Monty Hall’s Other Problems

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By John Tierney

Tags: Cognitive dissonance, mathematics, psychology

Now that so many Lab readers have brilliantly mastered the Monty Hall Problem discussed in my Findings column, I’ve got several new problems for you. (If you’re not yet clear on the original one, you can still play the Monty Hall game.)

Now, for extra credit, here are three new related challenges — which will be more of a challenge if you resist the temptation to look at the answers from other readers in the comments section:

1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?

2. John Allen Paulos, a mathematician at Temple University, has proposed a 10-door version of the Monty Hall Problem. Once again, there’s a car behind one door and goats behind all the others. You pick one door, but before it’s opened Monty will always open seven other doors to reveal goats. Then he’ll give you a chance to stick with your original door or switch to either of the remaining two unopened doors. Suppose you originally chose Door 1, and Monty opens every other door except Door 5 and Door 8. Should you switch to 5 or 8? And what are the odds the car is behind either 5 or 8? [Update: I originally wrote “the goat” instead of “the car” in that previous sentence, but I’ve corrected it. Thanks to Harris for noting the mistake.]

3. One unconvinced Lab reader, Scott, insisted yesterday that the best way to play the classic three-door Monty Hall game was to simply flip a coin. (I told him I’d be glad to let him bring his coin, and some bills, so we could play the game for money — with him flipping a coin and me using the switching strategy.) He’s wrong about the best way to play the game, but his strategy does raise a question: Suppose, when you’re confronted with the final choice to stick or switch doors, you flipped a coin — heads, you stick with your original door; tails, you switch. What would be your odds of winning the car?

I’ll give the answers tomorrow, along with the results from experiments that provide a psychological explanation for our tendency to get fooled by the Monty Hall Problem. I’m impressed that so many readers figured it out — disbelievers like Scott were a tiny minority among those who responded yesterday — and I think much of the credit goes to my Times colleagues who produced the online Monty Hall game: Ken Chang, Sarah Graham, Viktor Koen and Michael Lindsay.

You just don’t want me to get any work done, do you.

†Posted by Steve

It seems the first choice is not really a choice let alone one with one in three odds. You can not loose on this choice so on one level it doesn’t matter what you choose as every time a goat will appear no matter what is behind your door.

The real choice is the second one with 50/50 odds — the only ones that matter in this game.

†Posted by Judith

I haven’t had to think mathematically in 5 years, so please don’t laugh if these are all way off the mark:

1. 2/3 I think. Sounds wrong but it seems most logical.

2. So there is a 1/10 chance of the car being behind Door 1. Odds the car is behind Doors 2-10 are 9/10. Once the 7 goats are revealed, the odds of the car being behind Door 1 increase to 1/3. Odds the goat is behind 5 or 8 are 9/10. So odds the car is behind 5 are 9/20 and odds it’s behind 8 are also 9/20. Those are better odds than 1/3 so switching is still a good idea. But less of a good idea than in the 3-door version.

3. 1/2 chance heads comes up. 1/3 chance the original door has a car behind it. So 1/6 chance are your odds to win a car by flipping the coin. Hope you won some money off that kid.

†Posted by cognitively dissonant lawyer

1. 1/3 (ignoring slight deviations from an ideal 50:50 sex ratio). Three equally likely combinations are BG, GB and BB, and only one of them has the (rather ill-defined) “other” child being a boy.

2. You *really* want to switch. It’s a 10% chance the car is behind the door you originally chose, and thus a 45% chance for each of doors 5 & 8.

3. Exactly 50/50. You’re essentially ignoring the extra information, and picking one of the two closed doors at random — one having the car and one having a goat.

†Posted by Mike Scott

Okay, I tried the Monte Hall game. I switched every time. After 20 game, I had 44% victories. After 90 games, that has risen to 60%, but never higher. Then I had a winning streak and end 100 games with a 64% success rate, the highest percentage I saw in the whole series.

So why did it take 100 games to come close to the expected percentage of wins?

†Posted by B. J.

1. 1/3 2. Switch to door 5 or 8, though each has a 55% chance of having a goat 3. 50%

†Posted by Jim

You have to be careful about exactly what information you’re giving. Suppose you’re Mr. and Mrs. Smith’s OB-GYN, and the midwife went to their house yesterday evening. You know that the Smiths’ first child is a boy, but the ultrasound technician has kept it confidential what sex the second child is. Furthermore, you know that the midwife would have called you if there were problems with the delivery or if the labor was lasting too long, but just emailed you with the weight, length, sex, and APGARs if everything is ok. What you know fits the description that “Mister Smith has two children, at least one of whom is a boy.” But you don’t know anything about the sex of the younger sibling.

It’s a different situation than if Mr. Smith was selected at random from among everyone with two children, at least of whom is a son. The question needs to distinguish.

†Posted by Dan

1. The odds are 1/3: there are four possible permutations for the kids, and eliminating one (the double female) leaves two mixed-gender permutations and one combination that’s double males. The only way to get a second son is to have the double son permutation; one of three possible outcomes.

†Posted by Kenneth

1. Assuming boys and girls are equally likely (p=1/2) each of BB, GG, BG, GB has the same probability (p=1/4) For clarity, assume the first character refers to the oldest child. Conditional on having at least one boy we are left with BB, BG, GB and the probability of the other child being a boy is 1/3.

This is different than asking “given that Mr. Smith has a boy, what is the probability that his next child is a girl?” because in this case the GB possibility does not exist and the probability is 1/2.

2. You should switch. The probability that the car is behind door 5 is 45%, behind door 8 is 45%, behind door 1 is 10%.

†Posted by Don

#1. Given that there are more boys born than girls (the by survival rate is lower) odds favor a boy as the second child.

†Posted by TRS

The “rationalization effect” could be rational.

Suppose you have a complicated process for evaluating things. Half the time it provides some benefit, and enables you to choose the better of two options. The other half it responds to background noise, and is equivalent to flipping a coin. You have no preference between different colors of M&Ms, but you prefer the ones made at their Ohio factory over the ones made at their other factory in Pennslyvania. The Ohio ones taste better.

Now, you’ve been given a choice between a red and a blue M&M and spent some time and effort spinning them, rocking them, and tapping them, in an effort to determine which is more likely to have the yummy stuff inside and which is more likely to have the mediocre contents. You have no vocabulary to describe the resilience, resonance, and moments of inertia that your examinations rely on to help you guess. Or even if you do, that’s a lot of details to remember. On the other hand, it’s easy to remember that you just decided that the other one was better, so you can save yourself some time and effort by doing only a cursory examination of the newly-offered alternative.

If situations that abstractly resemble that scenario are common in our lives, “rationalization effect” could be a rational heuristic for efficient decision-making.

†Posted by Dan Wylie-Sears

1) 50% (or thereabouts… acutally I think it’s something like 52% if both children are under age 1 and due to a natural sex ratio). Of course this doesn’t consider factors like if he adopts and has a preference for having two kids of different sexes (I wonder what the statistics on that are).

2) 5 and 8 are indifferent, but you should switch. The probability of getting a goat in Door 1 is 90%, Door 5, 55% and Door 8 55%.

3) Flipping heads or tails to decide evens the odds, since one is always right. So if you flip, your odds flatline at 50%. (The same would be true regardless of the game if there is a binary outcome that a coinflip will invert. If there’s a game where if there’s a 99% chance of outcome A or a 1% chance of outcome B but you have a 50% chance of inverting that, your overall outcome is 50% even.)

†Posted by Cww

Very much like the Monty Hall problem, it is a question of the universe of possibilities that exist after the additional information is recognized. The original universe of possibilities is: GG,GB,BG,BB. However once the condition of at least one of the children is a boy the the universe of outcomes is reduced to GB,BG,BB. The probability of the other child being a boy is then 1/3.

The Monty Hall problem is similar but different and I think presents a paradox that you might call dissonance. I’ll try to put in words in another entry.

Regards,

Bob Nicholls

†Posted by Robert Nicholls

C’mon now, these are way too easy. I’ll work through the first one and leave the others for someone else to explain- I don’t want to ruin *all* of the problems…

If you have two children, there are four possibilities, each with probability 1/4: 1. boy, boy; 2. boy, girl; 3. girl, boy; 4. girl, girl

We can eliminate the fourth possibility from the list. That leaves us with the first three. Given that we are in cases 1,2 or 3 with equal probability, we can see what the chances are that the second child is a boy. In cases 2 and 3, the other child is a girl; only in case 1 is the other child a boy. That means that the odds are 1/3 that the second child is a boy.

†Posted by C

33%; should switch / 66.67%; ~41.17%

†Posted by RH

An interesting twist on Mr. Smith’s two children:

As you ponder the odds of both of Mr. Smith’s children being boys, one of the children comes down stairs, and you see that he is a boy. Now, what are the odds that both of Mr. Smith’s children are boys?

You already knew one child was a boy. Did seeing that one kid change the odds? And if so, why?

†Posted by David W.

1. There is a 1/3 chance that the other child is a boy. A priori there are four equally likely possibilities: two boys, two girls, an older boy with a younger sister, and an older girl with a younger brother. Knowing that there is at least one boy rules out the two girls possibility, which leaves the other three alternatives, only one of which is two boys — hence the answer is 1/3.

2. A priori, you have a 1/10 chance of picking the right door, and a 9/10 chance of picking a goat. When Monty shows you seven goats, your 9/10 a priori chance of missing is now concentrated in the two unopened doors. So, the final chances are that you have the car behind your initial choice with probability 10%, while the car is behind door # 5 with probability 45%, or behind door #8, also with probability 45%. You should switch, but it doesn’t matter whether you switch to door #5 or door #8. (In your question you actually asked “What are the odds the *goat* is behind either 5 or 8.” I presume this is a typo, but if not, then there is 55% chance there is a goat behind each of door #5 and door #8, along with a 90% chance of a goat behind the door you initially selected — and note that .55 + .55 + .9 = 2, which is the number of goats left when you are down to three doors, one of which has the car.) [John: You’re right, it was a mistake, and I’ve now corrected it to “the car” instead of “the goat.” Thanks for catching my carelessness — and kudos on calculating the probabilities both ways.].

3. If you toss a coin in the original problem to decide whether to keep or switch, then you have a 50% chance of winning the car. Via the coin toss, you have a 50% chance of keeping your original pick, and conditional on that you have a 1/3 chance of winning; plus you have a 50% chance of switching doors, and conditional on that you have a 2/3 chance of winning. So, your probability of winning if you flip the coin is equal to .5*1/3 + .5*2/3 = .5. Note that this would be true for *any* random mechanism by which Monty could place the car either behind your initially chosen door or the remaining unopened door — if “p” is the probability Monty puts the car behind your door (after you choose your door), and “1-p” is thus the chance he puts it behind the other unopened door (since he opens one to show a goat), and you toss a coin, then your chance of getting the car equals .5*p + .5*(1-p) = .5. The original problem has p=1/3. Note that if Monty always hid the car in the other door (p=0), or if he always initially gave you the car (p=1), or anything else, if you toss a coin, you have a .5 chance of winning for both scenarios (or for any other value of p as previously shown).

†Posted by Harris

For number 2, you should definitely switch to a different door. Since you’ve got a 1/10 chance of correctly guessing that the car is behind door number 1, and the all-knowing Monty then informs you that the car is certainly behind either 1, 5 or 8, it follows that there is a 9/10 chance that the car is behind either door 5 or door 8. In particular, if you choose to switch to (say) door 5, you’d be riding home in a a shiny new car with a probability of 9/20. If you stick with door 1, you’ll be riding home on a goat with probability 9/10.

John Riverside, CA

†Posted by John

And an addendum:

If you really want to confuse people here’s a variation:

Suppose that Monty decides to change up the game a little: 10 doors, one car, 9 goats (or lady and tigers)

You get to pick a door. That’s yours. Then Monty will randomly pick doors to open. Here’s the catch. If he picks a door with the car. Game over, you don’t get the car. If he picks a door with a goat, you can stop, trade for any of the remaining doors or keep your own and reveal OR you can keep going, letting him pick doors. (You also get to make this choice before he opens any doors.)

Is there a strategy that will improve your odds of getting the car beyond 10%? Obviously if you reveal at the very beginning, it’s 10% and if you let him open all 9 doors and get all goats then there’s a 100% chance that your door has the car… but your odds of making it this far are slim.

(Ok, I obviously know the answer to this question, but I don’t know what the INTUITIVE answer is… What do people feel is the best strategy? What does that say about us?)

†Posted by Cww

I tried to figure out the Mister Smith problem, but got distracted by thinking about the Red Queen’s job offer to Alice, which included jam every other day: “It’s jam yesterday and jam tomorrow, but never jam today.”

So, it seemed to me that there’s a 100% chance that Mister Smith’s other child is a boy… because of one child is a girl, then the other has to be the boy, and if they’re both boys, well, there you are.

But then, I’m the one who “solved” the Monty Hall problem by reasoning that I already have a car, but don’t have any goats, and they’d probably be fun… so I’m a winner whichever door I pick!

†Posted by The Supreme Dalek

1) 50% 2) switching pays off 45% of the time; staying pays off only 10% of the time 3) if you flip a coin your odds are a third plus a sixth, that is 50%.

†Posted by jon w

1. 50/50, or whatever the current ratio is between male and female births.

2. If you don’t switch, you have a 1/3 chance of winning the car. If you do switch, you still only have 1/3 chance of winning the car (1/3 x 0 if you abandoned the car, and 2/3 x 1/2 if you abandoned a goat, for a total chance of improvement of 1/3). It doesn’t matter if you switch or not. Doors 5 and 8 each have a 2/3 chance of being a goat.

3. 1/3 x 1/2 (the odds of winning when you flip heads) plus 2/3 x 1/2 (the odds of winning when you flip tails) = 1/6 plus 1/3 = 50/50, which is, of course, worse than the 2/3 expectation of winning by switching doors all the time.

†Posted by Geoff, Ohio

2. The odds of the car being behind doors 5 OR 8 is 9/10. This is because the odds of you picking the right door from the outset being correct is 1/10, and Monty is introducing information that alters the odds since now one of the three MUST have the car. We’re still working with the inital odds that you picked the wrong door (highly probable) and Monty knowing that must include the correct door in your choices.

I’m pretty certain this is one case where choosing from the remaining doors is a random choice and results in an equal chance of 45% winning the car. Someone more clever than I will probably show me wrong, though.

3. Adding an even weight random chance gives you exactly a 50/50 chance of winning, which is still losing to always swapping. Map all the outcomes: you’ve got a 66% chance of winning the car if you swap. You got a half chance of getting that door swap by coin flipping: 33%. Switch to the other possible door configuration. You now have a half chance of flipping to stay on the door and win: 16.5% This is a rounded probability, combined, of 50%.

Let’s check by calculating the odds of losing: flip a coin and stay to get a goat, which occurs 33% of the time. In the minority scenario, flipping a coin to switch to the goat is 16.5% This, again, is 50%.

†Posted by Kenneth

1. The first question is not phrased as artfully as it could be. I assume you mean “in a two child family, given that at least one child is a boy, what is the probability that the both children are boys?” The answer is one in three. Assuming 50-50 birth rates, 75% of two child households have at least one boy, but only 25% have two boys. .25/.75 - .33

2. This works like the original Monty Hall game. The opening of doors and revealing of goats provides information about the doors that you did not select, but not about the door that you did select. The backhanded way to solve this problem is to “credit” the remaining unopened, unselected doors with the original probabilities assigned to the doors that were opened. In our example, the 7/10 probability from the doors that were opened are added to the 2/10 probability of doors 5 and 8. Thus, the odds that the car resides behind either door 5 or door 8 is 9/10, or 45% for each of those doors individually.

3. Your odds of winning by flipping a coin are 50/50. You have a 50% chance of choosing a 1/3 winning strategy and a 50% chance of choosing a 2/3 winning strategy. Mathematically, 1/2 x 1/3 plus 1/2 x 2/3 - 1/2.

†Posted by Holt from New York

Jim (6) and Mike Scott (4) are right on all three counts.

†Posted by Jon

I guess that you might like to play against the coin-flipper but remember that random walks can move in either direction and both of you better have a limit on money or flips. Apiori, the selection of a door with no change gives the participant a 1/3 chance of getting the car. However, if we use the notation of /S for selected door and /O for opened door and C & G for cars and goats then we have the universe of outcomes of: C/S,G,G/O C/S,G/O,G C,G/S,G/O C,G/O,G/S With this we see that the winning strategy of changing doors has a probability of 2/4 or 1/2. Now if the player flips a coin then the probability of success will be 1/2*1/3+1/2*1/2=5/12 which is clearly less than the 1/2 or the pure strategy.

Underlying this paradox is the heuristic argument that there is a re-sampling with a restricted set once the door with the goat is revealed. Regards, Bob N-

†Posted by Robert Nicholls

The sexes of ones children are not independent (ask Henry VIII); there is no answer to the first question.

†Posted by bkp

1. A well known problem and another example of how the phrasing of a question determines the nature of the game, meaning the confusion comes from the way the problem “hides” the real question. The answer is 2/3 because there are 4 possible combinations of boy/girl and only the girl/girl one is excluded. The answer is, of course, different if birth order is considered. 2. Why is this a question? The first door has a 10% chance and the others have a 90% chance. The more of the other doors you open, the more that 90% becomes concentrated in the remaining doors. Since there are two doors left open of those nine, then you have a 45% chance - so switch, dude, switch. The important point, again, is that Monty is not a second player choosing at random but is providing you with information about the probabilities related to your choice. That fact determines the nature of the game. If Monty were picking at random and by chance chose seven doors with a goat, then you’d gain nothing by switching because Monty’s choices would be independent of your choice and would not be providing information about the probabilities of your choice. 3. You have a 50% chance of picking a door with a 1/3 chance and a 50% chance of picking a door with a 2/3 chance. I don’t remember my elementary probability rules very well, but I believe you multiply independent events and thus you’d halve each chance.

†Posted by jonathan

Maybe I’m dense, but I still don’t get the original problem.

Specifically, why does Monty Hall’s reduction in possibilities only influence probability of the non-chosen set?

(Restated: If one starts with three choices and one is eliminated, why does probability of the original choice remain at 1/3? Why does that elimination only affect probability of the unchosen door? After all, both are still unknown outcomes.)

I suspect that many who claim to understand this are unwilling to admit ignorance. But I could be wrong again. :)

†Posted by fred

I spent quite a lot of time explaining this to my boyfriend yesterday, so I think I’ve got a handle on it. The crux is that you have a limited number of correct answers, in a rapidly depleted pool of possibilities. So, for today’s questions: 1. 33.3%. Statistically speaking, if the odds of each child are 50/50 for either sex, with two children, you will end up with the mixes Boy-Boy, Boy-Girl, Girl-Boy, and Girl-Girl. The terms of the puzzle remove Girl-Girl from consideration. Therefore, of the remaining mixes, there is a one out of three chance the remaining child is a boy, and a two out of three chance the remaining child is a girl. (This is different in nature from knowing how many boys are in a family, and trying to guess if the oldest or youngest is a boy, which would more closely match the Monty Hall problem.)

2. Switch. 45%. Your first guess has a 10% chance of being right, and a 90% chance of being wrong. With 7 doors opened, that 90% chance is now spread across two unopened doors.

3. Just shy of 50%. Which is better than sticking with your original choice (33.3%) but not as good as switching every time (66.6%).

†Posted by Jennifer

If the genders of one’s children are independent, then the probability of two boys is 25%. Knowledge that the one of the children is a boy does not change the fact that the probability of two boys occurring is 25%. Have you people learned nothng from the Monty Hall game?

†Posted by bkp

1. 50% (assuming we take the probability of any random person being male/female to be 50/50). The fact that one child is male is extraneous. (By way of example, the chances of a coin coming up tails six times in a row is one in sixty four; if it has already come up tails five times in a row, however, the chance of the next flip being tails is one in two–it’s unaffected by previous flips.)

2. A number of people have already tackled this, but you’re much better off switching, and there’s no reason to prefer door 5 to door 8 (45% chance for each door).

3. Adding the coin toss renders all other information irrelevant: it reduces the problem to two doors, one winning and one losing, with the coin having even odds of picking either one. 50/50. (You would, of course, do better by always switching.)

†Posted by Evan

You are all being duped. It is a statistical problem and the “decider” has no influence on the outcome at all. Put the “Monty Hall” problem in a small room with Uri Geller and you will get your bent spoon. This sort of behavioral research belongs in an art gallery, not a laboratory.

†Posted by Schwannee Howiluvya

There are other factors involved in the boy-or-girl issue: supposedly some research suggests that acid/alkaline factors in the womb, viability of X sperm vs Y sperm, etc. vary by individual. So you can’t figure the odds on that question - a woman might have only boys with one partner, only girls with another, a mix with a third. As #27 said, ask Henry VIII (who had an illegitimate son well before he had a legitimate one).

†Posted by ACW

The answers to all three are 50:50

1. What gender is the other child. Options = boy or girl with a more or less equal chance that it is either.

2. With three doors left unopened, you know there is a goat behind two of them. It goes back to the original game. With one door chosen, the odds of a goat behind door 5 is 50% or door 8 is 50%.

3. Assuming you have a fair coin, the odds of heads or tails is 50:50.

†Posted by DM

I wish you would stop saying “odds” when you mean “chances”, or “probability”. “Odds” does not mean “chances”. Look it up in an elementary math or stat book, or try Wikipedia.

†Posted by Marjorie Ireland

OK Fred, let’s simplify and expand Paulos’ 10 door problem: Suppose there were a million scratch-off lottery tickets, and one of them held a million dollar prize. You pick one of them and don’t do anything to it.. I pick all the others. I scratch off all the 999,999 tickets that I have. I show you 999,998 of them. What are the odds that the one I didn’t show you is the million dollar ticket? Now, suppose I offer to switch the one I didn’t show you for the one you hold in your hand? Would you take that offer?

†Posted by Andrew Berman

Fred: Because the situation changes with the interference of a NON-random element, that is, Monty picking a door and revealing information about it.

Take the original choice. Knowing nothing about the doors, you pick one. The odds of winning is 1/3. Monty knowingly reveals a loser. The odds that you picked right initially did not change because you could not have incorporated that information at the start. The probability that the other two doors hides the car combined is 2/3, right? Well, now there’s only ONE door, but the combined odds are still 2/3. The information is that IF the winner is in the other two doors, it MUST be the one left. Those odds never changed, but your choices are now constrained, which is why switching is the answer.

The best way to illustrate this principle is if we add an absurd number of doors to the problem. You pick one from a thousand doors. Your odds of getting it right is 0.1%. With a wave of a hand, Monty causes 98 other doors to open, revealing a herd of goats. The act of eliminating those doors wasn’t random, however your initial choice was, so we carry that probability over and compare that with the new scenario and choice. The remaining door contains the 99.9% probability that you chose wrong initially, and it is almost certainly the right choice if offered the chance.

In other words, your choice isn’t really between two doors of uncertain provenance; it’s between two *sets* of outcomes, one choice having the luxury of eliminating all possible “wrong” outcomes.

†Posted by Kenneth

To Fred and others confused by original game: In bridge, they explain the Monty Hall problem as restricted choice: If you picked the car Monty could have picked either door with a goat, but if you picked a goat, Monty had only one choice. In the 2/3 of the cases that you picked the goat, you are certain to win if you switch. To bkp - You got the boy question confused. Yes, there is a 25% chance orignally of two boys. Once you have eliminated the 25% possibility of two girls, though, the probability that you have two boys is 25%/75% or 1/3. To Jennifer: flipping a coin is exactly 50%. The car is behind one of the two doors and you are randomly making your choice.

†Posted by Marc

All you folks who are figuring that there are 3 possible permutations of the boy/girl question (BG, GB, BB) are wrong. If each of those letters is a child, then one of the children’s sex has already been determined. So there is neither a GG nor, if we count the already-established boy as Child #1, a GB. There are only BB and BG. So the odds that the other child is a boy are 50%. QED.

†Posted by Ken

bkp - I’m afraid you haven’t learned as much from the game as you thought. One of the lessons of the game is that you have to take into account all of the information at your disposal.

It’s true that, assuming equal sex ratios, and knowing nothing else, 2 boys have a 25% probability.

But we do know something else — the 2 girls possibility has been excluded. This leaves us with three possibilities, all equally likely:

boy girl girl boy boy boy

Given our new information, the probability is 1/3.

†Posted by Dave

To everyone who is saying that the odds of the first problem is 25%, remember that the problem explicitly excludes the two girls outcome (if one of the kids is a boy, how are you getting an outcome that both kids are girls? Androgeny doesn’t count!). That leaves three logical possibilities, only one of which is two boys.

†Posted by Kenneth

Hi John -

Problem 1 is another problem that provokes all sorts of arguments, and the right answer is - it depends!

It depends on HOW you know at least one child is a boy.

Suppose you’re sitting in Mister Smith’s lounge. You ask him how many children he has, and he replies “two.” You look out of the window, notice a boy playing in the garden, and ask “is that one of your children?” Mister Smith answers “Yes.”

Well, now you know there’s a 50% chance the other child is a boy, assuming boys and girls are split 50:50.

However, suppose your visiting the Smiths, and ask your wife in a cellphone call what present you should bring. She answers “I know they don’t have two girls,” and at that moment the call is cut off.

Now you reason that the three remianing possibilities (elder child first) are BB, BG, and GB. In only one out of three cases do they have two boys, so your best estimate now is a 33% probability that Mister Smith has two sons.

Like the original Monty Hall problem, this gets people into all kinds of arguments. Most people intuitively think the answer is 50%, but many people who pose the question do so on the basis of the second scenario. I’ll be interested to read your answer tomorrow.

†Posted by Andrew

re: #16 — if you randomly select one of the two kids, knowing at least one is a boy, and the one you select is a boy, the odds of both being boys are, I think, 50%

re: #19 — the intuitive answer is there is no strategy that gets you better than 10% and your odds are, I think, 10%, no matter what your strategy is.

†Posted by Leonard Kim

bkp, yes, the p of two boys occurring is 25%, but the question was not “what is the p of two boys occurring in a family”–there was a condition set on it which changes the pool of outcomes. One gets 25% from the total set of possible outcomes:

BG GB GG* BB

*But GG is not allowed as an outcome in this question. And so we have just BG, GB, or BB…1/3.

†Posted by cm

fred,

You asked, “Specifically, why does Monty Hall’s reduction in possibilities only influence probability of the non-chosen set?”

Answer: it doesn’t. The non-chosen set is always equal to 2/3. By Monty narrowing it down to just one choice, that one choice has the 2/3rds chance.

Imagine there are 1 million doors. You choose one. Obviously, you have a poor chance of winning with your choice (1 in a million). Now Monty, who knows where the car is and will never accidentally reveal it, commands that all the other doors except one be revealed as the goat-doors. Now, you have the chance to switch. Would you say that your original choice is equal in probability to the one remaining door? No. The 999,999 out of million odds have been collapsed down onto that one remaining door.

Note that if Monty DIDN’T know where the car was, he would have high odds of accidentally revealing it as he showed more and more doors. But he does know where the car is, and that’s the point of why you should switch.

The same is true with three doors, but it is just much less intuitive with such a small set of doors.

†Posted by cm

I don’t understand why Monty’s knowledge makes a difference. In other words, I’ve read that, “if Monty chooses at random, and it happens to be a goat, then it doesn’t matter whether or not you switch.” I don’t understand that. Chances are, when you first select, you landed on a goat. If Monty then chooses randomly, and it happens to be a goat, then I still believe you should switch since you are still likely sitting on the other goat. Can anyone help me with this?

Additionally, I’m not sure why, in the first problem of this column, the sex of one child affects (or is dependent upon) the other. Suppose Mr. Smith had a child in 1995, and then another in 2002. Do you think that because Mr. Smith had a boy in 1995 that that affects the birth in 2002? It seems to me that the probably of a boy in 2002 is still 1/2. I grant that it’s unlikely that would have two boys versus all the other combinations, but just because you see a 13-year old boy walk down the stairs doesn’t mean that there is less than a fifty percent chance that 6-year old who follows down the stairs will be a boy.

†Posted by Erik

1. Given that one of the children is a boy there are three equally likely possibilities: GB, BG, and BB. BB happens 1/3 of the time so the probability is 1/3. 2. The chance you chose the door right the first time is 1/10. The chance one of the remaining doors is right is 1/3. So you should always switch. 3. You have 1/2 chance of switching and then, if you switch, you have a 1/2 chance of getting the car, so 1/4; plus, you have a 1/3 chance of choosing the car in the first place and a 1/2 chance of not switching, so 1/6. The total is 1/4 + 1/6 = 5/12.

One thing you haven’t pointed out about the Monty Hall problem is that you’ll probably lose if you’re actually playing against Monty Hall and he’s trying to win. Because he’ll be more likely to offer you the option of switching when you have chosen correctly.

†Posted by Jon Webb

1) 50% - the probability of having two boys is equal to the probability of having a boy and a girl.

2) Switch - The probability that your first choice is a car is only 10%. Thus, there is a 90% probability the car is behind one of doors 5 and 8. More specifically, a 45% chance the car is behind door 5 and a 45% chance the car is behind door 8.

3) the answer is 50% -> 0.5(1/3) + 0.5(2/3) = 0.5

†Posted by Brian

The key for my understanding of almost every statistics problem is never to calculate the chance of something happening but the chance of something *not* happening.

For example in the ten door problem the chance of not picking the correct door intitially is obviously 90%. So then there is a 90% chance the prize is behind the two remaining doors. Then it is 50/50 split between the two doors or 45% prize for each. Then 100% minus 45% is 55% goat. Therefore

Door 1 10% prize, 90% goat Door 5 45% prize, 55% goat Door 8 45% prize, 55% goat It adds up to 100% = 1 prize and 200% = 2 goats

†Posted by BigMattP

Fascinating stuff! I know I’m right on #3, but now I’m at sea concerning #2. I understand that the previous approach to the MH problem would indicate that doors 5 & 8 now have a 90% probability of concealing the car, and thus a 45% chance for either door, but something about TWO doors being unopened is driving me nuts, telling me that the situation is somehow different. I just can’t put my finger on it. As for #1, I admit my mistake; the correct odds that “the other” is a boy are 1/3 when “at least one of them” is a boy.

†Posted by Geoff, Ohio

Props: Mike Scott (#4) was the first one to nail all three answers.

1. 1/3 2. Yes, you should certainly switch; the probability that the car is behind door 5 or door 8 is 90% (45% each) 3. 50%. I like Mike’s reasoning. Another way of looking at it is this:

The solution to the original problem was that if you stay put, and you’ve got a 1/3 chance of getting the car, and if you switch, you’ve got a 2/3 chance.

In the new problem, there’s a 50% chance of staying put, and a 50% chance of switching, so the chance of getting the car is half of 1/3 (=1/6) plus half of 2/3 (=1/3). 1/6 + 1/3 = 1/2, Q.E.D.

†Posted by Conrad Chaffee

1) 50% You already know there is one boy. Given that information, the chance that the other was a boy is 50%.

2) 9/20. There was a 9/10 chance you originally picked a goat, so there is a 9/10 chance that one of either 5 or 8 is a goat, and the other is a car. Hence there is a 4.5/10 or 9/20 chance that you’ll get the car if you choose one of those doors.

3) 50% For two reasons. The 1/3 and 2/3 probability would average out to 1/2 if you randomly choose each one 50% of the time using a coin. Also, since you know the car is behind one or the other, flipping a coin still gives you a 50% chance of getting the car for a given trial. Similarly, in the door problem the probabilities are 1/10 4.5/10 and 4.5/10, but if you flip a coin you’ll get a car about 1/3 of the time as you would expect (and those probabilities average out to 3.33/10 or 1/3).

†Posted by DavidSTVZ

Ah, I screwed up. If a friend told me I have two children and at least one is a boy. I should have said, that’s not surprising, that would happen 3 times out of 4. And if he said what’s the chance that they’re both boys? I’d say, well, out of those 3 times I mentioned, that happens 1 time out of 3. Oops. What can I say, it’s been at least 6 six years since I took theory of probability and I only got a B.

†Posted by DavidSTVZ

Your article is mistaken. It depends on an ambiguity in language, not logic. Consider:

Sequence 1. You pick a door (at 1:3 odds). Monty opens a goat door and asks “Do you want to switch?” Insofar as this continues the same game, the odds of your original pick remain 1:3, and switching picks makes your odds of winning 2:3.

Sequence 2. You pick a door (at 1:3 odds). Monty opens a goat door and says “OK, it’s now a new game, with just 2 doors. Which door do you pick?” In this new game, there is 1 goat door and 1 car door, and thus the odds are 1:2.

But how can the odds change depending on how Monty phrases the question? The answer is that they don’t. In Sequence 1, Monty doesn’t continue the original game but begins a new one, just as in Sequence 2, even though he uses words that imply continuation. The question “Do you want to switch” requires you to choose one door or the other, at a time when there are only two possibilities.

Thus in either sequence your odds are 1:2.

†Posted by KRS

Erik: Monty is compelled to reduce all the remaining doors to a choice between two, one of which must contain the prize. Let’s take the extreme version with a million doors to pick — odds are if he’s acting randomly, he’ll not be able to get it down to two doors in one try, at least not without making it moot by accidentally revealing the car. Because he’s not interested in embarrassing himself, he’s going to choose to reveal only losers, and the rest of the problem is reasoned out in other posts above.

Also, the gender of the first child is not necessarily as important so much as remembering that kids aren’t marbles drawn from a bag; they have ages and a birth order. This adds a second implied dimension to the problem, and because the question states the gender but not age, you must consider the possibility of a younger OR older sibling, and their potential genders.

†Posted by Kenneth

Ken - there is no grounds to assume that the already established child is child #1.

The problem simply states that there is at least one boy. So you cannot exclude either BG or GB because you don’t know that it is Child #1 that has been determined. So there really are 3 possibilities.

†Posted by Dave

Have you seen the less known problem as follows:

There are two envelopes with money in them. You are told that one of them has double the money of the other. You can choose one of the envelopes, examine how much money is in it, and decide to keep it or switch to the other envelope.

Problem 1) Can you calculate the expected value of switching vs. staying (and justify your answer as making sense?).

Problem 2) Can you figure out a strategy that will work better than random guessing?

-Ben Weiss

John: Good questions. Answers, anyone?

†Posted by Ben Weiss

The first question asks, “What is the probability that the other [child] is a boy?” As phrased, the answer is 1/2. Those who say the answer is 1/3 either misinterpret or have rephrased the question to ask, “What is the probability that both children are boys?” That is a very different question indeed. The probability of an independent event is not affected by the outcome of other events. So the probability of a child being a boy is 1/2 whether Mr. Smith has 1, 10 or 100 other boys.

†Posted by Bill

1. 50% 2. Doesn’t matter if you switch 3. Must get some work done

†Posted by LL

Bill — You’re right that if it was an independent event, the answer would be 1/2. But it’s not independent, because we don’t know which child! If our information was ‘the first child is a boy’ then the sex of the second would indeed be independent. But in this case, ‘other child’ may refer to the first or the second child, *depending* on the sex of the original. You can’t assume more information than the problem provides.

†Posted by Dave

I’m with Scott for the answer to the original problem. 2/3 of the time your initial choice will be wrong, and since you are switching 1/2 the time, you will end up with 1/3 wins. The other 1/3 of the time your initial choice is correct, and since you switch 1/2 the time, you end up with another 1/6 wins, so you get wins 1/3 + 1/6 = 1/2 of the time.

†Posted by Michael Spivak

1. I misread the question in my answer at 28 above, because the usual statement is the odds of the other kid being a girl. The question above actually asks the odds of the other child being a boy when the first kid is a boy but there is no birth order specified. There are 3 possible combinations and two of those have girls - not g/b, b/g but maybe b/b - so the odds are 1/3 that the second and thus both children will be boys. If birth order were specified, then you’d eliminate another possibility - i.e., if the boy were first, then only two choices would be possible, b/g and b/b, which means the odds of a second boy would be 1/2.

The two envelopes problem in 58 above requires a longer discussion and my opinion is there is no advantage to switching in the pure case of two unopened envelopes. If you do simple math, you find that the other envelope always has an advantage so you end up in a switching loop, going back and forth. That is, if you assign a values to the money “in the other envelope” you can get a neat equation - like 1/2(2A) + 1/2(A/2), which says the other envelope is worth 5/4A, which is more . . . so you switch. (The hard part in that equation is understanding that A is the money in your envelope. That means the other envelope either holds twice A or 1/2 A; the equation is not 2A and A but 2A and 1/2A.) The issue is “in the other envelope” means that as soon as you switch, the other envelope is worth 5/4A so you switch back, etc., etc., etc.

†Posted by jonathan

Question 1: People are mixing up the logic here. It’s not normal mendelian genes. The wife always gives a X, and the husband either gives an X or a Y. So your choices are only XX and XY, with a 50% probability. Obviously you ignore the independent event of the first child.

†Posted by Michael K

At the risk of sounding too pedantic, I have a few comments on your proposed problems.

In the first problem, may we assume that using sibling boys and girls is merely a way of expositing the problem? That is, may we assume that Mister Smith is equally as likely to have a boy than a girl? and that the sex of any child is independent of the sex of previous children? If so, we can restate the problem as so:

Mister Smith flips two distinct, fair coins. Let’s call the coins A and B. He then shows you that one of the coins was heads, but you do *not* know if the coin he showed you–which was heads–was coin A or coin B. What is the probability that the other coin is also heads?

A major source of confusion results when someone fallaciously, implicitly assumes that the coin Mister Smith shows–which is heads–is coin A, and that the question is asking for the probability that coin B is also heads. Once this confusion is cleared up, the solution is fairly straight forward:

There are four possible outcomes, each of which are equally likely, viz. HH, HT, TH, TT, where the first H or T is coin A and the second is coin B. Of the four, only the first three can be considered, since Mister Smith showed a coin that was heads and thus TT could not have happened. Now, of the first three possible outcomes that we are considering–which are all equally likely–only the first one results in the other coin also being heads. Therefore, the answer is 1/3.

Let’s now consider the often confused version, namely falsely assuming that Mister Smith showed us coin A and that we are trying to determine the probability of coin B also being heads. Then, we must only consider the first two possible outcomes, namely HH and HT. Therefore, the answer to this confused version is 1/2.

As you can see, finding the solution to this problem has more to do with clearing up a confusion rather than needing any real creativity. The issue I have with your second problem is the same one I have with the standard Monty Hall problem.

The standard problem (and not the version that Monty Hall actually used on the show) is almost never stated in a complete way. ( There was a comment to your blog from yesterday noting this omission, and I think it’s worth restating.) Once this omission is clearly stated, the solution to this problem, as with the previous one with Mister Smith, becomes much clearer.

One time in three will the contestant initially correctly choose the winning door. Since Monty Hall must open one losing door of the remaining two, one time in three will Monty’s choice be unrestricted as to which door to open. Two times in three Monty’s choice will be restricted, i.e. he has no choice but to a open a particular door. Those two times in three, by opening a door, Monty is implicitly pointing to the correct door. But what about that one time in three when Monty must make a choice as to which door to open? How does Monty decide?

If Monty’s always has no preference between which of the two doors to open, and it’s equally likely that he chooses either of the two, then the contestant will win by switching with a probability of 2/3. However, let’s say that Monty, whenever he has a choice, will always open the lowest numbered door. Then if the contestant initially picks door 1 and Monty opens door 3, then contestant will know with certainty (assuming no cheating of course) that the prize is in door 2. For if it wasn’t, Monty would have opened door 2. Now if Monty instead opens door 2, then the probability of winning if the contestant switches is only 1/2.

In order to state the problem in a complete, and thus correct way, you must say that when the contestant happens to initially choose correctly, Monty is equally likely to open either of the remaining two doors. The rub is that by adding this clause, you’re kind of giving away the answer. That is, the solution is merely a matter of fully understanding the problem, and not one of needing genuine creativity.

†Posted by William Hanisch

re: 58. It’s funny. I did the calculation on a post-it, immediately got an expected payoff of 1.5 x the lower value (or 0.75 x the higher value) regardless of strategy. That result seemed so easy and intuitive, I went internet surfing thinking there had to be a catch and immediately got thousands of hits about the “two envelope paradox” which I had managed to avoid because of my calculation method.

†Posted by Leonard Kim

The new problems are fun, but for many NYT readers the main issue now is what is wrong with Scot’s logic (http://tierneylab.blogs.nytimes.com/2008/04/07/monty-h all-meets-cognitive-dissonance/#comment-123531) on the original Monty Hall’s problem. Here is one way to expalin. If the door for placing the car was picked by Monty with equal probability of 1/3 (which everyone assumes to be so by default), then, the probability that the car is behind one of the door not picked by the player at first is 2/3. Therefore, if Scot wants to toss a coin to decide on switcing or not, then he should toss a BIASED coin with heads having 2/3 chance (for switching) and tails 1/3 chance, because, according to the information available to the player at that point, the odds are not 50/50 percent for the two doors, but the odds are 1/3 and 2/3. Just to make this more clear, imagine how this problem changes if Monty told the palyer at the begining that he has placed the car with uneven probabilities of 10%, 20%, and 70%, behind doors 1, 2, and 3, respectively. Then it pays to pick door 3 and stick with it. This is the situation that the player faces when he has to decide whether to switch or not. He has to decide between 2 doors with known uneven probabilities of 1/3 and 2/3. Murali

†Posted by Murali

1/3 is not correct for the first one. Yes, we know there is no GG possibility, but GB and BG are not distinct outcomes, but rather they are the same choice; birth order is irrelevant for this question. We know only that one of the two children is a boy, and want to know the sex of the other - the rest of the question is a red herring. The coin toss analogy is apt.

†Posted by Mark Sowul

Answer to Question #1: the probability that other child is a boy equals 1/3. Because: there are three ways that at least one child is a boy (GB,BG, and BB); of these, in only one (BB) is the other child a boy.

†Posted by Leonard Katz

An easier way of thinking about the Monte Hall problem which several of the commenters have mentioned difficulty understanding is to assume the car is behind Door #1. It’s pretty obvious that you have only a one-in-three chance of picking that door. But what happens if you pick one of the other doors? If you pick Door #2, then Monte will open up Door #3 to reveal a goat, and you should switch. If you pick Door #3, Monte opens Door #2 and you should switch.

On the money question in #58, I’d guess for problem 2 if the amount of money you discover in the wallet is an odd number, like $5.67, you should switch; if it’s even you should probably stick. Of course, it is possible that they put amounts in both envelopes that ends in an even number, in which case you’re 50-50. So my guess is using that method you’d be 75% likely to get more money, and only 25% likely to lose out. Of course, this assumes that they don’t get really tricky and cut a penny in half.

†Posted by Pat Curley

Regarding question #1: The way I see it, the chances are 1/4. The odds of a child being a boy is 1/2. The odds of another child being a boy is 1/2. 1/2 multiplied by 1/2 = 1/4.

In other words, the odds of two boys in a two child household is 1/4. The possibilities of a two child family are, obviously: GG, BB, BG, GB. So, BB is 1/4. Just because the first child is not a girl doesn’t eliminate the possibility that it could have been a girl.

†Posted by maya

I look forward to tomorrow’s column on why so many people have trouble with the original Monty Hall problem. Most people think it doesn’t matter whether you switch or not after Monty reveals a goat as your chances are fifty-fifty. They fail to spot the logical/time fallacy. Namely, Monty flipping the door AFTER you have already chosen cannot possibly change the original odds which were 1 out of three. I try to use this analogy to explain it: You and I go hiking near some cliffs. Despite my warning of the danger of falling, you approach the edge and fall 300 feet to your death on the rocks below. The next week, the park service builds a fence to prevent future accidents. Does that fence save your life? No! Why? Because it was built after you fell. Monty flipping the door AFTER you choose, can NOT change your odds. But switching now can and does change your odds from one in three to two in three. Counter intuitive problems are tough to get.

†Posted by Kenneth Kaye

The two-child problem is quite old and a lot of fun, and I’ve given it to gifted math students. Here are 5 different takes on it…

The Two-Child Problem

Problem 1: A man tells you, “I have two children. One of them, Fred, is a boy.� You ask yourself what the probability is that Fred has a brother. You reason as follows: It is equally probable that a child is a girl or a boy, and so it is equally probable that Fred’s sibling is a girl or a boy. Hence, your answer is 1/2.

Problem 2: Again: A man tells you, “I have two children. One of them, Fred, is a boy.� You ask yourself what the probability is that Fred has a brother. You reason as follows: There are 4 types of 2-child families, each one equally probable:

I. Boy, then Boy II. Boy, then Girl III. Girl, then Boy IV. Girl, then Girl

The man has eliminated type IV, and so each of the remaining three are equally probable, and only type I gives Fred a brother. Hence, your answer is 1/3.

Problem 3: You meet a boy on the street. Fred (for some reason, always his name) says, “I have one sibling.â€� You ask yourself what the probability is that Fred has a brother. You reason as follows: Assume the street was filled with the children of, say, 400 2-child families. Then there are 800 children on the street, 400 boys and 400 girls. ¼ of those families (100 families) were of type I, contributing 2 boys each, so a total of 200 boys are from type I. Therefore your chance of meeting a boy from a type I is 200/400. Hence, your answer is 1/2.

Problem 4: You meet a boy on the street. Fred (for some reason, always his name) says something you don’t understand in Italian. Seeing your confusion, he gets his father, who speaks English. The father tells you, “I have two children. One of them is Fred, here.� You reply, “I believe that I’ve met you before, sir, and I can say, without hesitation, that the chance that your son, Fred, has a brother is either 1/2, if he is the one asking, or 1/3 if you are the one asking. Please see Problems 2 and 3 for explanations.� The father says something in Italian that sounds suspiciously like, “crazy academic.�

Problem 5: Let us let the world creep into this problem a little. Suppose you are in the People’s Republic of China (PRC) and a native Chinese parent tells you, “I have two children. One of them, Fred (Anglicized), is a boy.� What is the probability that Fred has a brother? You reason as follows: In the recent past, the PRC had a strict policy of allowing only one child per family (we are simplifying here), with sanctions against families with detected multiple children, and with increasing penalties with increasing family size. But there is a strong cultural and economic incentive to have at least one male child in the family. In this model, we will assume that parents will continue to have children until a boy is born, and then stop. Therefore, the chance that Fred has a brother is 0.

†Posted by Peter Brooks

I’m amazed how many people who responded got this wrong. I would have thought/hoped that the people who read math problems on the NYT’s website for fun would be a bit smarter!! (Or perhaps the smarter folks have better things to do with their time, not that I can talk!!)

Well done Mike Scott for being first to get it right. (And amazing how many folks got it wrong even with his correct answer posted!!)

Taking #1 a step further, a US Dept of Labor study found 51% of first children in the study were boys, a 50/50 split of 2nd children after a boy, and a 54.5 boy /45.5 girl split of 2nd children after a girl. If for fun one assumed that the study were 100% predictive (forgetting about the margin of error), the odds of BB would be 32.816%, not 1/3. (www.in-gender.com/XYU/Odds/Gender_Odds.aspx)

†Posted by rich

I continue to believe that the boy odds are 50%, but no longer for the reasons I stated in #40. I agree with Mark Sowul, #68.

†Posted by Ken

Mark (comment 68): You can consider GB and BG the same outcome, but then the possible outcomes are not equally likely. Among two-child families, “one of each” is twice as common as “two boys.” In many situations, it’s simpler to find the likelihood of each different sequence of events, even if you have to add some of the probabilities up to answer the original question.

†Posted by Steve Kass

Well it seems question 1 is ambiguous, depending on how the reader interprets the question. So I can’t state that 1/3 is wrong.

Anyway, http://www.wiskit.com/marilyn/boys.html

†Posted by Mark Sowul

There is indeed a flaw in Monty Hall. Flaw which everyone seems to ignore.

If you _chose_ door #1 then he should open door #1, not #2 or #3. Opening another door just makes you doubt your initial choice.

You are thus forced to re-evaluate your decision even though you don’t have any new substantial data upon which you could base that decision.

Assuming your initial choice was good, asking you to make another choice actually reduces your chances of winning the game.

Think of the Monty Hall game this way — would you prefer to pull the trigger in the game of Russian Roulette once or twice?

As for boy or girl, what are the chances of another child being identical twin or a hermaphrodite? Nobody mentioned that.

As for those monkeys, how about picking green M&M’s just because green is the primary color which is detected with more photo-receptors than the other two primaries used in that completely flawed test?

It is sad that taxpayer’s money goes into such poorly thought out “studies”.

†Posted by Igor Levicki

Thank you Kenneth (post #39). You’re the first one to clearly and adequately explain for me why the best decision is to “switch” in the original 3-door brain teaser. Of course, now I feel stupid (yet enlightened), because I was among the readers muttering the author of the article was an idiot.

†Posted by Saty

MarkSowul - #1 is pretty unambiguously phrased.

A lot of people are introducing additional assumptions into the problem (like assuming it was the first child), but that isn’t warranted by the problem statement.

GB and BG are not the same event, just ask the children whose sex would be changed if you picked one instead of the other.

Igor - that’s not a flaw, that’s just the rules of the game. And the Monty Hall problem as described isn’t psychological at all. You could replace Monty with a computer and it wouldn’t make any difference. And you do have new substantial data - you now know what’s behind one of the 2 doors you didn’t pick.

†Posted by Dave

#1. The four outcomes are BB, BG, GB and GG, each with a theoretical 25% probability. Just because we know that one child was a boy doesn’t change the probability that having two boys is 25%.

Now, if the question were worded “the FIRST child was a boy, what are the odds the second is a boy” it would be 50%. But it’s “the OTHER” child, and thus we must leave all options over. Just because GG didn’t happen doesn’t mean it is eliminated as a possible outcome. It’s just an outcome that didn’t happen.

Now… Monty Hall is simple for those who don’t believe it.

You are given the choice of 3 doors. You pick one. What if, at that point, Monty asks: “can I trade you my two doors for your 1?” you’d say yes. The odds are in your favor.

This is what he is doing when he opens one door and then offers you the other. He’s still offering you BOTH doors. Take them!

Deal or No Deal works the same way. You always, always have to assume that your suit case DOESN’T have a million dollars. The act of choosing it makes the odds stink for you. So you continue to play the game until the offer from the bank is one you would be happy with, take it and call it a day. Don’t ever doubt yourself on this. And even if it’s down to 2 suitcases, yours and one other, take the deal. Odds are 19/20 that you don’t have the million, but odds are 100% that you’ll get the money the bank offers you… ;)

†Posted by Mikey

BJ wondered why his/her results lagged the expected 67% winning percentage.

This is basically a binomial test with each trial having a 67% chance of success, essentially a special coin that comes up heads 2/3 and tails 1/3.

Given 20 trials, the probability that you would be at exactly 45% cumulative is about 2.5%, and the probability that one falls short of the expected 2/3 success rate is about 52.1%. (BJ must have been mistaken about being at 44% at that point; after 20 trials, that is just not possible.)

After 90 trials, the probability of having 54 successes (or 60%) is about 3.6%, and the probability that one falls short of the expected 60 is about 45.1%.

After 100 trials, the probability of having 64 successes (or 64%) is about 7.1%, and the probability that one falls short of the expected 2/3 is about 48.1%.

Had BJ repeated this experiment several times, i.e., run several dozen 100-game trials, the results from each experiment would begin to take on an approximately normal distribution. There is nothing particularly special or troubling about BJ’s result.

†Posted by Patrick

On the Monty Hall choice, you really have a 50% chance of choosing the correct door if you DON’T switch, and, indeed you have a 66.6% chance if you DO.

By choosing one of the three closed doors, you’re really only choosing one of two doors, because you know Monty will eliminate the second “goat” door for you, thus increasing your odds from the start.

†Posted by Eric P

Dealing with problem 1, I do not understand why people are looking at BG and GB as distinct alternatives. While it would quite naturally matter, in real life, which child were older it seems to be insignificant in this problem.

Unfortunately this particular phrasing of the problem seems to invite controversy. I believe an adequate rephrasing of the intended problem would be that two marbles are selected simultaneously or in sequence by a perfect randomization procedure to be either blue or green from a pool composed of an infinite number of identical green marbles and an infinite number of identical blue marbles. You are informed that one of these two marbles is blue and asked to guess at the color of the second.

As can be seen in this example, BG and GB are not distinct choices and further the selection process for one marble does not affect that of the second. The chance that any individual marble, or child in problem 1, is a girl or boy is determined independently and is therefore always 1/2.

Provided that the gender of each of Mr. Smith’s children is determined independently of that his other children this holds equally true if Mr. Smith has a thousand children (or marbles) and 999 were known to be boys. The 1000th child would still have a 1/2 chance of being a boy or girl (or a blue or green marble).

On the other hand the result would be far different if the position of the whole were determined such that each outcome were equally possible: there would be a 1/1000 possibility of the last element of an all B set being G.

Proper representation of the possibilities in problem 1 should be:

Position 1: B Position 2: B or G

The idea that there is somehow a choice between BB/GG/GB and, if the way the possibilities are written matters, BG states is a mistake.

This problem does change significantly when asked to guess the probability of gender/color for multiple positions. In other words, if one knew the gender of neither child (or the color of neither marble), there would be a 1/4 chance of two boys (blues), 1/4 chance of two girls (greens), and 1/2 chance of a girl and a boy/boy and girl (green and a blue/blue and a green). Again, order of birth (selection) should not be regarded as important.

If order of birth (selection) matters there is always a 1/2 chance that any nth child (marble) is a boy (blue) and a 1/2 chance of being a girl (green).

†Posted by Francis

i just don t get what all this fuss is about. These problems are just (elegantly) repackaged elementary Bayesian statistics problems. Or do i miss something? Can somebody help?

†Posted by S.E.H.

John Tierney, are you trying to be the M.C. Escher of grammar?

Question #1 should be rephrased one of two ways:

“Mr. Smith has two children, one of whom is a boy. What is the probability that the other is a boy?” (answer: 1/2)

or

“Mr. Smith has two children, at least one of whom is a boy. What is the probability that both children are boys?” (answer: 1/3)

This resolves the grammatical nightmare created when Mr. Tierney asks us about “the other” child without establishing who the first child is.

†Posted by Gilad

Question #1 is ill-formed. Here’s why:

When you use the word “the,” you are implicitly claiming that there is only one of the thing that “the” modifies. So when you talk about “the other child,” you are implying that there is one child to which that phrase refers. However, in the first part of the question, you say that there might be two girls. In that case, there is no clear referent for “the other child.”

We might rephrase the question this way: At least one of the two Smith children is a boy. What are the odds that both of them are boys? Now the question is easy to answer. There are four equally probable cases: Both are girls, only the older child is a boy, only the younger child is a boy, or both are boys. We’ve ruled out the first case, so the odds of the fourth case are 1/3.

Or we might phrase the question this way: At least one of the Smith children is a boy. Suppose we pick a random Smith boy (there might be only one, but we can still pick him randomly :-) ). What are the odds that the other one is a boy? Again, we have the four equally probably cases, and we’ve ruled out one. In only one of the other three cases is the other child a boy, so the odds are still 1/3.

In other words, even though I can see two significantly different ways of interpreting the question, those two ways happen to have the same answer.

But that still doesn’t excuse asking an ill-formed question.

†Posted by Andrew Koenig

Submitting my answer before I read anything!

1) 75% If child A is a girl, then 100% chance of B being a boy.If child A is a boy, 50/50 chance of the other being a boy.

2) Swap 1/10 chance of choosing the car right away. Remove 7 goats and there’s a 90% chance that one of the two others is the car. Swap to one of the other two, 45% chance each.

3) 50% 50% chance of getting a car 2/3 of the time. 50% chance of getting a car 1/3 of the time.

†Posted by Darshanand

Help!!

Mr. Smith HAS two children. Their sex is already determined. If at least one is a boy, as we are told, then the only permutations are BB and BG. In the scenario as presented, how is GB statistically relevant?

†Posted by Mark

I read the explanations over and over and none of them made sense to me. Here is my explanation to the Monty Hall game:

From the start, you have a choice of 3 doors: A, B or C, one of which has the car. Assume A and C are the goat and B is the car.

Let’s keep track of the score for switching and not switching in each possible situation.

If you pick A, Monty takes away C and let’s you switch. If you switch, you get the car; otherwise, you get the goat. Switch: 1. Don’t switch: 0.

If you pick B, Monty takes away A or C and let’s you switch. If you switch, you get the goat; otherwise, you get the car. Switch: 1. Don’t switch: 1.

If you pick C, Monty takes away A and let’s you switch. If you switch, you get the car; otherwise, you get the goat. Switch: 2. Don’t switch: 1.

So, if you take all of the possible situations (there are 3), in 2 of those situations switching will get you the car, whereas in only 1 situation will not switching get you the car. Therefore, switching will be successful 2/3 of the time.

So my way of understanding it is that when you start, you have more opportunities for picking the wrong answer and, therefore, more opportunities for picking the right answer later when a goat is taken away and you can switch.

†Posted by Mark

Depending on how you interpret #1, the answer could be 1/3, 2/3, or 1/2. The reason why is because if there are two boys, then there are two “other” children who are eligible to be boys, not just one “other” child in the case of girl/boy. #24 hit it right on the head by rephrasing it unambiguously. If we accept that interpretation of the question, the answer is obviously 1/3.

†Posted by luke

#58) The enveloppes problem. Not sure about the first one. Do we assume the amount is integer?

If you count the money in your enveloppe and it is an odd number, then you must switch, as it can’t be the the “double” of anything.

If we don’t assume the amount is an integer, let’s call the minimum amount A.

We have 2 enveloppes, A and 2A. Pick one randomly (expected value 1.5 A) Counting the amount does not give any new information if we do not assume an integer amount.

Switching as two possible outcomes: I Lose 2A and win A (-A) 50% I lose A and win 2A (+A) 50%

Switching doesn’t change anything.

For question 2, let’s assume we’re talking about integers and the amount is random.

Count your amount: if it’s an Odd number, you know you picked A, and you must switch. if it’s an even number, don’t switch.

If A is an odd number (50%), your expected value following this strategy is 2A

If A is an even number (50%), the strategy doesn’t change anything and your EV is 1.5A

Total EV of the strategy is 1.75 A (slight improvement)

†Posted by Pierre Pinguet

1. I don’t know. I keep thinking 50%. 2. Doesn’t matter anymore. You’ll always have 67% chance of getting a goat. You’ve just reduced it to the first step of the Monty Hall problem. 3. Don’t remember how to do that one.

†Posted by Tek

Ooh… My guess for #1 is 33% - there’re four ways you can have two kids. B/B G/G B/G G/B We know it’s not G/G. There are three other possibilities, one of which is B/B.

†Posted by Tek

1. 1/2. The fact one is a boy doesn’t affect whether ‘the other’ is a boy.

2. Switch. (1/10, 9/20, 9/20)

3. 1/2 odds to win the car, no matter what the original probability distribution is.

†Posted by Sunny Wei Zhao

How about this variation of the monty hall?

http://magpietangent.blogspot.com/2008/04/nhl-playoffs- variation-to-monty-hall.html

†Posted by mt

1) The probability the other child is a boy is 50%. The argument of BB, GB, BG is flawed. Let’s name the first boy Fred. There are four possibilities: Fred, Boy; Boy, Fred; Fred, Girl; Girl, Fred 2) Your original door will have a 10% success rate, the other two will have 45% success rates. The explanation is long and covered in previous posts anyway. 3) You would get the car (1/3)*.5+(2/3)*.5 percent of the time by flipping a coin, which comes out to 50%. You would get the car 2/3 of the time by switching doors.

†Posted by Rob

Here is a neat little twist to question 1 that I heard in a class once.

First of all, I believe that the answer to Q1 is 1/3. But what if the question is changed to this:

Mr. Smith has two children, one of whom is a boy named John. What is the probability that the other is a boy?

(Assume they are not twins, and assume that if the other child is a boy, he is NOT named John)

My answer: 1/2. Yes, changing the question simply by giving the boy a name, John, changes the answer from 1/3 to 1/2!

†Posted by mt

I understand the logic of the 1-5-8 problem being the same as the original Monty proposition: 1/10 remains 1/10. BUT I can’t get my mind around this (paradox, maybe):

After he opens the seven dummies, you have 5 and 8 remaining. There are only three possibilities: 5 has the car or 8 has the car, or neither. Therefore 1/3. Where am I going wrong?

†Posted by Donald Kahn

The two events are mutually exclusive. The fact that the first baby was a boy has no bearing on the gender of the second baby. Thus the probability of the second baby being a boy is 50%. It’s the same as roulette and the reason casinos put the board up to show the previous results. If someone sees a lot of red had previously come up they will bet on black thinking they have a greater chance when in reality the have the same chance as another red showing up. The previous result does not affect the future/current result.

†Posted by Joe Taliuaga

It seems the best way to overcome the ‘counter-intuitive’ nature of the problem is to realize that because Monty always opens a ‘losing’ door he avoids giving the contestant any information additional to what was known at the outset. The contestant already knew that, no matter if he picked a winner or loser, at least one other door would also be a loser. When Monty reveals a goat behind a door the contestant did not choose, he is only reiterating what was already known.

†Posted by Tom S

After going back and forth on the Mr. Smith problem, I think the answer you’re looking for is 1/3, but I don’t like the question.

The difference is in how the question is phrased. Since we don’t know whether the son is Mr. Smith’s older or younger child, the possibilities appear to be BB, GB, and BG — 33%.

If the son was defined as the 1) older or 2) younger child, the possibilities would be 1) BB/BG or b) GB/BB — 50%.

On the other hand, we know that the son must be either the younger or older child… and the outcome of both those scenarios are the same — 50/50. So I’m going with 50%, because it makes me feel better.

†Posted by Aaron

For the Monty Hall problem dissenters, the bottom line is this:

When you made your first choice, you probably chose a goat, since there are two goats and one car. That’s why it will always be better to switch; because, 2/3 of the time, you’ll choose a goat to start.

If you just disregard the doors and the opening of the door, which really has no bearing on the problem, it makes more sense.

It’s like the lottery, or poker, or blackjack. Most people see those as games of chance, not games of skill. Most of the time for most people, it will be a losing proposition.

However, there are some people with the skill to increase their odds.

Most people would say this is not true with the lottery. What if I told you there was a way to very slightly increase your odds in some lotteries?

Any game like Powerball (or other smaller individual state games) which has an odd number of balls can give certain players an advantage.

If anyone can post why, you’ll be super awesome.

†Posted by Jake

As is often the case, a problem properly defined is half the solution.

Let’s rephrase the question in an equivalent form: “Would you rather choose a door before or after Monty opens a losing door?”

If you choose before the door is opened (and stick with your choice), your chance of being correct is 1/3.

If you wait until he opens a goat door to make your choice, then you have 1/2 chance of being correct. You abandon your original 1/3 probability, switching to a 1/2 probability.

†Posted by Kash

Boy/girl question: 50-50. BG and GB are the same outcome for this proposition, so assuming there are 4 distinct outcomes (GG, GB, BG, BB) is incorrect. There are only 3 possible outcomes, one of which (GG) is eliminated, leaving BB and BG/GB.

†Posted by Scott Graham

The answer to Question #1 is “it depends”. The problem with the answer “1 out of 3″ is Monty Hall all over again!

We need additional information to answer definitively. Specifically we need to know how the speaker knows his information.

Suppose the extra information was something like : “>> Mr. Smith has two children, at least one of whom is a boy. I know because I was there when his first son was born. My neighbor told me he had a second child but didn’t specify the sex”

In this case, the odds would clearly be 50/50 for the other child being a boy.

†Posted by Barry Levinson

And to clarify on my last comment, I’m saying that birth order doesn’t matter, but even if you account for it, you should still get an answer of 50%.

†Posted by JasonK

In response to Igor: Your analogy to Russian Roulette depends entirely on the number of loaded versus empty chambers.

I think the most common mistake people make in justifying sticking with their original choice is that they believe that the choice between the 2 doors is independent of their original choice between 3 doors which is simply not the case.

†Posted by Ben

I understand the Monty Hall problem, but there has always been something that I was unsure of. That is, I think that one should switch ONLY if the person who reveals the goat knows that there was a goat in there. In other words, your chances of winning increasing is dependent on the person purposely revealing the goat. For example, say I chose door number 1. If lets say I am allowed to randomly choose another one of the two doors and it contains a goat, does that still mean I should switch? I think in this case my chances are 50/50. Any ideas? Sorry if this isn’t appropriate in this article.

†Posted by martintxo

Can I ask a slightly different vaiation on the Monty Hall question? Update it to apply to Deal or no Deal. It’s the final decision… there are 2 cases left, one worth 1 million and one worth a dollar. Do you switch or keep your original choice? This seems a similiar situation and I feel one would do best to switch. The method of getting to the final result differs, but I suspect the odds are the same.

†Posted by Dale Taylor

There is a 1/3 chance that you chose the car. There is a 2/3 chance the car is behind a door you have not chosen. Monty opens one door revealing a goat. Knowing your original odds are 1 in 3, and knowing that the other 2 doors have 2 of 3, you switch to the remaining door as it has 2 of 3 chance of having the car. The fact that you now know Monty’s door has a goat just forces you to choose the remaining door with the 2/3 chance of having the car. Remembering that your unchosen doors have a 2/3 chance of having the car. Monty turning over a goat forces you to switch to the door set with the 2/3 chance. At this point - the set of doors with the 2/3 chance includes just one door.

†Posted by Mark Stevens

With respect to problem #1, I have to agree with the people above that are pointing out that there is no difference between (BG) and (GB). The question did not state anything about birth order, so (BG) and (GB) are the same, one boy, one girl. Thus, the options are (BG) and (BB), so the probability that the other child is a boy is 50%.

†Posted by The Tim

1) There are four possibilities with two children (ignoring the transgendered, of course):

BB, BG, GB, GG (yes, you have to have both BG and GB in there, they’re referring to different kids).

We are told that GG is not true, so the remaining ones are equally probable. It’s a 1 in 3 chance.

2) We have a 10% chance for our original guess, but a 90% that one of the remaining two are correct. The remaining two have the same chance of being correct (it doesn’t matter which door numbers he leaves behind unless one of them is fixed and the other is not). So the remaining two split that 90% chance evenly and we have a 45% chance with either of them.

3) I saw someone answer 1/6. This is half right, but you forgot to add the other chance! Both doors COULD be right, though they’re not as probable. We have a 50% chance of a 1/3 and a 50% chance of a 2/3 chance. So 1/6 + 1/3 = 1/2. Yup, we’re back to 50-50 by flipping a coin. Better to take the 66% (2 in 3) chance by changing.

Of course, this means that you shouldn’t challenge him for money, you’d both come out even, statistically…

Fun exercise, though. It’s been ages since my last class on game theory :)

†Posted by Joe

Another good probabilistic problem is the one involving rare disease testing. Assume a test for a disease has a 99.9% accuracy rating. Assume that you are among a population of 1 million that are tested for this disease. Assume that roughly 1,000 people in this population have the disease. What is the probability that you have the disease if you test positive? The surprising answer is %50. The number of people falsely identified by the test is the same as the number of people who actually have it.

†Posted by Jon Flerchinger

The odds are 10% that the car is behind door 1, and 90% that it is behind door 5 or 8. 90%/2 = 45%, which is much better than 10%. You should switch to door 5 or 8.

†Posted by Darel Rex Finley

Reply to David W. (16):

50%

If the first (random) kid we see is a boy, the the possibility of girl,boy is eliminated and we are left with boy,girl and boy,boy

Reply to Cww (19):

Every strategy will give you 10% chance of winning. Monty doesn’t add any information to the game.

†Posted by alex roemcke

1) 75%

2) Switch - 10%, 45%,45%

3) 50%

†Posted by Chris MacDonald

I’ll bet you my next drink that I can flip a quarter 10 times and get 9 heads, how about it?

If you take my bet, you’ll be buying me a drink.

RE: Coin toss

What would you say the odds of flipping a quarter are? Almost everyone says 50/50.

I can PROVE experimentally that it is not 50/50 under average circumstances.

I define average as standing in an open area, no wind, and doing a standard thumb flip with open palm catch.

Now, based on that situation, I can force the statistical outcome to follow a 90/10 split on a large sample. (100 or more flips)

First, you need a quarter. Flip it a few times. Get used to catching it. Now, instead of catching it at random, make sure that your catch hand is at the same height your hand was when you flipped the coin.

If you can get within a small variance of the right height, you can manipulate the outcome. Staying at EXACTLY the same height for flip and catch means that you can approach 100%.

Going a small amount up or down should reverse the outcome. (1/2 the height of a quarter on it’s edge)

In simple terms, the rotational velocity of the coin is constant and therefore if you catch at the same level as you flipped from (equal up and down arcs) then the outcome is predictably opposite the starting position.

†Posted by zac

For question one: 33%

The question is deceptive, however, because it depends on the nuanced way the question is asked. Consider this alternative:

Mr. Smith has two children. The first born was a boy. What is the probability that the second born was a boy?

Then, both GG and GB are removed, leaving the BB and BG options — so the more intuitive 50%

Verses the question as stated, which only eliminates GG — so 33%

†Posted by Forestgomp

I’m blown away by how many people think that boy/girl and girl/boy are the same thing, and therefore the probability is 50%. It’s true that boy/girl and girl/boy are the same, but the point is that there’s two different ways to get one girl and one boy, but only one way to get two boys, so girl/boy is twice as likely as boy/boy. So the answer is 1/3.

†Posted by luke

The chances for a boy or a girl might be influenced by nation of origin, proximity to power lines, sun spot activity, RGBH in milk. Furthermore, one cannot limit the options to male and female. Some of my best friends could be hermaphrodites.

†Posted by Dr. Park Rogers

For those of you who’ve got correct answers to all the above (including the two envelopes and the one where you get to choose whether Monty randomly opens a door), here’s another one: Do boys or girls have more sisters on average?

†Posted by Chris

1. 1/3 assuming boy:girl ratio is 1:1. There are 4 equally likely choices in the choice tree BB, BG, GB, and GG. Three of those have at least one boy, and only one has two boys.

2. 4.5/10 or 45% so I would switch. There is a 90% chance it is behind one of the two doors left and they’re as likely as one another, because at the beginning your chances were 10% of picking the car.

3. 50/50 odds assuming a fair coin. half the time you will get 1/3 odds, the other half of the time you will get 2/3 odds. 1/3*.5+2/3*.5 = 1/2

†Posted by Madison M.

In response to query 1. The answer is based on information not present in the question.

Armenia : 58.5% chance that it will be a male child. Virgin Islands : 53% chance that it will be a male child Faroe Islands : 50/50 China : 56% chance that it will be male. USA : 52.5% chance that it will be male. http://en.wikipedia.org/wiki/List_of_countries_by_sex_r atio

†Posted by Zac

1) The 1/3 answers count the same answer twice; BG and GB are identical in terms of the question given, since age does not play a role. If you want to consider an older sister and a younger sister two separate options, you need to do the same with brothers, so your choices become (using lower case for younger and upper for elder): 1) gB 2) bG 3) bB 4) Bb

This, again, leaves you with 50% chance that the other is a boy.

†Posted by Bob Bobson

Actually to be precise the correct answer for #1 is It Depends.

In order for the 1/3 answer to be correct, the revelation of the fact that one child is a boy must be forced and not random.

Let’s assume I move in across the street from the Smiths. My real estate agent tells me the Smiths have two children. While getting the mail one evening, I meet a young man who tells me he and his family just moved in.

I now know that Mr Smith has two children and at least one of them is a boy. There is NOT a 1/3 chance that his other child is a boy.

Again, for that to be the case, the information that was revealed must have been selective, not random.

So without additional information, I have to argue that 50% is also a correct answer to #1. In the MH scenarios, there is no such problem, since we know the process by which information is revealed (it is that very process that creates the wrinkle in the problem). Everyone is making a leap regarding the information in #1, but to be precise, such a leap is not necessarily justified by the information provided. This would be an exceptionally bad test question.

†Posted by ARS

It’s 50% the other is a boy. It’s like asking me to flip a coin. A second coin has already been flipped. It’s glued to the table. It’s Heads. There’s a 50% chance the first coin is flipped Heads. Order doesn’t matter. It’s like looking at both coins from the north side of the table and then looking at it from the south side. The vantage point won’t influence the outcome.

†Posted by Toddius Zho

Well, the only interesting problem here is the two child problem. Its amusing how many here seem to think 1/3 is the correct answer. Its 50% (disgregarding the biological fact that chances of having a male child are slightly larger). If the order were given, everybody would surely agree at first glance. “Our first child is a son. Is the other (unborn or not) a boy”? 50% Obviously. Independantly “This is our second child, a boy. Is the younger child a boy?” Obviously this is also 50%. If i only see one of two children, the chances of seeing the younger child is x (and seeing the older child is 1-x). x*50%+(1-x)*50%=x*50%+50%-x*50%=50%

The problem arguing with the 25% distribution of GG,BG,GB,BB and dropping the GG case doesnt create a 33% probability for the other cases. By gaining the additional information “theres a boy in the familiy” the probabilities are now 50%,25% and 25% for BB,BG and GB respectivly.

†Posted by Syren Baran

1/2 - the sex of the second child is no way related to the sex of the first.

9/10 - 1/10 it was behind the first, so 9/10 it was behind one of the others.

1/3 - half the time you stand a 1/3 chance, the other half a 2/3 chance.

†Posted by David Irwin

1)1/3 2)45% 4)50%

†Posted by Josh L.

1. already stated above, but it’s 1/3. 2. 45% 3. 50%

In “Deal or No Deal”, if you pick a case and get to the end and you still have the million dollar case on the board, should you switch?

†Posted by Jake

Question 1 is 50%. Here’s why: Replace the kids with two marbles in a bowl. Marbles can be either red or green. If one marble is red, what’s the probability the other is green? Answer: 50%. The reason is that there is no differentiation between R/G and G/R, and the probabilities for either marble are not tied together. You can even rephrase it differently: if you cover one marble, what’s the chance that the other marble is green? Again, it’s 50%, because the color of one marble has no impact on the color of the other marble.

The difference between this problem and the door problem is that the probabilities of one kid being a boy is not tied to the probability of the other kid being a boy. In the door problem, the probability of the doors having the prize are all tied together. That’s why adding information about one door changes the overall probability of the other doors hiding the prize.

†Posted by Rene

You all are missing a point.

Monty Hall problem is not logical in the first place. The only reason to temp you with another door was to lengthen the game or lower the cost of game.

Since he has knowledge of the game and what is where, he can offer and reoffer so that switching or not is always in the house’s interest.

†Posted by Jack

#1 is interesting. Most readers seem to see a 1/3 chance for a second boy based on the premise that the possibilities are BB BG GB GG. However I think the introduction of sequencing is wrong so BG, GB is the same. When you have 2 kids the possibilities are: - 2 boys - 1 boy, 1 girl - 2 girls Since we know 2 girls is ruled out we have 2 choices left, which makes the odds 50%. (This of course ruling out the imbalances of probability introduced by mother nature). :-) stw

Do we have an official answer somewhere?

†Posted by Stephan H. Wissel

I dunno what’s wrong here but for #1 the answer is 1/2 (that’s 50%). since when GB is different from BG? GB and BG means he has 2 kids. there’s nothing to differentiate between BG and GB (like age, IQ or whatever). so there are only 2 outcomes possible BB and BG. out of 2 only 1 is good (that’s BB) so there you go: 1/2 chance the other one’s a boy.

for those who still don’t get it please continue this explanation and when you get stuck you’ll see what I mean: “BB means one is a Boy and one is a Boy” “BG means one is a Boy and one is a Girl” “GB means [..continue here..]”

#2: 1/10 for door 1 that’s for sure. for the other 2 doors 4.5/10 seems pretty accurate.

#3: tossing a coin brings down by 2 your chances of winning so 2/6 if you switch and 1/6 if you don’t. it may not sound right but it is: you waste half your chances so the other 3/6 (that’s 1/2) are gone

†Posted by amnezick

For all those people who are trying to calculate the answer to (1) using combinations of girls and boys, think carefully about what you’re listing as choices:

1) boy, boy 2) boy, girl 3) girl, boy 4) girl, girl

What do you notice about choices (2) and (3)? That’s right, they’re identical. Since the question does not say anything about, for example, which sibling is older, or any other way of ordering the two children, having “one boy and one girl” is exactly the same situation as having “one girl and one boy”. Thus for any father of two children, one of whom is a boy, there are exactly two choices:

1) Two boys 2) One boy and one girl (in any order you like)

Only one of those two situations includes two boys, therefore the probability is 1/2.

Another way to think about this: the gender mystery of one child has already been eliminated, therefore the question is only asking the probability for the gender of one remaining child, which is just the probability of any given person being male (about 50%).

†Posted by Marcus Polus

About the difference between #3 and the classic Monty Hall problem. Suppose that you enter the room after the door has been opened. Obviously, the 2 doors are the same to you and you chances of being right are 50:50, same as a coin flip. Now how could it be 1/3:2/3 for someone who was in the room before? What information are you missing?

Once you realize that a door was locked by the contestant’s choice and that you see that Monty’s choice wasn’t free, then you’ll understand why you may be wrong, even if you don’t believe it. Cognitive dissonance is hard to dispel.

About #2, let’s reverse the question. What are the chances all 9 other doors have goats? And when I open 7 of them to show you, will the car magically move to another door?

About #1: “Do you have kids?” “Yes, 2″ “Any daughters?” “No”. This dialog sets the same situation without involving any mathematical meaning.

So, strictly speaking: 1/3; 9/10; 1/2. But some people will try to interpret the question to match the answer they want to give.

†Posted by SLC

In 129 i miss posted that it’s 1/3, it is 1/2 for the same reason that’s listed.

†Posted by David Irwin

I think the reason so many people got it, is because a lot of people were introduced to the Monty Hall problem by the TV series “Numbers” about a year ago.

†Posted by Russ

Response to #116 - It is fine to say “If the first kid is a boy” but then you must remember to include when the first child is a *girl*.

†Posted by Kamal

Re: #80 - Ambiguity

Dave,

There is still some unclarity in the problem statement. The problem, to me, comes from the word _other_ in the question.

1) Mr. Smith has two kids. At least one is a boy. What is the probability that the _other_ kid is a boy?

2) Smith has two kids. At least one is a boy. What is the probability that _both_ kids are boys?

To me, these are not the same question. #1 is looking at the kids separately, where #2 looks at them collectively. Let’s answer them.

Solution 1 - “Other” X1 is gender of child 1. X2 is gender of child 2. Gender of siblings are assumed to be independent. Then: P(X2 = B | X1 = B) = 1/2 P(X1 = B | X2 = B) = 1/2 The question asks about the gender of the _other_ child, which I have shown above is a boy with prob 1/2, regardless of any assumption about birth order.

Solution 2 - “Both” X is the number of boys among Smith’s children. P( X = 2 | X >=1 ) = 1/3 (by the BB,BG,GB,GG argument.)

So, depending on your interpretation, the answer changes. The two different solutions go with two different problems.

I admit that this is getting very particular in the wording of the problem. Probability questions often hinge upon their wording, so I don’t think it’s unreasonable to accept both solutions as long as it is made clear by the person working it out what exactly is being answered.

†Posted by Brendan

Yeah, the people claiming #1 is 33% are falling for the Gambler’s fallacy. If you flip a coin 20 times in a row and they are all heads, what are the chances that the next flip will be heads?

Or to phrase it like the child one: I have 21 flipped coins. 20 have landed heads, what side is the last.

†Posted by BA

1. 1/2 The question is identical to, “Smith has one child. What is the chance that that child is a boy?”.

2. By staying with the original, the chance would be 1/10, by switching the chance would be 1/2.

3. It is the average between the chance of winning by staying (1/3) and the chance of winning by switching (1/2). I think that comes out to 5/12. Not as good as always switching (1/2).

Goats are smarter than sheep.

†Posted by Gordon Fogus

Edit: 2. 1/10 for staying, 2/3 for switching. 3. the average between 1/3 and 2/3 = 1/2.

†Posted by Gordon Fogus

re:#16 After actually seeing a boy, the chances that they are both boys is 50%. Given three possible families of BG, GB, and BB, there are four one-child samples that are boys, and two of those samples come from a family of two boys. This is different than just knowing that there is a boy in the family because that is a chosen sample, not a random sample.

re:#19 Yes, there is a strategy with better than 10% success. Obviously, opening the door you originally chose can’t be successful more than 10% of the time, but intuitively the more information you have before switching, the better. Monty’s door opening isn’t completely random–he won’t open the door you have selected, so you want him opening doors that have the least probability of being the car. After he opens one door, the door you hold has a 10% chance of being the car, and each of the eight other doors has a (9/10)/8 or 11.25% chance of being the car. So you switch, and then have Monty randomly pick another door. You then switch, have him open a door, switch, and so on until Monty has opened eight doors. At which point you switch and open that door. I don’t feel like doing the math to get to that point, but if you don’t bother switching until Monty has opened eight doors, then there is an 80% chance that Monty has found the car before you did and a 10% chance that the door you open will be the goat. Since of the 20% of the time you get to open the door, you find the car 90% of the time, your overall chances of getting the car are 18%. But if you switch every time your chances will be even better.

†Posted by curril

1) A vast majority of the posters above have chosen 1/3rd as their answer. I’m going to go ahead and be the black sheep here:

Everyone has assumed that child order actually matters, and thus the four choices are BB, BG, GB, and GG. The problem is, the question is clearly worded as a combination, not a permutation. It’s not specifying *which* child’s probability we’re after, just “a” child’s. Thus there are only three possible outcomes. The wording below might clear it up for those in doubt:

Option 1: We have 1 boy and another boy Option 2: We have 1 girl and another girl Option 3: We have 1 boy and 1 girl.

There is no fourth option. Thus the probability that “the other” is a boy is 75%.

2) Probability of original guess being right: 1/10. Probability of the remaining doors Monty knows: 9/10. Thus each of the two mystery doors not opened by Monty are 45%.

3) The coin flip problem doesn’t interest me. It’s just 50/50.

†Posted by Andypro

I see a lot of you saying 1/3 for the sex of the second child being a boy. This is incorrect. Unlike the other scenarios, these two events are independent of each other. You can’t use the outcome of the first child to calculate odds for the second one. Just like if you flip a coin and get heads, that doesn’t make it any more likely that you will get tails the next time. The odds of the next coin flip being heads or tails is still 50%.

So, the odds that the second child is also a boy is 50% (or i think it’s closer to 48% in actuality)

†Posted by Jeff

Jack: What knowledge of the game can he possibly leverage? He can’t choose the door with the car (to show you), and he can’t make you choose the door with the car originally. If you don’t originally choose the car (which happens 2/3 of the time if you just choose randomly), he has NO option but to choose the other door with the goat. So, you switch and win 2/3 of the time. You lose when you pick the car originally, which only happens 1/3 of the time.

Rene, sorry, but that is incorrect. This is a simple application of Bayes Formula:

P(A|B) = P(B|A)*P(A) / P(B) P(BB|1 or 2 boys) = P(1 or 2 boys|2 boys)*P(BB) / P(1 or 2 boys) = (1)*(1/4) / (3/4) = 1/3

You are confused about the concept of mathematical independence. We are concerned with sets of 2 marbles (in your example) - not single marbles. Yes, the probability of the next marble, taken alone, to be R or G is 1/2 for both. But if we accept what you said, we’ve ignored the possibility of a GG combination (you have said that P(RG) = P(RR) = 1/2, meaning P(RG or RR) = 1 and P(not RG or not RR) = P(GG) = 0).

†Posted by Casey

I see a lot of 1/3 and 50/50 answers for #1. Maybe I am missing something, but shouldn’t the answer to be 75%??

†Posted by Confused

Reply to alex roemcke (116):

You’re right - every strategy has the same 10% chance of success, but the statement “Monty doesn’t add any information to the game” is a bit misleading. The opening of any door with a goat does add information - all the remaining doors are (equally) more likely to hide the car - but at a cost (risk that the car is behind the opened door, ending the game in a loss) exactly equal to the advantage gained by the new position.

†Posted by Mark Jackson

Looks like I was wrong. You can treat the BG and GB combination as identical, but then you have to assign different probabilities: 0.25 for BB, 0.5 for BG|GB and 0.25 for GG. Same thing for my marble example. Removing the GG probability means that there’s a 2 to 1 probability that the BG|GB scenario is true - which means it’s a 2/3 probability to 1/3 for BB.

Well, I learned something new today.

†Posted by Rene

The best way I’ve found to convince someone of the Monty Hall solution is to consider an extreme variant of the game. I secretly choose any real number, say in this case (Pi-3)/11. You then make a guess, say 0.4. I then tell you something like “The number is either 0.4 or (Pi-3)/11.” At that point I’m pretty sure nobody would think they had a 50/50 chance that 0.4 was actually the number I chose.

A chain of related games can then be played to establish the relationship between this and the original game. Try choosing an integer from 1 to 1,000,000, then an integer from 1 to 100, then an integer from 1 to 3, which is equivalent to the original problem. For the person to balk only when accepting the final game they have to consciously break their mental schema of the continuum of games, but there isn’t a strong enough stimulus to justify this mental step (i.e., reducing from 1..4 to 1..3).

†Posted by Mark

Sample Space = { BB, BG, GB, GG }

P( BB | BG or GB or BB) = P( BB and (BG or GB or BB) ) / P( BG or GB or BB ) = P( BB ) / ( P(BG) + P(GB) + P(BB) ) = (1/4) / (3/4) = 1/3

ps. thanks for pointing out publicly what us probabilists (ie, mathematicians) have known for years: psychology is crap since they don’t understand the numbers they get

†Posted by some guy

#1 is badly written; but I think the answer is 50%. I think “at least” is what’s throwing off a lot of people. The questions is not what are the odds of both children being boys; rather. It identifies one child as a boy (child A) then asks the odds of the “other child” (child B) being a boy. The odds of child B being a boy is independent of the sex of the child A hence 50%. It’s like flipping a coin. The odds of you flipping heads on any individual toss is 50% regardless of the outcome of any previous tosses. On the other hand, the odds of you flipping 2 heads in a row on your next 2 tosses is 25%.

†Posted by rayson

Ahhh, on the drive home realized that the poor wording of question 1 indeed does not matter.

Dang, it is 1/3 the other will be a boy. You can pick either one child or the other to present first as “the boy”. This allows you to switch the order and constitutes extra information. It has to do with the difference of saying “the first child is a boy” vs. “one of the two is a boy”.

†Posted by Matt

The first question has never been worded satisfactorally in all the years I’ve seen it.

Here, I fixes it:

1) I have a pool of 100 families, each with 2 kids. I choose a family at random, and find that one child is a boy. What are the odds that his sibling is a boy?

2) I have a pool of 100 families, each with 2 kids. I choose a boy at random. What are the odds his sibling is a boy?

Now 4 sum credit, ahh…

†Posted by Owen R

You’re not being fair to Scott with regard to his probability analysis. Scott’s assumption is that if Monty can or could swap the prize between the closed doors after he had opened the goat door, AND did so randomly, so that the player has an equal chance of the prize being behind either closed door.

Your assertion is that no such swap is allowed. 1) IF a swap happens randomly, Scott’s analysis of the probability is correct. 2) IF If no swap ever happens, your analysis of the probability is correct. 3) IF a swap can happen but it is non-random, neither analysis is correct and you’d have to analyse the outcomes to try to model the swapping process to predict the probabilities.

However, even if Scott’s assertion is true (the swap can happen), AND the swap is random, the switch strategy is just as valid as tossing a coin, IF the position of goat and prize are random. If not, you’re back to modelling.

†Posted by Willlllllll

Problem 1 is whats really interesting. If given this problem in a vacuum, statistically the information provides for 1/3 as many have pointed out.

Alas, we don’t live in a vacuum, and there is plenty of real world data that shows the vacuum statistical answer is useless.

To be useful we would really need to consider male/female birth ratio which can be skewed even slightly off the purely vacuum statistical answer by any or all of the following: -Birth order of known boy -Number/order of siblings/sexes -family tree history of child sex ratio -socioeconomics of family -genetic predispositions -time in history (born in 20BC or 2 years ago) And oh yeah…… -method of conception hahaha

I’m sure there are other skewing factors that make the vacuum statistical answer inaccurate in real world application.

†Posted by Bryan

As others have pointed out, the child question is a bad question. “Ask Marilyn” asked this question a few years ago, and Marilyn stuck to her guns about it being 1/3 even though I (and several people I know of) wrote to her explaining it was a bad question. She published some disagreeing responses, but not any which properly explained the conundrum.

It is socially irresponsible for popular sites (and even more so for “Ask Marilyn”) to publish questions like this, with the “correct” answer designed to confuse people. It either makes lay people think they could never understand math (when, in fact, their intuitive understanding of the answer to this question is ABSOLUTELY CORRECT for the manner in which they are likely to encounter the question, e.g. two mothers meeting at a park, one mother sees the other’s boy playing and a baby in a stroller, and inquires as to the gender of the baby), or it confirms their suspicions that educated people are, well, stupid. Neither of these is a useful outcome.

†Posted by Patrick Maupin

Sorry Jack, you’re wrong. Opening the goat door can only increase your information about the system, and improve your odds of getting the prize. Even if the prize can be swapped between the closed doors after you’re shown the goat, you’re still down from 1/3 to 1/2 odds.

The odds get longer again if Monty knows where the prize is and is skilled enough to make you choose the wrong door (whichever one it is) without making it obvious.

†Posted by Willlllllll

1. Based on the way you defined this question, the probability of the *other* child being a boy is independent of the fact that one of the two children *is* a boy. So, it’s whatever the rate of boy vs. girl births is. If you wanted to turn this into the Monty Haul problem, you should have said, “What is the probability that if you randomly choose one child, that it’s a boy?”

2. This is actually an intermediate step to explaining the Monty Haul problem to people who “don’t get it.” The way I explain it it to use the million-door example. It goes like this: Say, instead of three doors, there are a million. You get to pick one door. Then Monty opens the other 999,998 doors (man, that’s a lot of goats…). So, either you picked the right door (one in a million), or the other door contains the car (999,999 out of a million). Once you use a sufficiently large number of doors, the solution becomes painfully obvious.

But in the 10 door example, with 2 doors remaining instead of 1, there’s a 1/10 chance you picked the right one and 9/10 chance you didn’t, and that one of the remaining doors contains the car. The way you worded the question, the answer is that “either of the doors” has a 9/10 chance that it contains the car. If you meant to say, “What is the chance that one of those two doors contains the car?” then you just divide that in half, and its

(9/10)*(1/2) = 9/20 9/20 = 0.45

3. There are six possible outcomes:

Original Door Coin Toss Outcome ————- ——— ——- Car Switch Goat Car Keep Car Goat1 Switch Car Goat1 Keep Goat Goat2 Switch Car Goat2 Keep Goat

The interesting thing, is that this extends to the n-door (where n>3) versions as well. By using a coin toss to choose to keep or switch, you always have a 50% chance to win. Now, while this may seem marginally worse that always switching in the 3-door version, think about the million-door version: coin toss gives you 500,000/1,000,000 but always switching gives you 999,999/1,000,000.

†Posted by Duane

It seems that the first question is confuddling lots of people, so let me clear it up:

Say you’re planning on having 2 kids. What is the probability of having two boys in a row? Obviously, these are the outcomes (assuming roughly equal chance of having a boy or girl):

Kid 1 Kid 2 —– —– Boy Boy Boy Girl Girl Boy Girl Girl

So, 1/4.

But this is not the question asked in problem 1. The question is *not* “What are the odds of having 2 boys in a row?” Think of the unknown-gender child as unborn. Now, let’s go back to the above table. We can eliminate the possibilities of the first child being a girl, so that leaves us with:

Kid 1 Kid 2 —– —– Boy Boy Boy Girl

So, 50%.

†Posted by Duane

1. a bit more than 50% (depending on the ratio of boys to girls) 2. switch 3. 5/12

†Posted by AdrianTM

If you randomly select from a group of families with two children, at least one of whom is a boy you will have a 1/3 chance that it is a older boy/younger boy family, a 1/3 chance that it is a older girl/younger boy family and a 1/3 chance that it will be an older boy/younger girl family.

The reason people keep picking 50/50 odds is they aren’t taking into account that you have actually already selected out the girl/girl families and that the boy could be either the younger or the older sibling. The odds of having 2 kids, at least one of them being a boy is 75%. If you narrowed the question to ‘is the younger child a boy’… oh, now my head hurts…

†Posted by Nathan Smith

#1) Oh come on, If you know that child “A” is a boy, the OTHER child (”B”) has about a 50% chance of being a boy. If “B” is the known boy, then there is still about a 50% chance that the OTHER is a boy. The fact that one child is a boy has no effect on the sex of the other.

†Posted by Dave C.

People are overthinking #1, I think. Let’s apply #2 to it for a minute. Mr. Smith has ten children, at least nine of whom are boys. What is the probability that the tenth one is a boy? Using the logic most people are presenting, in this example, the probability the tenth one is a boy is 10%? Therefore, a 90% likelihood that it’s a girl? No. It doesn’t work that way. If he has 19 boys, does the probability drop to 5%? This problem is not mathematically the same as the Monty Hall problem. You don’t get to swap the second child for a third child of the opposite gender. Hermaphrodite/intersexed people notwithstanding, there are only two possibilities, boy or girl. Ergo, it’s 50/50. Everyone has the right idea for the other two problems.

†Posted by Anne Nonymous

1) 1/3rd

Create a random set of 4,000,000 families that each have 2 children and you will end up with approximately 1,000,000 families with two girls, 2,000,000 families with one boy and 1,000,000 families with two boys.

Now take those 3,000,000 families that have two children with at least one being a boy.

Ask your self how many of them have two boys and it is clear that there are 1,000,000.

Which leads back to the correct answer to problem #1 of 1/3rd odds that both children will be boys.

—————-

Now let’s imagine that you are very ambitious and go ring the doorbell of each of those 3,000,000 families that have two children with at least one being a boy.

In each case the door is answered by a son named Fred.

What are the odds that Fred has a brother?

1/3rd

You didn’t randomly select the person answering the door, Fred (a boy) always answered.

———

Now let’s say that in each of case of the 3,000,000 families the door was answered by a daughter named Jill. What are the odds that Jill has a brother?

Those who argue that the answer to #1 is 50% because the sex of the 2nd sibling is independent of the first must use similar logic and declare that Jill only has a brother 50% as well.

Even though it is obvious that the odds that Jill has a brother is 100%.

†Posted by William Burr

Rene,

Let’s see if I can clear up your confusion. You have a giant bowl of marbles, 1/2 red and 1/2 green.

You randomly select two marbles.

There is a 1/4th chance of getting two green 1/4th of getting two red 1/2 of getting one of each

Do it 1000 times and keep track of the results.

Now ignore all the times you got two greens (two girls), about 250 times.

You would have two reds about 250 times You would have one of each about 500 times

Given only the cases where you have at least one red marble (one boy) what are the odds that you have two red marbles?

750 cases where you picked at least one red marble and 250 cases where you picked two red marbles

1/3

Or, to phrase it like problem #1, if you randomly select two marbles (have two children) out of your pool of 1/2 red, 1/2 green marbles and at least one of the marbles you picked is red (a boy) what are the odds that the other marble is red (a boy)?

1/3

†Posted by William Burr

People are right to say that #1 (Mr. Smith’s boy[s?]) is ambiguous. The essence of the ambiguity is this: Did you find out that Mr. Smith has at least one boy in a way that could have led you to instead find out that he has at least one girl? Or, did you find out in a way that could have instead led you to find out that he does not have any boys? The answer is 1/3 in the latter case, but not in the former. If you could just as easily have learned that it was a girl, then the answer is 1/2. Other people have explained this 1/2 answer using scenarios that involve different ways you could have learned from Mr. Smith (or his family) that he has a son.

There’s another way of demonstrating the ambiguity, looking at the issue of how John Tierney made the decision to pose the challenge in terms of boys rather than girls. Consider these two procedures by which John could have crafted this challenge by selecting the parent, Mr. Smith, out of the population of all adults with exactly two children.

Procedure A: John walks down the street, wanting to find an example person for a neat puzzler. He asks each adult he passes “do you have exactly two children?” If they say yes he asks the follow-up “is at least one of them a boy?” The first person to answer “yes” to both questions becomes the subject of John’s puzzler (that person happens to be named Mr. Smith). In this case, you should bet that there’s a 1/3 chance that Mr. Smith has 2 sons, since only 1/3 of the people who answer “yes” to both questions have two sons (only BB out of BB, BG, and GB).

Procedure B: John walks down the street asking passing adults “do you have exactly two children?” and chooses as the subject of his puzzler the first person who answers “yes” to this question (this person happens to be named Mr. Smith). Then he asks Mr. Smith how many sons and daughters he has. If Mr. Smith has 2 sons, John poses the puzzler exactly as he has above. If Mr. Smith has 2 daughters, John poses the puzzler slightly differently, informing us that at least one of the children is a girl, and asking the probability that the other is a girl. If Mr. Smith has a son and a daughter then John uses some gender-neutral method to decide between the two ways of posing the puzzler (such as randomly choosing between the two formulations or focusing on the gender of Mr. Smith’s elder child). In this case, you should bet that there is a 1/2 chance that Mr. Smith has two children of the same gender, regardless of whether John asks about boys or girls.

One way to see that the answer with Procedure B is 1/2 is that half of the people who answer “yes” to John’s screening question have two different-gendered children, and since the cases where John focuses the puzzler on boys are symmetric to the cases where John asks about girls, there must be a 1/2 chance of two different-gendered children in each case. A more analytical way of seeing the answer is to realize that there is only one way in which Mr. Smith can have 2 sons while there are two ways to have a son and a daughter, but if he has a son and a daughter then there is only a 50% chance that John will ask us about boys. This means that there are four equally likely possible outcomes of this procedure: 1) Mr. Smith has 2 daughters & John asks about girls, 2) Mr. Smith has 2 sons & John asks about boys, 3) Mr. Smith has one of each & John asks about girls, and 4) Mr. Smith has one of each & John asks about boys. By posing the question about boys, John rules out 1) & 3), leaving a 50-50 chance of two boys.

(Note how this fits the format I described in my first paragraph: we the readers had an equal chance of instead finding out Mr. Smith has at least one girl.)

If this seems convoluted or unclear, here’s a simpler way to consider the difference between Procedures A & B. The question, in a nutshell, is this: what would John have done if Mr. Smith had two daughters? Would he have asked about girls instead of boys, or would he have found another person who did have at least one son? The argument for the 1/3 answer assumes that John only asks about boys, so it only holds up if we assume that the latter is true.

The same analysis applies (more clearly, I think) to William Hanisch’s example in #65. In that example, Mister Smith flips two fair coins, shows you that one landed heads, and asks you the chances that both landed heads. The question is, what would he have done if he flipped 2 tails? Would he have flipped again, or abandoned his game? Or, as seems more plausible, would he have shown you one of the tails instead of showing you a head? In the latter case, there is a 1/2 chance that the other coin is heads when he shows heads, unless we know something about how he would choose which coin to show if the coins came up 1 heads and 1 tails.

(To make things more complicated, if you did have some idea of what approach Mister Smith would use in deciding which coin to show you if he flipped a head and a tail, then you could argue for any probability from 1/3 to 1. If you are sure that he will show you a head if at all possible, and only show a tail if forced to by flipping 2 tails, then if you see a head there is a 1/3 chance that both are heads (since there are 3 equally likely live possibilities: HT, TH, and HH). On other other hand, if you are sure that he’d show you a tail if at all possible, then the probability of two heads would be 1 once he’d shown a head, since the fact that he chose to show a head would rule out the HT, TH, and TT possibilities, leaving HH as the only option. And if you suspect that he has a coin-side preference, but are not entirely sure, then your answer would fall somewhere in between. This analysis can also apply to my Procedure B, if John does not necessarily use a gender-neutral procedure to decide whether to pose the puzzler in terms of boys or girls. In that case the four possible outcomes listed a few paragraphs ago would not be equally likely; outcomes 3) and 4) would have different probabilities.)

With the Monty Hall problem, the question retains most of its simplicity and trickiness after eliminating ambiguities (Monty always opens an unchosen door with a goat and then gives the contestant the option of switching to the other unopened door). I don’t think that’s quite as true with the Mr. Smith question, although I invite anyone to prove me wrong by posting a simple, tricky, and unambiguous version of it.

†Posted by Vince

The questions have been answered, but in some of the above comments people referred to some order of the children (first, second, or youngest, oldest). If you specify which is the boy, then the other’s sex is independent (1/2); when the situation is /unordered/ is when the possible states are BG GB BB, giving rise to the 1/3.

†Posted by Andrew H

For the 1st question, conditional probability tells us that the answer is 50%, provided the probabilities of getting a boy for the 2nd child is independent of that of the 1st child.

P(B given A) = P(B and A) / P(A)

P(B and A) = .5 x .5 = .25 P(A) = .5 Hence P(B given A) = .5

Too lazy to think about the other 2 questions..

And to Jack: Did you actually read about the Monty Hall problem? Switching is always NOT in the house’s interest for Monty Hall’s game. Point taken about ‘lengthen the game’, not so for ‘lower the cost of game’ since if there is no offer to switch at all, the expected cost will definitely be lower…

†Posted by Yijin

I know the results have been announced, but… I still say the answer to #1 is 1/2, and here’s why: The “correct” answer (1/3) depends on permutations - ordered pairs. But there is no information in the problem to indicate that the order of the children matters. Without any such information, there are only 3 possible combinations of children - 2 boys, 2 girls, or 1-and-1. Knowing that there is at least 1 boy eliminates the 2 girls choice, leaving only 2; thus there is a 1/2 chance that the other child is a boy. But let’s assume that permutations matter; in which case we should take care of the order of children. There is a 50-50 chance that the boy we already know about is either the older or younger child. If the known boy is older (50%), that leaves 2 possible permutations: (b,g) and (b,b), giving a 50% chance that the other child is a boy. If they known boy is younger, that also leaves 2 choices: (b,b) and (g,b). 50% x 50% = 25%, for both possibilities: 25% + 25% = 50%, so the possibility that the other child is a boy (or girl) remains 50%. If you want the answer to come out to 1/3, you have to specify: ask the possibility that the Older (or younger) child is a girl, and then that’s the answer.

†Posted by Charley

@ 128

You’re really adding far more information than is in the original problem.

Your argument comes in three parts: If the order were given, everybody would surely agree at first glance. 1. “Our first child is a son. Is the other (unborn or not) a boy�? 50% Obviously. 2. Independantly “This is our second child, a boy. Is the younger child a boy?� Obviously this is also 50%. 3. If i only see one of two children, the chances of seeing the younger child is x (and seeing the older child is 1-x). x*50%+(1-x)*50%=x*50%+50%-x*50%=50%

1 and 2 are technically correct. If you see a child and you are told whether it is the first born or second born, it is easy to see that the other child is a boy 50% of the time.

Your point 3 is misstating the original problem. What you are calculating with your equation is the probability of seeing the younger or the older child BUT ALSO being told whether they are oldest or youngest (as per 1 and 2). This is simply not the original question — in that question, you don’t know if the boy is the eldest or the youngest.

We know there are two children, at least one of which is a boy. We don’t know if the boy is the eldest or not, therefore the probability of the other child (whether the boy is younger or older) is 1 in 3. Each outcome of BB, BG, GB is equally likely.

†Posted by Ivan

You write “I think much of the credit goes to my Times colleagues”. Fair credit and all, but I was struck by the intuitive simplicity and force of your own formulation: “But when you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.”

That’s clarity.

(So many of the arguments about Monty Hall come down to just how you describe the reasoning you’re doing, and whether you can get someone to admit that it’s valid.)

†Posted by tikitu

the chance of getting the car in door1 is 1/10…so 9/10 chance of it in either door5 or door9.

†Posted by akao

This thing is as old as the invention of the wheel, better said the invention of the game show.

This is more funny: If tree people are in a room and five go out. How money people have to go in to have an empty room?

†Posted by Jack

Since people seem to be posting answers…

1. 1/3 2. switch (which other door is irrelevant). 9/20 success. 3. 1/2 (1/2 * 1/3 + 1/2 * 2/3 = 3/6 = 1/2)

†Posted by Tim

1.) 1/3, bg, gb, bb

2.)Switch, odds = 1/3. It is in one of the three remaining doors.

3.) 1/2, (2/3 * 1/2 + 1/3 * 1/2).

What is interesting to me about probability is that it is fundamental to common sense reasoning, and yet most people cannot do even the simplest of problems.

How about an inverse problem? Say that man flips a potentially biased two sided coin, and the first flip is heads. Using only that information what probability should you assign to heads showing up in the second flip? I’ll give you a hint, it is not 1/2.

I took that problem from this book Probability the Logic of Science, which is a fantastic book.

†Posted by Jonathan

Too many people have been tricked by #1. The question is designed to be extremely short to have your mind wander.

Known facts. 2 children, 1 is a boy. 50% chance the other is also.

Having kids is essentially flipping a coin. If you have 12 kids, each one of them are boys, the next child will STILL have a 50/50 chance of being a boy.

Having one child or twelve who are already male mean nothing.

†Posted by Mike

1) 50% chance, Chance of the next child being a boy is always 50/50

2) door 1: 10% chance of car, door 5 and 8: 45% chance of car each

3)50%

†Posted by Lisa

People who insist on GB and BG being different forget that in that case there are also two equally distinct BB cases… there are 4 possible cases if you consider it that way:

1) he has an older sister == GB 2) he has an younger sister == BG 3) he has an older brother == BB 4) he has an younger brother == BB

So BB is still 2/4 = 50%, as would be if you took GB==BG.

†Posted by L

TRS - “Given that there are more boys born than girls (the boy survival rate is lower) odds favor a boy as the second child.”

Why????? Each time someone gets pregnant you stand the same chance of giving birth. Plus, why are there more boys born if the boy survival rate is lower??

Made no sense at all.

†Posted by Rob

Jonathan (178)

I think you have underspecified your biased coin problem because I don’t think you can solve it unless you make an assumption about the probability distribution of the biased heads probability P.

In effect, you are asking for the distribution function of the event A = Pr(p

†Posted by Phil Koop

The first problem is flawed, in fact there are two possible answers: 1/4 and 1/2. That depends on whether you know that Mr Smith has a boy after you knew that he has two children or not. The second case is trivial, the answer is 50% (~50% to be pedantic). In the first case you have four possibilities, with probability 1/4: a)gg b)gb c)bg d)bb After the sentence “at least one of whom is a boy”, this is the scenario: 1)the sentence is true (75% of the cases: bg,gb,bb): the probability that the second child is a boy is 1/3. 2)the sentence is false (25% of the cases: gg): the probability that the second child is a boy is 0. So the probability that the second child is a boy is (3/4)*(1/3)=1/4 This problem is quite different from Monty’s one: Monty’s problem is always valid: you’ve got always two goats and a car; this one not: you can’t affirm a priori that you have at least one boy, because in that case the answer is 1/2. In Monty’s problem you make a choice, then you get a new information and then you can choose again; in this problem you have only one chance: a priori is 1/4, a posteriori is 1/2.

Sorry for my bad english.

†Posted by Stefano Guidoni

Years ago when I first thought about the Monty Hall problem, I found the best way was to take it to extremes. !0 doors is helpful, but suppose there are infinite doors. Your probability of picking the one car is zero. After Monty unveils infinity-1 goats, should you switch to the one remaining door? You bet!

†Posted by Tony Russo

The chances of having two boys, in this problem, is 1/3.

Also, I believe the question is NOT ambiguous.

To arrive at the answer of 1/2, you must know if the mentioned son is an older child or a younger child. The question gives us no information on whether he is older or younger: we just know that at least one of two children is a son.

Given we have a two-child family, and we know one child is a boy, we are limited to a world of two-child families with at least one son. There are twice as many of these families with a son and daughter as there are these families with two sons. In other words, in two-child families with at least one son, there is a 2/3 chance their other child is a daughter and only 1/3 chance their other child is a son.

Compared to boy-and-boy, a combination of boy-and-girl is twice as likely.

If we were told that the one son is an older child, then we could restrict our world to two-child families with an older child who is a son. In that case, there are as many boy-then-boy families as there are boy-then-girl families, and the answer would be 1/2.

However, in this problem, there is no ambiguity - we can NOT know that the one child is older or younger, so we must consider three types of families that occur with relatively equal frequency: boy-then-boy families, boy-then-girl families, and girl-then-boy families.

Only 1/3 of these families will have two boys.

†Posted by asr

The fact that one child is male is extraneous. The fact that the phrase “other child” is used to determine which child we are talking about, however, makes it relevant - making two distinct, equally likely possibilities (boy then girl, girl then boy) produce the same result (girl).

†Posted by David Thomas

Reply to L

BG is definitely not the same as GB. However, older brother is equal to younger brother under the parameters of the question (which is important).

†Posted by NP

For number 1, which seems to have confused the most people. Forget for a moment that there are not 2 girls so you could have BG, GB, BB, GG. There’s a 0.25 propbability of a couple having any given combination. What are the probabilities for each of the 4 possible child pairings that you could pick a boy? BG - 0.5; GB - 0.5; BB - 1; GG - 0. This gives overall probablities of a boy coming down the stairs from each pairing of: BG - 0.125; GB - 0.125; BB - 0.25; GG - 0. Given that there is no GG combination, the chances that a boy from BB comes downstairs is 0.25/(0.25+0.125+0.125) = 0.5 The chances of a boy comgin downstairs from GB or BG is therefore also 0.5.

Basically, there may be 2 ways of GB or BG, but there 2 ways of getting a B form BB, which balances them out probability-wise when a B is sighted. If a child was seen coming down the stairs, rather than a boy, this wouldn’t be the case, but that fact makes all the difference.

This is actually the Monty Hall, but with boys as goats and girls as prizes. Imagine that there are 3 children: 2 boys and a girl. 2 of them are Smiths, 1 is a Brown. What are the chances that the Browns have a girl i.e. the Smiths have 2 boys? 1/3. But if you see one of the Smith children is a boy, what are the chances that the browns have a girl? It becomes 1/2.

†Posted by Jonathan Boyd

I haven’t read all of this, and I’m just restricting myself to correcting one question, but even so:

@89: You aren’t told which child is older. Therefore you must still take GB into account.

@97: But you don’t know which boy is older. So the chance of the first two outcomes is 1/2 as much.

@98: Same mistake. The chance of John being the older boy is 1/2, so you need to factor that in. Then you’ll get the right answer.

@105: They are the same outcome, but they are equally likely. Therefore combining them is twice as likely as BB.

@134: Buth there are two ways of getting one boy and one girl, hence twice the chances.

@136: Two choices, but _not_of_equal_probability_.

@179: No, not true. Assume chance of boy is 50%, and that gender of first child doesn’t influence gender of second child. Then we have four possible outcomes for two children:

BB = 25% BG = 25% GB = 25% GG = 25%

Because of the question, we know that there is at least one boy, but we _do_not_know_if_he_is_older_. Therefore we can only eliminate the last option, leaving three equally likely possible outcomes, only one of which has a second boy. Your analysis would be true if we knew that the _older_ child was a boy, as your reasoning shows: “If you have 12 kids, each one of them are boys, the next child will STILL have a 50/50 chance of being a boy.” This is correct, but not quite what the problem was asking.

†Posted by Timotiis

@Dale RE: Deal or No Deal

The difference between “Deal or No Deal” and “Monty Hall” is that in “Monty Hall” the revealed door is not random–Monty MUST open a losing door. In “Deal” the cases being opened are not constrained, no one knows where the $1,000,000 is. Therefore, each selection is a new game. At the end with two cases left it’s 50% to switch and (unless the case values are close together) the best move is to take the deal and go home. – JimFive

†Posted by JimFive

in ref. to #1, the reader comments seem to address two separate questions: Prob (child1 is male given child2 is male) Prob (having two male children)

just wanted to make the distinction that they are separate probabilities and have…well…separate probabilities.

†Posted by Elcheecho

1. How can so many people bring ordering into this problem? It doesn’t ask anything about sequences of boys / girls it asks about distributions. You have two urns labeled B and G and two balls. You’re told one ball is in urn B, what is the probability the other ball is in urn B…

its 50%

†Posted by Devin

#1 is 1/3, but you will think it is 1/2 if you aren’t careful. After explaining this to a friend, here are some common reasons you might think it is 1/2, and what is wrong with those reasons. I apologize if I duplicate any previous explanations; I just wanted to capture all the issues that my friend had run into.

* With no other information, the odds of a boy are 1/2. This leads some to believe that the odds are always 1/2. However, if I tell you truthfully that I am a boy, the odds of me being a boy jump to 1/1. So information affects the odds. In this case we have the information that one child is a boy. This means we either have BB, BG, or GB (GG has been eliminated as an option). Since the choices were restricted by the information, the odds are now 1/3 for BB (1/3 for BG, 1/3 for GB).

* Some people get hung up on whether BG and GB should be considered the same. You can do this either way, but you have to watch your math. The odds of BB, BG, GB, and GG are all 1/4. If you preserve the order, you get 1/3 BB (out of equal BB, BG, and GB). If you decide to ignore the order, that is fine, but remember that now you are working with BB (1/4), BG (2/4), and GG (1/4), so BG is twice as likely as BB, resulting again with 1/3 BB. The common mistake here is to think that by ignoring order you get 3 equally probable options, when that isn’t the case.

* Some people mistakenly read this question as “If your FIRST child is a boy, what are the odds of your SECOND being a boy?”. The answer to that question is 1/2 (since previous results don’t affect subsequent results). However, this question is actually “If ANY of your two children is a boy, what are the odds of the OTHER being a boy?”. The difference is subtle but crucial. This trap is especially common since the FIRST+SECOND version of the question is hammered into us in probability classes, so it distracts us here.

-JD

†Posted by JD

I had the following exchange with John Tierney:

———————–

One reason why this problem causes so much controversy is that the way Monty plays the game is not completely specified, and the details matter (I too have spoken with Monty).

The critical question (usually not specified) is whether Monty _must_ reveal an open door once the contestant has made a choice. If he is constrained in that way, then switching is the optimal strategy.

But in the way the game was actually played, Monty _may_ reveal an open door, or he may not.

In game theory terms, given that Monty is not constrained, he adopts a “mixed strategy,” playing Reveal with some probability and No Reveal with some other probability. The wrinkle is that this probability is different, depending on whether the contestant’s initial guess is correct or not: he should Reveal twice as often when the contestant’s initial guess is correct compared to when it is not.

[In other words, Monty optimally offsets the signalling effect of Revealing, by inducing the contestant to Switch more often in the event that the contestant ought to Stay.]

When Monty adopts this mixed strategy, he ensures himself the payoff (no prize giveaway) at least 2/3 of the time.

In response to Monty’s mixed strategy, the best response for the contestant is to randomize equally (i.e., to also adopt a mixed strategy) between Switching and Staying.

In other words, the optimal strategy is, in fact, a coin flip.

The other wrinkle is that, as long as Monty is not too constrained (meaning that, in this case, he must Reveal less than 2/3 of the time), it doesn’t make any difference what the percentage of Reveals is, as long the ratio is 2:1. If Monty must Reveal more than 2/3 of the time (say, because the ratings people demand more winners), then Switching becomes the dominant strategy for the contestant. Of course, that would completely remove any uncertainty about what the contestant would do, which would make the show much more boring, which would probably reduce ratings.

In the context of the “rational choice” research by Chen and others, this raises the larger question of what one means by “intuition.” An “uneducated” person may intuit that, in a zero-sum game, Monty is not actually trying to help him when he reveals information. That strong intuition turns out to be correct if Monty is free to behave as an adversary.

Thus, the interesting social phenomenon is a bunch of “educated” people trying to convince a bunch of “uneducated” people why their intuition is incorrect, which the “educated” people can prove only by assuming that one of the contestants (Monty) behaves in a way that is suboptimal for him. The “uneducated” person, who merely has street smarts, correctly thinks that that assumption is hooey. —————- To which John responded:

Yes, Monty’s freedom, or lack thereof, to choose is crucial. I wrote about that in my original 1991 article about it, and Monty himself pointed this out. —————– To which I replied:

Since the outcome depends critically on whether and how often Monty has to reveal an open door, I think this should be made an explicit condition of the problem, as well as any explanation for it. Otherwise, people are arguing over (and feeling stupid for not understanding) an indeterminate set of conditions.

John: Yes, that’s why, when I posed the problem, I said that Monty will always open a door to reveal a goat.

†Posted by Jon Putnam

#1. 1/3 “Mr. Smith has two children” Had a boy then a boy = 25% Had a boy then a girl = 25% Had a girl then a boy = 25% Had a boy then a girl = 25% So BB = 25%, BG = 50%, GG = 25%

“at least one of whom is a boy” Remove the GG case ==> BB 33%, BG = 66%

It’s funny how people can so easily be tricked by the Monty Hall problem even after reading an article on how the Monty Hall problem can trick people. Let me rephrase the original problem so that the 50% answers can be right. “Mr. Smith has two children, the eldest one is a boy. What is the probability that the other is a boy?” OR “Mr. Smith has two children, the first one you met is a boy. What is the probability that the other is a boy?”

I will break it down another way: “Mr. Smith has two children, the eldest one is a boy or the yougest one is a boy. What is the probability that the other is a boy?”

Older Boy: 50% having a younger brother

†Posted by Will

Several posters have been fixated on the idea that GB and BG are identical outcomes. Specifically, #162 added the additional information that the boy given in the problem was not only first born, but that the other sibling is not yet born. This is a completely different problem. It’s true that ordering is not mentioned, but there are human beings in the problem, and human beings are ordered by age naturally. It’s implicit that the boy *has* to be either the younger or older sibling. Knowing that, the possibilities of GB and BG really are different, and as such, the probability of having a boy and girl is twice as likely as having two children of the same sex. So you can say “GB and BG are the same event”, but you have to assign that event 50% probability in this case.

It’s also true that the problem can (accurately) be read to not be a question of conditional probabilities at all, and so those posters that say “the sex of your second child is unrelated to the first” are correct. However, anyone who’s taken an intro probability course has probably seen this problem before, and knows what’s *really* being asked.

†Posted by MattM

#181: That’s the answer to the question, “Fred has a sibling. What the probabilities that his sibling is a boy?” That’s not the question that’s being asked here (though some people might quibble with that). In the problem as given, we haven’t actually seen a boy, so in that scenario, there is no bB or Bb, i.e., we can’t know *which* boy was being referred to. Reference some of the other posts about how giving the boy a name changes the probability. You’re right on the money about why this is true: we now have to add in the two viable options that Fred is either older or younger. I hadn’t realized that until I saw your post, so thanks for indirectly clearing that up for me!

†Posted by MattM

“1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?”

This reminds me of the following. A class of 200 students is told, “raise your hand if you have a brother”. A largish number of students raise their hands. They are then told “raise your hand if you have a sister”. The students are shocked because the number of hands which go up this time is obviously much smaller.

This is not a fluke. Why are the numbers so different? Wouldn’t we expect about the same numbers of students to have sisters as have brothers?

†Posted by Richard Weber

@ 58 Ben Weiss…

I can’t believe you slipped the 2 envelope paradox in there you disturbed sick man. It is like watching lambs to the slaughter.

Ben Carr

†Posted by Ben Carr

Regarding the question posed by the commenter in post #19: This problem you posed doesn’t lend itself to a “intuitive” answer, rather you have to calculate the odds. I believe the answer is to have monty keep opening doors until you either lose (he shows you the car) or he has opened 8 doors. At this point, you ought to switch to the other unopened door and take your chances (which I think are 60%) given that Monty has opened 8 doors and shown you eight goats. If you have him open one more door, you either will reveal the last goat (you win) or the car (you lose) each with probability 50%

†Posted by Jeff

Hi guys. Already posted #135 but will correct some things I got carried on at first sight of the problem. After carefully reading the problem and a comment that shed some light on some thoughts I unwillingly must have “firewalled” I have come to the conclusion that 1/2 is definetely the answer because:

BB, BG, GB, GG are irrelevant. Quote: “What is the probability that the other is a boy?”. Logics tell me that the quantifier “the other” refers to 1 (ONE) child. So now we have 1 child that can be either a BOY or a GIRL. We know the first one is a BOY but it has no relevance whatsoever on what gender the other one might be. So now that we’ve made it clear there’s only one child involved we assign it two possible outcomes: BOY or GIRL (we’re not considering recent results on cow-human hybrids). 1 outcome, 2 options => 1/2 change that one of the options will make it so there’s 1/2 chance the we have a BOY on our hands.

John, or whoever posted please shed some light since it’s your post.

†Posted by amnezick

#1 is indeed 50% as worded. In order to get the 33% answer, you need to make the assumption that the question is the same as “of all 2 child families with one boy, how many have two boys”. This is NOT the same as the question asked.

The question that was asked comes in two logical parts: (1) one child is a boy, and (2) knowing that, what is the other child’s sex?

By pure logic, section (1) is irrelevant to (2), and immediately leads to an answer of 50%.

If you want to run this out with the 4 possible ordered pairs (MM,MF,FM,FF), you need to be sure you’re actually using said pairs correctly.

There are two ways to go about this:

1) Ordered pairing: Most people assume this means MF = male is older, female is younger. However, the order should be considered not the age of the children (as this is irrelevant), but rather the ORDER IN WHICH YOU LEARN INFORMATION. In this case, you learned information about one child, and remove the cases where it does not match (FM,FF), leaving you with 2 equally likely outcomes (MM,MF) - 50%.

2) Unordered pairings: In the case of unordered pairings, you MUST assume that your information could apply to EITHER half of any possible pairings. As such, you have one piece of information which could apply to either the first or second child. As such, you get a 50% chance that it means the older child (leaving MF,MM), and a 50% chance that it means the younger child (MM,FM). This leaves 4 possibilities (MF,MM,MM,FM) - once again 50%.

†Posted by Andrew R.

I’m not doing this work for you. You’re just trying to get me to do your work for you!!!

†Posted by Mohib

For #1, we are talking about the father. If you took all the fathers in the world who have exactly 2 children, you will find that 1/3 of those fathers have 2 boys, and the other 2/3 have a boy and a girl.

However, if you talk to one of all the boys in the world who have exactly one sibling, 1/2 of the boys will have a sister and 1/2 will have a brother. That’s because the 2-boy families contribute twice as many boys to the pool as the boy-girl families.

Look at it another way: there are 100 fathers, each with 2 children. If you eliminate the 25 fathers who have two girls, there are 75 fathers left. 25 of them have 2 boys, and the 50 have one of each. But, now look at the children. There are 200 of them. The 25 BB families contribute 50 boys to the mix. The 50 one-of-each families contribute another 50. Ignore the all-girl families. That’s 50 boys who have a brother and 50 who have a sister.

Since the problem is about Mr Smith, the father, and not young Fred, the answer is 1/3.

†Posted by DR

To everyone who says #1 is 50%, a question:

Why do you think that 2 boys will occur as often as 1 boy and 1 girl?

It’s like flipping a coin twice. Is the probability that two heads will come up the same as the probability that one head and one tail will show up? Not at all - as long as you don’t use a trick, you’ll get a head and a tail about 50% of the time… meanwhile you’ll only get 2 heads 25% of the time.

The difference comes when you realize that the only way to get two boys if for the first to be a boy and the second to be a boy. Meanwhile, to get a boy and a girl, we have two cases: (1) the first is a boy and the second is a girl (2) the first is a girl and the second is a boy Hence even though we end with the same result, there are more ways to get there.

†Posted by Mo

#3: Switching gives a 2/3 chance of winning; not switching gives a 1/3 chance, as explained in the article. So, flipping a coin between these two just averages 33% and 67% to give a 50/50 chance on the final choice.

†Posted by Greg Conquest

The older child is a boy: 50% The younger child is a boy: 50% One of the 2 children is a boy: 75%

When the older child is a boy, the younger is a boy: 50% - independent When the younger child is a boy, the older is a boy: 50% - independent When one of the 2 children is a boy, the other is a boy: 33% - correlated

The sex of the other child is not independent, because “one of the 2 children” isn’t an actual kid.

If you cannot believe you are wrong, I suggest you look at your neighbors, family and friend. Between your children, your siblings, your in-laws, … you can come up with 20+ families. Pick the eldest 2 children of each and you’ll get close to 33%, give or take 5% for real-life sex variations.

†Posted by SLC

Oops! Edit my above post #205; the first paragraph has a sentence which should read:

If you took all the fathers in the world who have exactly 2 children OF WHICH ONE IS A BOY, you will find…

†Posted by DR

In response to # 206:

Not true. The odds of any flip of a coin coming up heads are 50% (.5). The probability of the first flip coming heads and the second flip coming heads is .5 X .5 = .25 HH, HT, TT, TH all have the same odds. The events are independent of each other. The outcome of the first flip does not affect the second flip. The same is true of children. Having a boy does not change the 50% odds of having a boy or a girl, so the odds that the second child is a boy is still 50%. There are only two possibilities for the second child… boy or girl… both are equally likely, so the odds of child #2 being a boy are 50%.

†Posted by Jeff

There’s an issue I’ve never seen addressed in discussions of the door (or other similar) problems. If you switch your choice then, yes, you’ll win a lot more often IF you play the game repeatedly. But if you get to play it only once, then why is what would happen if you played it repeatedly of any relevance whatever? In other words, why should long-run relative frequencies apply to a single unrepeated event?

†Posted by Frank Williams

Let’s rephrase the first problem in a way that the BB, BG, GB, GG people will understand:

Consider “b” to be the child whose gender we ALREADY know. Therefor, the possibilities REALLY are:

bB, Bb, Gb, bG, and the GG/GG that cannot happen.

We are therefor left with bB, Bb, Gb, and bG, leaving a 50% chance the other child is male (assuming a coin-flip chance determines gender). If we accept that birth order is relevant for BG/GB, then we must accept birth order for Bb/bB. If we DON’T accept birth order as relevant, then we have the possibilities of:

Bb, Gb

Which still gives us a 50% chance of the other child being a boy.

†Posted by Sam

#210, the issue of dependence of events can be tricky.

There are two assumptions implicit in the question: 1. The probability that a child is a boy is 1/2, and that it is a girl is also 1/2. 2. The sex of a child is independent from the sex of any other child.

[Two events, say A and B, are independent exactly when P(A|B) = P(A). In words this means that the probability of A occuring given that B is known to have occured, is the same as the probability of A occuring.]

Lets refer to Mr. Smith’s children as child 1 and child 2.

2) implies, by definition of independent, that the probability that child 1 is a boy given that the sex of child 2 is known, is equal to the probability that child 1 is a boy (which by the first point is 1/2).

So, if the question were to state: child 2 is a boy; what is the probability that child 1 is also a boy? Then the previous paragraph tells us that the answer is 1/2.

However, the question states something different: at least one of the children is a boy. Lets write E for this event. Then the probability of E is 3/4. Write F for the event that the other child is a boy. These events, E and F, are -not- independent of each other. We can see this by showing that P(F|E) is not the same as P(E).

Indeed, the probability of F given E (which also answers the original question) is, by definition of conditional probability, P(F and E)/P(E). But F and E both occuring is the same as having two boys, which has probability 1/4. So the conditional probability is (1/4)/(3/4) = 1/3, and this is not the same as P(E) = 3/4.

I’m sorry that I don’t have a precise explanation that also does not involve formulas.

†Posted by Sasha

Thank you Sam, You saved me the time of writing this out. Well put!

50/50

If BG and GB is viable then so is Bb and bB

†Posted by Vyoletblu

can you apply this to “deal or no deal” and why so many people lose the money or stay in the game too long?

†Posted by Adrian VanEss

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John Tierney always wanted to be a scientist but went into journalism because its peer-review process was a great deal easier to sneak through. Now a columnist for the Science Times section, Tierney previously wrote columns for the Op-Ed page, the Metro section and the Times Magazine. Before that he covered science for magazines like Discover, Hippocrates and Science 86.

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