### Puzzle Corner

With a hat tip to the mathematician John Baez, who in turn tips his hat to the science fiction author Greg Egan, who in turn credits the journalist Alex Bellos, who got this from the puzzle designer/collector Gary Foshee (who seems to have no website):

(For those who want more precision: We gather all those women in the world who have exactly two children, tell each of them to “go home unless you have a boy born on a Tuesday”, and select a woman randomly from those who remain. Assume that births are equally likely to occur on any day of the week, and that on any given day, boys and girls are equally likely.)

#### 58 Responses to “Puzzle Corner”

1. 1 1 EricK

There are 14 possibilities for each child (boy born on Monday, girl born on Monday, boy born on Tuesday, girl born on Tuesday etc). So imagine a 14×14 grid of all pairs of children. 27 of them have a boy born on a Tuesday (14 from one column, 14 from a row, but 1 is double counted). And of those 13 have 2 boys (7 from the column, 7 from the row, but again 1 is double counted).

This method easily generalises to cultures with weeks of different length, and even to species with more than 2 possible genders (assuming the probability of each is the same)

2. 2 2 ErikR

I’m not sure that your rephrasing of the problem is equivalent. The original phrasing states “one of whom is a boy born on a Tuesday”. One might imply “exactly one”, so the other child is NOT a boy born on a Tuesday. In which case I think the answer is 12/26.

If one implies one or two (as your rephrasing seems to assume), then EricK’s answer looks correct.

3. 3 3 Jeff Semel

EricK’s answer looks right to me. As an aside, Greg Egan (whom Steve tipped his hat to above) wrote a wonderful short story called “Luminous,” about the more-than-Platonic reality of mathematical truths. I was reminded of this story when I first read Steve’s position on the nature of mathematics in The Big Questions, the book.

4. 4 4 Sam

I saw a largish amount of comments regarding this on reddit, with many proclaiming 1 in 3 (*BB*, BG, GB), some saying 1/2 (BB,BG) and the others responding with the official answer of 13/28 (as mentioned above).

My feeling is that this is a wording confusion.

I say, 1 in 2, because no two boys are the same == (B + (G/B) is 50/50). The ‘double counting’ adjustment removes any uniqueness between the two boys both born on the Tuesday. I conceptualise the both-boy/both-Tuesday occurrence as B1B2 or B2B1, each distinct from each other, in that we know one is a boy but not which one so both instances need to be counted. If understood like this 13/28 -> 14/28 or 1/2.
That is, gender is linked to unique people and not treated like a red or blue coin returned to a selection pot after each pick.

The ‘unintuitive’ thing with this problem (so I’ve read), is that the more specifiers added the question, the more the probability asymptotically approaches 1 in 2. So, what is ordinarily a 1/3 situation becomes 13/28 with a day ‘property’ added; and, along with gender and day, adding a hour-born property will bring the probability closer again to 1 in 2 (335/672) than 13/28.

Mind you, I haven’t done probability, so I probably missing something.

5. 5 5 Al

I’ve encountered this problem before and every time I think about it I get more confused.

On the face of it EricK’s answer looks right to me, but then thinking about the apparently simpler problem of:

“A woman has two children, one of whom is a boy. What is the probability that they are both boys?”

In this case there are only 4 possible combinations of children: B,B B,G G,B or G,G and the 4th of these options has been eliminated. This would lead to the conclusion of the probability of both children being boys as 1/3.

What I don’t understand is HOW the probability changes by knowing the woman has a son born on a Tuesday.

Every child has to have been born on some day of the week; so even if you don’t know which day the boy was born on, we could simply assign the characters A, B, C, D, E, F, G to days of the week, assume the child was born on day A and then deduce A = Tuesday if we ever learn which particular day of the week he was born on.

The day of the week a child was born on can’t alter the probability that their sibling is a brother rather than a sister.

6. 6 6 Ben

@Al:

The probability ONLY changes when you learn something new.

That’s because probability is about knowledge, not about the physical universe itself (I’m ignoring quantum mechanics here).

For example:

I take a pack of cards, remove the jokers, and shuffle it well. I look at the top card, and ask you: “What is the probability that the top card is a spade?” The answer is 1/4.

Then I show you the card. I ask you “what is the probability that the top card is a spade?” You will not answer 1/4 – because now you know. You will answer either 1 or 0, depending on whether it was a spade.

The top card has not changed – it is still the same one I saw. Your knowledge changed.

That’s because probabilities are about knowledge, i.e. about what we know of the world, not about the world itself, which continues to exist regardless of whether we know about it.

7. 7 7 William

When Tim Harford did a segment on this question for “More or Less”, they phrased it in a way that made the answer unambiguous.

“All the people in the world are standing up. An announcement is made: “Please sit down if you are not the mother of exactly two children.” Then: “Please sit down if you are not the mother of at least one boy.” Finally, “Please sit down if you do not have a son born on a Tuesday.” From this group of women, you will get the answer you are looking for, 13/27

To answer this question, I need to know how you found this woman and how you chose “boy” and “Tuesday” as the important factors. You could have just said to any mother of two, “write on this piece of paper the gender and weekday of birth of one of your children,” which wouldn’t affect the probability at all.

8. 8 8 Martin

The answer is simply 50%. The question asks what the odds are that given that she already has a boy, the second child is a boy. The fact that she however already has a boy is irrelevant.

Think of this problem as a roulette-table without the zero or doublezero in it. What are the odds that the next number will be black given that the previous number was black? The fact that the previous number was black, that a \$100 was bet, won or lost, that it was tuesday 5 pm or friday 7 pm is irrelevant. The problem though is that the question is phrased in such a way to imply that it is dependent, although it is independent.

What matter is the sex of the next child and that is independent from all these factors (in the question that is).

That’s my 2 cents.

9. 9 9 Martin

To put in it in the terms of BB BG GG GB is confusing the issue. We already know that there is B? and the ? can either be G or B, so we are picking between BB and BG.

10. 10 10 Harold

I have heard this before, and it is was then also boy born on Tuesday, not any of the other possible combinations. This still confuses me, so forgive me if I work it through a bit.

From Al, if we say “A woman has 2 children, one of whom is a boy”, then the probability of the them both being boys is 1/3. If asked to give a quick answer, I would probably have said 1/2. I would have been confusing the question with the un-asked question: “If a woman has a son, what is the chance that her second child will also be a son?” The original question is asked in an unusual way.

Using EricK’s method, we have a 2×2 grid, giving 4 with “boy” (2 from row and 2 from column), but one is double counted, leaving 3. Of these, 1 has 2 boys in it (one from row, one from column, but 1 is double counted) Giving 1/3.

If we add in more information, say one child is a boy born in the first half of the year (2 divisions). Using the grid system we have a 4 x 4 grid, 7 have a boy born in 1st half of year. Of these, 3 have 2 boys, giving 3/7. If we say first 1/3 of year, we have a 6 x 6 grid, and an answer of 5/11. For 1/4 of a year, we have 7/15. For each additional piece of information we add 2 to the numerator and 4 to the denominator. Probability = (2x-1)/4x-1) where x is the number of “divisions”. So for 7 days of the week we get 13/27. So if we say one is a boy born between 12:00 and 12:01, we get probability of both boys of 2879/5759, or 0.4999.

Mathematically, this makes sense. I don’t yet see “why” it makes sense. How does knowing more specific details of the time of birth change the probability?

If we gather all those women with 2 children, the probability of selecting a woman with 2 boys from these is 1 in 3. If we send home all those without a boy born in the first half of the year, we will be sending away more girls than boys, because there are more girl / boy combinations than 2 boy combinations. The more specific the question, the closer we are getting to the “un-asked” question: “if a woman has one boy, what is the probability that the second will be a boy”. The answer to this is of course 1/2. I think I have it clearer in my mind now.

11. 11 11 Steve Landsburg

Harold:

The more specific the question, the closer we are getting to the “un-asked” question: “if a woman has one boy, what is the probability that the second will be a boy”. The answer to this is of course 1/2.

Yes. You’ve got it exactly.

12. 12 12 Alan Gunn

Some people have a fairly strong preference for having children of both sexes (at least in the US; likely not in some cultures). If that’s the case, someone whose first two children are boys (or girls) would be more likely to have another child, in hopes of getting one of a different sex, than someone whose first two kids are one of each. Since we are looking at a sample of people with two and only two children, one of whom is a boy, wouldn’t that group (if Americans, anyway) be made up disproportionately of people whose other child is a girl?

I realize that taking this into consideration takes the mathematicsl fun out of the problem.

13. 13 13 Steve Landsburg

Alan Gunn: Agreed on all counts (that this would change the answer, and that it takes away the mathematical fun).

14. 14 14 Steve Landsburg

Martin: Your analysis would be correct if we were told that her *first* child is a boy. But that’s not what we’re told. We’re told that *at least one* child is a boy, which makes it a different problem.

15. 15 15 Steve Landsburg

William: You could have just said to any mother of two, “write on this piece of paper the gender and weekday of birth of one of your children,” which wouldn’t affect the probability at all. On the contrary, it would have eliminated all mothers of two girls, and therefore changed the probability. I agree that the *way* in which it changes the probability depends on how you got the information, which is why I added the clarifying information in the original post.

16. 16 16 Michael

The day of the week isn’t relevant to the question asked–it’s noise added to obfuscate. If we assume equal probability between child being a boy or a girl, then the answer is 1/2 * 1 (we know with 100% certainty that one is a boy) = 1/2.

I’m not sure I understand why the questions is difficult. My math skills aren’t that rusty, are they?

17. 17 17 thedifferentphil

@Alan: the math fun can be restored by altering the setup as follows:

“We gather all those women in the world who have two or more children, tell each of them to “go home unless one of your first two children is a boy born on a Tuesday”, and select a woman randomly from those who remain and ask whether the other of her first two children is also a boy.”

That correction means that whether you stop having kids or go on, we only examine the first two. The decision to continue with a third or fourth child has no impact on the sex distribution of first births.

18. 18 18 Henry

The permutations are “two boys born on Tuesday”, “a boy born on a Tuesday and a born not born on a Tuesday”, “a boy not born on a Tuesday and a boy born on a Tuesday” a “boy born on a Tuesday and a girl” and a “girl and a boy born on a Tuesday”.

Let:

T = probability of having a boy born on a Tuesday (1/14)
B = probability of having a boy (1/2)
Thus, B-T = probability of a boy not born on Tuesday (6/14)
G = probability of having a girl (1/2)

Pr(Both are boys) = (TT + T(B-T) + (B-T)T)/(TT + T(B-T) + (B-T)T + TG + GT)

((1/14)*(1/14) + (1/14)*(6/14) + (1/14)*(6/14))/((1/14)*(1/14) + (1/14)*(6/14) + (1/14)*(6/14) + (1/14)*(1/2) + (1/14)*(1/2))

0.06633/0.13776

= 0.48148

Note that we can simplify the above expression by dividing by T:

(2B – T) = (2B – T + 2G)

We can also use the above formula to find generic solutions, if T = “the probability of a boy being born given a certain independent qualifier”. For example, for probability 0.5 (e.g. “the probability of a baby being born a boy on any day of the week), we can see the answer is 1/3, the same for the original problem without qualifiers. As T approaches 0, we can see the answer approaches 1/2, which is the same as the girl named Florida or when we know the identity of the first boy.

Why does this happen? Well, one way to think of it is the “boy and a boy” combination gets two opportunities to have a “Tuesday boy”, whereas the “girl and a boy” and “boy and a girl” permutations only have one opportunity each. Another is that it introduces a “third type”, so just like there can be “GB and BG”, there can now be “BT and TB”.

If the qualifier is very common (say, one child is a boy with two legs), then “boy with two legs and boy with two legs” makes up almost all the probability of there being two boys. Close to 1/4 of all families will have two boys who each have two legs. Close to 1/4 of all families with two children including a boy will have their older child a girl and the younger a boy with two legs, and close to 1/4 of all families will have the older child a boy with two legs and the younger child a girl. Thus, around 1/3 of all families with a boy with two legs will have two boys.

However, if the qualifier is very rare (such as a boy with less than two legs), then there are now two permutations of “boy with two legs and boy with less than two legs” which come to dominate the probability that that both children are boys (since two boys who both have less than two legs is extremely unlikely). Since these are about as likely as the two “girl and a boy with less than two legs” permutations, the probability that both children are boys are about 1/2.

The answer is difficult to find intuitive because we don’t really deal with conditional probabilities like these. We see the answer and think something like “Hey, I was born on Tuesday, how could that possibly affect the probability of my sibling’s sex?” But of course, it doesn’t. However, it’s unlikely that you’ll ever hear anyone say “I have two children and at least one boy.” The only exception would be if one was a fetus in utero, in which case we can now eliminate the possibility of the first child being a girl.

19. 19 19 Martin

Mr. Landsburg, I don’t think I understand the difference between my analysis and that of Harold.

I don’t see whether it really matters for my analysis that the boy is born first or second. For the roulettetable it is not relevant whether we ask if 1 = black, then 2 = ? or that we ask if 2 = black then 1 = ?. The probability remains 1/2.

Same goes for ?B vs B?. I don’t see what time or the sequence of events has to do with it here. The probability seems to me to be independent of any of the information provided.

I don’t really see my error here. I take it though that the explanation of this post’s question will be the subject of a subsequent post.

20. 20 20 William

@Steven:

You could have just said to any mother of two, “write on this piece of paper the gender and weekday of birth of one of your children,” which wouldn’t affect the probability at all. On the contrary, it would have eliminated all mothers of two girls, and therefore changed the probability.

My grasp of Monty Hall was never too firm, so let me make sure I’m understanding correctly. Here’s the situation: You find a mother of two. You say, “write on this slip of paper the gender and weekday of birth of one of your child. I don’t care which one.” You get the slip. It says, “[gender], [weekday]“.

You then ask yourself, “What are the odds that the child the mother wasn’t describing on this slip of paper is a boy?” Are you telling me the answer is different depending on whether [gender] was “boy” or “girl”?

21. 21 21 Ken B

Whenever it seems puzzling go back to the population space of the sample. This is what EricK did in the first (and I think correct) response.

The number you get looks funny. What happened? Why did the Tuesday affect things? Because you actually changed the “shape” of the sample space by adding the Tuesday caveat. You excluded mothers with daughters born on Tuesday but sons on another day. That changes the sample space. (And this is that extra bit of information you have: the mother in NOT a son-friday/daughter-tuesday mother.)

22. 22 22 Douglas Bennett

Michael-

The day of the week does matter, as does the method for weeding out the families. After gathering all mothers of 2 children, asking the mothers of only girls to leave leaves 1/3 having BB, 1/3 having BG, and 1/3 having GB. Within this new group, ignoring the Tuesday part, randomly picking a mother would yield a probability of picking a family with BB of 1/3. HOWEVER, randomly picking a BOY would yield a probability of picking a family with BB of 1/2, which is the answer at which you arrived. You can’t overlook the fact that there are more ways to get a boy and a girl than there are to get 2 boys.

Now the Tuesday part: Each of those 1/3 groups (BB, BG. GB)has the same probability of having /at least one child/ born on a Tuesday. However, for a lot of the BG or GB families, that child will be a girl and they will need to leave anyway (more specifically, approximately half of each of those groups will have to leave because the child born on a Tuesday is of the wrong gender (even more specifically, it’s slightly less than half because some will have both children born on a Tuesday)). Every BB family will stay, becasue no matter which child(ren) was born on a Tuesday, it was a boy.

In terms of probabilities, once you have only families with boys, 1/7 of the BG families and 1/7 of the GB families will have a boy born on a Tuesday (when the girl is born is irrelevant). 1-(6/7*6/7)=13/49 of the BB families will have a boy born on a Tuesday, because they have more 7-sided “boy” die to roll to try to land a Tuesday. You are now left with 13x, 7x, and 7x for BB, BG, and GB where x/49 = 1/4 of the total initial families (including GG). Thus, the answer is 13/27, as the first poster pointed out.

23. 23 23 Steve Landsburg

William:

In your experiment, let’s ignore the days of the week and just have mothers write down genders. Then 1/4 of all mothers (those with BB) will write “boy” for sure, and 1/4 (those with GG) will write “girl” for sure.

That leaves 1/2 of the mothers. We don’t know what they’ll do. If they all write “boy”, then 1/3 of the mothers who wrote “boy” have two boys. If they all write “girl”, then 100% of the mothers who wrote “boy” have two boys. If they split equally, then 1/2 of the mothers who wrote “boy” have two boys. To reach a conclusion, you need an explicit assumption about what these mothers do.

24. 24 24 Jonathan Campbell

Harold: Here is a way to think about how it makes sense intuitively (to add to your analysis). Imagine an extremely specific question e.g. “A woman has 2 children, one of whom is a boy struck by lightening more than once. What is the probability they are both boys.” In this case, the probability of a woman having 2 boys, both struck by lightening more than once, is very small. Thus, it is nearly true that if a woman has 2 boys rather than 1, her chances of having 1 struck by lightening more than once are twice as great – so the original 1:2 ratio of families with 2 boys to families with at least 1 boy is adjusted almost to 2:2 (doubling the first figure).

Within the context of EricK’s analysis, this is the same as saying that the “double counted” cell is very insignificant.

In general, the less significant the double counted cell is, the closer to 1/2 the answer is (if you weren’t careful, and actually did double count it, the answer you’d get would be 14/28 = 1/2). For very specific questions like the one I mentioned, this double counted cell is very insignificant indeed.

25. 25 25 William

@Steven: To reach a conclusion, you need an explicit assumption about what these mothers do.

Cool. No disagreement here. I can see that I was assuming that every mother just picked one of her kids at random and described him. Not necessarily true.

26. 26 26 Michael

Douglas, thank you for the response, but…that isn’t what the question is asking (that I can discern, anyhow).

Here’s the text: “A woman has two children, one of whom is a boy born on Tuesday. What is the probability they are both boys?”

I took “a woman” to mean one specific woman. If I’m reading you and Steve Landsburg correctly, neither of you are. I’m not sure why. If it had said, any woman I would agree with you one hundred percent. But if it means one specific woman, we need not weed out any families.

27. 27 27 nobody.really

A woman has two children, one of whom was a boy born on a Tuesday. What is the probability they are both boys?

It’s a word problem: time to torture the pronouns!

“They” is a plural pronoun, so it can refer only to some aforementioned plural item. “Children” is the only candidate, so “They” refers to “children.” That’s no fun.

But “One of whom” is singular, so it could refer to any prior noun. Yes, it could refer to “children.” But could it also refer to “woman”? Thus, when calculating probabilities, we need to account for the possibility that the woman was a boy born on a Tuesday who subsequently had a sex change operation. Now the fun starts!

With this interpretation, any of the following mutually-exclusive scenarios would fulfill the problem’s requirements:
1A. Woman born female, B born on Tuesday, B not born on Tuesday
1B. Woman born female, B not born on Tuesday, B born on Tuesday
1C. Woman born female, B born on Tuesday, B born on Tuesday
2. Woman born female, B born on Tuesday, G
3. Woman born female, G, B born on Tuesday
4. Woman born male on Tuesday, B, B
5. Woman born male on Tuesday, B, G
6. Woman born male on Tuesday, G, B
7. Woman born male on Tuesday, G, G

Calculate the probabilities now!

Never underestimate the power of a committed mind to miss the point.

28. 28 28 Harold

It is good to see Henry’s and EricK’s methods both give the same result. It is enough to restore my faith in numbers, which I was starting to doubt the existence of.

29. 29 29 wkw

There are two possibilities, depending on how one interprets the question:

1. The gender of a child is independent of the gender of his sibling, or P(A|B) = P(A). So, roughly (not exactly, empirically) 1/2. This interpretation can basically be rephrased as “What is the probability that any given child is a boy, irrespective of other extraneous factors?”

2. The gender of a child is not independent of the gender of his sibling, or P(A|B) /= P(A), and the answer is roughly 1/3 (because [B,B], [B,G], [G,B]). This interpretation is equivalent to “What is the probability that two children are both boys, given that at least one of them is?”

Either one is correct depending on underlying assumptions about how the question is created, which depends on beliefs about how families are being sampled. Personally, I find #1 to be the best answer since #2 requires some semantic gymnastics to make sense. Occam’s Razor, baby.

30. 30 30 Al

Harold’s explanation has helped me get my head around this.

My intital stab at this was right, but I started doubting myself, which is incredibly easy to do with questions like these, which led me down the wrong path.

31. 31 31 Dave

I am so perplexed about why I don’t get that it’s 1/2. I did EricK’s exercise by drawing up the grids and agree that’s what comes out but I don’t get why.

Intuitively, it should be 50/50 or 14/28.

For some reason we remove 1 from both the numerator and denominator and get 13/27 because we are double counting or something?

Would I be right in saying that if the question was: “A woman has 2 children, one of whom was a boy born in January. What is the probability they are both boys?”

Assuming each month has 30 days, could I take the same tack and think, intuitively it’s 24/48 but deduct 1 from both the numerator and denominator and land at 23/47?

What if the question was: “A woman has 2 children, one of whom was a boy born in a leap year. What is the probability they are both boys?”

Could I take the same tack and think, intuitively it’s 8/16 but deduct 1 from both the numerator and denominator and land at 7/15?

If I’ve found some magical shortcut (ie double the possible number of outcomes for birth event, phrase 50% as a fraction using that number as the numerator and deduct 1 from both numerator and denominator), then it’s a pure fluke.

I don’t get it.

Am I right in that shortcut? Can anyone explain it to me? I still think it’s 50% even though I’ve done the visual exercise probing it’s not.

32. 32 32 Dan

I got tired of trying to reason out the solution to this, so I went to Excel. I produced a sheet with 100,000 rows.

Column A: Sex of oldest child. Uniform random variable. 0=girl; 1=boy.
Column B: Day of week of oldest child’s birth. Uniform random variable. 0=Sunday, 1=Monday, 2=Tuesday, and so on.
Column C: Same as A, but for younger child.
Column D: Same as B, but for younger child.
Column E: TRUE if at least one of the two children was a boy born on a Tuesday.
Column F: TRUE if both children are boys.
Column G: TRUE if columns E and F are both TRUE.

12,032 rows have TRUE in column E.
25,105 rows have TRUE in column F.
5,837 rows have TRUE in column G.

The answer to this problem, as I understand it, is approximated (subject to the true randomness of Excel’s random number generator) by dividing the number of TRUEs in column G by the number of TRUEs in column E. This number is 0.485123. As many times as I run the simulation, I never get a number greater than 0.5. Rather, it seems to bounce around near 13/27 (0.481481), the answer proposed by other commenters.

33. 33 33 EricK

With problems of this type it is vital to realise that how the information was arrived at is as important as the information itself. In particular, freely volunteered information is not the same as forcibly revealed information.

Take the simpler problem: A woman has two children, at least on of which is a boy. What is the probability that both her children are boys?

If we ask random women the questions “Do you have two children?” and “Is at least one of them a boy?” and consider the population of women who answer “yes” and “yes”, then of these women, 1/3 will have two sons and 2/3 will have one child of each gender. This is because our sample space eliminates all the women with two children both of which are girls, but leaves in all the other women with 2 children.

On the other hand, if we ask random women “do you have 2 children?” and “tell me the gender of one of your children” and consider the population of women who answer “yes” and “boy”, then of these women we would expect 1/2 would have 2 sons and 1/2 will have one of each. This is because our sample space has eliminated not only all the women with 2 duaghters but ALSO (approximately) half the women with one son and one daughter i.e. those who happened to answer “yes” and “daughter”.

34. 34 34 EricK

Dave,

I suspect the reason you intuitively think it is a half is that you say to yourself “Right, one child is a boy born on Tuesday, what about the other one? Well it’s either a boy or a girl and that is 50/50″. The trouble with that reasoning is that “The other one” is not always well defined. If only one of the children is a boy born on a Tuesday then “the other one” is unambiguous”, but if both happen to be boys born on Tuesday then it is not clear which is “the one” and which is “the other one”. Essentially, this is the source of the double counting, and why the probability differs from 1/2. And it is also why the rarer the auxiliary event (eg being born on a Tusday, being born on January 1st etc), the less likely it is that both he and his brother will qualify, and so the smaller the deviation from 1/2.

35. 35 35 Dave

EricK – aha!

thanks! that is the perfect explanation!

36. 36 36 John Faben

I like this problem in that it seems counterintuitive to different people for different reasons. For some people it seems hard to grasp that the answer *isn’t* 1/2 – they’ve obviously never encountered the simpler, and older version “I have two children, one of whom is a boy, what is the probability both are girls” – which is mentioned above. To which the answer (when the problem is specified in the same way at Steve’s version) is 1/3.

For people who have heard the “1/3″ problem, it’s often counter-intuitive that learning the birthday of one of the boys could move the probability away from 1/3. After all, any boy is born on *some* day, right? (Steve’s careful wording of the Tuesday Boy problem is necessary so that we can give the answer “yeah, but we didn’t ask them about any other day” – careful wording of the original is needed so we can answer “yeah, but we didn’t ask them about any other gender”).

If you’re still struggling with an intuition which tells you the answer should be 1/3, consider the following version – “a woman has two children, one of whom is a boy whose birthday is the 9th of January, what is the probability both are boys?”.

37. 37 37 wheninrome15

I just applied Bayes rule and got 13/27.

This class of questions is so frequently posed, and while I would never impose an expectation that people know how to solve them, it does surprise me that so many haven’t yet realized something funny is going on in them. I don’t care if you don’t remember the way to solve it from last time it showed up, but don’t you at least have the meta-awareness that the “obvious solution” is not correct?

The Law of Two-Sentence Probability Puzzles clearly states that the answer is never 1/2.

38. 38 38 David

@John Faben

The probability is 1/3

What’s the difference between specifying the day of the year and specifying the day of the week?

I see nothing in either your question or the original which alters the likelihood of the sex of the second child. I am supposed to conclude that hair color or the month when the child first walked also change facts about the second child?

39. 39 39 Steve Landsburg

David:

I am supposed to conclude that hair color or the month when the child first walked also change facts about the second child?

What if I tell you that one child was elected president of the United States in the year 1828?

On the one hand, this seems no more relevant than hair color or the month when the child first walked. On the other hand, do you see how it changes the probability from 1/3 to 1/2?

40. 40 40 Steve Landsburg

David: Let’s do this another way —-

We have in front of us all the world’s mothers of two children.

We say to them: Stick around if you have a boy; otherwise leave. 1/4 of the mothers (those with two girls) go home. Of those that remain, 1/3 have two boys. I think you and I agree on this.

Now we say to those that remain: Stick around if you have a boy born on a Tuesday; otherwise leave.

Just before we say this, 2/3 of the remaining mothers have a boy and a girl. Of those, 1/7 have a boy born on Tuesday. They all stay.

The other 1/3 of the remaining mothers have two boys. Of these, MORE than 1/7 have a boy born on a Tuesday, because each has two chances to have such a boy.

So—of the 2/3, 1/7 stick around. Of the 1/3, MORE than 1/7 stick around. Therefore the proportion of remaining mothers who have two boys must grow.

41. 41 41 Thomas Bayes

Let’s have some fun with Bayes’ Rule . . .
(B,B) = two boys
B_T = boy born on Tuesday
C_2 = two children total

Pr[(B,B) | B_T, C_2] =
Pr[B_T | (B,B), C_2] Pr[(B,B)|C_2] / Pr[B_T|C_2]

The easy one is the probability of two boys given she has two children: Pr[(B,B)|C_2] = 1/4.

The probability of a boy born on Tuesday given that she has two children is a little more complex, but not too tough. There are 14×14 = 196 possibilities. There are 14 ways to have the first child be a boy born on Tuesday, and 14 ways to have the second child be a boy born on Tuesday. One of these is counted twice, so there are 14+14-1 = 27 ways to have at least one boy born on Tuesday. So,
Pr[B_T|C_2] = 27/196.

The probability of a boy born on Tuesday given that there are two boys and two children is equal to the probability of a boy on Tuesday given that there are two boys. There are 7×7 = 49 possibilities. There are 7 ways to have the first boy be born on Tuesday, and there are 7 ways to have the second boy be born on Tuesday. One of these is counted twice, so there are 7+7-1=13 ways to have a boy born on Tuesday given she has two boys:
Pr[B_T|(B,B),C_2] = 13/49.

The desired probability is:

Pr[(B,B)|B_T,C_2]
= (13/49)*(1/4)/(27/196)
= (13/49)*(1/4)*(196/27)
= 13/27

Of course EricK’s approach is a little quicker to the point.

42. 42 42 Jeffrey

David,

Suppose you have two possibilities, A and B, and event E is more likely in case B than in case A. It’s intuitive to see that learning about event E raises the likelihood of possibility B. (Mathematically, it’s a Bayes factor.) This is the situation with the Tuesday data.

If someone has only one son, there’s a 1/7 chance that they have a son born on Tuesday. If they have two sons, there’s a 13/49 chance that they have a son born on Tuesday. “A son was born on Tuesday” is data that is more likely to be true of people with two son than of people with one. Thus, this Tuesday data increases the chances that someone has two sons.

43. 43 43 Michael

@wkw

Exactly. I’m not sure why the second is being assumed on this thread….

44. 44 44 Evan P.

It’s really important to think about what you actually know in this problem (which can be a function of how you gathered the information). Consider two scenarios in which you know a woman has two children, one of whom is a boy born on a Tuesday:

However, suppose you are interviewing a woman about whom you know only one fact: she has exactly two children. You are about to ask her the following question: “do you have a son born on Tuesday?” Before you ask you can think about the likelihood of a yes or no answer given the genders of her two children. You know the following:

-There is a 1/4 chance that she has two girls, a 1/2 chance that she has a girl and a boy, and a 1/4 chance that she has two boys (based on the assumption that having a girl is equally as likely as having a boy).
-If she has two girls there is no chance she will answer your question yes. If she has one boy and one girl there is a 1/7 chance that she will answer your question yes. If she has two boys, there is a 13/49 chance she will answer yes (since there are 49 possible day-of-birth combinations for her two boys, 13 of which include at least one boy being born on a Tuesday).

Therefore the probability of a yes answer is 1/4 x 0 + 1/2 x 1/7 + 1/4 x 13/49. The probability that she has two boys and gives a yes answer is 1/4 * 13/49. So, if she answers “yes” (in which case, again, you “know a woman has two children, one of whom is a boy born on Tuesday”) then the probability of her having two sons is P(yes + 2 boys)/P(yes) = (13/196)/(13/196 + 1/14) = 13/27.

The key information difference in the first and second scenarios is that in the first you know something about one specific child. This gives you no information about the second child. In the second, you know a fact that applies to the woman’s children as a pair. Because you know something about the pair of children you can use that information to figure out the likelihood of the pair being made up of any of the three possible gender combinations.

When you first read this riddle it’s easy to think in terms of the first scenario (e.g. “Ok, I know something about one of this woman’s kids; what does that tell me about her other kid?). But when you pay careful attention to Prof. Landsburg’s disambiguation, you realize you know something about her children as a pair.

45. 45 45 Thomas Bayes

Here is a simpler* version of this problem. I just flipped two fair coins. You ask if at least one of them showed heads. I answer yes.
What is the probability that they both showed heads?

*Simpler in the sense that the sample space is smaller and you can try it for yourself.

46. 46 46 Cos

Thomas Bayes:

Ahh, but what if you lived in a world in which billions of people had flipped a small number of fair coins and saved the result, and then you selected from that world all of those who had flipped exactly two coins, at least one of which was heads?

The initial question is ambiguous and open to different interpretations that yield different probabilities. To get an answer, you need to specify the interpretation you want to think about (which this blog post does).

47. 47 47 Thomas Bayes

I believe the coin problem helps illustrate the confusion about the ‘conditions’ that define these problems.

My formulation of the coin problem requires that we solve for the probability of two heads showing conditional on the compound event of two tosses and at least one head. To solve this problem, we don’t need to know the marginal probability of two tosses. In your formulation, the question is the same. We still don’t need to know the marginal probability of making two tosses, only the conditional probability of two heads given that two coins were tossed with at least one head showing.

Bayes’ Rule is beautiful because it provides a systematic way to solve all of these problems. For the coin flip:
T_2 = two tosses
H_1 = at least one head shows

Pr[(H,H)|H_1, T_2]
= Pr[H_1|(H,H), T_2] Pr[(H,H)|T_2] / Pr[H_1|T_2]

All of the probabilities are conditioned on two tosses, so we never need to know the probability that the person tossed two coins. Provided everyone made independent and fair coin tosses, it doesn’t matter which one we select.

To solve this problem . . .
Pr[(H,H)|T_2] = 1/4
Pr[H_1|T_2] = 3/4
Pr[H_1|(H,H),T_2] = 1

Pr[(H,H)|H_1, T_2] = 1/3.

How we get to the condition of two tosses doesn’t matter in the answer.

The important part of the clarification for the birth problem is to ensure that the original statement didn’t imply that EXACTLY one boy was born on a Tuesday.

48. 48 48 Clifford Nelson

2/7 chance of having a boy born on Tuesday for women who have two kids .2857

.500
.500
.500
.500
.500
.500
.500 (for each day of the week odd of having a boy)

Then average to get odds for both are boys.

3.5 plus .2867 3.7867

47.33 percent chance

Right?

49. 49 49 Clifford Nelson

Wait … the odds of having a boy on Tuesday when you have two kids is 1/7 = .1428

The odds of having a boy on any day of the week is 1/2 = .500

Thus, the chance is 45.535% chance

.500 odds for child 1 being boy born on Monday
.500 odds for child 1 Tuesday
.500 …. Wednesday
.500 …. Thursday
.500 …. Friday
.500 … Saturday
.500 …. Sunday
.1428 … odds for boy born on Tuesday

3.6428/8 = .45535

50. 50 50 I.G. Noranz

I hate to brag about how dumb I am, but I see the answer as:

[The population of boy-children in the world, minus 1] divided by [the population of boy-children in the world, minus 1, plus the population of girl-children in the world].

Where: boy-children = all the boys in the world belonging to women with two children only, one of which was born on a Tuesday, and: girl-children = all the girls in the world belonging to women with two children only, one of which was a boy born on a Tuesday.

Also, it seems to me that EricK makes a counting mistake (response #1) when he says: “…27 of them have a boy born on a Tuesday (14 from one column, 14 from a row, but 1 is double counted).” Aren’t 2 boys double counted here, the BB boy and the BG boy?

51. 51 51 Ryan JL

[i]“The day of the week isn’t relevant to the question asked–it’s noise added to obfuscate. “[/i]

On its surface, it does seem to be a red herring, but it’s an important piece of information because of what it implies implicitly about the other child. What it’s really saying is that child #2 (the mystery child) is NOT a Tuesday-boy. It could still be a Tuesday-girl, or a boy born on a different day, but there are now more possibilities of girls (7) than boys (6), and hence the funny probability calculated by Post #1.

52. 52 52 Thomas Bayes

Ryan: This problem does not rule out the possibility of having two boys born on Tuesday. The ‘funny’ probability arises because the question and corresponding answer do not specify which of the two children is a boy born on Tuesday. If we had learned that the oldest child was a boy born on Tuesday, then there would be a 1/2 probability that the other child was a boy. If we had learned that the youngest child was a boy born on Tuesday, then there would be a 1/2 probability that the other child was a boy. Our knowledge in this problem, however, is that one of the children is a boy born on Tuesday, but we don’t know if it is the younger or older child. This is the reason for the “funny” probability.

Think about about flipping two coins. If I tell you that at least one of the coins showed heads, then three things could have happened: Heads followed by Heads; Heads followed by Tails; or Tails followed by Heads. Each of these is equally likely to have happened, but only one of them gives two Heads. Hence, the probability that both showed Heads given that I tell you one of them did is 1/3.

If, however, I tell you that the first toss showed Heads, then two things could have happened: Heads followed by Heads; or Heads followed by Tails. Each of these is equally likely, and one of them gives two Heads. Hence, the probability that both showed Heads is 1/2.

The reason this simple coin toss example gives an unintuitive answer is exactly the same reason that the original problem does.

For more fun with the coin toss problem, imagine that I toss two coins and hold one in each hand. If my hands are closed to prevent you from seeing, we can go through a series of questions:

2. I can tell you that at least one showed heads (if that is true), and ask you the probability that both show heads. The answer is 1/3.

3. I can show you one of the coins that is showing heads, and ask you the probability that the other shows heads. The answer is 1/2.

When confronted with a version of question 2 (which we are in the ‘boy born on Tuesday’ problem), most people think about an answer that is more applicable to question 3. That, I believe, is the source of confusion on this problem.

53. 53 53 Thomas Bayes

Suppose you first gather all of the women in the world with exactly two children.

Then you ask all women that do not have at least one boy to leave.
What is the probability that a woman who remains has two boys?

Now you ask all women that do not have at least one boy born on Tuesday to leave.
What is the probability that a woman who remains has two boys?

Now you ask all women that do not have at least one boy born on Tuesday in the AM to leave.
What is the probability that a woman who remains has two boys?
(It is probably not true, but assume all hours of the day are equally likely for a birth.)

Now you ask all women that do not have at least one boy born on Tuesday between midnight and 6AM to leave.
What is the probability that a woman who remains has two boys?

Now ask all women that do not have at least one boy born on Tuesday between midnight and 1AM to leave.
What is the probability that a woman who remains has two boys?

These probabilities are increasing from 1/3 toward 1/2. Do you see why?

54. 54 54 Clifford Nelson

As I noted above, to get the odds you add all the odds of the necessary events together and take the average. However, I again made a mistake with the odds for the boy born on Tuesday.

For a given week:

Boy born on Tuesday – .5/7 .0714
Boy born any other day – 1/2 .500

Since we are working with a seven-day period (which of course continues, but that is irrelevant):

We use .5/7 once because we are told that we have that boy so we need to find the odds – given what we know, that both are male.

We use 1/2 seven times since the other child can be born on any day of the week (think about how many more kids one should expect) we account for that by multiplying the odds by each day a child in the relevant period can be born.

So once again I revise my answer: .446

55. 55 55 Stephen Coy

Why does everyone seem to assume that the sex and/or birthdate of one child affects the other?

The odds of one children being a boy born on Tuesday — 1.0 (This is one of the “facts” given in the puzzle)
The odds of the other child being a boy — 0.5

Final answer : 1 * 0.5 = 0.5

Another way to look at this: “A woman has a child and a dog. The child is a boy born on Tuesday. What is the probability that the dog is male?” Still get 13/27ths?

56. 56 56 Steve Landsburg

Stephen Coy: Here’s why your example is not like the original problem:

With the boy and the dog, it is given, as part of the problem, which of the two was born on Tuesday (namely the boy).

In the original problem, it is not given, as part of the problem, which of the two boys was born on Tuesday.

So your problem is not the same as the one that was posed.

57. 57 57 Stephen Coy

@Stephen: I completely fail to see the difference that makes. The point I’m making is that the sex of the ‘other’ child (or dog) is completely independent of the sex of the first, or the day of the week the first was born on. You’re implying that knowing which of the two was born on Tuesday is relevant information. I’m saying it’s not. I have a set with two “beings” in it. I know that one of the two beings is a boy born on Tuesday. The other member of the set can be a human or a dog, it doesn’t change the fact that the odds of this being being male is 1/2.

The beauty of this puzzle is that it leads people to think that there’s some subtle, complex answer when in fact the obvious solution is the right one.