# Probability puzzles

post by johnclark · 2011-04-22T21:11:48.313Z · LW · GW · Legacy · 10 commentsThere are 2 probability puzzles that I like:

1) Suppose I tell you that I have 2 children and one of them is a boy, what is the probability that I have 2 boys?

The correct answer is not 1/2 but 1/3. How can that be? Well there are 4 possible combinations, BB,GG,BG and GB but but at least one is a boy so you can get rid of GG. So all that's left is BB,BG and GB; and in only one of those 3 possibilities do I have two boys.

2) Now I tell you that I have 2 children and one of them is a boy born on a Tuesday. What is the probability that I have 2 boys?

You may think that Tuesday is not useful information in this matter so the answer would be the same as the previous example, but you would be wrong. The correct answer is 13/27. How can that be?

Well there are 14 possibilities for EACH kid:

B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su

G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su

But I told you the one of my kids (the first or the second) was a boy born on a Tuesday so that narrows down the field of possibilities to:

First child: B-Tu, second child: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.

Second child: B-Tu, first child: B-Mo, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.

No need to put B-Tu in the second row because it's already accounted for in the first row.

So now just count them out, 14+13= 27 possibilities. How many result in 2 boys? Count them out again 7+6=13. So 13 out of 27 possibilities give you 2 boys.

John K Clark

## 10 comments

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## comment by gjm · 2011-04-22T21:45:15.780Z · LW(p) · GW(p)

The correct answer to puzzle 1, as posed, is not in fact simply 1/3, because you've got to factor in Pr(you say "I have two children and one is a boy" | each scenario) and it's not at all clear that these are equal in the BB, BG, GB cases. For instance, if you say that in the ordinary course of events (rather than, e.g., to pose a puzzle) I think BG and GB are very much more likely than BB, because if you had two boys why on earth would you say "I have two children and one is a boy"?

There's a general principle here: when you discover something new (by, e.g., being told something, or seeing something), the correct information to update on is not *what you've been told, or seen* but *the fact that you've been told, or seen, it*. In some cases this doesn't matter. In many others (including, but not limited to, those where you have reason to doubt the accuracy of what you've been told or seen) it does.

## ↑ comment by steven0461 · 2011-04-22T21:56:12.019Z · LW(p) · GW(p)

A more general lesson is that whenever the answer to a puzzle causes you to go, "oh how wondrous that this question could have such a strange answer", you were probably tricked into accepting an anti-helpful framing of the problem, and one of the reasons why the puzzle-poser didn't guide you into a helpful framing instead was probably exactly that such anti-helpful framings cause people to feel that way.

Replies from: gjm, Martin-2## ↑ comment by Martin-2 · 2014-08-12T01:04:55.803Z · LW(p) · GW(p)

This post is not evidence for that lesson. When OP's puzzle is stated as intended it indeed has a wonderful and strange answer. The meta-puzzle: "Are these two puzzles essentially the same?" referring to the puzzle as intended and as presented also has a wonderful and strange answer; in fact, John Baez and maybe all of his commenters have been getting it wrong for several years. Our intuition is imperfect, and whether the puzzles you come across tend to use this fact or just trick you with sneaky framing depends on where you get your puzzles.

## ↑ comment by ata · 2011-04-22T22:05:29.663Z · LW(p) · GW(p)

The correct answer to puzzle 1, as posed, is not in fact simply 1/3, because you've got to factor in Pr(you say "I have two children and one is a boy" | each scenario) and it's not at all clear that these are equal in the BB, BG, GB cases. For instance, if you say that in the ordinary course of events (rather than, e.g., to pose a puzzle) I think BG and GB are very much more likely than BB, because if you had two boys why on earth would you say "I have two children and one is a boy"?

Indeed, and this exact malformed problem is also discussed in the post "My Bayesian Enlightenment":

In the

correctversion of this story, the mathematician says "I have two children", andyouask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys.

## comment by ArisKatsaris · 2011-04-23T17:34:25.070Z · LW(p) · GW(p)

If this post wasn't already at -6 I'd downvote it, for failing to provide the *correct* form of the puzzles, which has the *second* person ask the first about this additional info. Otherwise you need to consider the *reason* the person is providing you this information.

For the "Suppose I tell you that I have 2 children and one of them is a boy" we have the following priors:

P(two girls) = 25%

P(two boys) = 25%

P(one of each) = 50%

Consider however the following possibilities.

Scenario A) The guy is a sexist who would loudly proclaim the existence of sons, and avoid discussing the existence of daughter. Then the probabilities he has two boys is 0%. If he had two boys, he would have told you both of them are boys. By mentioning only one, he unintentionally revealed he had only one. Putting differently P(mentions only one boy|two boys) ~= 0

Scenario B) The guy thought to himself "I'll randomly pick a child and mention its gender". Then with no gender bias we have

P(mentions a boy) = 50%

P(mentions a girl) = 50%

and we go to:

P(mentions one boy|two girls)= 0%

P(mentions one boy|two boys) = 100%

P(mentions one boy|one of each) = 50%

Therefore P(two boys|mentions one boy) = P(mentions one boy|two boys) * P(two boys)/P(mentions one boy) = 100% * 25% / 50% = 1/2

So if the guy randomly picked a child who's gender he picked to mention, possibility of two boys is 50% given the info he provided.

## comment by alcha · 2019-11-09T19:10:06.793Z · LW(p) · GW(p)

The way the puzzle is posed is ambiguous, as it does not state exactly how the trial is carried on. It may be reworded in (to my understanding) 2 different reasonnable ways, leading to 2 different answers.

1) Out of the set of 4 boys and 4 girls, that is 4 families representing the 4 combinations BB,BG,GB,GG, you choose at random a boy (rejecting the trial if you get a girl). Then the probability of the other member of the family being a boy is 1/2. This is the same as Ariskatsaris scenario B) below.

2) Out of the 4 families, you chose at random one that has at least one boy (rejecting the family with 2 girls), then the probabilty is is 1/3.

The way the puzzle is worded appears to me closer to scenario #2 than #1, but this my biased interpretation.

## comment by gjm · 2011-04-22T23:58:07.430Z · LW(p) · GW(p)

Here's a possibly useful intuition-adjuster for the "Tuesday boy" problem: replace "born on a Tuesday" with something much less probable. Pr(two boys | two children, one a boy who will one day be President of the USA) is "obviously" about 1/2 rather than about 1/3, because now you can (almost) meaningfully talk about "the other child", which you can't in the case of Pr(two boys | two children, one a boy). The less uniquely-identifying the extra information, the nearer you are to the original "two children, one a boy" scenario.

If you want to do the actual calculation, you might want this picture in your head: a 14x14 table of possibilities, boys in the first 7 rows/columns, Tuesday in the first and 8th row/column. "Two children, one a boy" excludes the bottom-right quadrant. "Two children, one a Tuesday-boy" excludes all but the first row and column.

## comment by engineeredaway · 2011-04-24T21:09:50.784Z · LW(p) · GW(p)

a lot of the comments here are critical of the way these scenarios are presented, but I don't believe there is in fact any deep issue. the fact of the matter remains, if you are in a situation where you have two things which you do not yet have the information to differentiate, and you know they have a state in some binary property, and that this state for one is independent of the other. Then, if you learn, by any means, that one of these objects has a specific state (call it A) with regard to the binary property, your probability needs to adjust to p(both are in state A|one is in state A) = 1/3 and p(only on object is in state A|one is in state A) = 2/3. johnclark just used a classic example to demonstrate the importance of how interchangeability has a large and unintuitive effect on probability. It is my contention that the sentiment this scenario is unintuitive because of the way johnclark is incorrect. I have seen this question posed in other ways to classes of smart college students studying probability and most of them getting it wrong. External knowledge of the way people tend to provide information isn't really a relevant factor here. This article is well written and important, and by no means deserves the very low karma score it has received (-6 as of this writing, -7 before my vote)