Probability puzzles

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.

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.

The Tuesday boy problem was discussed here.

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.

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.

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)