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by HWR_14
1027 days ago
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I think now I understand. "I will make you make exactly one of two statements and each results in a 2/3 chance of BG" doesn't make sense. "You freely made one of two statements and each results in a 2/3 chance of BG" does. I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works. So it comes down to if we decide on the question (however we do that) before we look at the kids or after. |
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The point is that original problem says "I tell you I have two children and that (at least) one of them is a boy". It doesn't say "I tell you I have two children and [when you ask me to confirm whether (at least) one of them is a boy] that (at least) one of them is a boy".
Reasoning from the cases "BB", "BG", "GB", "GG" - and discarding the last one to get p(BB)=1/3 - is implicitly using the cases "BB and I tell you that at least one of them is a boy", "BG and I tell you that at least one of them is a boy", "GB and I tell you that at least one of them is a boy", "GG and I tell you something else like at least one of them is a girl".
That breaks the symmetry between the "I tell you that at least one of them is a boy" and the "I tell you that at least one of them is a girl" problems. Using the cases in the previous paragraph the answer for the former is p(BB)=1/3 but the answer to the latter is p(GG)=1.
If you want to have the same solution when you switch girl <-> boy ["one of them is a boy, what is the probability that I have two boys" <-> "one of them is a girl, what is the probability that I have two girls"] you should treat them equally.
A quite natural way to do so would be to consider the eight equiprobable cases (some of them repeated)
but then the answer to both problems is 1/2.You can make the answer to both problems 1/3 but the eight cases that you would need to consider for that are quite unnatural: