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You unfortunately fall right into the trap that he's warning about: just because you can enumerate the possible outcomes doesn't mean that they are all equally
by kmod 16y ago
You unfortunately fall right into the trap that he's warning about: just because you can enumerate the possible outcomes doesn't mean that they are all equally likely (cf: the Monty Hall problem).
The conclusion that BB/BG/GB are all equally likely follows from the assumption that the man would definitely state that he has a male child if and only if one of those conditions were true. But what if instead we add the fact that the man would only say that he has a son if he has no daughters? This isn't contradictory with anything else in the problem, but now the answer is clearly "P(BB) = 1", which makes it hard to state that certainly "P(BB) = 1/3" when there is a consistent interpretation of the problem that gives a different answer.
You should reread the way he set up the problem again: he makes the subtle distinction between things that are given to us by the omniscient problem writer and things that are given to us by characters inside the problem, for whom we have to apply Bayes Law an additional time.
- Tichy 16y agoI think you are confusing the likelihood of something happening with the estimate based on a given set of facts. The man doesn't have quantum children that are 1/3 boy or girl. So trivially P(the gender of the man's children is the gender of the man's children) = 1. They don't change gender depending on your guess. The only thing that changes is the trueness of your guess of the other child's gender (it changes when you change the guess) The way to get clear hear is to simulate meeting the man 1000 times, and counting how many times you would have guessed right if you said he has two boys. This yields 1/3 because you don't even guess when he has 2 girls.