Think you understand Monty Hall? Try the Tuesday boy problem.

Re:Think you understand Monty Hall? Try the Tuesday boy prob... by rubergly on Tuesday November 29, 2011 @06:38AM (#3289928)

    Let's try a simpler problem. Suppose we know that a 
    certain man has two children and we also know that the
    older one is a boy. In this case we would say that the
    probability that the other child is a boy is 1/2. After
    all, the sex of one child is independent of the sex of
    the other child. That the older child is a boy has no
    bearing on the sex of the younger child.

    Now suppose we know simply that a man has two children
    and that one of them is a son. This time we would reason
    that there is no possibility that the person has two
    girls. It follows that the sexes of his two children,
    ordered from oldest to youngest, are either BB, BG or GB.
    Since these cases are equally likely, and since only one
    of them involves having two boys, we would say the
    probability that the man has two boys is 1/3.

Maybe I'm missing something, but this seems fundamentally wrong. Pr(two boys | older child = boy) is not equal to Pr(two boys | one+ child = boy)? Why is time so special? Could we not order them based on their height, and assert Pr(two boys | taller child = boy) = 1/2. Or, if there were a scale of masculinity, order them based on that and assert Pr(two boys | manlier child = boy) = 1/2, where manlier child = boy <==> one+ child = boy, so Pr(two boys | manlier child = boy) = Pr(two boys | one+ child = boy) (contradiction).

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Re:Think you understand Monty Hall? Try the Tuesday boy prob... by incremental on Tuesday November 29, 2011 @06:53AM (#3289960)

For a randomly selected family with two children, there are four possible boy/girl combinations: B B, B G, G B, G G

In the first case we are told that the older child is a boy. This leaves only two cases: B B, B G Therefore, there is a 50% chance the second child is a boy.

In the second case, we are told only that [at least] one child is a boy. This leaves three possibilities: B B, B G, G B Therefore, the probability that both children are boys is 1/3.

Enumerating possible states of the world like this is the fundamental insight you need to have to be able to understand these types of problems - but it does take a while to get used to!

[ Parent | on HN ]

Re:Think you understand Monty Hall? Try the Tuesday boy prob... by rubergly on Tuesday November 29, 2011 @07:06AM (#3289985)

But you are assuming that each of those three possibilities has equal probability; can you explain the rationale for that? (it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case, but less clear why BB, BG, and GB have equal probability in this restricted case)

Besides, this is just a rephrasing of the original article's argument, and doesn't counter mine at all. I am open to the possibility that there is a flaw in my argument, but where is it?

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Re:Think you understand Monty Hall? Try the Tuesday boy prob... by incremental on Tuesday November 29, 2011 @07:53AM (#3290096)

So you say: "it is clear why BB, BG, GB, and GG have equal probability in the unrestricted case"

The restricted case is just the unrestricted case + one additional bit of information, that is, you're told that GG is not an option. This eliminates GG from the unrestricted case, but says nothing more about the probabilities of the other options. So the probabilities stay equal, although they now equal 1/3 each (if you eliminate options, the remaining options all become more likely).

What you're missing is this: The statement "the older child is a boy" has more information than the statement "one of the children is a boy". The first statement allows you to eliminate two options (GB and GG), while the second statement only allows you to eliminate one option (GG).

The "older" part is not fundamental to the problem. Equally, the statement "the taller child is a boy" has more information than "one of the children is a boy". The problem with this is that probabilities for height are not so friendly like the 50/50 probabilities for birth order (e.g., boys are likely to be taller than girls, older children are taller than younger, etc), which introduces unnecessary complexities to a logic problem. So that's why birth order is used for these types of puzzles.

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Re:Think you understand Monty Hall? Try the Tuesday boy prob... by ajanuary on Tuesday November 29, 2011 @07:48AM (#3290089)

Using age as the ordering is not important. What matters is that when enumerating possibilities you count the probability of the first and then the second, and the second and then the first.

If manlier child = boy <==> one+ child = boy then surely in your final equation one of the sides boils down to Pr(two boys | impossible event) as the probability of either manlier child = boy or one+ child = boy must be 0?

The problem comes from the fact you've asked an impossible question, not from applying an ordering. If the ordering is used consistently, then it should all work fine.

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Re:Think you understand Monty Hall? Try the Tuesday boy prob... by rimantas on Tuesday November 29, 2011 @07:40AM (#3290073)

  > Why is time so special?

Because it is a part of the condition? How else can you have "older child" in your condition, if time is not important? Stating that first child is a boy is the same as opening the door in Monty Hall problem: it eliminates particular combination.

If you don't know the order of the kids you have three ways to have a situation where two boys are possible: BB, BG, GB. When you know that first one is a boy, you eliminate the GB case and are only left with BB and BG.

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