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probability

Posted By: Daniel Murphy
Date: Wednesday, 30 July 2008, at 6:12 p.m.

In Response To: probability (Tom Keith)

There is enough information to answer Christian Munk-Christensen's question. The knowledge we have of the children's sex is what C M-C has told us: that "at least one of them is a girl." C M-C has not told us that we know whether Child A is a girl or whether Child B is a girl. That's a different problem with a different answer.

For A, you can substitute "the older child" (and B is the younger), or "the one vacationing is France" (while B is in Germany), or "the one Tom sees in the street" (while the other child is home).

In C M-C's problem, since of A and B at least one of them is a girl, there are three possibilities: A is a girl and B is a girl, or A is a girl and B is a Boy, or A is a boy and B is a girl. The probability that both are girls is 1/3.

In Tom's problem, Tom doesn't know merely that at least one of them is a girl; he knows that Child A who he saw in the street is a girl. In which case (assuming that A has only one sibling), either A and B are both girls, or A is a girl and B is a boy. The probability that both are girls is 1/2.

Again with the dice:

I roll two dice and don't look at them. What is the probability that both are 6s? The answer is 1/36.

I roll two dice and look at them and tell Tom that at least one of them is a 6. What is the probability that both are 6s? Of 36 dice rolls, 11/36 contain at least one 6, and 1 of them contains two 6s. So the answer is 1/11.

That's similar to C M-C's problem. Because we know that "at least one child is a girl" we can ignore two-child combinations in which neither child is a girl. Because we know that "at least one die is a 6" we can ignore the 25/36 dice rolls in which neither die is a 6.

I roll two dice and one -- megacocked dice -- rolls out into the street and Tom sees that it is a 6. What is the probability that the die Tom sees in the street and the die resting on the right-hand side of my backgammon board are both 6s? The answer is not 1/11, but 1/6.

"If a man who cannot count finds a four-leaf clover, is he lucky?" - Stanislaw Jerzy Lec

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