Re^7: Marilyn Vos Savant's Monty Hall problem (odd odds)

Your possible sets of kids that we assume are equally likely:

son, son
son, daughter
daughter, son
daughter, daughter

If you ask the first pair of questions, the answers will be "2" and "Yes" for the first 3 possibilities, and "2" and "No" for the last. So with the first pair of questions you reduce down to 3 possibilities, of which 2 have a daughter. Note that your initial odds of not getting the second answer that you got are 1/4.

If you ask the second pair of questions, the mother told you a random story, your possible situations of interest double:

son, son, story about first son
son, son, story about second son
son, daughter, story about son
son, daughter, story about daughter
daughter, son, story about daughter
daughter, son, story about son
daughter, daughter, story about first daughter
daughter, daughter, story about second daughter

Now your answers are, "2" and "story about son" for possibilities 1, 2, 3, and 6. So out of 4 equally likely situations, in half of them the other child is a son. Note that your initial odds of having the second answer not being the actual answer that you got are 1/2. The extra situations eliminated now that weren't before are the cases where she had a son and a daughter, and decided to talk about her daughter rather than her son.

Looking at that explanation, do you see how important it is for the probability calculations to not just be clear on what did happen, but also on what could have happened?

Comment on Re^7: Marilyn Vos Savant's Monty Hall problem (odd odds)

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Re^8: Marilyn Vos Savant's Monty Hall problem (odd odds) by bollen (Novice) on Aug 24, 2004 at 12:41 UTC
I don't get it. Why include both "son,daughter" and "daughter,son" in the equation, when the order is - as far as I can see - ireelevant? To me it's 50/50 anyhow, but thanks for a particularly well documented answer :)	[reply]
Re^9: Marilyn Vos Savant's Monty Hall problem (odd odds) by zby (Vicar) on Aug 24, 2004 at 14:40 UTC
To start with an even probability distribution. If you would take unordered pairs that is "two sons", "two daughters" and "son and daughter" the last case would have two times bigger probability that the first two and in my opinion that would make it more difficult to think about.	[reply]
Re^9: Marilyn Vos Savant's Monty Hall problem (odd odds) by tilly (Archbishop) on Aug 24, 2004 at 14:23 UTC
Pedagogical reasons. While you're right that there is a symmetry that makes the order not matter, I wanted to avoid all potentially confusing shortcuts. Being explicit may bore, but it is less likely to confuse.	[reply]