Your possible sets of kids that we assume are equally likely:
  1. son, son
  2. son, daughter
  3. daughter, son
  4. 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:

  1. son, son, story about first son
  2. son, son, story about second son
  3. son, daughter, story about son
  4. son, daughter, story about daughter
  5. daughter, son, story about daughter
  6. daughter, son, story about son
  7. daughter, daughter, story about first daughter
  8. 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?

In reply to Re^7: Marilyn Vos Savant's Monty Hall problem (odd odds) by tilly
in thread Marilyn Vos Savant's Monty Hall problem by mutated

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