in reply to Re^4: Marilyn Vos Savant's Monty Hall problem (odd odds)
in thread Marilyn Vos Savant's Monty Hall problem

The textbook fell into the common mistake of stating a standard probability problem with insufficient clarity. Textbooks are unfortunately especially prone to this because they crib a lot from each other, and it is very easy to miss an important condition. Unfortunately for the textook, it seems to have messed up the important condition very badly. More on that in a second.

Consider the following situation: You meet someone and ask how many kids she has. She says 2. You ask whether she has a son. She says yes. Assuming that she told the truth, what are the odds that she has a daughter? That question is stated unambiguously. And the reasoning is exactly what is in the textbook, and the answer comes out at 2/3. (Actually it is slightly below 2/3, children are slightly more likely to be male than female.)

Now consider the following alternative. You meet someone and ask how many kids she has. She says 2. You ask her for a random recent story involving one of them and she tells you something about her son Ralph. What are the odds that she has a daughter? That question is stated unambiguously as well, but the correct reasoning is very different. There is no reason to believe that she'd be more likely to come up with a story involving her daughter or her son, and now your reasoning applies and the odds come out at 50%.

Unfortunately for the textbook, the way that it stated the problem is the latter situation. In this case you are right and "the textbook answer" is wrong.

The moral is that when you state a probability problem, you need to think very hard about how you are stating it and be sure to specify not just what did happen, but what could have happened. When you ask the mother a direct question about whether she has a son and she says, "yes", you've eliminated the 1 chance in 4 that she had 2 daughters. When you ask the mother for a story, you've eliminated the 1 chance in 2 that she'll give you a story about a daughter, and that extra 1 chance in 4 that you eliminate is the possibility that she has a son and a daughter, and chose to tell you about the daughter instead of the son.

The questions were different. The possible answers were different. And therefore the appropriate probability calculations are different.

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Re^6: Marilyn Vos Savant's Monty Hall problem (odd odds)
by Aristotle (Chancellor) on Aug 23, 2004 at 23:42 UTC

    When you ask the mother for a story, you've eliminated the 1 chance in 2 that she'll give you a story about a daughter, and that extra 1 chance in 4 that you eliminate is the possibility that she has a son and a daughter, and chose to tell you about the daughter instead of the son.

    I can almost but not quite wrap my head around that. Can you expound all the possibilities you're taking into account here and which you have been eliminated?

    Makeshifts last the longest.

[reply]
      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?

[reply]
        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]

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