in reply to Re^3: Marilyn Vos Savant's Monty Hall problem
in thread Marilyn Vos Savant's Monty Hall problem

No, your chances go from 1/3 to 2/3. For your chances to be 1/2, you'd have to assume no bias in what is behind the two (remaining) doors, which means you'd win as often by picking the same door you previously picked (which you appear to agree isn't the case).

You start with these odds for each door having the car behind it: A=1/3, B=1/3, C=1/3. You pick "A" (it doesn't matter). Then, if the car is in A, the odds that Monty reveals B are 1/6, and that Monty reveals C are 1/6. If the car is in B, then Monty reveals C (1/3) and changing doors means you win. If the car is in C, Monty reveals B (1/3) and changing doors means you win.

So keeping the same door means you win 1/6+1/6 = 1/3 of the time. Switching doors means you win 1/3+1/3 = 2/3 of the time, or twice as likely.

You can't have the choice of "switch doors or not" result in you winning 1/3 of the time one way and 1/2 of the time the other way. You'd have 1/6 of the odds unaccounted for.

The human mind appears not well suited for dealing with probabilities (unlike calculating parabolic trajectories).

A similar probability "paradox" was taught to me:

You know that your stats professor has two children. Once you heard him mention his son. What are the odds that his other child is a girl?

Our text book said that the odds are equal of boy-boy, boy-girl, girl-boy, girl-girl and knowing that one is a boy makes the girl-girl case impossible and therefore we have 1/3 chance that the other child is also a boy and 2/3 chance that the other child is a girl.

Despite this being in a text book and being taught without challenge in my statistics class, I disagreed with this result.

My argument is that if mentioning a boy can make the odds of the girl-girl case go to zero, then why should it not affect the odds of the other cases as well. It was my assertion that the professor is more likely to mention a son when he has two sons (or is relatively more likely to have had two sons by virtue of having mentioned one of them) and so the odds are 50-50 just like instinct tells us.

You can structure the problem a little more strictly as:

You know that your stats professor has two children and research shows that boy and girl children are equally likely and that the sex of one's first child does not have a statistically significant impact on the likelihood of the sex of one's next child. You flip a (fair) coin and if heads, you ask your professor the sex of his younger child, but if tails, you ask your professor the sex of his older child. If your professor (truthfully) answers "male", then what are the odds that the other child is female?

Odds that the first child is a boy 1/2, girl 1/2. Odds that the second child is a boy 1/2, girl 1/2. So the odds are boy-then-boy 1/4, boy-then-girl 1/4, girl-then-boy 1/4, girl-then-girl 1/4. If you flip heads (it doesn't matter), and the professor answers "male", then we have the odds b-b 1/4, b-g 1/4, g-b impossible, g-g impossible. Odds that the other child is a girl: 50%.

But in the restructured problem, you not only know that the professor has a son, but also whether the son in question is the older or younger child. So perhaps this difference changes the odds and we've over-thought the problem to come up with an incorrect answer just to justify our initial, instinctive answer.

There is also a check we can perform.

If we go by the text book's interpretation but don't specify whether the sex that we heard mentioned by the professor was "boy" or "girl", then we have the following matrix of possibilities: 

 children's sexes
 ---------------- heard
 b-b b-g g-b g-g  -----
 1/3 1/3 1/3  0   "boy"
  0  1/3 1/3 1/3  "girl"

And since there is no reason to bias for "boy" or "girl" being heard, we assign each of those odds of 1/2 so that combined odds are: 

      children's sex
      ---------------  heard
total b-b b-g g-b g-g  -----
 1/2  1/6 1/6 1/6  0   "boy"
 1/2   0  1/6 1/6 1/6  "girl"
  1   1/6 1/3 1/3 1/6  total

which violates our initial common-sense assertion. That is, having heard the sex of one of his children suddenly makes it twice as likely that his children are not the same sex. Sounds rather magical.

So, was my text book wrong?

I think it was, but I'm even more sure that this demonstrates that humans aren't very good a analyzing probabilities.

- tye        

  • Comment on Re^4: Marilyn Vos Savant's Monty Hall problem (odd odds)

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Re^5: Marilyn Vos Savant's Monty Hall problem (not so odd odds)
by bmann (Priest) on Aug 23, 2004 at 18:13 UTC
    So, was my text book wrong?
    Yes, your textbook was wrong. Marilyn expounded on this in one of her columns, and she was wrong too.

    The logical mistake she made is that we don't know if it the professor was talking about the older or the younger child! Knowing this, there are two cases of boy-boy (or girl-girl) - one where he is talking about the younger and one where he is talking about the older.

    Let's pretend you hear "boy" and you don't know the relative ages. You have a 1 out of 2 chance the child is the oldest or the youngest. I'll put a star by the child we are assuming to be a boy and I'll leave out the girl-girl red herring.

    total  gender
      b*-b b*-g (assume prof talking about the older)
 1/2  1/4  1/4  
      b-b* g-b* (assume prof talking about the younger)
 1/2  1/4  1/4   
  1   1/2  1/2

    [download]

    Let's change boys/girls to heads/tails and assume that we are talking about flipping a coin. Toss a coin and we know that one of the tosses is heads, but not which one. Using the same logic your textbook presented, there are 4 possibilities - HH, HT, TH, and TT - therefore the odds are 2/3 that the other one is tails??? This is easily tested and refuted. Toss two coins without looking. Look at either one, then the other. Just knowing what one of the coins is has no impact on the probability that the other coin is heads or tails.

    The bottom line is that calculating probablility is not intuitive, and the original problem is like tossing a coin. The fact that someone flips a coin and it came up heads (or girls or boys ;) has no effect of the odds of the second toss.

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[d/l]
Re^5: Marilyn Vos Savant's Monty Hall problem (odd odds)
by tilly (Archbishop) on Aug 23, 2004 at 20:28 UTC
    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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      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.

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        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?

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