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This is good reasoning. Unlike other probability puzzles, the new information isn't useful in this one. P(my first choice was right) = P(my first choice was r

This is good reasoning. Unlike other probability puzzles, the new information isn't useful in this one. P(my first choice was right) = P(my first choice was right | given that I've been shown a goat in another door) It's my second-favorite probability puzzle; my favorite is the one in which the new information has no relevance at all but still changes the probabilities, the "not both daughters" problem.

pklausler 11y ago
I should have just given the other problem, sorry for the laziness. Your friend tells you that she has two children. They're not both boys, she adds. What are the odds that they're both girls? (Obviously 1/3). When she tells you that one child is named Mary, the odds that they're both girls change (to 1/2)! Even though you knew that at least one of the children was a girl, and the name might as well be arbitrary, and the new information seems useless... the odds have changed. It's nontrivial to work this one out; exercise for the reader.
- logfromblammo 11y ago
  After first datum (there are two children): Presuming A and B to be independent events (first child is a girl, second child is a girl) P(A) = 1 - P(!A) = 1/2 P(B) = 1 - P(!B) = 1/2 P(A and B) = P(B|A) * P(A) = 1/2 * 1/2 = 1/4 P(A and !B) = P(!B|A) * P(A) = 1/2 * 1/2 = 1/4 P(!A and B) = P(B|!A) * P(!A) = 1/2 * 1/2 = 1/4 P(!A and !B) = P(!B|!A) * P(!A) = 1/2 * 1/2 = 1/4 Good so far. But the probability later on depends on how you find out that at least one of your friend's children is a girl. If your friend tells you a random fact about her whole family, the probability is 1/3. If she tells you a fact about a randomly chosen child, the probability is 1/2. You can presume from "they're not both boys" that the statement is equivalent to "there is at least one girl". When she tells you the name, you now know "a specific child of mine is a girl". "Not both boys" gives you the precondition "!(!A & !B)" which equates to "A or B". P(A or B) = P(A and B) + P(A and !B) + P(!A and B) = 3/4 P(A and B|A or B) = P(A or B|A and B) * P(A and B) / P(A or B) = 1 * 1/4 / 3/4 = 1/3 "Mary is a girl" gives you either the precondition "A" or the precondition "B" P(A and B|A) = P(A|A and B) * P(A and B) / P(A) = 1 * 1/4 / 1/2 = 1/2 P(A and B|B) = P(B|A and B) * P(A and B) / P(B) = 1 * 1/4 / 1/2 = 1/2 These are both the same number, so that works out fine.
  - logfromblammo 11y ago
    Now see if you can work it out if your friend told you this beforehand: Between pregnancies, I had an accident that resulted in an odd medical condition. The doctors said that after it happened, 50% of boy fetuses that otherwise would have resulted in pregnancy would simply fail to implant, and I wouldn't even know about it. Now there's an all new ambiguity. How do you decide whether your friend was planning on having two children all along, or if she has two children because that's how many times she got pregnant?
    - pklausler 11y ago
      Here's an epistemological head-scratcher: do the odds still change if you don't hear the daughter's name accurately? Why?
      - logfromblammo 11y ago
        Remember, those odds are the level of certainty you have that a given statement may be true based only on what you already know. You can also assign a level of certainty to the facts you think you know. Given the number of ways to mishear "Mary" and the dialectical variant in pronunciation characterized by the Mary-marry-merry merger, I'd tentatively assign the following values: 25% One child is a girl. (Heard correctly) 25% One child is a boy. (Heard incorrectly) 50% No new information. (Incomprehensible or ambiguous) In the first case P(A and B|A) = P(A and B|B) = 1/2 In the second, P(A and B|!A) = P(A and B|!B) = 0 In the third, P(A and B|A or B) = 1/3 In any case, you're just guessing at the probability you might mishear something, so there's no definitive answer. The mere possibility that I might have heard a boy's name means that the overall probability could be anywhere between 0 and 1/2. By the above, my new odds are 5/12. So the odds do change, based on your confidence in what you heard (and your confidence that your friend isn't the kind of parent to name her son "Meriadoc" or something similar).

pklausler 11y ago

I should have just given the other problem, sorry for the laziness. Your friend tells you that she has two children. They're not both boys, she adds. What are the odds that they're both girls? (Obviously 1/3). When she tells you that one child is named Mary, the odds that they're both girls change (to 1/2)! Even though you knew that at least one of the children was a girl, and the name might as well be arbitrary, and the new information seems useless... the odds have changed. It's nontrivial to work this one out; exercise for the reader.

logfromblammo 11y ago
After first datum (there are two children): Presuming A and B to be independent events (first child is a girl, second child is a girl) P(A) = 1 - P(!A) = 1/2 P(B) = 1 - P(!B) = 1/2 P(A and B) = P(B|A) * P(A) = 1/2 * 1/2 = 1/4 P(A and !B) = P(!B|A) * P(A) = 1/2 * 1/2 = 1/4 P(!A and B) = P(B|!A) * P(!A) = 1/2 * 1/2 = 1/4 P(!A and !B) = P(!B|!A) * P(!A) = 1/2 * 1/2 = 1/4 Good so far. But the probability later on depends on how you find out that at least one of your friend's children is a girl. If your friend tells you a random fact about her whole family, the probability is 1/3. If she tells you a fact about a randomly chosen child, the probability is 1/2. You can presume from "they're not both boys" that the statement is equivalent to "there is at least one girl". When she tells you the name, you now know "a specific child of mine is a girl". "Not both boys" gives you the precondition "!(!A & !B)" which equates to "A or B". P(A or B) = P(A and B) + P(A and !B) + P(!A and B) = 3/4 P(A and B|A or B) = P(A or B|A and B) * P(A and B) / P(A or B) = 1 * 1/4 / 3/4 = 1/3 "Mary is a girl" gives you either the precondition "A" or the precondition "B" P(A and B|A) = P(A|A and B) * P(A and B) / P(A) = 1 * 1/4 / 1/2 = 1/2 P(A and B|B) = P(B|A and B) * P(A and B) / P(B) = 1 * 1/4 / 1/2 = 1/2 These are both the same number, so that works out fine.
- logfromblammo 11y ago
  Now see if you can work it out if your friend told you this beforehand: Between pregnancies, I had an accident that resulted in an odd medical condition. The doctors said that after it happened, 50% of boy fetuses that otherwise would have resulted in pregnancy would simply fail to implant, and I wouldn't even know about it. Now there's an all new ambiguity. How do you decide whether your friend was planning on having two children all along, or if she has two children because that's how many times she got pregnant?
  - pklausler 11y ago
    Here's an epistemological head-scratcher: do the odds still change if you don't hear the daughter's name accurately? Why?
    - logfromblammo 11y ago
      Remember, those odds are the level of certainty you have that a given statement may be true based only on what you already know. You can also assign a level of certainty to the facts you think you know. Given the number of ways to mishear "Mary" and the dialectical variant in pronunciation characterized by the Mary-marry-merry merger, I'd tentatively assign the following values: 25% One child is a girl. (Heard correctly) 25% One child is a boy. (Heard incorrectly) 50% No new information. (Incomprehensible or ambiguous) In the first case P(A and B|A) = P(A and B|B) = 1/2 In the second, P(A and B|!A) = P(A and B|!B) = 0 In the third, P(A and B|A or B) = 1/3 In any case, you're just guessing at the probability you might mishear something, so there's no definitive answer. The mere possibility that I might have heard a boy's name means that the overall probability could be anywhere between 0 and 1/2. By the above, my new odds are 5/12. So the odds do change, based on your confidence in what you heard (and your confidence that your friend isn't the kind of parent to name her son "Meriadoc" or something similar).