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Two girls on an island problems

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COUNTERINTUITIVE PROBABILITIES REDUX

islnd

Drawn in by our post Tuesday’s child is full of probability problems, author and Professor of Operations Research and Probability Henk Tijms writes in with two new puzzles:

Problem 1:

An isolated island is ruled by a dictator. Every family on the island has two children. Each child is equally likely a boy or a girl. The dictator has decreed that each first-born girl (if any) in the family should bear the name Mary Ann (the name of the beloved mother-in-law of the dictator). Two siblings never have the same name. You are told that a randomly chosen family that is unknown to you has a girl named Mary Ann. What is the probability that this family has two girls?

Problem 2:

The dictator has passed away. His son, a womanizer, has changed the rules. For each first-born girl in the family a name must be chosen at random from 10 specific names including the name Mary Ann, while for each second-born girl in the family a name must be randomly chosen from the remaining 9 names. What is now the probability that a randomly chosen family has two girls when you are told that this family has a girl named Mary Ann? Can you intuitively explain why this probability is not the same as the previous probability?

If you need a hint, he adds this postscript:

P.S. As you know, the wording in this kind of problems is crucial. I found that the best approach to attack this kind of problems is to use Bayes’ rule in odds form. This specific form of Bayes forces you to make transparent the assumption you are (implicitly) making in solving the problem. I take the liberty to mention that in the recent third edition of my book Understanding Probability (Cambridge University Press, 2012), I advocate the use of Bayes’ rule in odds form (and Bayesian thinking in general).

Who can solve it first?

7 Comments

  1. Karim says:

    Well, I’m not sure, so maybe I need the book. I got 1/3 for Problem #1, and 1/2 for Problem #2. The second one is higher because whereas in P1, the probability of having a girl named Mary Ann is the same for families with 1 or 2 girls (p = 1), for P2, the probability of having a girl named Mary Ann is much higher for families with 2 girls.

    One remaining question, is Henk’s mother-in-law actually named Mary Ann??

    February 26, 2013 @ 10:40 am

  2. Nate says:

    For me, the intuition on P2 is that the Mary Ann in question is equally likely to be either the first or second born, regardless of whether the family is a BG or GG family.

    February 26, 2013 @ 12:40 pm

  3. Eliezer Yudkowsky says:

    Am I missing something? I’m a pretty experienced Bayesian and I’m getting 1/2 for both cases. In case 1, being told that the family has a girl named Mary Ann gives a posterior probability of ~1 the first child was a girl (this name is far more likely to be chosen for first-born girls than any other) so we’ve got a 50% probability the other child is a girl. In Bayesian terms, families with two girls are twice as likely to have a first-born girl as families with 1 girl, hence twice as likely to have a Mary Ann, hence prior odds of 1:2 for GG vs. GB times a likelihood ratio of 2:1 for “Mary Ann” yields posterior odds of 1:1.

    In the second case, a family with 2 girls is again twice as likely to have a girl named Mary Ann as a family with 1 girl (there are 2 girls instead of 1 who might randomly be assigned that name) so prior odds of 1:2 times a likelihood ratio of 2:1 again equals posterior odds of 1:1 that both children are girls.

    Am I misunderstanding the problem? This looks almost exactly like a practice problem that I designed for one of my own sessions on Bayesianism so I might be misinterpreting it to say the same thing my practice problem did.

    February 26, 2013 @ 11:58 pm

  4. Henk Tijms says:

    Eliezer , the wording in the first problem may be not perfect. It is said that the dictator has decreed that the first girl (if any) who will be born in the family should bear the Mary Ann. The family has no choice: the girl MUST bear the name Mary Ann.

    February 27, 2013 @ 2:10 am

  5. Eric says:

    (Sorry if this posted twice)

    Eliezer, you are correct given how you interpreted the problem. This is why the postscript was added about the wording being crucial and Bayes’ rule forcing you to make your assumptions transparent.

    Karim (and I) interpreted the phrase “first-born girl” to mean the first girl born to a couple, even if it is the second child to be born. Thus we interpreted the dictator’s law in the first problem as saying, “Mary Ann shall be the name of the first girl a family has, even if they have a boy first, and if they have a girl first, the second girl shall be given another name,” whereas you interpreted the law as saying, “Mary Ann shall be the name of a girl only when it is the first child of a couple.”

    Your interpretation means that only half of all non-GG families with girls will have a girl named Mary Ann. Under our interpretation, the first problem is just asking you to figure out the probability a family is GG given the fact they have a girl, since all families with at least one girl have a girl named Mary Ann. And that probability is 1/3. As you said, GG families are twice as likely to have a girl named Mary Ann in the second problem, so the probability is 1/2.

    One problem with interpreting the “first-born girl” naming rule as you did is that you are forced to assume all BG families in the first problem never name their second children Mary Ann, although we are not given any information about a rule against naming second-born children Mary Ann if they are the first girl born to a family. I think the second problem also must imply our interpretation, since the second-born girl’s list is shortened by the selection of the first-born girl, (“for each second-born girl in the family a name must be randomly chosen from the remaining 9 names,”) which would not be possible in BG families.

    February 27, 2013 @ 2:40 am

  6. Henk Tijms says:

    Eric, thanks for your clarification. On second thought about the first problem, I think that it is not relevant for the asked probability whether any second girl in the family may also bear the name Mary Ann after the first girl born in the family has received the name Mary Ann. This follows from a closer examination of the derivation of the probability using Bayes’ rule in odds form.

    February 27, 2013 @ 3:09 am

  7. Eric says:

    Henk, you are right, it is not relevant to the first problem if families with two girls have a second girl named Mary Ann.

    However, it is relevant if families that first had a boy cannot name their first girl Mary Ann because she was not the first born child. That would imply that non-Girl,Girl families with at least one girl are half as likely to have a girl named Mary Ann as Girl,Girl families. And since there are twice as many non-Girl,Girl families, half of the first-born Mary Ann’s are with them (in the Girl, Boy families) and half are in the Girl, Girl families.

    February 27, 2013 @ 4:23 am

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