[math-fun] Schroedinger's daughter
OK, I thought that I understood this, but here's a variation: A friend writes a girl's name on a piece of paper and puts it (unopened) in your pocket. A random parent of two children comes up to you and says, "I have a daughter named Mary." At this point, the probability that Mary's sibling is a girl is 1/3 (I think). However, if you look at the piece of paper, and it says "Mary", the probability changes to 1/2 (approximately). Is this correct? Bill C.
No. The point that is generally overlooked in this kind of problem is that how the information is obtained is as important, sometimes more important, than what that information is. In this case, you haven't told us enough to really determine how we obtained the information. On what basis did this parent decide to divulge to us Mary's existence? Without knowing this, we have no way to decide what the probability is after the original statement. To make it more definite, we need to go up to this parent, and ask a strange question like "if you have a daughter, please tell me her name; if you have more than one, please choose one at random and tell me her name". Now the probability is 1/3, both before and after we look at the paper. One way of looking at the reason for this is that we have not, in this general procedure, determined if there is daughter named Mary. There might be, but we could still get her sister's name instead. Franklin T. Adams-Watters -----Original Message----- From: Cordwell, William R <wrcordw@sandia.gov> OK, I thought that I understood this, but here's a variation: A friend writes a girl's name on a piece of paper and puts it (unopened) in your pocket. A random parent of two children comes up to you and says, "I have a daughter named Mary." At this point, the probability that Mary's sibling is a girl is 1/3 (I think). However, if you look at the piece of paper, and it says "Mary", the probability changes to 1/2 (approximately). Is this correct? Bill C.
At 10:38 AM 6/21/2006, you wrote:
No.
The point that is generally overlooked in this kind of problem is that how the information is obtained is as important, sometimes more important, than what that information is.
Ok, here's a concrete proposal. Chose a city with at least 50,000 inhabitants, say Dubuque, Iowa, population 57,686. Get access to census data. I predict 1.Among all families having at lest two children the two eldest will have opposite sexes in about half the cases. 2. Among all Mary's who are the eldest or second eldest child, the other sibling will be a brother in about half the cases. dg
In this case, you haven't told us enough to really determine how we obtained the information. On what basis did this parent decide to divulge to us Mary's existence? Without knowing this, we have no way to decide what the probability is after the original statement.
To make it more definite, we need to go up to this parent, and ask a strange question like "if you have a daughter, please tell me her name; if you have more than one, please choose one at random and tell me her name". Now the probability is 1/3, both before and after we look at the paper.
One way of looking at the reason for this is that we have not, in this general procedure, determined if there is daughter named Mary. There might be, but we could still get her sister's name instead.
Franklin T. Adams-Watters
-----Original Message----- From: Cordwell, William R <wrcordw@sandia.gov>
OK, I thought that I understood this, but here's a variation:
A friend writes a girl's name on a piece of paper and puts it (unopened) in your pocket. A random parent of two children comes up to you and says, "I have a daughter named Mary." At this point, the probability that Mary's sibling is a girl is 1/3 (I think). However, if you look at the piece of paper, and it says "Mary", the probability changes to 1/2 (approximately).
Is this correct?
Bill C.
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Right, assuming not too many families stop at one if and only if the first is a boy - that would throw off 2. Note 2': among all girls who are the eldest or second eldest child, the other sibling will be a brother in about half the cases; in contrast with 3: among all families with at least two children, with at least one of the older two being a girl, one-third have two girls. The difference is that 2' double counts families with two girls - once for each girl. Franklin T. Adams-Watters -----Original Message----- From: David Gale <gale@math.berkeley.edu> Ok, here's a concrete proposal. Chose a city with at least 50,000 inhabitants, say Dubuque, Iowa, population 57,686. Get access to census data. I predict 1.Among all families having at lest two children the two eldest will have opposite sexes in about half the cases. 2. Among all Mary's who are the eldest or second eldest child, the other sibling will be a brother in about half the cases.
participants (3)
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Cordwell, William R -
David Gale -
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