[math-fun] Error in Scientific American math puzzle.
I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-1270-8 4C783414B7F0000 is incorrect. But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors. Andy Latto andy.latto@pobox.com
I think the solutions online already have the correction (that the problem as stated should have a probability of 1/2, not 1/3). I like the analysis incorporating "which kid is outside" as a third letter in the string. --Joshua Zucker On 6/12/06, Andy Latto <Andy.Latto@gensym.com> wrote:
I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-1270-8 4C783414B7F0000 is incorrect.
But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors.
Andy Latto andy.latto@pobox.com
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You're quite right, Andy. I think I've seen this puzzle stated incorrectly more often than stated correctly; it's hard to say it right. But if you walk up to the house of a family which you know has two kids and there's a girl playing in the yard, the probability that her sibling is a girl is indeed 1/2, not 1/3. "The child playing outside" has distinguished the two kids. I'm surprised; Dennis Shasha is usually more careful than that. --Michael Kleber On 6/12/06, Andy Latto <Andy.Latto@gensym.com> wrote:
I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-1270-8 4C783414B7F0000 is incorrect.
But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors.
Andy Latto andy.latto@pobox.com
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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
I had a scarier experience just recently. Scientific American published a Domino sudoku by me, but the artist left out the domino placements. After looking at an issue in terror, I spent some hours trying to solve the puzzle as published. Fortunately, the puzzle was still solvable. In fact, it was a much improved puzzle, going back and forth between a domino tiling problem and a sudoku problem. Page 3 of of the following PDF gives the published and intended puzzle: http://sciam.com/media/pdf/sudoku_solutions_REV.pdf Ed Pegg Jr
Andy Latto wrote:
I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-1270-8 4C783414B7F0000 is incorrect.
But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors.
Andy Latto andy.latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I completely concur with you, but that this is notoriously tricky AND that the article got it wrong. The probability you seek is NOT: P{both are girls | at least one is a girl} but rather P{both are girls | the one outside is a girl} In the author's examples, it's like the penny and dime where the dime is known. David
Actually, the answer to question 3 also contains incorrect statement. 8/15 is not four times 2/5 (although it is greater). Additionally, to get the 1/2 result for question 4 implicitly depends on another assumption: boys and girls are equally likely to be playing in the yard. If, for example, we assume instead that two girls will always play together, then the probability that the other child is a boy is 1. Franklin T. Adams-Watters -----Original Message----- From: Andy Latto <Andy.Latto@gensym.com> I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010& articleID=000E3F5B-04C7-1270-84C783414B7F0000 is incorrect. But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors. Andy Latto andy.latto@pobox.com ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com
As already pointed out, the article has the "1/3 should be 1/2" correction; and its statement that 8/15 = 4 * 2/5 is a bit of an exaggeration. But in the boys and girls problem, knowing that boys and girls are equally likely is not sufficient information to solve the problem. It might be that the probability of having same-sex siblings is significantly greater than (or less than) 1/2. In actuality, I suspect it is significantly greater than 1/2. The determination of the gender of a child includes a number of factors that are specific to its parents, including such things as genetics and sexual practices. For example, didn't Henry VIII's wives have some problem with that? --ms Andy Latto wrote:
I think the solution to the last of the probability puzzles on the Scientific American web site at http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-1270-8 4C783414B7F0000 is incorrect.
But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors.
Andy Latto andy.latto@pobox.com
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Actually, the probability of having same-sex siblings is very close to 1/2. The only deviation of greater significance than the actual variation of birth distribution from 1/2 is that due to identical twins. And that's pretty small. Anecdotal evidence greatly exaggerates the frequency of same-sex siblings. Franklin T. Adams-Watters -----Original Message----- From: Mike Speciner <speciner@ll.mit.edu> As already pointed out, the article has the "1/3 should be 1/2" correction; and its statement that 8/15 = 4 * 2/5 is a bit of an exaggeration. But in the boys and girls problem, knowing that boys and girls are equally likely is not sufficient information to solve the problem. It might be that the probability of having same-sex siblings is significantly greater than (or less than) 1/2. In actuality, I suspect it is significantly greater than 1/2. The determination of the gender of a child includes a number of factors that are specific to its parents, including such things as genetics and sexual practices. For example, didn't Henry VIII's wives have some problem with that? --ms Andy Latto wrote:
I think the solution to the last of the probability puzzles on the Scientific American web site at
http://www.sciam.com/article.cfm?chanID=sa010&articleID=000E3F5B-04C7-127 0-8
4C783414B7F0000 is incorrect.
But since this sort of puzzle is notoriously tricky to get right, I thought I'd check with fellow math-funsters rather than risk embarrassing myself with an incorrect letter to the editors. Andy Latto andy.latto@pobox.com
___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com
Mike Speciner speciner@ll.mit.edu said:
It might be that the probability of having same-sex siblings is significantly greater than (or less than) 1/2. In actuality, I suspect it is significantly greater than 1/2. The determination of the gender of a child includes a number of factors that are specific to its parents, including such things as genetics and sexual practices. For example, didn't Henry VIII's wives have some problem with that?
--ms
This brings up a point that has always troubled me a bit. It's challenging to state all of the assumptions required to solve artificial "real-world" word problems. But in my view, stating these assumptions should be left to the problem solver, and not the problem poser. Problems would be too dry if every such problem clearly stated every assumption. It's best to leave it to the reader to make some reasonable assumptions for the purpose of fleshing out an otherwise underspecified problem. While only the most sophisticated of solvers will correctly identify these assumptions, a good solution should include them. In the case of this problem, I don't think it's unreasonable to make the usual (if inaccurate) assumptions even if not all surviving births are independently XX or XY with 50-50 probability. But Mike is dead-on in pointing out that a complete solution explicitly includes these assumptions.
participants (7)
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Andy Latto -
David Wolfe -
franktaw＠netscape.net -
Joshua Zucker -
Michael Kleber -
Mike Speciner -
Xeipon