June 2006 - mathfun@mailman.xmission.com

[math-fun] Re: Error in Scientific American math puzzle.
by James Propp 12 Jun '06

12 Jun '06

> "A family has two children, at least one of whom is a boy; what > is the probability that both are boys?" > > and > > "A family has two children, at least one of whom is a boy named > Bartholomew; what is the probability that both are boys?" > > have different answers. Well, I could be wrong; these matters are notoriously slippery. But let me reconstruct my understanding (it's been a few years since I taught probability) and then you can tell me if you buy it. I'll start by taking probability theory out of the problem and turning the two questions into demographic questions: "Of those families that have two children, at least one of whom is a boy, what fraction are two-boy families?" "Of those families that have two children, at least one of whom is a boy named Bartholomew, what fraction are two-boy families?" Among four million two-child families, we'll have three million that have at least one boy, and one million of those families are two-boy families, so the answer to the first question is 1 million divided by 3 million, or 1/3. For the second question, let's assume (for definiteness) that one child in a thousand is named Bartholomew. So, of the two million families with one boy and one girl, there are 2000 families with a boy named Bartholomew, and of the one million families with two boys, there are 1000 two-boy families in which the older boy is named Bartholomew and 1000 two-boy famlies in which the younger boy is named Bartholomew. (There may also be 1 family in which both boys are named Bartholomew --- cf. the Dr. Seuss poem "Too Many Daves" --- but they are too few to skew the demographics by much.) That makes 2000 two-boy families in which at least one of the children is a boy named Bartholomew. So the answer to the second question is 2000 divided by 2000+2000, or 1/2. If you think my reasoning is wrong (not just dependent on unstated assumptions that might shift the answers slightly from 1/3 and 1/2, but seriously wrong), please let me know where my mistake is. Jim Propp

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[math-fun] Re: Error in Scientific American math puzzle.
by James Propp 12 Jun '06

12 Jun '06

The way I usually teach this topic in introductory probability classes is that (under assumptions that need to be made explicit but are very reasonable), the question "A family has two children, at least one of whom is a boy; what is the probability that both are boys?" and "A family has two children, at least one of whom is a boy named Bartholomew; what is the probability that both are boys?" have different answers. I learned this gimmick (giving the boy a name) from Charles Grinstead or Peter Doyle (I'm not sure which). Jim Propp

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Re: [math-fun] Error in Scientific American math puzzle.
by dasimov＠earthlink.net 12 Jun '06

12 Jun '06

Thanks, Franklin: << Oh, yes, the following should be easy assuming a society where boys and girls are born in roughly even numbers. If you know a family has two kids and you see one playing outside and the kid is a girl, what is the chance that the other is a girl? >> Interesting that even if a parent informs you that they have two children, and flips a coin to decide which child's sex to divulge to you (and it's a girl), the probability is still 1/2 that the other child is a girl. --Dan

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[math-fun] Error in Scientific American math puzzle.
by Andy Latto 12 Jun '06

12 Jun '06

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(a)pobox.com

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Re: [math-fun] Error in Scientific American math puzzle.
by dasimov＠earthlink.net 12 Jun '06

12 Jun '06

Could someone please restate the original flawed question at sciam.com ? (The link provided is now defunct.) Thanks, Dan

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[math-fun] Tower of Hanoi problem
by David Wilson 10 Jun '06

10 Jun '06

In the n-plate Towers of Hanoi problem, consider the fixed base of each peg to be plate 0, and number the largest through smallest movable plates 1 through n. In the solution of the problem, how many times is plate p placed on top of plate q (0 <= p,q <= n)?

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[math-fun] domino question
by David Wilson 09 Jun '06

09 Jun '06

For any n, does there exist a polyomino with exactly n distinct domino tilings?

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RE: [math-fun] Latin square question
by dasimov＠earthlink.net 09 Jun '06

09 Jun '06

Christian writes of Mike Reid's nice example: << However, you are using the particularity that Dan allows to not analyze all the lines, if they are empty (why, Dan?). >> I agree with your implication that it's a more consistent condition if it's applied to *every* row or column, regardless of how many squares are filled initially. (I.e., it should apply to any row or column with k squares filled, 0 <= k <= N for an NxN square.) Your example is very nice too, Christian, especially considering that you posted it within about one minute of my question! << If we extend the Dan's condition to all the lines, empty or not, we can directly see that the last row (and the last column) cannot be filled. >> --Dan

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RE: [math-fun] Latin square question
by Michael Reid 09 Jun '06

09 Jun '06

> I'm afraid I may not have expressed myself clearly. What I meant was, suppose a matrix is partially filled. > These entries will never change in this question. > > CONDITION (*): Suppose that any *individual* row that already contains at least one filled square satisfies the condition that its unfilled squares can be filled without repetition so that there is also no repetition in any column. *And* the same thing after switching the roles of rows & columns. > > This is clearly a prerequisite to extending the originally partially-filled matrix to a Latin square. > > Question: Is this sufficient to know the original partially-filled matrix can be extended to a Latin square? how about putting an (n-1) x (n-1) latin square in the upper left corner? if i parsed it correctly, that appears to satisfy your condition (*), but obviously (well, for n > 2 ) cannot be extended to an n x n latin square. mike

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Re: [math-fun] domino question
by Michael Reid 09 Jun '06

09 Jun '06

> For any n, does there exist a polyomino with exactly n distinct domino tilings? yes, consider the following sequence of polyominoes: +--+ | | +--+ | | +--+ +--+--+ | | | +--+--+ | | | +--+--+ +--+--+ | | | +--+--+ | | | +--+--+ | | | +--+--+ +--+--+ | | | +--+--+--+ | | | | +--+--+--+ | | | | +--+--+--+ +--+--+ | | | +--+--+--+ | | | | +--+--+--+ | | | | +--+--+--+ | | | +--+--+ +--+--+ | | | +--+--+--+ | | | | +--+--+--+--+ | | | | | +--+--+--+--+ | | | | +--+--+--+ +--+--+ | | | +--+--+--+ | | | | +--+--+--+--+ | | | | | +--+--+--+--+ | | | | +--+--+--+ | | | +--+--+ +--+--+ | | | +--+--+--+ | | | | +--+--+--+--+ | | | | | +--+--+--+--+--+ | | | | | +--+--+--+--+ | | | | +--+--+--+ ... (this sequence is surely well-known.) mike

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