October 2009 - mathfun@mailman.xmission.com

Re: [math-fun] best published explanation of the Monty Hall Paradox?
by Dan Asimov 29 Oct '09

29 Oct '09

Victor wrote: << . . . However, another possibility is that Monty chooses a door at random in which case b = 1/2, yielding the answer of 1/2. So the correct answer depends on what procedure Monty uses (which we don't know). . . . >> Huh? If the original pick hides a goat, the door Monty opens is forced to be the only other goat, of course. If the original pick hides the car, then it makes absolutely no difference what procedure Monty uses to pick one of the goats. Because the probability that the original pick hides a goat remains 1/3, and so the remaining probability of 2/3 covers both other doors. Since Monty has eliminated one of them, that 2/3 applies to the door the player is allowed to switch to. OR, maybe you're referring to some *different* game from how I defined the game in a previous post? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] best published explanation of the Monty Hall Paradox?
by Dan Asimov 29 Oct '09

29 Oct '09

<< In August 2001, I argued stridently on the fledgling Wikipedia that switching was irrelevant, and wrote code to prove it, only to be proven wrong by my own code. I posted the code as an "empirical example" subpage off the main article. Unfortunately, the current article no longer includes an executable demonstration. >> The key point in the explanation of why switching is better is that the showing of a goat does not affect the probability of 1/3 that the original door is correct. If you agree to that, the other steps are pretty routine and easy to accept. So a good explanation should address this issue. The reasoning is simple: After the initial guess, the two remaining doors play identical roles with respect to what the player knows. The player knows that at least one of them hides a goat. So when a goat is shown, that gives the player no new information about the original guess. This is true regardless of what algorithm may be used to choose the goat door, as long as the player doesn't know the algorithm. And so the probability of 1/3 cannot change. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] best published explanation of the Monty Hall Paradox?
by Dan Asimov 28 Oct '09

28 Oct '09

<< What book or magazine article or website has the best (in your opinion) written explanation of the Monty Hall Paradox for the lay reader? >> Question: --------- Behind one of three doors - A, B, and C -- is a car; behind each of the other two is a goat. The player, P, gets to keep what's behind the door he chooses to open. First P picks one door that remains closed at first, say door A. Now one of the other doors is opened to show a goat -- say door B -- and P gets to make their final choice: open the original choice of door A, or the other door -- call it C -- that's still closed. Which is more likely to get the car? ------------------------------------ Answer: ------- The player should pick door C for the best chance of getting the car. Explanation: ------------ Player P has 1/3 chance of guessing the car with their original choice of door A. Showing the goat sheds no light on whether the initial choice of door A is correct, so its probability remains 1/3. Hence the probability the car is not behind door A must be 2/3. Since door B has been excluded from hiding the car, that 2/3 probability belongs to door C. QED --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] fractal sponges
by Jeannine Mosely 28 Oct '09

28 Oct '09

Some of you may know that I built a large origami model of Menger's sponge a few years ago out of 66,048 business cards. Recently I have been playing around with a different fractal sponge and I'm wondering if anyone else has studied it or named it. I have tried some web searches, but if this other sponge is out there I cannot find it among the thousands of sites that mention Menger's sponge. To make Menger's sponge, take a cube and divide it into 3x3x3 smaller cubes, then throw away the cubes in the centers of the faces and in the center of the body, keeping only the corner and edge cubes, then repeat this process with the remaining cubes an infinite number of times. To make my sponge, at each iteration throw away the corners and the body center, keeping the edge and face cubes. Some people protest that there is no point in throwing away the body center cube because you cannot tell that it is missing. But this is only true at the first iteration -- after that the missing corners allow you to see into the holes. I have modeled it in Mathematica and it looks pretty cool. When viewed along one of the original cube's long diagonals, this fractal is riddled with holes that have a cross section shaped like a snowflake curve. Because of this I have been calling it a Snowflake Sponge. If it already has a name I'd like to know. You can create other fractals by keeping different subsets of the sub-cubes. I wrote my Mathematica program to take as arguments a list of numbers from 0 to 3, indicating whether or not to keep the corner, edge, face or body cubes. If you want it to be connected then obviously the numbers have to be consecutive. Menger's sponge is {0,1}, my sponge is {1,2}. {2,3} makes an interesting fractal but it's more a tree than a sponge. -- Jeannine Mosely

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Re: [math-fun] Interesting gap in OEIS
by N. J. A. Sloane 23 Oct '09

23 Oct '09

> ... it's not on OEIS ... and I'm not sure it should be. ... Well, A098720 + 1 (now called A152301) comes pretty close! (Superseeker found the connection) Neil

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[math-fun] Interesting gap in OEIS
by Allan Wechsler 23 Oct '09

23 Oct '09

I found a sequence that many participants in this forum must have brushed into in the past. I think I'm not the only one here who calculated the first dozen or twenty entries in this sequence back in the 1970s. And yet it's not on OEIS ... and I'm not sure it should be. I don't want to give the definition in this message ... I want to see how many other math-funsters recognize it. It's zero-origin, and a[5] is the place OEIS gives up. Here are the values of a[0] through a[20]; let's have the first few replies to this message merely affirm that they recognize the sequence, or express bafflement, and we can give the answer a little later. 0 1 1 0 2 6 12 14 48 20 2 15 15 24 28 40 32 24 20 25 20 After we do the reveal, I'd like to know how much is known about the asymptotic behavior of the sequence.

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Re: [math-fun] linked spheres
by Dan Asimov 23 Oct '09

23 Oct '09

Fred wrote: << On 10/20/09, Dan Asimov <dasimov(a)earthlink.net> wrote: << Linking is a fascinating phenomenon in topology, but the only chance of proving anything about it will require the use of a definition of linking. >> Which is not supplied; although at a later stage (ex 5.8) a definition is requested! It turns out that for standard k-spheres, a simple and intuitive criterion is available. >> Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order. (In other words, L intersects the spheres in two linked 0-spheres.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] linked spheres
by Dan Asimov 23 Oct '09

23 Oct '09

***Spheres Rule*** (for separating unlinked circles in 3-space) Rich wrote: << (b) Fred's separability test seems to work if we allow spheres as separators (instead of just allowing planes). >> As I previously posted, the family of all planes and spheres is sufficient to separate any two unlinked circles in 3-space. (This after initially believing that spheres alone are not sufficient). But Fred kindly mentioned a proof he has that spheres alone are sufficient. And now I see that -- since the family of planes and spheres is sufficient -- it follows that spheres alone must be sufficient. (For, any plane can be approximated arbitrarily well, in an arbitrarily large finite region of space, by spheres.) Finally, this might be what Rich was saying all along. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] An a-Rouse-ing question
by Thane Plambeck 22 Oct '09

22 Oct '09

Anyone know the answer to this? ******************** Hi Thane, I read books onto cd for the blind at the Canadian National Institute for the Blind, and I am looking for a site that will give me the correct pronunciation of (Walter William) Rouse Ball. Given your comprehensive site, I wonder if you might know whether the pronunciation is ROUSS (a kind of hissing s) or ROWSE where s sounds like z...? I have found both options, and am left clueless. Thank you for any help you might give me. The book I am working on is Alan Turing:The Enigma. I know little on the subject, but he is a fascinating, heartbreaking man. Regards, Jennifer -- Thane Plambeck tplambeck(a)gmail.com http://thaneplambeck.typepad.com/

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Re: [math-fun] An a-Rouse-ing question
by N. J. A. Sloane 22 Oct '09

22 Oct '09

ROUSS (a kind of hissing s) is correct! Neil

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