March 2007 - mathfun@mailman.xmission.com

[math-fun] today's Jeopardy! outcome: nonzero three-way tie
by Thane Plambeck 17 Mar '07

17 Mar '07

The first three-way nonzero tie in the history of the program occurred in the program aired today. I copied this interesting (if slightly flawed) analysis from http://newsfromme.com/ * * * **snip * * * ** No Winner, No Losers I know we've been paying too much attention to this one Jeopardy! episode but I wanted to write one more post about how it's unlikely to happen again. The three-way tie was an anomaly in large part because one player didn't play the game to win. Going into Final Jeopardy!, here's how the totals stood... * Scott: $13,400 * James: $8,000 * Anders: $8,000 So how much do you wager if you're one of those contestants? If you're James or Anders, you'd presume the following: That you can't win unless you're right and Scott is wrong. In theory, you could win the game with the wrong answer if Scott was also wrong and he wagered so much that his total dropped to below where you wound up, but you have to operate from the assumption that he's not dumb enough to do that. James and Anders made the right wagers. Neither one could afford the risk that the other would bet more than he did. You'd also assume that the odds are that it's unlikely you will get the answer right and the other two will both get it wrong. That's possible but it's much more likely that it either stumps everyone or no one. So you'd wager as each of them did: The full eight thousand. Now, let's say you're Scott. You'd assume from the above that at least one of the other two players — and probably both — is wagering $8,000. A person who does that will beat you unless you're also right and you wind up ahead of them. Therefore, the correct bet for you is at least $2,601. That way, one or both of them could wind up with $16,000 and you'd have $16,001. You'd also win if all three of you were wrong. The only way you could lose is if you're wrong and one or both of the others is right...but that's going to be true no matter what you bet. So at least $2,601 is the proper bet here. The trouble is that Scott didn't bet that way. I'm assuming the Game Theory expert opined that this wouldn't happen again because future players will recognize that Scott blew his chance to win and they won't make that mistake. But is it really much of a mistake? If Scott had bet $2,601, he'd take home $16,001 and come back on Monday to play against two new opponents. Since he bet $2,600, he takes home $16,000 and comes back on Monday to play against the same two opponents again. Not much difference from his standpoint. Scott didn't really lose anything except being able to say he won. * * * Also, this comment from Roy Leban on the puzzlers.org list: * * * Not the greatest analysis. How could anyone think that a computer science professor who had already won twice, once betting $1401 once, wouldn't know exactly what it would take to win. I figured it was a three-way tie early on when Squonk [ie Scott] didn't sweep the board (and, wow, that first board was tough for me too! I think I might have had a $0 at the end, if I was lucky). When I saw the twin $8,000's, it seemed inevitable. Very cool! > Yeah, very exciting! Infinity noticed that the guy in the middle must > have said, "Thank you," since during Squonk's close-up he shakes the > dude's hand and says, "You're welcome." My wife and I thought Squonk also said "You're welcome" earlier. He turned to the player next to him right before his answer was unveiled and said something. And "You're welcome" is right -- he could have won and had some more money, but he essentially gave those two guys $16,000 each. Plus, it should be an interesting rematch on Monday. Squonk, I hope you'll fill us all in at some point about what went through your mind and what the other players and Alex thought about it. Is there another three-way tie on Monday?! -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] Centrist Groupoids, Universal Equation, Somos Sequences & Modular Theta functions, ExpMath slides
by Richard Schroeppel 16 Mar '07

16 Mar '07

Centrist Groupoids Some of you will recall Central Groupoids, which are systems with a binary operator (magmas, shudder) that satisfy the equation AB.BC=B (or, (AB)(BC)=B), and have the curious property that the finite systems have a square number of elements. I decided to look at the relaxed equation AB.BA=B, which, after 30 seconds of deep thought, I have decided to call Centrist Groupoids. (Medial Magmas?) Of course, any Central Groupoid is also Centrist. The rule is anti-commutative. How Many of Order N? There are no 2- or 3-element systems. There are 58 4-element systems, and 6634 5-element systems, and no/nada/nil/0/zilch 6-element systems. Universal Equation Is there an equation that is true for all 2-element binary operators? (Besides A=A. :-)) PDP10 aficionados will recall the 16 boolean instructions (ANDCA, etc.). There are 10 non-isomorphic tables, or 7 if we allow argument swapping. (Zero, Left, And, Xor, Nor, ~Left, Andcm.) All of them satisfy the identity A:.A.AA:A = A:.AA.A:A or A{[A(AA)]A} = A{[(AA)A]A}. Puzzle: Show that similar equations exist with more letters, and for systems with more than 2 elements, and for ternary operators (tragmas). Problem: Find some. Somos Sequences etc. Gosper has come up with some interesting determinant identities for Somos sequences. I've finally updated my web page. Gosper's results, along with my foray into defining theta functions mod 43, are on the new page. I've also put up the slides from my Experimental Math talks, which include contributions from several Funsters. (counting polyhypercubes, ...) The page is at http://www.cs.arizona.edu/~rcs Rich rcs(a)cs.arizona.edu

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[math-fun] bad pun
by R. William Gosper 15 Mar '07

15 Mar '07

http://gosper.org/rasta.png

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[math-fun] Probably dumb question about primes
by Daniel Asimov 14 Mar '07

14 Mar '07

Goldbach's conjecture is that every even number >= 4 is the sum of two primes. But what's known about this question: << Is every even number the difference of two primes? >> --Dan

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[math-fun] Today is the picentennial (correction)
by R. William Gosper 14 Mar '07

14 Mar '07

of 14-15 Jan 1693

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[math-fun] Today is the picentennial
by R. William Gosper 14 Mar '07

14 Mar '07

of 27-28 Feb 1693. Various sites claim for 1693: Newton's 2nd mental breakdown, 1st published description of fluxions, 1st publication of Wallis's Algebra, Leibniz rediscovers determinants, ..., Not clear what happened on 27-28 Feb, or 16-17 Feb if we count the missing eleven days. --rwg

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[math-fun] Group representation question about PSL(2,p)
by Daniel Asimov 13 Mar '07

13 Mar '07

Let p be an odd prime, and let G denote PSL(2,F_p). (Cf. < http://en.wikipedia.org/wiki/Projective_special_linear_group >.) (Each such G is simple, of order #G = (p^3-p)/2.) 1. What's the least n for which the orthogonal group O(n) contains G as a subgroup ? Denote this n by Or(p). 2. What's the least n for which the symmetric group S_n contains G as a subgroup ? Denote this n by Sy(p). (It's clear that Or(p)+1 <= Sy(p) <= #G since any group permutes itself faithfully, and any S_n embeds in O(n-1) via isometries of the regular (n-1)-simplex.) --Dan

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Re: [math-fun] Re: Simplest Ovals
by Marc LeBrun 13 Mar '07

13 Mar '07

Fascinating and surprisingly long-lived thread! My "natural" choice would be to make ovals by interpolating or blending two ellipses (perhaps from too much computer graphics): Standardize the graphs of the ellipses by aligning them with their major axis coincident with the X axis and scale them so their ends are at x=0 and x=1. These graphs form a family indexed by "eccentricity parameter" [e]: y = ellipse[e](x) Now define oval[e0,e1] := (1-x) ellipse[e0](x) + x ellipse[e1](x) (having "eccentricity vector" [e0,e1]) So if ellipse(x) is quadratic in x then oval(x) will be some cubic. Alternatively, we might define oval[e0,e1] := ellipse[(1-x) e0 + x e1](x) For what parameterizations of eccentricity are these the same?

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[math-fun] Hopefully trivial question about triangles
by Henry Baker 09 Mar '07

09 Mar '07

If you inscribe a triangle within another triangle, so that the inscribed triangle vertices actually touch the 3 sides of the larger triangle, what is the shape of the inscribed triangle of _minimum perimeter_ ? Is it the tangent points of the inscribed circle?

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Re: [math-fun] Hopefully trivial question about triangles
by Michael Reid 08 Mar '07

08 Mar '07

> If you inscribe a triangle within another triangle, so that the inscribed triangle vertices actually touch the 3 sides of the larger triangle, what is the shape of the inscribed triangle of _minimum perimeter_ ? Is it the tangent points of the inscribed circle? for acute triangles, it is the "pedal triangle", obtained by joining the feet of the altitudes of the triangle. for right and obtuse triangles, the answer degenerates to the "triangle" which has two vertices at the right/obtuse angle, and whose other vertex is the foot of the altitude from the right/obtuse angle. mike

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