should have said "never 4 or fewer" instead of "never fewer than 4" On 8/28/07, Thane Plambeck <tplambeck@gmail.com> wrote:
I offer this Toss Up scoresheet, in which I'm proud to say I figure as "T", the winner of a three-way game against my two sons:
http://www.flickr.com/photos/thane/1263170453/
I play a conservative game, never rolling 3 or fewer dice, and when in the lead, never fewer than 4. After I have the lead, the boys feel compelled to "catch up" after their rash bankruptcies and are then led into repeated subsequent bankruptcies.
In most games I plod inexorably to victory. * * On the "original" rules pointed out by Michael Kleber: suppose two players both have 99 points. I think both players would want to "pass", since it's got to be better to let the other guy cross 100 first, and see where he lands up before rolling your own dice? Or maybe not?
On 8/27/07, Schroeppel, Richard <rschroe@sandia.gov> wrote:
The original paper wants x to be a vector, A to be a matrix, and b,b' to be vectors. Two obvious methods of solving it are (a) assume Left or Right for each component of Max, solve the resulting linear system, and select out the combinations which match the Left/Right choices; (b) assume a starting guess for the vector x, and iterate, hoping for convergence.
I actually did something similar once-upon-a-time, solving 2-checker Backgammon. I assigned upper and lower bounds for the probability of winning, to each Backgammon position. These were initialized to 1.0 and 0.0, and then a routine calculated updated probabilities based on the Backgammon rules. I had to leave out the doubling cube (memory constraint). The loop converged very quickly, with upper & lower probabilities being equal or adjacent. But then I faced the problem of making sense of the data, which I never really solved.
Rich
-----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Monday, August 27, 2007 1:57 PM To: math-fun Subject: Re: [math-fun] Toss Up Komi
As Neller and Presser say, "There is no known general method for solving equations of the form x = max(Ax+b, A'x+b')."
Assuming A, A', b, b' are fixed given numbers, the graph of y = max(Ax+b, A'x+b') is continuous, piecewise linear, and consists of two segments. Its intersection with y = x consists of 0, 1, 2 or points or a continuum, and these are the solutions.
Gene
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-- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm