I'd like to see the official rules of Toss Up, just to be sure of the facts this time, but I can't seem to find them on-line.
I take it back: they're at http://www.patchproducts.com/pdf/7367_TossUprules.pdf and it looks like my version is the official one: * Once your points are entered on the score sheet, they are safe, and you cannot lose them. Here "entered on the score sheet" is what happens to points when you voluntarily end your turn. It's okay, Thane, I won't tell your kids if you won't. --Michael Kleber
From the mathematical point of view, it actually makes a big difference. The game is hard to solve because it is potentially loopy: to decide whether to stop or roll again, you need to compare P(win|stop) with P(win|roll) = P(going bankrupt)*P(win|state after bankruptcy) + other terms. In the small-penalty version that I was thinking about, state-after-bankruptcy is not so far from current state; you can do an induction on the total number of "banked" points of both players and only need to solve a handful of piecewise-linear equations each time.
But in Thane's big-penalty version, every single decision depends on the probability associated with a bankrupt state in which your total score has reverted to zero, so there's no stratification to help tame the dependencies. In this case solving formally really is hopeless, and the iterative methods we were talking about earlier probably are your only hope.
--Michael Kleber
On 8/29/07, 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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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.