warren has it right. that's exactly what i did. for example, in the last round, if you have 2 points and only need 1 more, your matrix of probabilities of success look like: 0 1 1 0 which has value 1/2 (both rows and columns should be used with prob 1/2). which means if you are in round 3 of 4, and you have only 1 point, the matrix is: 0 1 1/2 0 which has value 1/3. further backtracking gives the probability of success at any stage. folks who want to explore this in more generality can use the following mathematica code. value[a_, b_, c_] := Module[{}, If[a == b, a, p1 = (a - b)/(a + c - 2 b); p2 = (c - b)/(a + c - 2 b); p1 c + p2 b]] f[x_, y_, 0, t_] := If[t >= 1, 1, 0] f[x_, y_, n_, t_] := value[f[x, y, n - 1, t + x], f[x, y, n - 1, t], f[x, y, n - 1, t + y]] g[x_, y_, n_] := f[x, y, n, 0] g[x,y,n] gives the probability of winning a game with n stages, if the two possible numbers written down are proportions x and y of the goal. so the answer to my original question is g[1/3,2/3,4]. those interested in a phase space diagram showing the probability of success g[x,y,4] can find it here: http://www2.stetson.edu/~efriedma/temp/prob.gif the vertical lines are at x=0, 1/4, 1/3, 1/2, and 1. by the way, in the freshman course in game theory i teach regularly, we cover repeated matrix game questions like this the last week of class. erich On Dec 23, 2011, at 12:34 AM, Warren Smith wrote:
RCS:
But Erich's version has complications that make it more challenging. First is the finiteness, and our goal is not to maximize our expected point total, but our chance of surmounting the three-point threshold. Second is the chance for both players to alter their play based on the state of the game so far.
--indeed. It would be possible to solve the game by working backwards from the end state, using the result of the game-solve at the penultimate stage as the payoffs for the game 1 stage earlier, then solve that game, and so on.
This is kind of in miniature the sort of problem faced by those building computerized poker players.
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