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The computer solves the three tower problem. (English) Zbl 0778.60050
Consider the following probability problem: We have \(n\) piles (towers) of \(x_ 1,\ldots,x_ n\) chips, respectively. Each second a pile \(X\) is selected at random, then another pile \(Y\) is chosen at random and a chip is moved from \(X\) to \(Y\). Stop, when a) one pile becomes empty, b) \(n-1\) piles become empty. Find the expected times \(f(x_ 1,\ldots,x_ n)\) and \(g(x_ 1,\ldots,x_ n)\) until stop, and the probability \(p_ i\) that pile \(x_ i\) wins.
The case b) was solved for any \(n\in\mathbb{N}\). The result is \[ f(x_ 1,\ldots,x_ n)=\sum^ n_{x_ i<x_ k}x_ ix_ k,\quad p_ i=x_ i/(x_ 1+\cdots+x_ n). \] The case a) was solved for \(n=3\) with the result \[ f(a,b,c)=3abc/(a+b+c). \] Exact computations for small \(a,b,c,d\) indicate that the case \(n=4\) is hopeless.
The paper touches a related problem: Players \(1,\ldots,n\) start with \(n\) chips. In one round each player stakes a chip. Then an \(n\)-sided symmetric die labeled \(1,\ldots,n\) is rolled and the winner gets all the chips staked. Stop, when a) any player is broke, b) all but one player are broke. Find the times \(f\) and \(g\) until stop.
For \(n=3\) the case a) was found 1966. It is \[ f(a,b,c)=abc/(a+b+c-2). \] The case b) was found to be \[ g(a,b,c)=ab+bc+ca-2abc/(a+b+c-2). \] For \(n>3\) there is no solution up till now. Exact computations for small pile sizes indicate that the results do not grow as fast as in the case of the tower problems.

MSC:
60G50 Sums of independent random variables; random walks
Keywords:
games of chance
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