This game consists of a sequence of rounds. In a round three players \(A_ 1\), \(A_ 2\), \(A_ 3\) toss a symmetric coin. If one player gets a result different from those of the others, he receives one dollar from each. If they all obtain the same result they toss their coins again. The initial amount of money of \(A_ i\) is \(a_ i\) dollars, \(i=1,2,3\). If T is the number of rounds until one of the players is ruined, it is proved that \[ E(T)=a_ 1a_ 2a_ 3/(a-2),\quad a=a_ 1+a_ 2+a_ 3. \] This result is obtained by a martingale method. More precisely, if \(Z_ i(n)\) is the capital of player \(A_ i\) after n rounds, \(i=1,2,3\), and \[ S_ n=Z_ 1(n)Z_ 2(n)Z_ 3(n)+n(a-2),\quad \sigma_ n=\sigma (Z_ i(k),\quad i=1,2,3,\quad 0\leq k\leq n) \] the author proves that \((S_ n,\sigma_ n)\) is a martingale, and that the optional sampling theorem holds for \((S_ n)\) and T, i.e. we have \(E(S_ T)=E(S_ 0).\)
Since \(S_ 0=a_ 1a_ 2a_ 3\) and \(S_ T=T(a-2)\), the main result is deduced easily.