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Three players, $A$, $B$ and $C$, repeatedly toss a standard six-sided die. If the die lands 1 or 2 on a particular toss, then $B$ and $C$ each give one dollar to $A$. If the die lands 3 or 4 on a particular toss, then $A$ and $C$ each give one dollar to $B$. If the die lands 5 or 6 on a particular toss, then $A$ and $B$ each give one dollar to $C$. Players $A$, $B$ and $C$ start the game with respective numbers of dollars $a$, $b$ and $c$. Let $X_n, Y_n$ and $Z_n$ denote the respective amounts of money that players $A,B$, and $C$ have after the $n$-th toss, $n \geq 1$, and let $X_0 = a$, $Y_0 = b$, and $ Z_0 = c$.

a) I'm trying to prove that $W_n = X_nY_nZ_n + n(a + b + c - 2)$ is a martingale.

b) And the expected # of tosses required to finish the game.

For part a, we're essentially looking for a fair game where no knowledge of past events can help to predict future winnings. So we can condition on what?

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What part(s) of my answer is (are) not clear to you? –  Did Apr 20 '12 at 16:05
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Hint for (a): Let $S=\{(2,-1,-1),(-1,2,-1),(-1,-1,2)\}$. Assume that $(\xi,\eta,\zeta)$ is uniformly distributed on $S$. Then, for every $(x,y,z)$, $$ \mathrm E((x+\xi)(y+\eta)(z+\zeta))=xyz-(x+y+z)+2. $$

  • Sub-hint for (a): For every $n\geqslant0$, $X_n+Y_n+Z_n=a+b+c$.

Hint for (b): The game stops at time $T=\inf\{n\geqslant0\mid X_n=0\ \text{or}\ Y_n=0\ \text{or}\ Z_n=0\}$, hence $T=\inf\{n\geqslant0\mid X_nY_nZ_n=0\}$ and $W_T=(a+b+c-2)\cdot T$ on $[T\ \text{finite}]$.

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