I am attempting the following question from Grimmet & Stirzaker which I’ve attached for context:
Q) A large casino contains infinitely many gamblers Gl, G2, . . . , each with an initial fortune of \$1. A croupier tosses a coin repeatedly. For each n, gambler Gn bets as follows. Just before the nth toss he stakes his \$1 on the event that the nth toss shows heads. The game is assumed fair, so that he receives a total of \$(1/p) if he wins, where p is the probability of heads. If he wins this gamble, then he repeatedly stakes his entire current fortune on heads, at the same odds as his first gamble. At the first subsequent tail he loses his fortune and leaves the casino, penniless. Let Sn be the casino's profit (losses count negative) after the nth toss. Show that Sn is a martingale. Let N be the number of tosses before the first appearance of HHH; show that N is a stopping time and hence find E(N).
The solution provides the following expression for the casino profit after the Nth toss: $S_{N} = N – (p^{-1}+ p^{-2} + p^{-3})$ but I am having real trouble understanding this. Intuitively I understand that at N-3, we have observed a tail, thus the casino will have taken a total \$(N-3) at this point, then we have a new gambler appearing at each time step betting on heads, so at t=N-2, the casino will have profit: $(N-3)-(p^{-1})$ and at t=N-1 the casino has profit: $(N-3)-(p^{-1}) – (p^{-2} + p^{-1})$. Continuing in this way I don’t see how I can arrive at the solution in the textbook.
Further to this I cant quite figure out why $\mathbb{E}[N]= (p^{-1}+ p^{-2} + p^{-3}) $? Any help on the above would be greatly appreciated.
Best Answer
If the game is to be fair, and we have a payoff of $-1$ with probability $p$ and a payoff of $x$ with probability $1-p$, then $-p+x(1-p)=0$, so $x=(1-p)/p=p^{-1}-1$. So if we start with $\$1$, we end up with $\$0$ with probability $p$ and $\$p^{-1}$ with probability $1-p$, but the casino has only paid us $p^{-1}-1$. Similarly, if we win twice in a row, we now have $p^{-2}$, but the casino has only paid us $p^{-2}-1$. If we win 3 times, we have $p^{-3}$, but the casino has only paid us $p^{-3}-1$.
Suppose that $N$ is the stopping time of the game. Everybody except the last 3 players have bet and lost, paying the casino $1$ each. The last player has won 3 times in a row, second to last has won twice in a row, and the third to last player has won one. Therefore, we have switched from $N$ players all having $\$1$ to $N-3$ players having $\$0$, and the others having $p^{-1}, p^{-2}, p^{-3}$. Thus, the casino has paid out $p^{-1}+p^{-2}+p^{-3}-N$. Alternatively, they have paid out $-(N-3)+(p^{-1}-1)+(p^{-2}-1)+(p^{-3}-1)$.
However, all of the games are martingales, and we satisfy the conditions of the optional stopping theorem, so the stopped game is also a martingale, which means it has expecctation $0$. Thus, $\mathbb E(p^{-1}+p^{-2}+p^{-3}-N)=0$. Expanding and rewriting, this give $\mathbb E(N)=p^{-1}+p^{-2}+p^{-3}$.