The casino game (European) Roulette works as follows. The wheel is divided into 37 equally spaced regions numbered $0$–$36$. The wheel is then spun and a ball is released on the wheel and players bet on what numbered region the ball will end up.
One possible bet in roulette is to bet on the ball landing on the $0$ spot. The payout for this bet is $35$-to-$1$. That is, you either win $35$ dollars or lose $1$ dollar.
Suppose a gambler decides to play roulette by repeatedly betting on the $0$ spot and playing until he wins for the first time. Let $W$ be his net winnings from all his plays when he finally quits. Compute $E[W]$
and $P(W > 0)$.
Let $\tau$ be the number of games he plays before quitting( when his net winnings are positive). Explain why $τ$ is the stopping time and why $E[τ]$ = $\infty$
I am thinking :
If he wins – he quits. He can win maybe in the first try(the profit is $34$ dollars), maybe in the second and so on. If he wins in his 35th try – it won't matter anymore because he won't make any money after that point.
So, $E[W]$ should be P(win in 1st try) * 35 + P(win in 2nd try) *35 + …. + P(win in 34th try)*35?
And $P(W > 0)$ should be the first 34 tries divided by (the entire set of tries which can go up to infinity)?
How can I prove that $τ$ is the stopping time and the expected value of it will be infinity?
Best Answer
If the gambler ends up playing $\tau$ rounds, including the winning one, he has gained $36-\tau$ dollars and the sequence has a $(36/37)^{\tau-1}(1/37)$ chance of occurring. We have $$E(W)=\sum_{k=1}^\infty(36/37)^{k-1}(1/37)(36-k)=-1$$ and $$P(W>0)=P(\tau\le35)=1-P(\tau>35)=1-(36/37)^{35}$$ Now $\tau$ is a stopping time because the gambler stops when he has a positive net profit. Its expectation is infinite because $E(W)<0$ (cf. gambler's ruin).