I have found online this pdf which treats the gambler's ruin problem. However, in the first page the writer implicitly assumes that the gambler must reach one of the barriers (state $0$ or $N$) in finite time. It seems that it uses this implicit assumption to justify that the probability of ruin is $1-P_i(N)$. Is there a way to justify it without using that assumption? I think it has to do with that the probability of never reaching a barrier is $0$, but how can I prove it?
Gambler’s ruin – Why is the probability of never reaching the barriers 0
gamblingmarkov chainsprobabilityrandom walkstochastic-processes
Best Answer
Note that if there were ever to be a sequence of $N$ winning bets in a row, the game must have ended at some point in that sequence (even if the gambler started at \$1, they would have ended up at \$N. If they started below \$1, the game ended at some earlier point). We can show the probability of ever getting a streak of $N$ wins is $1$.
Let $A_0$ be the event that we get a streak of $N$ heads starting from the $0$-th toss, $A_1$ be the event that we get a streak of $N$ heads starting from the $N+1$-st toss, and so on to define $A_n$. Note that $P(A_n) = p^N > 0$, and each $A_n$ is independent. The probability of getting a streak of $N$ heads is at least \begin{align*} \mathbb{P}(\bigcup_{n=1}^\infty A_n) = 1-\mathbb{P}(\bigcap_{n=1}^\infty A_n^c) = 1-\prod_{n=1}^\infty \mathbb{P}(A_n^c) = 1-\prod_{n=1}^\infty (1-q)^N = 1, \end{align*} so we conclude that the probability that the gambler's wealth hits a barrier at some point is $1$.