Gambler's ruin in the limit (only stopping rule ruin)

Question

Imagine a classical Gambler's ruin with winning probability p and losing probability q = 1-p. You start at 1\$ and lose once you reach 0\$. The only stopping rule is that the game is over when the gambler loses. No restriction on p besides the usual $0\leq p\leq1$. My question is, what is the probability of the gambler losing.

My intuition would be that $P(lose) = 1$, or at least very close to one. The question, I believe, is whether there are values of $p$, where even when considering an infinite time horizon, the Gambler never reaches 0.

Answer 1

Your conclusion that $\ \frac{1-p}{p}\ $ is the probability of the gambler's ruin is correct if $\ p>\frac{1}{2}\ $. If $\ p\le\frac{1}{2}\ $ then the probability of his ruin is $\ 1\ $.

Let $\ \pi_k\ $ be the probability that the gambler is ruined if he starts with $\ k\ $ dollars. Then $\ \pi_0=1\ $ and $$ \pi_k=p\pi_{k+1}+(1-p)\pi_{k-1} $$ for $\ k\ge1\ $. If $\ p\ne\frac{1}{2}\ $ the general solution of this recursion is $$ \pi_k=A+B\left(\frac{1-p}{p}\right)^k\ , $$ and if $\ p=\frac{1}{2}\ $ it is $$ \pi_k=A+Bk\ . $$ The initial condition $\ \pi_0=1\ $ gives $\ A+B=1\ $ when $\ p\ne\frac{1}{2}\ $ or $\ A=1\ $ when $\ p=\frac{1}{2}\ $. If $\ p\le\frac{1}{2}\ $, the condition $\ 0\le\pi_k\le1\ $ for all $\ k\ $ implies that $\ B=0\ $, and in that case $\ \pi_k=1\ $ for all $\ k\ $.

If $\ p>\frac{1}{2}\ $, we have $\ \lim_\limits{k\rightarrow\infty}\pi_k=A\ $. The plausible conjecture that this limit is zero turns out to be true, although I don't know any simple way of establishing that fact. If we take it as given, however, then we have $\ A=0\ $ and $$ \pi_k=\left(\frac{1-p}{p}\right)^k\ , $$ which gives $\ \pi_1=\frac{1-p}{p}\ $ as the probability of the gsmbler's ruin if he starts with $\ \$1\ $.

Answer 2

There is a slick way to answer this question, in the case where $p\neq 1/2$, using Doob's martingale convergence theorem. I shall prove that

• if $p\le 1/2$, then the gambler certainly goes broke.

• if $p>1/2$, then the probability the gambler goes proke is $q/p$.

Let $X_n$ be the gambler's wealth after $n$ games, with the convention that as soon as $X_t=0$ for some $t\in \mathbb N$, then $X_{t+s}=0$ for all $s\in \mathbb N$. Then, let $$ M_n=(q/p)^{X_n} $$ I claim that $M_n$ is a martingale. This is true because $$ \begin{align} \mathbb E[M_n\mid M_{n-1}] &=\;\;\;\,(q/p)^{X_{n-1}+1}\cdot \mathbb P(\text{win $n^\text{th}$ game})\\ &\;\;\;\;+(q/p)^{X_{n-1}-1}\cdot \mathbb P(\text{lose $n^\text{th}$ game}) \\ &=M_{n-1}\cdot (q/p)\cdot p+M_{n-1}\cdot (q/p)^{-1}\cdot q\\&=M_{n-1}(p+q)=M_{n-1} \end{align} $$ In this computation, I am ignoring the part that the process stops when $X_t$ hits zero, but this is OK, because it is well known that a stopped martingale is also a martingale.

Since $M_n\ge 0$, we can apply the martingale converge theorem to conclude that $\lim_{n\to\infty}M_n$ exists almost surely. Let $M:=\lim_n M_n$ be this limit.

First, suppose $p<1/2$. Since $q/p>1$, the only way that $M_n=(q/p)^n$ can converge is if it is eventually constant, which only happens when $X_n$ hits zero. We conclude that $X_n$ must hit zero with probability $1$. We have shown the gambler certainly loses in this case, as expected.

When $p>1/2$, we instead have that $M_n=(q/p)^n$ is a positive exponent of number between $0$ and $1$, so there are two possibilities. Just as before, $X_n$ could eventually hit zero, in which case $M=\lim_n M_n=(q/p)^0=1$. However, it is also possible to have $X_n\to +\infty$, which would mean $M=\lim_n M_n=0$.

Since the random variable $M$ is either equal to zero or one, we have $\mathbb E[M]=P(M=1)$, which is exactly the probability the gambler loses. On the other hand, since $0\le M_n\le 1$ surely, we know the original martingale is uniformly integrable, so Doob's second martingale convergence theorem implies $$ P(\text{gambler loses})=\mathbb E[M]=\lim_{n\to\infty} \mathbb E[M_n]=\mathbb E[M_0]=(q/p)^{X_0}=q/p $$

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Finally, let us deal with the case where $p=1/2$. In this case, the martingale method does not tell us anything, since $M_n=1^n=1$ is trivial. However, there is also a clever argument to prove the gambler certainly loses.

The gambler's wealth starts at one. Let us divide the random walk of the gambler's wealth into "phases," as follows.

• Phase 1 starts at the beginning, and ends the first time that $X_n$ hits either $0$ or $2$. If $X_n$ hits $0$, then the experiment is over. Otherwise, phase $2$ starts.

• Phase $2$ ends the first time that $X_n$ hits either $0$ or $4$. Just as before, the whole experiment ends if $X_n$ hits zero. If instead $X_n$ hits $4$ first, we move to phase $3$.

• Phase $3$ ends the first time that $X_n$ hits either $0$ or $8$.

• Phase $4$ ends the first time that $X_n$ hits either $0$ or $16$.

• $\quad\vdots$

The point is, by symmetry, each phase has a $50\%$ chance to end with $X_n=0$, independently of previous phases. The only way that the gambler could avoid ruin was if the gambler ended every phase by hitting $X_n=2^n$, instead of $X_n=0$. But, the probability that this $50\%$ chance event occurs infinitely many times is clearly zero.

To be complete, you need to show each phase is almost surely finite. This is not hard to do. For example, phase $4$ ends if at any point there are $16$ heads in a row, so if you divide all of the flips into disjoint blocks of $16$, there will certainly be a block of $16$ heads.