Eventual Ruin in the Fair Infinite-Capital Game

The Fair Walk With No Upper Boundary

In a finite gambler’s-ruin game, a fair gambler eventually reaches one of two boundaries: zero fortune or a fixed upper target. With no upper target, those finite games do not directly describe the whole future. A path could first reach a high fortune and then return to zero later. The key is to let the upper target increase and track the events of ruin before that target is reached. Their probabilities approach one, and the events together account for every path that is eventually ruined.

Let $(\Omega,\mathcal{F},\mathbb{P})$ carry independent increments $(\xi_n)_{n\ge1}$, with $\mathbb{P}(\xi_n=1)=\mathbb{P}(\xi_n=-1)=\tfrac12$. For a starting fortune $i\in\{0,1,2,\ldots\}$, define the discrete-time process $(X_n)_{n\ge0}$ on the state space $S=\{0,1,2,\ldots\}$ by $X_0=i$, and by $X_{n+1}=X_n+\xi_{n+1}$ when $X_n>0$, while $X_{n+1}=0$ when $X_n=0$. Thus zero is absorbing: play ends at ruin. For every positive state $k$, the transition probabilities to $k-1$ and $k+1$ are both $\tfrac12$, so that row sums to $\tfrac12+\tfrac12=1$; at zero, the only transition is to zero with probability $1$, so its row also sums to $1$.

Write $\mathcal{F}_n=\sigma(X_0,\xi_1,\ldots,\xi_n)$ for the information available by time $n$, and define $T_0=\inf\{n\ge0:X_n=0\}$ and $T_N=\inf\{n\ge0:X_n=N\}$, with $\inf\varnothing=\infty$. The probability of eventual ruin from $i$ is $\mathbb{P}_i(T_0<\infty)$. The finite-target quantity $\mathbb{P}_i(T_0<T_N)$, for an integer $N>i$, asks whether ruin happens before the walk first reaches $N$. These are not the same event for a fixed $N$.

Theorem (Finite-target ruin probability): For an integer $N\ge1$ and $0\le i\le N$, the fair walk started at $i$ reaches one of $0$ and $N$ in finite time almost surely. Its probability of reaching zero before $N$ is $$ \mathbb{P}_i(T_0<T_N)=1-\frac{i}{N}. $$

Proof. If $i=0$, then $T_0=0<T_N$, so the probability is $1$. If $i=N$, then $T_N=0$, so the probability of $T_0<T_N$ is $0$. Now suppose $1\le i\le N-1$. From any interior state $k$, a sequence of $k$ consecutive downward steps reaches zero within the next $N$ steps; its probability is $2^{-k}\ge2^{-N}$. Consequently, conditional on not yet reaching either boundary at the start of any block of $N$ steps, the probability of reaching a boundary during that block is at least $2^{-N}$. Independence of future increments therefore gives $\mathbb{P}_i(T_0\wedge T_N>mN)\le(1-2^{-N})^m$ for every integer $m\ge0$. The right side tends to zero, so $T_0\wedge T_N<\infty$ almost surely.

Put $h_k=\mathbb{P}_k(T_0<T_N)$ for $0\le k\le N$. The boundary values are $h_0=1$ and $h_N=0$. For an interior state $k$, conditioning on the first step gives $h_k=\tfrac12h_{k-1}+\tfrac12h_{k+1}$. Thus $h_{k+1}-h_k=h_k-h_{k-1}$: the consecutive differences are constant. Since their sum from $k=0$ through $N-1$ is $h_N-h_0=-1$, each difference is $-1/N$. It follows that $h_k=1-k/N$, which satisfies both boundary values and the recurrence. The constant-difference argument also shows that no other solution can satisfy the same recurrence and boundary values. Since the walk reaches a boundary almost surely, these probabilities describe the finite-target exit event. $\square$

Taking the Upper Target to Infinity

The finite-target formula alone does not say that ruin happens before any particular fixed $N$ with probability one. Instead, consider all upper targets larger than the starting fortune. For each integer $N>i$, let $A_N=\{T_0<T_N\}$. If $N_2>N_1>i$, a nearest-neighbor path starting below $N_1$ must pass through $N_1$ before it can reach $N_2$. Therefore, a path that reaches zero before $N_1$ also reaches zero before $N_2$, and $A_{N_1}\subseteq A_{N_2}$.

Theorem (Eventual ruin in the fair infinite-capital game): For the fair walk defined above, from every positive integer starting fortune $i$, $$ \mathbb{P}_i(T_0<\infty)=1. $$

Proof. The events $A_N=\{T_0<T_N\}$, for integers $N>i$, increase as $N$ increases. Their union is exactly $\{T_0<\infty\}$. To see one inclusion, $A_N$ entails $T_0<\infty$. For the other, suppose $T_0<\infty$. The path up to time $T_0$ has only finitely many values, so its maximum $M=\max_{0\le n\le T_0}X_n$ is finite. Choose an integer $N>\max\{i,M\}$. The path has not reached $N$ by time $T_0$, and hence $T_0<T_N$; this path belongs to $A_N$. Continuity of probability for an increasing sequence of events now gives $$ \mathbb{P}_i(T_0<\infty) =\lim_{N\to\infty}\mathbb{P}_i(A_N) =\lim_{N\to\infty}\left(1-\frac{i}{N}\right) =1. $$ This proves the claim for every $i\ge1$. If $i=0$, then $T_0=0$, so the same conclusion holds directly. $\square$

The increasing-event argument is the essential step. For a path that is eventually ruined, the segment from its start to its first visit to zero is finite and has a finite maximum. A sufficiently high target lies above that entire segment, so the path is counted as ruin-before-target for that target and for every still higher one. No assumption is made that the fortune stays bounded over the infinite future.

Worked Example: Ruin Before a Target of Eight

Start with fortune $i=3$ and set the upper target to $N=8$. The positive-state transition rows each have two entries, $\tfrac12$ downward and $\tfrac12$ upward, summing to $1$; the boundary rows at $0$ and $8$ each have a single entry $1$, also summing to $1$. The finite-target formula gives

\mathbb{P}_3(T_0<T_8)=1-\frac{3}{8}=\frac{5}{8}.

One can check the formula directly from the finite recurrence. The values at states $0,1,\ldots,8$ are $1,\tfrac78,\tfrac68,\tfrac58,\tfrac48,\tfrac38,\tfrac28,\tfrac18,0$. The endpoints are $1$ and $0$. At every interior state the value is the average of its two neighbors: for example, at state $3$, $\tfrac12(\tfrac68)+\tfrac12(\tfrac48)=\tfrac{5}{8}$; the same constant spacing of $\tfrac18$ verifies the recurrence at each other interior state.

Check: substituting $i=3$ and $N=8$ gives $1-\tfrac38=\tfrac58$, as above. This is the probability of ruin before the first visit to eight, not yet the probability of ruin at any later time.

Worked Example: Increasing the Target

Start with $i=2$, and compare targets $N=4,8,16$. For each positive state the two possible moves have probabilities $\tfrac12$ and $\tfrac12$, hence row sum $1$; the absorbing boundary rows have sum $1$. The finite-target probabilities are, one at a time,

\begin{aligned} \mathbb{P}_2(T_0<T_4)&=1-\frac{2}{4}=\frac{2}{4}=\frac12,\\ \mathbb{P}_2(T_0<T_8)&=1-\frac{2}{8}=\frac{6}{8}=\frac34,\\ \mathbb{P}_2(T_0<T_{16})&=1-\frac{2}{16}=\frac{14}{16}=\frac78. \end{aligned}

These probabilities increase because the events are nested: reaching zero before four guarantees that zero is reached before eight, and reaching zero before eight guarantees that it is reached before sixteen. The values approach $1$ as $N$ grows. The limit theorem identifies the union of these events—not any one of them—with eventual ruin.

Check: for general $N>2$, the formula gives $1-\tfrac{2}{N}$. As $N\to\infty$, $\tfrac{2}{N}\to0$, so the finite-target probabilities tend to $1$, consistent with the three exact values.

What a Fixed Target Does—and Does Not—Tell Us

For a fixed upper target, the event $T_0<T_N$ excludes paths that first reach $N$, even if those paths later return to zero. In particular, its probability is generally smaller than the probability of eventual ruin. The limit works because increasing the target eventually includes every path that reaches zero in finite time; it does not work by identifying eventual ruin with the event for one chosen target.

Worked Example: A Path Missed by the First Target

Start at $i=1$ and set $N=2$. Each positive-state row has downward and upward probabilities $\tfrac12,\tfrac12$, and the boundary rows at $0$ and $2$ each sum to $1$. The finite-target calculation is

\mathbb{P}_1(T_0<T_2)=1-\frac12=\frac12.

Consider the particular sequence of three moves $+1,-1,-1$. The fortunes are $1,2,1,0$, so this path reaches $2$ before zero and is not in $\{T_0<T_2\}$, although it is eventually ruined. The probability of this specified sequence is $\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$. This gives a concrete reason not to equate the finite-target probability $\tfrac12$ with the eventual-ruin probability.

Check: at target $N=4$, the same starting fortune gives $\mathbb{P}_1(T_0<T_4)=1-\tfrac14=\tfrac34$; at target $N=8$, it gives $1-\tfrac18=\tfrac78$. These values increase, and their limit is $1$, as the theorem requires.

The same argument also gives a useful finite-target bound. If the walk never reaches zero, then it must reach $N$ before zero with probability one: the finite-boundary absorption proof shows it cannot remain forever between zero and $N$. Therefore, for $N>i$, the probability of never being ruined is at most the probability of reaching $N$ before zero, which is $i/N$. As $N$ can be chosen arbitrarily large, this upper bound can be made arbitrarily small. This is another way to see why the probability of nonruin is zero.

Common mistake. A tempting but incorrect calculation is to set $\mathbb{P}_i(T_0<\infty)=1-i/N$ for one fixed $N$. For instance, from $i=1$ with $N=2$, that would give eventual ruin probability $\tfrac12$, incorrectly treating paths that hit $2$ first as if they could never later hit zero. The correction is to use the identity only for $\mathbb{P}_i(T_0<T_N)$, then take the limit over increasing targets. The events increase to eventual ruin, which is why their limiting probability gives the desired answer.

The conclusion depends on fairness. With a systematic drift upward, the finite-target ruin probabilities need not tend to one; a walk may escape to arbitrarily large fortunes without ever reaching zero. In the fair case, however, each finite upper boundary leaves a positive probability of ruin before that boundary, and the probabilities of these increasing ruin events converge to one. The result concerns probability, not the expected number of steps until ruin; an almost-surely finite time need not have finite expectation.

Check Your Understanding

Use the finite-target formula and the increasing-event argument to answer each question.

For a fair walk starting at $i=4$, calculate the probability of reaching zero before $N=10$.
Why does $\{T_0<T_{N_1}\}\subseteq\{T_0<T_{N_2}\}$ hold when $i<N_1<N_2$?
Explain why every path with $T_0<\infty$ belongs to $\{T_0<T_N\}$ for some sufficiently large integer $N$.
For $i=1$, calculate $\mathbb{P}_1(T_0<T_N)$ for $N=5$, then state its limit as $N\to\infty$.
What is the error in using $1-i/N$ as the eventual-ruin probability for one fixed $N$?

Eventual Ruin in the Fair Infinite-Capital Game

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