The Infinite-Capital Gambler's Ruin Model

From a Finite Barrier to an Unbounded Fortune

In a finite-capital gambler’s ruin model, the game ends when the gambler reaches zero or a fixed upper fortune. The infinite-capital model removes that upper boundary: the gambler can continue playing at every positive fortune, however large. This change makes it important to distinguish two questions. Does the gambler eventually reach zero? And does the gambler ever reach one particular finite target? Reaching a target does not end play in the unbounded model.

We use a discrete-time process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). At each play, a positive fortune increases by one with probability \(p\) and decreases by one with probability \(q=1-p\). Throughout, assume \(0<p<1\), so both outcomes have positive probability. The state space is \(S=\{0,1,2,\ldots\}\), and the initial fortune is a fixed integer \(i\in S\).

For conditioning, take \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). The independent next increment determines the next fortune whenever the current fortune is positive. Thus, for \(j\in S\), the one-step transition probabilities are \(\mathbb{P}(X_{n+1}=j\mid\mathcal{F}_n)=p_{X_nj}\) almost surely, where \(p_{00}=1\), \(p_{i,i+1}=p\), and \(p_{i,i-1}=q\) for \(i\ge1\); all other entries are zero. Each row sums to one: at zero its only nonzero entry is \(1\), and at every positive state its nonzero entries sum to \(p+q=1\). This is a Markov chain, but its state space has no largest state.

Ruin and Finite Targets

For a set of states \(A\), the hitting time is \(T_A=\inf\{n\ge0:X_n\in A\}\), with \(\inf\varnothing=\infty\). Write \(T_0=T_{\{0\}}\). For a positive integer \(b\), write \(T_b=T_{\{b\}}\). The eventual ruin event is \(\{T_0<\infty\}\). If the initial fortune satisfies \(0<i<b\), the event of reaching \(b\) before ruin is \(E_b=\{T_b<\infty,\ T_b<T_0\}\). The finiteness condition makes the comparison unambiguous even when a hitting time is infinite.

For this nearest-neighbor process, when \(0<i<b\), reaching \(b\) at any finite time is equivalent to reaching it before ruin. Indeed, once the chain reaches zero it remains there forever, so it cannot subsequently reach positive \(b\). The finite target is not itself an absorbing state in the infinite-capital model: after reaching \(b\), the next transition is still governed by the ordinary rule for a positive fortune.

Proof. Let \(b_2>b_1>i\). On \(E_{b_2}\), the chain reaches \(b_2\) at a finite time before reaching zero. Because each upward step increases the fortune by only one, it must visit \(b_1\) before its first visit to \(b_2\). That visit to \(b_1\) is also before ruin, so \(E_{b_2}\subseteq E_{b_1}\). A path belongs to every \(E_b\), for integer \(b>i\), exactly when it reaches every such level before ruin, which is the stated unbounded-level event. If \(T_0<\infty\), only finitely many steps occur before ruin, and each step changes the fortune by at most one. The fortunes before that finite time therefore have a finite maximum. Such a path cannot reach arbitrarily large levels before ruin. Hence the intersection is contained in \(\{T_0=\infty\}\). \(\square\)

This set relation is deliberately not a formula for the probability of eventual ruin. It separates the questions without assuming an answer about which paths occur with positive probability. In particular, reaching one fixed target before ruin is not the same event as avoiding ruin forever.

Why a Finite Target Is Not a New Boundary

A finite target \(b\) can be used to ask whether the process reaches \(b\), but it does not change the transition law. By contrast, a finite-capital game with upper boundary \(b\) stops when it reaches \(b\). The two models can be compared using the same sequence of increments: in the finite game, stop at the first boundary hit; in the infinite-capital game, continue applying the transition rule after every visit to \(b\).

Proof. Let \(T=\min(T_0,T_b)\). From any interior state \(x\in\{1,\ldots,b-1\}\), \(b\) consecutive losses would reach zero within at most \(b-1\) plays. The probability of those losses is \(q^b>0\); using \(b\) losses gives a uniform lower bound even if the boundary has already been reached earlier. Thus, conditional on the chain still being in the interior at time \(kb\), its probability of remaining in the interior for the next \(b\) plays is at most \(1-q^b\). Repeated conditioning gives \(\mathbb{P}_i(T>kb)\le(1-q^b)^k\) for every integer \(k\ge0\). Since \(0<q^b<1\), the right-hand side tends to zero as \(k\) tends to infinity. Therefore \(\mathbb{P}_i(T=\infty)=0\), as claimed. \(\square\)

A Common Modeling Error

A frequent mistake is to treat a chosen finite target as absorbing in the infinite-capital model. For example, with initial fortune \(2\), target \(b=4\), and the parameters in the previous example, the first two wins have probability \(\tfrac25\cdot\tfrac25=\tfrac4{25}\). That is the probability of reaching \(4\) on the first two plays. It is not a reason to set the process permanently at \(4\): on the four-play path just computed, the infinite-capital fortunes continue from \(4\) to \(3\) to \(2\).

The correction is to distinguish the process from the event being studied. The process keeps its original transition rule at every positive fortune. The event \(E_b\) records whether level \(b\) is reached before zero; it does not alter any transition. Conversely, using \(X_{n+1}=X_n+Y_{n+1}\) even at zero is also incorrect: a loss from zero would produce \(-1\), outside the state space. The absorbing boundary rule is part of the model.

These definitions establish the unbounded state space and the finite-target events without yet calculating their probabilities. Such probabilities require additional analysis; they cannot be read off from a single path calculation or from the finite-boundary model without checking how its stopping rule differs.