Boundary States and Absorption in Gambler's Ruin

From Capital Paths to Absorption

In a finite-capital gambling game, the gambler’s capital changes one unit at a time until it reaches zero or the upper limit \(N\). The two boundary states have a special role: once reached, the game ends, and the capital stays there if we continue to record it. This makes the game a Markov chain with absorbing boundary states. Its stopping time records when the path first reaches either boundary, and its final state records the exit outcome.

Formally, work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Fix an integer \(N\ge1\), a starting capital \(i\in\{0,1,\ldots,N\}\), and a probability \(p\) of winning one unit on each play; write \(q=1-p\). Assume successive plays are independent and \(0<p<1\). The capital process \((X_n)_{n\ge0}\) has state space \(S=\{0,1,\ldots,N\}\), discrete time \(n\), and initial state \(X_0=i\). Its filtration is the natural filtration \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\). At an interior state \(j\in\{1,\ldots,N-1\}\), it moves to \(j+1\) with probability \(p\) and to \(j-1\) with probability \(q\). At states 0 and \(N\), it stays put.

The definition includes the starting time \(n=0\). Thus if \(i=0\) or \(i=N\), then \(T=0\), not 1. If \(0<i<N\), each step changes the capital by exactly one. A path that first reaches 0 at time \(m\) has \(X_0,\ldots,X_{m-1}\) in the interior and \(X_m=0\); the corresponding statement for an upper exit replaces 0 by \(N\). The word “first” matters: after absorption, the continued chain remains at its boundary, but the game’s exit time has already occurred.

Proof. If \(i\) is a boundary state, \(T=0\), so both claims hold. Suppose \(1\le i\le N-1\). Consider any block of \(N\) consecutive plays begun before absorption. If all \(N\) plays are losses, then either the path has already reached 0 during the block or, if it were still interior, its capital would have decreased by \(N\), which is impossible from a capital at most \(N-1\). Thus the probability of avoiding absorption throughout such a block is at most \(1-q^N\). Independence of future plays from the past gives, conditional on \(\mathcal{F}_{kN}\) and on \(T>kN\), the same bound for the next block. Consequently, by induction, \(\mathbb{P}_i(T>kN)\le(1-q^N)^k\) for every integer \(k\ge0\). Since \(q>0\), the right side tends to zero as \(k\) tends to infinity. The events \(\{T>kN\}\) decrease to \(\{T=\infty\}\), so continuity of probability for decreasing events gives \(\mathbb{P}_i(T=\infty)=0\). Finally, from either boundary the transition stays at that boundary with probability 1. Applying this at each time after \(T\) shows that \(X_n=X_T\) for all \(n\ge T\), almost surely. \(\square\)

This argument proves termination; it does not calculate the probability of each exit. It also handles paths that revisit an interior state: revisiting a state does not restart the game or make the path absorbing. If \(p=0\) or \(p=1\), the block argument is unnecessary: from any interior starting state the capital moves deterministically to a boundary in finitely many steps. The theorem’s stated assumption \(0<p<1\) is used for its geometric bound, not because the endpoint cases fail to terminate.

Reading a Path and Its Probability

For a particular finite path \(x_0,x_1,\ldots,x_m\), its probability is the product of its transition probabilities, provided \(x_0=i\). Indeed, repeatedly applying the Markov property gives the probability of the successive transitions as a product. A path is an absorption path at time \(m\) only if \(x_0,\ldots,x_{m-1}\) are interior states and \(x_m\in\{0,N\}\). This condition is equivalent to \(T=m\), for that path. In particular, an event such as \(T=m\) uses an exact time, whereas \(T\le m\) includes every earlier boundary hit as well.

For the game’s transition matrix \(P=(p_{jk})\), the rows are indexed by the current capital and sum to 1. The boundary rows have probability 1 of staying in place. To calculate an exit probability, one can group absorption paths into exhaustive, disjoint families and sum their probabilities. If a family contains arbitrarily many possible paths, an infinite series may be needed. The geometric series below is simply a sum of path probabilities; no recurrence for ruin probabilities is needed.

Boundary Values and a Common Counting Error

When the chain is described beyond the end of the game, both boundary states are absorbing: \(p_{00}=p_{NN}=1\). For the stopped process, the boundary values are therefore fixed. If a quantity records whether the game exits at the upper boundary, its value at 0 is 0 and its value at \(N\) is 1, because lower exit and upper exit are mutually exclusive outcomes. This statement identifies the boundary values only; it does not provide the interior probabilities. Calculating those for general starting capitals is a separate problem.

The absorption theorem also clarifies the meaning of the two exits. Since \(T<\infty\) almost surely and \(X_T\in\{0,N\}\), the events \(\{X_T=0\}\) and \(\{X_T=N\}\) are disjoint and have total probability 1. In a path calculation, however, individual paths need not have probabilities summing to 1 until all possible paths in the relevant outcome have been accounted for. A path that returns to an interior state is not a new, independent game; its probability remains part of the original path.

Common pitfall: counting only the shortest path. In the \(N=3\) example starting at 1, the shortest upper-exit path is \(1\to2\to3\), of probability \(\tfrac13\cdot\tfrac13=\tfrac19\). Calling \(\tfrac19\) the total upper-exit probability omits paths such as \(1\to2\to1\to2\to3\). The correction is to include every possible number of return excursions. Their total is \(\sum_{k=0}^{\infty}(\tfrac29)^k\tfrac19=\tfrac17\), not \(\tfrac19\). The missing paths have total probability \(\tfrac17-\tfrac19=\tfrac{2}{63}\), which is positive.

There is a different error at a boundary start: assigning \(T=1\) because “one step is needed” contradicts \(T=\inf\{n\ge0:X_n\in\{0,N\}\}\). If \(X_0\) is already a boundary, membership holds at time zero. Always check the initial state before describing the path’s first play.