Ruin Probabilities in a Fair Finite Game

The Fair Finite Game and Its Two Outcomes

In a fair finite game, the player’s capital changes by one unit at each play, with equal probabilities of a win and a loss. The game ends when capital reaches either zero or a fixed upper boundary. The probability of reaching the upper boundary before zero is a natural measure of success; its complement is the probability of ruin. The first-step recurrence turns either probability into a boundary-value problem whose solution is linear in the starting capital.

Let \(N\ge1\) be an integer. On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \((X_n)_{n\ge0}\) be the player’s capital after \(n\) plays, with state space \(S=\{0,1,\ldots,N\}\) and discrete time \(n\ge0\). The states \(0\) and \(N\) are absorbing: once the capital reaches either boundary, it stays there. At each interior state \(i\), the next capital is \(i-1\) or \(i+1\), each with probability \(\tfrac12\). The filtration is \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\). Write \(\mathbb{P}_i\) for probabilities when \(X_0=i\).

For clarity, define \(\tau_0=\inf\{n\ge0:X_n=0\}\) and \(\tau_N=\inf\{n\ge0:X_n=N\}\), with the infimum of the empty set equal to infinity. Then \(h_i=\mathbb{P}_i(\tau_N<\tau_0)\) and \(r_i=\mathbb{P}_i(\tau_0<\tau_N)\). At a boundary, the corresponding hitting time is zero, so \(h_0=0\), \(h_N=1\), \(r_0=1\), and \(r_N=0\). For interior starting states, the two boundary hitting times cannot be equal: the chain can occupy only one state at any given time.

Why the Game Reaches a Boundary

Before identifying the two outcomes as complements, it is useful to verify that the game is absorbed with probability one. In any block of \(N\) plays, the event that all \(N\) plays are losses has probability \(2^{-N}\). From any interior capital, that event forces the chain to reach zero within the block: if the starting capital is \(i\le N\), then \(i\) consecutive losses suffice, and the remaining losses leave the chain at the absorbing state zero.

Consequently, conditional on still being in an interior state at the start of a block, the probability of remaining unabsorbed throughout the next \(N\) plays is at most \(1-2^{-N}\). Applying this bound successively gives \(\mathbb{P}_i(\tau_0\wedge\tau_N>kN)\le(1-2^{-N})^k\) for every integer \(k\ge0\). The right-hand side tends to zero as \(k\) tends to infinity. Thus the game reaches one of its two boundaries almost surely. In particular, from any interior state the upper-hit and ruin events are complementary, and \(r_i=1-h_i\).

First-Step Recurrence and Its Linear Solution

If the game starts at an interior state \(i\), its first play moves it to \(i-1\) or \(i+1\), each with probability \(\tfrac12\). Conditioning on that first play and using the Markov property gives the recurrence

Proof. First suppose \(N\ge2\). For any solution of the recurrence, multiply the interior equation by \(2\) and rearrange to obtain \(h_{i+1}-h_i=h_i-h_{i-1}\) for every \(1\le i\le N-1\). Define the successive differences \(d_i=h_i-h_{i-1}\) for \(1\le i\le N\). The rearranged equation says \(d_{i+1}=d_i\), so all \(N\) differences equal a constant \(c\). Summing them gives \(h_N-h_0=\sum_{i=1}^N d_i=Nc\). The boundary values give \(h_N-h_0=1\), hence \(c=\tfrac1N\). Since \(h_0=0\), summing the first \(i\) differences yields \(h_i=ic=\tfrac{i}{N}\). This argument also proves uniqueness: every solution must have those same differences and boundary values. Direct substitution verifies existence, since \(\tfrac12\frac{i-1}{N}+\tfrac12\frac{i+1}{N}=\tfrac{i}{N}\) for each interior \(i\). The ruin formula follows from almost-sure absorption and \(r_i=1-h_i\). If \(N=1\), there are no interior states; the two boundary values alone give \(h_0=0\), \(h_1=1\), which agree with \(h_i=\tfrac{i}{N}\). The corresponding ruin values also agree with \(r_i=\tfrac{N-i}{N}\). \(\square\)

The recurrence is linear because the two possible next states are weighted equally and the boundary values are fixed. The equality of the two transition probabilities matters: it makes the successive differences constant. This tutorial concerns that fair case; changing the win and loss probabilities changes the recurrence and its solution.

Worked Calculations

Interpretation and a Common Mistake

The starting capital affects the two probabilities in opposite directions. Increasing \(i\) by one increases the chance of reaching \(N\) first by exactly \(\tfrac1N\), and decreases the chance of ruin by the same amount. At \(i=0\), ruin is certain; at \(i=N\), ruin is impossible. These endpoint checks are useful for identifying whether a calculation is answering the question asked.

Common mistake: swapping success and ruin probabilities. A wrong calculation might label \(i/N\) as the probability of ruin. For \(N=4\) and \(i=1\), that gives a claimed ruin probability of \(\tfrac14\). But \(i/N=\tfrac14\) is the upper-hit probability: it has boundary values zero at \(i=0\) and one at \(i=N\). Ruin instead has boundary values one at zero and zero at \(N\), so its value is \(1-h_i=1-\tfrac14=\tfrac34\). Checking the two boundary values before interpreting a recurrence solution prevents this reversal.