Exact Ruin Probability Calculations

From a Game to an Exact Probability

In a finite gambler’s ruin game, a player’s fortune moves one unit at a time until it reaches either zero or a fixed target. The probability of reaching zero first can be found from a short recurrence. The arithmetic may be exact, but it is easy to reverse the win and loss probabilities or to impose the wrong boundary values. We will set up the model carefully, derive the solutions, and check each calculation directly in the recurrence.

Take a probability space \((\Omega,\mathcal{F},\mathbb{P})\) carrying independent game outcomes. Let \(N\) be a positive integer, and let the state space be \(S=\{0,1,\ldots,N\}\). The discrete-time process \((X_n)_{n\ge0}\) records the fortune after \(n\) plays, stopped when it reaches either endpoint. From an interior state \(i\), the fortune increases to \(i+1\) with probability \(p\) and decreases to \(i-1\) with probability \(q=1-p\). The boundary states 0 and \(N\) are absorbing. We assume \(0<p<1\), so \(0<q<1\), unless a deterministic boundary case is discussed separately.

For \(1\le i\le N-1\), conditioning on the first play gives \(r_i=q r_{i-1}+p r_{i+1}\): a loss moves the fortune to \(i-1\), while a win moves it to \(i+1\). This recurrence and its boundary values are the calculation’s starting point. The probability of reaching \(N\) first is \(1-r_i\) when the game reaches one of the two endpoints; here the recurrence concerns ruin specifically.

Solving the Recurrence

The formulas below require no limiting argument. We verify that each candidate satisfies the recurrence and boundary values, then use uniqueness to conclude it is the ruin probability. The uniqueness argument is included so that substitution is justified.

Proof. First suppose \(p=q=\frac12\). The proposed values \(r_i=1-i/N\) have \(r_0=1\) and \(r_N=0\). For each interior \(i\),

Now suppose \(p\ne q\), and put \(a=q/p\). Since \(p\ne q\), \(a\ne1\), so the denominator \(1-a^N\) is nonzero. The proposed values \(r_i=(a^i-a^N)/(1-a^N)\) give \(r_0=1\) and \(r_N=0\). For an interior \(i\), the constant term is preserved by the recurrence because \(p+q=1\). Also, because \(a=q/p\), we have \(q/a=p\) and \(pa=q\), and hence

Thus the term \(a^i\), and therefore the proposed expression, satisfies the recurrence. To prove uniqueness, let \(d_i\) be the difference between two solutions. Then \(d_0=d_N=0\) and \(d_i=q d_{i-1}+p d_{i+1}\). If the finite list \(d_0,\ldots,d_N\) had a positive maximum, it would occur at an interior index \(i\). Because \(p,q>0\) and sum to 1, \(d_i=q d_{i-1}+p d_{i+1}\) can equal that maximum only if both neighbors equal it. Repeating this implication reaches a boundary, contradicting \(d_0=d_N=0\). Applying the same argument to \(-d_i\) rules out a negative minimum. Therefore every \(d_i=0\), proving uniqueness and the claimed formulas. \(\square\)

In an actual calculation, it is usually most efficient to substitute the given \(p,q,N,i\) into the appropriate formula and simplify the fraction. The recurrence remains an important check: it detects an incorrect ratio or a misplaced boundary value.

Checks That Prevent Reversed Answers

A useful first check is structural: from an interior state, a loss must use \(r_{i-1}\) and a win must use \(r_{i+1}\). Since \(r_i\) is the probability of ruin, the lower boundary gets value 1 and the upper boundary gets value 0. Reversing either convention changes the problem being solved.

Common mistake. In the game with \(p=2/3\), \(q=1/3\), and \(N=4\), using \(p/q=2\) in place of \(q/p=1/2\) gives the incorrect calculation

This wrongly assigns the house-favored result from the other example to a game favoring the player. The recurrence confirms the correction: with \(r_1=7/15\), \(r_2=1/5\), and \(r_3=1/15\), the equation at state 2 is \(\tfrac13(7/15)+\tfrac23(1/15)=1/5\). The correct ratio is \(q/p\), because the recurrence weights the value one step down by \(q\) and one step up by \(p\).

The formula for \(p\ne q\) assumes both probabilities are positive. If \(p=1\), a player starting strictly between the endpoints moves upward at every play, so the ruin probability is 0. If \(q=1\), the player moves downward at every play, so the ruin probability is 1. These deterministic cases should be handled directly rather than by substituting into a formula that divides by \(p\) or \(q\).