Common Errors in Finite Ruin Probabilities

A Reliable Setup for Finite Ruin Probabilities

Finite gambler’s ruin calculations are short, but several conventions must line up: which probability means a win, which boundary is ruin, and whether the target fortune is counted as a state. A formula can look plausible and still answer a different question if one of these choices is reversed. This tutorial develops checks that expose those errors before they affect the answer.

Let $N\ge2$ be an integer, and let $(X_n)_{n\ge0}$ be a discrete-time Markov chain on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, with state space $S=\{0,1,\ldots,N\}$. The process describes a gambler’s capital. States $0$ and $N$ are absorbing: capital $0$ is ruin, and reaching $N$ is success. At each interior state $i\in\{1,\ldots,N-1\}$, the capital increases by one with probability $p$, and decreases by one with probability $q=1-p$, where $0<p<1$. Thus $P=(p_{ij})$ is row-stochastic: from an interior state $i$, $p_{i,i+1}=p$, $p_{i,i-1}=q$, and the other entries in that row are zero; the boundary rows have $p_{00}=p_{NN}=1$.

Write $T_0=\inf\{n\ge0:X_n=0\}$ and $T_N=\inf\{n\ge0:X_n=N\}$. The finite ruin probability from capital $i$ is $u_i=\mathbb{P}_i(T_0<T_N)$. The strict inequality matters: this event means ruin occurs before success. If the chain starts at $0$, then $u_0=1$; if it starts at $N$, then $u_N=0$.

Definition: In this model $p$ is always the probability of an upward step and $q$ is always the probability of a downward step. The finite ruin problem uses the boundary values $u_0=1$ and $u_N=0$. For each interior state, conditioning on the first step gives $u_i=p\,u_{i+1}+q\,u_{i-1}$.

That equation is a useful first diagnostic. A win leads to $i+1$, so its probability multiplies $u_{i+1}$; a loss leads to $i-1$, so its probability multiplies $u_{i-1}$. The order of the terms is determined by the model, not by the order in which a formula happens to be remembered.

Why the Recurrence Gives the Ruin Probability

For completeness, we first check that the process is absorbed at a boundary with probability one. Let $\mathcal{F}_n=\sigma(X_0,\ldots,X_n)$ be the history through time $n$. From any interior state $j$, $j$ consecutive losses reach $0$ within at most $N$ steps. The probability of those losses is $q^j$: all $j$ transitions occur while the chain is still in the interior, and the last one takes it to $0$. Since $j\le N-1$ and $0<q<1$, $q^j\ge q^{N-1}\ge q^N$. Consequently, from every interior state the conditional probability of absorption within the next $N$ steps is at least $q^N>0$.

Proof. Set $T=\min(T_0,T_N)$, and let $a_k=\mathbb{P}_i(T>kN)$, for any fixed initial interior state $i$ and integer $k\ge0$. On $\{T>kN\}$, the chain is in an interior state at time $kN$. By the Markov property and the bound just established, its conditional probability of remaining unabsorbed for the next $N$ steps is at most $1-q^N$. Therefore $a_{k+1}\le(1-q^N)a_k$. Since $a_0\le1$, induction gives $a_k\le(1-q^N)^k$. The right-hand side tends to zero because $0<1-q^N<1$. If $T=\infty$, then $T>kN$ for every $k$, so $\mathbb{P}_i(T=\infty)\le a_k$ for every $k$, and hence $\mathbb{P}_i(T=\infty)=0$. Thus absorption occurs almost surely from every interior state. $\square$

The recurrence follows by conditioning on the first step: from $i$, a win moves to $i+1$ and a loss to $i-1$. The boundary values follow from the definitions of $T_0$ and $T_N$. We can now solve the recurrence and verify that its solution is unique.

Theorem: For the finite gambler’s ruin model with $N\ge2$ and $0<p,q<1$, let $r=q/p$. Then $$ u_i= \begin{cases} \dfrac{r^i-r^N}{1-r^N}, & p\ne q,\\[6pt] \dfrac{N-i}{N}, & p=q=\tfrac12, \end{cases} \qquad 0\le i\le N. $$ These values are the unique solution to $u_0=1$, $u_N=0$, and $u_i=p\,u_{i+1}+q\,u_{i-1}$ for $1\le i\le N-1$.

Proof. Define differences $d_i=u_i-u_{i-1}$ for $1\le i\le N$. The recurrence at an interior state is equivalent to $p(u_{i+1}-u_i)=q(u_i-u_{i-1})$, so $d_{i+1}=r d_i$. It follows by induction that $d_i=d_1r^{i-1}$. The boundary conditions require $$ -1=u_N-u_0=\sum_{i=1}^{N}d_i=d_1\sum_{k=0}^{N-1}r^k. $$ The sum is positive because $r>0$, so these equations determine $d_1$ uniquely and thus determine every $u_i$ uniquely. For $r\ne1$, summing the first $i$ differences gives $u_i=1-\frac{\sum_{k=0}^{i-1}r^k}{\sum_{k=0}^{N-1}r^k}$. Since $\sum_{k=0}^{i-1}r^k=(1-r^i)/(1-r)$ and $\sum_{k=0}^{N-1}r^k=(1-r^N)/(1-r)$, this equals $(r^i-r^N)/(1-r^N)$. If $r=1$, all the differences are equal; their sum is $-1$, so each is $-1/N$ and $u_i=1-i/N=(N-i)/N$. The actual ruin probabilities satisfy the boundary conditions and recurrence, so uniqueness identifies them with these solutions. $\square$

The finite-sum form also makes the fair-game limit precise. For $r>0$, $$ u_i=\frac{\sum_{k=i}^{N-1}r^k}{\sum_{k=0}^{N-1}r^k}. $$ When $p$ tends to $1/2$, $r=q/p$ tends to $1$. Both sums contain finitely many terms, and each term tends to $1$, so their ratio tends to $(N-i)/N$. The fair formula is therefore the continuous limiting value of the biased formula, not a separate value obtained by substituting $r=1$ into a quotient with zero denominator.

Worked Examples: Diagnose the Common Errors

Worked Example: Matching Step Probabilities to Destinations

Suppose $N=4$, the gambler starts at $i=2$, and each step is a win with probability $p=\tfrac23$ or a loss with probability $q=\tfrac13$. Find the ruin probability, checking the recurrence directly.

The transition matrix, in the state order $0,1,2,3,4$, is $$ P= \begin{bmatrix} 1&0&0&0&0\\ \tfrac13&0&\tfrac23&0&0\\ 0&\tfrac13&0&\tfrac23&0\\ 0&0&\tfrac13&0&\tfrac23\\ 0&0&0&0&1 \end{bmatrix}. $$ Its row sums are $1$, $0+ \tfrac13+\tfrac23=1$, $\tfrac13+\tfrac23=1$, $\tfrac13+\tfrac23=1$, and $1$, respectively. Here $r=q/p=(1/3)/(2/3)=1/2$. The formula gives $$ u_2=\frac{(1/2)^2-(1/2)^4}{1-(1/2)^4} =\frac{1/4-1/16}{1-1/16} =\frac{3/16}{15/16} =\frac15. $$ For a recurrence check, the same formula gives $u_1=(1/2-1/16)/(15/16)=7/15$ and $u_3=(1/8-1/16)/(15/16)=1/15$. Thus $$ p u_3+q u_1=\frac23\cdot\frac1{15}+\frac13\cdot\frac7{15} =\frac2{45}+\frac7{45}=\frac15=u_2. $$ If the probabilities are swapped in the recurrence, the calculation instead describes a model with upward probability $\tfrac13$ and downward probability $\tfrac23$, not the stated game.

Worked Example: Keeping Ruin and Success Boundaries Straight

Let $N=3$, $p=q=\tfrac12$, and $i=1$. The state order is $0,1,2,3$, and the transition matrix is $$ P= \begin{bmatrix} 1&0&0&0\\ \tfrac12&0&\tfrac12&0\\ 0&\tfrac12&0&\tfrac12\\ 0&0&0&1 \end{bmatrix}. $$ The row sums are $1$, $\tfrac12+\tfrac12=1$, $\tfrac12+\tfrac12=1$, and $1$. Since ruin is hitting $0$, set $u_0=1$, not $0$, and set $u_3=0$. The fair formula gives $u_1=(3-1)/3=2/3$. To check, it gives $u_2=(3-2)/3=1/3$, and the recurrence at state $1$ gives $\tfrac12u_2+\tfrac12u_0=\tfrac16+\tfrac12=\tfrac23$.

A common reversed-boundary calculation sets a quantity $v_0=0$, $v_3=1$. That quantity is the probability of reaching $3$ before $0$, not the ruin probability. Indeed, for this fair chain $v_1=1/3$, and $u_1+v_1=2/3+1/3=1$. The boundary values must describe the event named in the question.

Worked Example: Counting the Target as a State

A fair gambler starts with capital $1$ and stops upon reaching $0$ or $4$. The target is capital $4$, so the state space is $\{0,1,2,3,4\}$: there are four unit steps from $0$ to the target, not three. The transition matrix is $$ P= \begin{bmatrix} 1&0&0&0&0\\ \tfrac12&0&\tfrac12&0&0\\ 0&\tfrac12&0&\tfrac12&0\\ 0&0&\tfrac12&0&\tfrac12\\ 0&0&0&0&1 \end{bmatrix}. $$ Its row sums are $1$, $\tfrac12+\tfrac12=1$, $\tfrac12+\tfrac12=1$, $\tfrac12+\tfrac12=1$, and $1$. With $N=4$, the fair formula gives $u_1=(4-1)/4=3/4$. The full values $u_0=1,u_1=3/4,u_2=1/2,u_3=1/4,u_4=0$ satisfy each interior equation: $$ \tfrac12u_0+\tfrac12u_2=\tfrac12+\tfrac14=\tfrac34,\quad \tfrac12u_1+\tfrac12u_3=\tfrac38+\tfrac18=\tfrac12,\quad \tfrac12u_2+\tfrac12u_4=\tfrac14. $$ Using $N=3$ by mistake would give $2/3$ from capital $1$. That calculation treats capital $3$, rather than capital $4$, as the success boundary.

Checks That Catch Errors Early

Before accepting a finite ruin calculation, write down the event, the state space, and the direction of each step. Then test the answer against the endpoints and the first-step recurrence. These checks address different mistakes: a correctly solved recurrence can still answer the wrong boundary problem, while correct boundary values cannot repair a recurrence with $p$ and $q$ assigned to the wrong destinations.

Common mistake and correction: For a fair game with $N=4$, a calculation may state $u_1=1/4$ because the starting capital is one quarter of the target. This is the success probability, not the ruin probability. Since the fair formula for ruin is $u_i=(N-i)/N$, the correct answer is $u_1=(4-1)/4=3/4$; the complementary success probability is $1/4$.

Use the following audit in order:

Check the model: $p$ must multiply the upward destination and $q$ the downward destination, with $p+q=1$.
Check the event: ruin has boundary values $u_0=1$, $u_N=0$; success has the reverse values.
Check the endpoint: a target capital of $N$ produces states $0,1,\ldots,N$, including both endpoints.
Check the equations: substitute the proposed values into $u_i=p u_{i+1}+q u_{i-1}$ at interior states.
Check the fair limit: the biased expression must tend to $(N-i)/N$ as $p$ tends to $1/2$.

These checks are useful because a finite recurrence has a unique solution under the stated conditions. Once the model and boundary conditions are correct, a proposed answer that satisfies the recurrence is fixed; if a check fails, the issue is usually a modeling or indexing choice rather than an algebraic detail.

Check Your Understanding

For each question, state the event and boundary values before calculating.

In a game with $p=\tfrac25$, $q=\tfrac35$, and target $N=3$, write the first-step equation for the ruin probability at state $i=1$.
For $N=5$, what are the two boundary values for the probability of reaching $N$ before $0$? How do they differ from the ruin-probability boundaries?
A fair gambler starts at capital $2$ and stops at $0$ or $4$. Calculate the ruin probability and state which number of unit steps separates the two boundaries.
Explain why the expression $(r^i-r^N)/(1-r^N)$ cannot be evaluated by direct substitution at $r=1$, and give its limiting value.
Why does the probability of $j$ consecutive losses from an interior state $j$ equal $q^j$, even though a block of $N$ steps was used to bound the chance of absorption?

Common Errors in Finite Ruin Probabilities

What You'll Learn