How Bias and Capital Affect Finite Ruin Risk

What Changes Finite-Game Ruin Risk?

In a finite gambler’s ruin game, the player’s fortune moves one unit at a time between two absorbing boundaries: zero, where the player is ruined, and a target fortune, where the game ends in success. The ruin probability depends not only on whether upward or downward steps are more likely, but also on where the player starts and how far away the target is. Comparing these effects is useful because a formula gives more than a single answer: it lets us see how risk changes when one feature of the game is altered while the others are held fixed.

Let \((X_n)_{n\ge0}\) be a Markov chain on \(S=\{0,1,\ldots,N\}\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Here \(N\ge2\) is an integer. At each step from an interior state, the fortune increases by one with probability \(p\) and decreases by one with probability \(q=1-p\), where \(0<p<1\). States \(0\) and \(N\) are absorbing. We use the filtration \(\mathcal{F}_n\) generated by the outcomes through step \(n\), and write \(\mathbb{P}_i\) for probability when \(X_0=i\). Define the ruin probability \(h_i=\mathbb{P}_i(T_0<T_N)\), where \(T_0=\inf\{n\ge0:X_n=0\}\) and \(T_N=\inf\{n\ge0:X_n=N\}\). The strict inequality means ruin must occur before the target is reached.

The transition probabilities are \(p_{i,i+1}=p\) and \(p_{i,i-1}=q\) for \(1\le i\le N-1\). Each interior row sums to \(p+q=1\); the boundary rows have probability one of remaining at their boundary. Thus this is a valid transition matrix. The question in this tutorial is how \(h_i\) varies with \(i\), \(N\), and \(p\).

The Formula and Why It Applies

Proof. First, absorption occurs with probability one. Starting from any interior state \(j\), the event that the next \(j\) steps are all downward has probability \(q^j\) and reaches zero during those steps. Since \(1\le j\le N-1\) and \(0<q<1\), \(q^j\ge q^N>0\). Consequently, conditional on the chain being in any interior state at the start of a block of \(N\) steps, its probability of absorption during that block is at least \(q^N\). If it is already at a boundary, it is already absorbed. Repeated conditioning at block endpoints therefore gives \(\mathbb{P}_i(T_0\wedge T_N>mN)\le(1-q^N)^m\) for every integer \(m\ge0\). The right side tends to zero, so \(T_0\wedge T_N<\infty\) almost surely.

Conditioning on the first step gives \(h_0=1\), \(h_N=0\), and \(h_i=q h_{i-1}+p h_{i+1}\) for \(1\le i\le N-1\). To verify uniqueness, suppose two solutions exist and let \(d_i\) be their difference. Then \(d_0=d_N=0\) and \(d_i=q d_{i-1}+p d_{i+1}\). If some \(d_i\ne0\), either the largest value of \(|d_j|\) is positive and attained at an interior state, or it is zero. In the positive case, change the sign of all \(d_j\), if needed, so that a positive maximum \(M\) is attained at an interior state \(i\). Both neighbors are at most \(M\), and their weighted average with strictly positive weights \(q,p\) equals \(M\). Therefore both neighbors equal \(M\). Repeating this argument along neighboring states forces \(d_0=M\), contradicting \(d_0=0\). Thus the solution is unique.

For \(p\ne q\), set \(r=q/p\) and consider \(g_i=(r^i-r^N)/(1-r^N)\). Its boundary values are \(g_0=1\) and \(g_N=0\). Also \(p r+q/r=q+p=1\), so for any constant \(c\), \(p r^{i+1}+q r^{i-1}=r^i\) and \(p c+q c=c\). These identities show \(g_i=qg_{i-1}+pg_{i+1}\), and uniqueness gives \(h_i=g_i\). The finite geometric-sum identity gives \(g_i=(\sum_{k=i}^{N-1}r^k)/(\sum_{k=0}^{N-1}r^k)\). When \(p=q\), the sequence \(g_i=(N-i)/N\) has the required boundary values and satisfies \(qg_{i-1}+pg_{i+1}=(g_{i-1}+g_{i+1})/2=g_i\). Uniqueness again identifies it with \(h_i\). At \(r=1\), the sum formula also gives \((N-i)/N\). \(\square\)

The sum form is especially useful for comparisons. It has only positive terms, applies on both sides of the fair case, and avoids interpreting a ratio with negative numerator and denominator when \(r<1\).

Comparing Initial Fortunes, Targets, and Bias

Proof. Write \(D_N=\sum_{k=0}^{N-1}r^k\), so \(h_i=(\sum_{k=i}^{N-1}r^k)/D_N\), with \(r=q/p>0\). Increasing \(i\) by one removes the positive term \(r^i\) from the numerator and leaves the denominator unchanged. Thus \(h_{i+1}=h_i-r^i/D_N<h_i\).

For a fixed \(i\), increasing the target from \(N\) to \(N+1\) adds \(r^N\) to both the numerator and denominator. The old numerator \(A=\sum_{k=i}^{N-1}r^k\) is strictly less than \(D_N\), since \(D_N-A=\sum_{k=0}^{i-1}r^k>0\). Hence \((A+r^N)/(D_N+r^N)>A/D_N\): cross-multiplication reduces this inequality to \(r^N(D_N-A)>0\). So raising the target strictly increases ruin risk.

Finally, regard \(h_i\) as a function of \(r\). Put \(L=\sum_{k=0}^{i-1}r^k\) and \(A=\sum_{k=i}^{N-1}r^k\), both positive. Then \(h_i=A/(L+A)\). The logarithmic derivative \(A'/A\) is a weighted average of the indices \(i,\ldots,N-1\), with positive weights proportional to \(r^k\); \(L'/L\) is a weighted average of \(0,\ldots,i-1\). Every index in the first set exceeds every index in the second, so \(A'/A>L'/L\). Differentiating gives \[ \frac{d h_i}{d r}=\frac{AL}{(A+L)^2}\left(\frac{A'}A-\frac{L'}L\right)>0. \] But \(r=(1-p)/p\) strictly decreases as \(p\) increases, since \(dr/dp=-1/p^2<0\). Therefore \(h_i\) strictly decreases as the chance of an upward step increases. The argument includes \(r=1\), that is, the fair game. \(\square\)

Reading the Formula Carefully

These comparisons are controlled comparisons: only one feature changes at a time. If the initial fortune increases while the target and step probabilities remain fixed, ruin risk falls. If the target rises while the initial fortune and step probabilities remain fixed, ruin risk rises. If the upward-step probability increases while the two fortunes remain fixed, ruin risk falls. Changing two or more features at once can produce competing effects, so the individual comparisons alone do not determine the answer.

A useful complementary quantity is the probability of reaching \(N\) before zero. Since absorption occurs almost surely, exactly one of the two boundaries is reached first. Its probability is \(1-h_i\), or, from the sum formula, \((\sum_{k=0}^{i-1}r^k)/(\sum_{k=0}^{N-1}r^k)\). The boundary values help keep the two questions distinct: ruin has probability \(1\) at state \(0\), whereas reaching the target has probability \(1\) at state \(N\).

Common mistake. The expression \((r^i-1)/(r^N-1)\) is not the ruin probability under the boundary convention used here. It has value \(0\) at \(i=0\) and \(1\) at \(i=N\), so it gives the probability of reaching the target before ruin. For example, with \(p=\tfrac35\), \(q=\tfrac25\), \(N=5\), and \(i=2\), it equals \(\tfrac{135}{211}\). The ruin probability is instead \(\tfrac{76}{211}\), and these add to \(1\). Always check the boundary values before assigning an interpretation to a formula.

The finite-game formula should not be carried over automatically to a game with no upper boundary. Here the target \(N\) is finite and absorbing, and the comparisons concern that finite setting. The formulas also require \(0<p<1\); if either step direction has probability zero, the chain no longer has the two-sided interior movement used in the recurrence argument, and the displayed ratio comparisons do not apply as stated.