The Infinite-Capital Game
In the infinite-capital gambler’s ruin model, the player starts with a positive integer fortune and plays a sequence of independent rounds. Each round changes the fortune by one unit: a win adds one and a loss subtracts one. There is no upper capital boundary. When the fortune reaches zero, the game ends. If losses are more likely than wins, the mean change in fortune per round is negative. The question here is whether that unfavorable drift implies that ruin eventually occurs with probability one.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge0}\) be the fortune after \(n\) rounds, with state space \(S=\{0,1,2,\ldots\}\) and discrete time \(n\). The state \(0\) is absorbing. From every state \(i\ge1\), the chain moves to \(i+1\) with probability \(p\) and to \(i-1\) with probability \(q\), where \(p+q=1\) and \(p<q\). The transition row at \(0\) assigns probability \(1\) to \(0\); every row at a positive state sums to \(p+q=1\). For conditioning, use the natural filtration \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\). Write \(\mathbb{P}_i\) for the law with initial fortune \(X_0=i\).
For \(0<p<q\), the one-step mean change before absorption is \(p-q<0\). This describes the direction of the average increment, but by itself it does not calculate the probability of eventual ruin. That probability is obtained from a first-step equation and the requirement that probabilities remain bounded. We treat \(p=0\) separately, because the recurrence solution below divides by \(p\).
The First-Step Equation
Starting from \(i\ge1\), the chain takes one step before the future ruin question is decided. A win leads to fortune \(i+1\), and a loss leads to \(i-1\). The Markov property says that after this step the probability of eventual ruin depends on the new fortune, so conditioning on the first transition gives the recurrence.
Proof. Fix \(i\ge1\). The events \(\{X_1=i+1\}\) and \(\{X_1=i-1\}\) are disjoint, have probabilities \(p\) and \(q\), and together have probability \(p+q=1\). Conditional on either event, the future chain has the transition law of the same process started at the state reached at time \(1\), by the Markov property. Thus \(\mathbb{P}_i(T_0<\infty\mid X_1=i+1)=h_{i+1}\) and \(\mathbb{P}_i(T_0<\infty\mid X_1=i-1)=h_{i-1}\). Both conditioning events have positive probability because \(0<p<q\). The law of total probability gives \(h_i=p h_{i+1}+q h_{i-1}\). Starting at zero means \(T_0=0\), so \(h_0=1\). \(\square\)
The recurrence alone admits solutions that grow without bound. The probability interpretation supplies an essential extra condition: for every \(i\), \(0\le h_i\le1\). We now use that boundedness to identify the probability solution.
Solving the Bounded Recurrence
Proof. We first solve the recurrence for any sequence satisfying it and \(h_0=1\). Define the successive differences \(\Delta_i=h_i-h_{i-1}\) for \(i\ge1\). Rearranging the recurrence gives $$ p(h_{i+1}-h_i)=q(h_i-h_{i-1}),\qquad i\ge1, $$ so, since \(p>0\), \(\Delta_{i+1}=(q/p)\Delta_i\). Set \(r=q/p\); the assumption \(p<q\) gives \(r>1\). Repeated substitution yields \(\Delta_k=r^{k-1}\Delta_1\) for every \(k\ge1\). Summing the differences from \(1\) to \(i\), $$ h_i=1+\Delta_1\sum_{k=1}^i r^{k-1} =1+\Delta_1\frac{r^i-1}{r-1}. $$ If \(\Delta_1\ne0\), then \((r^i-1)/(r-1)\) tends to infinity as \(i\) increases, so this expression is unbounded. Therefore every bounded solution must have \(\Delta_1=0\), which makes every \(\Delta_k=0\) and hence \(h_i=1\) for all \(i\). Conversely, the constant sequence \(h_i=1\) satisfies the boundary condition and the recurrence because \(p\cdot1+q\cdot1=p+q=1\). Thus it is the unique bounded solution. The probabilities \((h_i)\) satisfy the recurrence by the proposition and are bounded between zero and one, so they equal this solution. \(\square\)
This proves certainty of eventual ruin, not a statement about how quickly ruin happens. The proof uses boundedness to exclude the other recurrence solutions; it does not require a conclusion about expected ruin time. In particular, a probability equal to one for eventual ruin does not say that ruin occurs within any specified finite number of rounds.
Worked Examples
Worked Example: Starting with Two Units When \(p=\tfrac13\)
A player starts with \(i=2\), wins each round with probability \(p=\tfrac13\), and loses with probability \(q=\tfrac23\). Find the eventual ruin probability by solving the recurrence.
At zero the chain stays at zero with probability \(1\). At every positive state the transition row assigns probability \(\tfrac13\) to the state above and \(\tfrac23\) to the state below, so its row sum is \(\tfrac13+\tfrac23=1\). The recurrence is \(h_i=\tfrac13h_{i+1}+\tfrac23h_{i-1}\), with \(h_0=1\). Here \(r=q/p=(\tfrac23)/(\tfrac13)=2\). The general form obtained by summing differences is $$ h_i=1+\Delta_1(2^i-1). $$ At \(i=2\), this gives \(h_2=1+3\Delta_1\). If \(\Delta_1\ne0\), the term \(2^i-1\) grows without bound, which is impossible for probabilities. Thus \(\Delta_1=0\), and \(h_2=1\).
Check: Substituting the constant answer into the recurrence at any positive state gives \(\tfrac13(1)+\tfrac23(1)=1=h_i\). At state zero it gives the required boundary value \(h_0=1\).
Worked Example: Starting with Four Units When \(p=\tfrac25\)
Now let the initial fortune be \(4\), with win probability \(p=\tfrac25\) and loss probability \(q=\tfrac35\). Find the eventual ruin probability and identify why a nonconstant recurrence solution cannot represent it.
The absorbing row at zero sums to \(1\); each positive-state row sums to \(\tfrac25+\tfrac35=1\). The first-step equation is \(h_i=\tfrac25h_{i+1}+\tfrac35h_{i-1}\), and \(h_0=1\). The ratio is \(r=q/p=(\tfrac35)/(\tfrac25)=\tfrac32\), so $$ h_i=1+\Delta_1\frac{(\tfrac32)^i-1}{\tfrac32-1}. $$ For \(i=4\), \((\tfrac32)^4=\tfrac{81}{16}\), and \(\tfrac32-1=\tfrac12\). Therefore $$ h_4=1+\Delta_1\frac{\tfrac{81}{16}-1}{\tfrac12} =1+\Delta_1\frac{\tfrac{65}{16}}{\tfrac12} =1+\frac{65}{8}\Delta_1. $$ The same parameter \(\Delta_1\) makes the solution unbounded as \(i\) increases unless \(\Delta_1=0\). Hence \(h_4=1\).
Check: With \(h_i=1\), the right side of the first-step equation is \(\tfrac25+\tfrac35=1\), and the boundary value remains \(h_0=1\). The result concerns eventual ruin even though the win probability is positive.
Worked Example: Eventual Ruin Versus Ruin by Round Three
Use \(p=\tfrac13\), \(q=\tfrac23\), and initial fortune \(1\). Compute the probability of ruin by round \(3\), then compare it with the eventual ruin probability.
Again, the row at zero sums to \(1\), and every positive-state row sums to \(\tfrac13+\tfrac23=1\). Ruin at round \(1\) occurs exactly when the first round is a loss, with probability \(\tfrac23\). If the first round is a win, the fortune becomes \(2\). To reach zero by round \(3\), the next two rounds must both be losses: the path is win, loss, loss, with probability $$ p q^2=\frac13\left(\frac23\right)^2=\frac13\cdot\frac49=\frac4{27}. $$ There is no path with ruin exactly at round \(2\): after a first-round win, a loss in round \(2\) returns the fortune to \(1\), not to zero. Therefore $$ \mathbb{P}_1(T_0\le3)=\frac23+\frac4{27} =\frac{18}{27}+\frac4{27}=\frac{22}{27}. $$ The theorem gives \(\mathbb{P}_1(T_0<\infty)=1\), which is larger than \(\tfrac{22}{27}\).
Check: The paths not yet ruined by round \(3\) are win-win-win, win-win-loss, and win-loss-win. Their probabilities are \(\tfrac1{27}\), \(\tfrac2{27}\), and \(\tfrac2{27}\), respectively, for a total of \(\tfrac5{27}\). Thus \(\tfrac{22}{27}+\tfrac5{27}=1\), confirming the finite-time calculation.
Degenerate Case and Common Pitfall
If \(p=0\), then \(q=1\): every round is a loss until absorption. Starting from \(i\), the fortune visits \(i-1,i-2,\ldots,0\), so \(T_0=i\) with probability one and \(h_i=1\). This case has certain ruin as well, but the difference calculation that divides by \(p\) does not apply. Treating it separately avoids dividing by zero.
Common mistake: reversing the ratio. For \(p=\tfrac13\), \(q=\tfrac23\), one might incorrectly try \(h_i=(p/q)^i=(\tfrac12)^i\), since this is bounded and decreases with the initial fortune. It does not satisfy the recurrence or the boundary condition: \(h_0=1\), but at \(i=1\) the proposed value \(h_1=\tfrac12\) differs from the recurrence’s right side, $$ \frac13h_2+\frac23h_0 =\frac13\left(\frac14\right)+\frac23 =\frac1{12}+\frac8{12} =\frac34\ne\frac12. $$ The correct ratio in the difference recurrence is \(q/p\), because rearranging gives \(p\Delta_{i+1}=q\Delta_i\). Boundedness then forces the difference to be zero, rather than producing a decreasing geometric ruin probability.
The central conclusion is specific: with no upper boundary and a positive probability of a loss on each round, a player facing \(p<q\) is eventually ruined with probability one, whatever the positive starting fortune. The probability of ruin by a fixed time can still be less than one, as the three-round calculation shows. The recurrence and its boundedness condition provide the exact eventual probability without making an unsupported claim about the expected duration.
Check Your Understanding
For each question, use the boundary value and recurrence where applicable, and distinguish finite-time events from eventual ruin.
- For \(p=\tfrac14\), \(q=\tfrac34\), and initial fortune \(3\), what ratio \(q/p\) appears in the difference recurrence, and what is the eventual ruin probability?
- Explain why the recurrence has nonconstant solutions but the ruin probabilities cannot be one of them when \(0<p<q\).
- With \(p=\tfrac13\), \(q=\tfrac23\), and initial fortune \(1\), compute the probability of ruin exactly at round \(3\).
- What is \(T_0\) when \(p=0\) and the initial fortune is \(5\)?
- Why does a proof that eventual ruin has probability one not imply that ruin occurs by round \(10\) with probability one?