The gambler’s ruin describes a scenario where a gambler starts with an initial capital $x$ and repeatedly bets a fixed amount (e.g. 1 unit) on a game. In each game, there are two possible outcomes: - *Win:* With probability $p$, the gambler's capital increases by one unit. - *Lose:* With probability $q=1−p$, the gambler's capital decreases by one unit. The game continues until the gambler either goes broke (reaches $0$ capital) or reaches a target capital $a$, at which point the game ends. ## Probability of Ruin Let $Q_x$ be the probability of ruin for the gambler starting from capital $x$. After one game, the capital either increased or decreased by $1$, which affects the future probability of ruin. Thus, we can write $Q_x$ in recursive form: $ Q_x = pQ_{x+1} + qQ_{x-1}$ This equation is a *second-order linear difference equation*, since $Q_x$ is expressed as a linear combination of its two adjacent terms in the sequence, namely $Q_{x+1}$ and $Q_{x-1}$. ## Solutions of the Difference Equation For the above recursive formula, there are two solutions: $ Q_x=1 \quad \text{and} \quad Q_x =\left(\frac{q}{p}\right)^x$ In mathematics, when an equation is linear and you have two solutions $f(x)$ and $g(x)$ that satisfy the equation, then any combination of these solutions is a valid solution as well. $ Q_x=A\,f(x)+B\,g(x)$ Substituting our identified solutions into that linear form: $ \begin{align} Q_x = A*1+B*\left(\frac{q}{p}\right)^x\\[4pt] = A+B*\left(\frac{q}{p}\right)^x \end{align}$ ## Applying Boundary Conditions We know the probability of ruin $Q_x$ for the two boundary conditions: - At $x=0$: The gambler is already ruined - At $x=a$: The gambler has reached the target capital and the game ends. $ Q_0=1, \quad Q_a=0$ Using these boundary conditions, we setup a system of 2 equations to solve for $A$ and $B$: Condition 1: $Q_0=1$ $ 1 = A+B\left(\frac{q}{p}\right)^0 \implies A=1-B$ Condition 2: $Q_a=0$ $ \begin{align} 0 &= A+B\left(\frac{q}{p}\right)^a \\[2pt] 0 &= 1-B+B\left(\frac{q}{p}\right)^a \\[2pt] B-B\left(\frac{q}{p}\right)^a &=1\\[2pt] B &= \frac{1}{1-\left(\frac{q}{p}\right)^a} \end{align} $ Now that we have determined $A$ and $B$, substitute them back into the general solution: $ \begin{align} Q_x &= A+B*\left(\frac{q}{p}\right)^x \\[2pt] &= 1-B+B\left(\frac{q}{p}\right)^x \\ &= 1-\frac{1}{1-\left(\frac{q}{p}\right)^a}+\frac{1}{1-\left(\frac{q}{p}\right)^a}\left(\frac{q}{p}\right)^x\\[6pt] &=\frac{1-\left(\frac{q}{p}\right)^a-1+\left(\frac{q}{p}\right)^x}{1-\left(\frac{q}{p}\right)^a}\\[6pt] \end{align} $ Simplifying further, the probability of ruin $Q_x$ is: $ Q_x=\frac {\left(\frac{q}{p}\right)^a-\left(\frac{q}{p}\right)^x} {\left(\frac{q}{p}\right)^a-1} $ ## Generalize the Step Size If the gambler bets $b$ units each time (instead of 1 unit), the capital changes by $\pm b$ per game. The recursive equation now becomes: $ Q_x = pQ_{x+b} + qQ_{x-b}$ The two solutions for this linear equation are: $ Q_x=1, \quad Q_x = \left(\frac{q}{p} \right)^{x/b}$ > [!note:] > This also makes intuitive sense, since the probability of ruin only cares about the number of discrete steps until hitting the boundary. Increasing the bet $b$ e.g. by a factor of $2$, is similar of lowering the target capital $a$ by half. By following the same algebraic steps as before, we obtain the following for the probability of ruin: $ Q_x=\frac {\left(\frac{q}{p}\right)^{a/b}-\left(\frac{q}{p}\right)^{x/b}} {\left(\frac{q}{p}\right)^{a/b}-1} $