Question

Consider the gambler's ruin problem where on each bet the gambler either wins 1 with probability \(p\) or loses 1 with probability \(1-p\). The gambler will continue to play until his winnings are either \(N-i\) or \(-i\). (That is, starting with \(i\) the gambler will quit when his fortune reaches either \(N\) or \(0 .\) ) Let \(T\) denote the number of bets made before the gambler stops. Use Wald's equation, along with the known probability that the gambler's final winnings are \(N-i\), to find \(E[T]\). Hint: Let \(X_{j}\) be the gambler's winnings on bet \(j, j \geqslant 1\). What are the possible values of \(\sum_{j=1}^{T} X_{j} ?\) What is \(E\left[\sum_{j=1}^{T} X_{j}\right]\) ?

Short answer

The expected number of bets the gambler will make before stopping is: \[E[T] = \frac{(N-2ip)i}{2p - 1}\]

Step-by-step solution

Define the Random Variables and Find Possible Values

Let \(X_j\) denote the gambler's winnings on bet \(j\) with \(j \geq 1\). Since the gambler either wins or loses on each bet, the possible values for \(X_j\) are \(1\) with probability \(p\), or \(-1\) with probability \(1-p\). The sum of winnings after T bets: \[\sum_{j=1}^{T} X_j\] Since the gambler quits when his fortune reaches either \(N\) or \(0\), the possible values of this sum are: \[N-i \text{ (if he wins)}\] or \[-i \text{ (if he loses)}\]

Determine \(E\left[\sum_{j=1}^{T} X_{j}\right]\)

To find the expectation of the sum of winnings, we must account for both cases: when the gambler wins and when he loses. Let \(q = 1-p\). Since the known probability of the gambler's final winnings is \(N-i\), we have: \[E\left[\sum_{j=1}^{T} X_{j}\right] = (N-i)p + (-i)q\]

Apply Wald's Equation

Wald's equation states that: \[E[T] = \frac{E\left[\sum_{j=1}^{T} X_{j}\right]}{E[X_1]}\] where \(E[X_1]\) is the expected value of a single bet. We have: \[E[X_1] = 1 \cdot p + (-1) \cdot q = p - q\] Now, we can plug \(E\left[\sum_{j=1}^{T} X_{j}\right]\) and \(E[X_1]\) into the equation: \[E[T] = \frac{(N-i)p + (-i)q}{p - q}\]

Simplify the Result

Simplify the expression for \(E[T]\): \[E[T] = \frac{(N-i)p + (-i)(1-p)}{p - (1-p)}\] \[E[T] = \frac{(N-i)p - ip + i}{2p - 1}\] \[E[T] = \frac{(N-2ip)i}{2p - 1}\] Thus, the expected number of bets the gambler will make before stopping is: \[E[T] = \frac{(N-2ip)i}{2p - 1}\]