The expected time for the drunk man to leave the bridge can be determined using the martingale property and quadratic variation of the random walk. We use the martingale property to determine the respective probabilities of the drunk man exiting at the start ($W_t = 0$) or the end ($W_t = 100$) of the bridge, and the quadratic variation formula to determine the expected time. $$ $$ The martingale property of the random walk is $$ \text{E}[W_t - W_0] = 0 \text{ for all } t \geq 0. $$ Suppose we travel infinitely far in time so that in all scenarios the drunk man has exited the bridge. Then, the following must be true $$ \text{E}[W_t - W_0] = P(W_\infty = 0)\left( 0 - W_0 \right) + P(W_\infty = 100)\left( 100 - W_0 \right) = 0. $$ Note that mathematically we need to treat the random walk as finished at the ends of the bridge, with no further movement, so that at $t = \infty$ the value of the random walk must either be $0$ or $100$. This equation can be solved to get the probabilities $$ P(W_\infty = 0) = \frac{74}{100}, $$ $$ P(W_\infty = 100) = \frac{26}{100}. $$ Now, we use the quadratic variation property of the random walk, that is $$ \text{E}[(W_T - W_0)^2] = T $$ Using the same argument, if we travel infinitely far in time so that in all scenarios the drunk man has exited the bridge, then $$ \text{E}[(W_\infty - W_0)^2] = P(W_\infty = 0)\left( 0 - W_0 \right)^2 + P(W_\infty = 100)\left( 100 - W_0 \right)^2. $$ and we can plug in the numbers we have obtained to get the result $$ \text{E}[(W_\infty - W_0)^2] = 1924. $$ On average it takes the drunk man $1924$ seconds to exit the bridge.