A trader starts with 5 units and repeatedly wins 1 unit with probability 0.4 or loses 1 with probability 0.6, stopping at 0 or at 10. Set up the difference equation for the probability of reaching 10 first, solve it, and evaluate it at 5.

A trader starts with 5 units and repeatedly wins 1 unit with probability 0.4 or loses 1 with probability 0.6, stopping at 0 or at 10. Set up the difference equation for the probability of reaching 10 first, solve it, and evaluate it at 5.

Approach: Condition on the first step to get a second order linear difference equation, solve its characteristic equation, and fit the two constants to the values at the absorbing boundaries.

0.1164. Let p_k be the probability of reaching 10 before 0 from k units. Conditioning on the first step gives the difference equation p_k = 0.4 p_{k+1} + 0.6 p_{k-1} with boundary conditions p_0 = 0 and p_10 = 1. The characteristic equation 0.4 r^2 - r + 0.6 = 0 has roots r = 1 and r = 0.6/0.4 = 1.5, so p_k = A + B * 1.5^k. The boundaries give A + B = 0 and A + B * 1.5^{10} = 1, so B = 1/(1.5^{10} - 1) and p_k = (1.5^k - 1)/(1.5^{10} - 1), the standard gambler's ruin formula. With 1.5^5 = 7.59375 and 1.5^{10} = 57.665, the value at k = 5 is 6.59375/56.665 = 0.1164. A fair game would give 0.5 here, so a 10 percentage point edge against the trader cuts the chance of doubling from one half to about one in nine.

Follow-up: What is the expected number of steps before absorption from 5, and how does that expectation behave as the edge goes to zero?

Key concepts: difference equation, characteristic equation, gambler's ruin, boundary conditions.