In a random walk starting at A and then moving to any adjacent point with equal probability. Find P(reach B before E)

Question

It seems to me that for symmetric reasons the answer should be $\frac{1}{2}$. I tried to prove this by separating paths that did not include A and those that did. The probability of the former is $\frac{3}{8}$ (using the sum of an infinite geometric sequence) and the probability of the latter seems to me to be $\frac{1}{2}$$\frac{1}{3}$P where P is the requested probability.(the $\frac{1}{2}$ comes from going from A to C or D and the $\frac{1}{3}$ from going from those back to A). This "recursive" equation results in $P=\frac{18}{40}$ which is not expected. Will be grateful for any thoughts on this

Answer 1

The answer is "obviously" $\frac12$ by symmetry.

Let's try a longer way: let $P_i$ be the probability that you reach $B$ before $E$ when starting from point $i$. Then

• $P_A=\frac14 P_B + \frac14 P_C + \frac14 P_D + \frac14 P_E$
• $P_B=1$
• $P_C=\frac13 P_A + \frac13 P_B + \frac13 P_D$
• $P_D=\frac13 P_A + \frac13 P_C + \frac13 P_E$
• $P_E=0$

which is five equations in five unknowns. You can solve them by substitution or otherwise and get the unique solution

• $P_A=\frac12$
• $P_B=1$
• $P_C=\frac58$
• $P_D=\frac38$
• $P_E=0$