In an election, candidate A receives n votes and candidate B receives m votes, where n>m. Assume that in the count of the votes all possible orderings of the n+m votes are equally likely. Let Pn,m denote the probability that from the first vote on A is always in the lead. Find Pn,m. [solution] Pn,m = P(A gets n votes, B gets m votes | last vote to A)*(n/n+m) + P(A gets n votes, B gets m votes | last vote to B) * (m/m+n) = Pn-1,m * (n/n+m) + Pn,m-1*(m/m+n) From the formula, we have Pn,1 = Pn-1,1 *(n/n+1) + Pn,0*(1/n+1) = (n/n+1)*Pn-1,1 + 1/(n+1) P1,1 = 0 => Pn,1 = (n-1)/(n+1) and Pn,2 = Pn-1,2*(n/n+2) + Pn,1*(2/2+n) = Pn-1,2*(n/n+2) + 2*(n-1)/(n+2) P2,2 = 0 ==> Pn,2 = (n-2)/(n+2) Please explain this solution to me. The stuff in red is where I got lost.