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.(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Conditional Probability at it's finest

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