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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Conditional Probability at it's finest

**Physics Forums | Science Articles, Homework Help, Discussion**