
#1
Feb1508, 11:26 AM

P: 225

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) = Pn1,m * (n/n+m) + Pn,m1*(m/m+n) From the formula, we have Pn,1 = Pn1,1 *(n/n+1) + Pn,0*(1/n+1) = (n/n+1)*Pn1,1 + 1/(n+1) P1,1 = 0 => Pn,1 = (n1)/(n+1) and Pn,2 = Pn1,2*(n/n+2) + Pn,1*(2/2+n) = Pn1,2*(n/n+2) + 2*(n1)/(n+2) P2,2 = 0 ==> Pn,2 = (n2)/(n+2) Please explain this solution to me. The stuff in red is where I got lost. 



#2
Feb1708, 01:31 AM

P: 239

This is a common method for 'random walk' type problems which are solved by difference equations. Pn,m is called 'steady state probability' and the corresponding equation is called ballance equation of transition. If you understand the first expression of Pn,m then it will not be difficult to understand how Pn1,m and Pn,m1 are eleminated. Go through the method of solving difference equations in any text book. Hope this is not a homework.




#3
Feb1808, 07:33 AM

P: 225

Yes it's a homework but it has already been turned in and graded. Any more responses are appreciated.




#4
Feb1808, 08:08 AM

Mentor
P: 14,457

Conditional Probability at it's finestFrom this, the latter equality need this to make the latter part of the above statement. Pn,1 = f(n)/(n+1). Using this form in the recursive relationship, f(n)/(n+1) = (n/(n+1))*(f(n1)/n) + 1/(n+1) from which f(n) = f(n1) + 1 In other words, f(n) = n+k where k is some constant. Since P1,1=0, f(1)=0 and k=1, or Pn,1 = (n1)/(n+1) You can verify this with recursion. The expression is obviously correct for n=1. Assuming its correct for n1, you should be able to prove it is true for n. The case m=2 is similar. 


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