Probability question, cross post

[AaronBrown]

pzhon is correct but if everyone has the same chance of having a number from 50 to 100, and all the events are independent, it's not too bad. If there are k other players, each with chance p of being in 50 to 100, your chance of winning is the sum from i = 0 to k of:

C(k,i)*p^i*(1-p)^(k-i)/(i+1)

Which is 1/[p*(k+1)] times the sum from i = 0 to k of:

= C(k+1,i+1)*p^(i+1)*(1-p)*(k-i)

The sum is the sum from j = 1 to k+1 of:

C(k+1,j)*p^j*(1-p)^(k+1-j)
= 1- (1-p)^(k+1)

So the answer is:

[1 - (1-p)^(k+1)]/[p*(k+1)]

If the p's are not equal, but there are a lot of them without too much variation, you might get a decent approximation by putting an average p into the formula above; or take the average using the maximum and minimum p's.