Hey all

I have a problem concerning Condorcet's Jury Theorem.

First of all, for those of you that don't know, the theorem looks like this (given that n is even)

$\displaystyle P_n=\sum _{h=(n/2)+1}^n \left(\frac{n!}{h!(n-h)!}\right)p^h(1-p)^{n-h}$

This is a standard binomial argument, to calculate the probability that a jury will have a majority of members arriving at a correct decision, given an individual probability of p for each member voting correctly. The variables are as follows:

n is the amount of people in the jury
Pn is the total probability of the jury arriving at the correct decision
h is the amount of the majority

My question is now:

Assume that we have jury a that must have a majority to "win" with the correct decision. Each person has an individual probability of 65%, voting for the correct decision. How many members should the jury contain to achieve a total probability of 95% voting correctly?

I have placed the values in the formula, so it looks like this:

$\displaystyle 0.95=\sum _{h=(n/2)+1}^n \left(\frac{n!}{h!(n-h)!}\right)0.65^h(1-0.65)^{n-h}$

But how to isolate n? I am using Mathematica to do my calculations, but a standard Solve or Reduce can't do the job, and other operators yell a strange "root"-answer - any ideas?

I will appreciate any help.