Reverse solve chi-squre statistic to find number of observations I am trying to prove that two proportions are equal with 95% confidence. I have a total population Tp (10000), and wins in that population of Wp (2000). My sample size is Ts (1500). I need to solve for Ws, the number of wins that have to be observed in the sample to prove the sample is proportional to the population. 
I've got that I need a test stat of 3.842 (qchisq(0.95, df = 1)). I'm having trouble reversing the Pearson test to come up with Wp. Since I can't calculate the expected value with a missing number, I'm a little stuck. 
 A: You seem to be thinking that a failure to reject in a chi-squared test shows the two proportions are the same, but it doesn't (failure to show an effect doesn't mean the effect is absent, since you can have a small effect for which the sample size wasn't big enough to reliably pick up whatever effect you had)
You can't show that the two proportions are the same (the nearest you'd get to that would be an equivalence test, if you can specify an equivalence criterion in your situation).
I think your actual question is "what values of the number of wins in the sample (X) would not lead to rejection with a chi-squared test at the 5% level?"
If that's the case, we have the following table:
            Sampled   Not sampled   Total

Wins          X        2000-X        2000
Not-Wins   1500-X      6500+X        8000

Total       1500        8500        10000

The chi-squared statistic for an observed $X$ of $x$ is $(x-300)^2\cdot (\frac{_1}{^{300}}+\frac{_1}{^{1200}}+\frac{_1}{^{1700}}+\frac{_1}{^{6800}})$ (without continuity correction) so we want to see where that's less than $3.84146$. We can easily solve this algebraically - the quadratic formula gives the bounds as $300\pm\sqrt{\Delta}$ where $\Delta=\frac{_{3.84146}}{^{\frac{_1}{^{300}}+\frac{_1}{^{1200}}+\frac{_1}{^{1700}}+\frac{_1}{^{6800}}}}$ - but since you seem to be familiar with R, let's just use it to solve our problem:
f <- function(x) (x-300)^2*(1/300+1/1200+1/1700+1/6800)
x <- 250:350
range(x[f(x)<qchisq(.95,1)])
[1] 273 327

Alternatively
f <- function(x) {chisq.test(matrix(c(x,2000-x,1500-x,6500+x),nr=2),
               correct=FALSE)$p.value}
range(x[sapply(x,f)>.05])

gives the same result. 
So the largest and smallest $x$'s that would not lead to rejection are 273 and 327.
(If you apply Yates' correction - simply by removing "correct=FALSE" above, which is the default in R - then it changes to 272 and 328)

