Number of Free Throws to Separate Two Players A problem I've been thinking of and want to make sure I'm on the right track. It's somewhat similar to this question I think: Binomial Probability Question. But I was curious if  I could approach it from a different direction.
Say I'm picking between two basketball players for my team. I want to evaluate them based only on their make percentage from $n$ free throws taken during the tryout. How many shots should I have them take to have some level of certainty in my selection of who is the better shooter?
My initial thought is that I could answer this via power analysis. If I assume a probability of Type I error, $\alpha=0.05$, probability of Type II error, $\beta=0.2$, and assume that make% of player 1 $P_1 = 0.5$ and make% of player 2 $P_2= 0.6$, then what does my sample size $n$ need to be?
I plug this information into G*Power (Proportions: Inequality, two independent groups (Fisher's exact test), two-tailed) and get a sample size of $n=404$ shots.
Is this a valid way to approach this problem? Can you think of a better or more generalizable way?
 A: Comments: Your answer seems about right. Here are results taking a couple of different points of view, and giving answers similar to yours.
With $p=0.5,$ the margin of error of a 95% CI for $p$ based on $n = 400$ observations would be $1.96\sqrt{.5^2/400} \approx .05.$ 
Similarly, with $p=.6$ the margin of error with $n=400$ is about the same.
So with $n = 400,$ you should rarely have trouble distinguishing
$p_1 = 0.5 \pm 0.05$ from $p_2 = 0.6 \pm 0.05$ fractions of successful free throws.
Computations in R:
1.96*sqrt(.5^2/400)
[1] 0.049
1.96*sqrt(.6*.4/400)
[1] 0.04801


Minitab's Power and Sample Size procedure for a test of differences in two proportions gives the following result for distinguishing between $.5$ and $.6$ at the 5% level with power .8. (The test uses normal approximations.)
Power and Sample Size 

Test for Two Proportions

Testing comparison p = baseline p (versus ≠)
Calculating power for baseline p = 0.6
α = 0.05


              Sample  Target
Comparison p    Size   Power  Actual Power
         0.5     388     0.8      0.800672

The sample size is for each group.


