Binomial Probability Question Given two basketball players.
John made 38/50 free throws.
Mike made 80/100 free throws.
What is probability that Mike is better at free throws than John?
 A: This is very much a problem in Bayesian inference.  You appear to have taken the first step in realizing that the sample is not the same as the underlying probability distribution.  Even though Mike had a higher mean in his sample, John might have a higher mean for his true talent distribution.  When applying Bayesian inference, we start with a prior distribution, use the evidence to update our knowledge, and come out with a posterior distribution.
The point I alluded to in my comment is that the prior knowledge can matter a great deal.  If we happen to know that the "John" happens to be basketball Hall of Famer John Stockton and the "Mike" happens to be me, then 100 shots are not going to be nearly enough to convince you that I'm better.  You prior distribution for John Stockton's true talent would probably be somewhat tight with a mean near .8 or so (he was .8261 for his career), whereas who knows for me...it might be .3 for all I know.
If you are estimating the true talent probabilities of making a free throw, you at least know that the values must lie in (0, 1), so a minimal prior distribution for each would be the uniform probability distribution on (0, 1).  This is equivalent to using the Beta distribution with parameters $\alpha = \beta = 1$.  Through some calculations I won't get into (along with Bayes theorem), it turns out that the posterior distribution is also a Beta distribution with parameters $\alpha + i$ and $\beta + j$ (assuming the player made $i$ shots in $i + j$ attempts).
So our posterior distribution for John is $Beta(39, 13)$ and for Mike it is $Beta(81, 21)$.  So with these priors our posterior estimate of John's mean talent is $\frac{39}{52} = .75$ and for Mike it is $\frac{81}{102} \approx .794.$  I believe finding the answer to how likely Mike is better in a nice closed form is messy, and I don't have the tools handy to calculate the exact answer, but you would integrate/sum the portions of those curves where Mike is better to find the answer.  Nick's code looks like it will probably get you a nice approximation.  I can tell you that with the prior I chose Mike will be higher (since they had the same prior and Mike had the higher sample mean, it does pass the smell test that Mike will have a higher posterior mean).  
A: Okay. I think I figured out the general answer:
Given a sample of n/N made, the probability that the population success rate is greater than x is defined by the posterior probability distribution:

$\frac{\int_{x}^1 r^n(1-r)^{N-n}\ dr}{\int_{0}^1 r^n(1-r)^{N-n}\ dr}$

So for my example the chance that Mike's free throw percentage is greater than x is:

$\frac{\int_{x}^1 r^{80}(1-r)^{20}\ dr}{\int_{0}^1 r^{80}(1-r)^{20}\ dr}$

We then need to integrate that over the probability distribution for John's free throw percentages:

$\frac{\int_{0}^1 \int_{x}^1 x^{38}(1-x)^{12} r^{80}(1-r)^{20}\ dr\ dx}{\int_{0}^1 x^{38}(1-x)^{12}\ dx \int_{0}^1 r^{80}(1-r)^{20}\ dr}$

Which as whuber said is ~ 72.66%
For the WolframAlpha solution:

Integrate[y^38 (1 - y)^12 x^80 (1 - x)^20, {x, 0, 1}, {y, 0, x}]/(Integrate[y^38 (1 - y)^12, {y, 0, 1}] Integrate[x^80 (1 - x)^20, {x, 0, 1}])

A: I think what you want to do is compare the predictive distributions of both players. The predictive distribution describes the probability that Mike/John will make his next shot given the data (integrating out the parameters). 
Here is some Matlab code you can play with:
clear;
clc;
rng = linspace(0, 1, 100);

% data
j = [38 50];
m = [80 100];

% priors (I assume uniform, but you can integrate external knowledge here)
a0 = 1;
b0 = 1;

% posterior distribution for each player
post_j = [a0 + j(1), b0 + j(2) - j(1)];
post_m = [a0 + m(1), b0 + m(2) - m(1)];

% visualize
postj = betapdf(rng, a0 + j(1), b0 + j(2) - j(1));
postm = betapdf(rng, a0 + m(1), b0 + m(2) - m(1));
figure(1); plot(rng, postj, 'r',...
                rng, postm, 'k');
title('Posterior Distributions');
legend('Jon', 'Mike');
xlabel('Theta');

%% SAMPLING FROM PREDICTIVE 

SAMPLES = 5000;

predj = zeros(SAMPLES, 1);
predm = zeros(SAMPLES, 1);
for i = 1:SAMPLES
    pj = betarnd(post_j(1), post_j(2));
    lj = binornd(1, pj);
    predj(i) = lj;


    pm = betarnd(post_m(1), post_m(2));
    lm = binornd(1, pm);
    predm(i) = lm;
end

% Comparison of (sampled) predictive distributions:
fprintf('P(d_john | Dj) > P(d_mike | Dm) = %.3f\n', sum(predj > predm) / SAMPLES);
fprintf('P(d_john | Dj) < P(d_mike | Dm) = %.3f\n', sum(predj < predm) / SAMPLES);
fprintf('P(d_john | Dj) = P(d_mike | Dm) = %.3f\n', sum(predj == predm) / SAMPLES);

pred_j = sum(predj) / SAMPLES;
pred_m = sum(predm) / SAMPLES;

fprintf('Probability that John will make his next shot: %.4f\n', pred_j);
fprintf('Probability that Mike will make his next shot: %.4f\n', pred_m);

