function p = BR_2aces(num_trials)
%
% 10^6 trials
% Shows that our numbers are reasonable
% calculating probabilities of atoms, then product of atoms is expected
%   value of second ace
%   ap(1)*ap(2) = 0.769*0.766 = 0.059

p = 0;
counts = zeros(1,2);

for t = 1:num_trials
    pe = randperm(52);
    if pe(1)<5
        counts(1) = counts(1) + 1;
        if pe(2)<5
            counts(2) = counts(2) + 1;
        end
    end
end
p = counts(2)/counts(1);
