What is the probability of having 13 wins and 3 losses in 16 games?

Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is

16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)

However, the assumption of constant and independent probability of a win is rubbish.

Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is

16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)

However, the assumption of constant and independent probability of a win is rubbish.

Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is

16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)

However, the assumption of constant and independent probability of a win is rubbish.

Suppose the result for each game is independent of the results of earlier games and the probability of a win in any game is the constant, p, then the answer is

16C3*p13*(1-p)3 where 16C3 = 16*15*14/(3*2*1)

However, the assumption of constant and independent probability of a win is rubbish.