- #1

#### member 428835

I am wondering the differences between the discrete and continuous case for expected value of minimum of 3 integers uniformly distributed from 1 to 13 vs 3 reals from 1 to 13.

The real case is direct: ##F = ((x-1)/12)^3 \implies f = 3(x-1)/12)^2## for CDF ##F## and PDF ##f##. Thus the expected value for the max of 3 reals in this range is ##\int_1^{13} x f \, dx = 10##. But now for the discrete case: the probability a random variable ##X_i## is less than some integer ##k## I think should be ##P(X_i \geq k) = (13-k+1)/13 \implies P(X \geq k) = ((13-k+1)/13)^3## but I really don't know how to proceed. Is there a direct way to arriving at the CDF that I'm missing?