Running hands twice

[icewizard008]

Suppose you are all in on the flop and can only catch one out to win (for example, set over set). Is it true that running it twice decreases your expected value?

Pot size = P
Your chance of winning once = x
Expected value of running it once = Px

Your chance of winning twice = 0 (x^2 in most cases, but 0 now because you only have one out)
Your chance of winning once and losing once to split the pot = 2(1-x)x
Expected value of running it twice = 0*P + 2(1-x)x*(P/2) = Px - Px^2.

Is the analysis correct?