[scdavis0]

Here's a reasonable formulation in the case of ONE out.

Let S = size of the pot

If you run it once your EV = S/44

If you run it twice, one of 3 things will happen

1. You'll miss twice
2. You'll hit the first board but miss the second
3. You'll miss the first board but hit the second

To get our overall EV, we simply sum the product of each probability by the percentage of the pot we will receive should they occur.

Scenario 1 -- miss twice:

We will miss on the first card 43/44. Given that we missed the first card, we will miss the second card 42/43. The product of these probablities is 21/22. We will miss both boards 21/22 and receive zero.

Scenario 2 -- hit 1st and miss 2nd

We will hit on the first board 1/44. Given that we hit the first board, we will miss the second board 43/43. The probability is therefore the product -- 1/44. In this case we get half of the pot.

Scenario 3 -- miss the first but hit the second

We will miss the first board 43/44. Given that we missed the first, we will hit the second 1/43. The product of these is again 1/44.

Our overall EV is therefore:

EV = (S * 21/22 * 0) + (S * 1/44 * 1/2) + (S * 1/44 * 1/2)
= S/44

Thus running it twice with one out does NOTHING to EV. For fun, solve it generally for any number of outs!

(If anyone cares to see the general solution I'll be happy to provide)