[Lost Wages]

OK, I've figured out how to do it using probability (I think).

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If you have X outs on the flop and Y outs on the turn then:

P(hitting 2 outs) = P(out on turn) * P(out on river)
= (X/47)*((Y-1)/46)

P(hitting exactly 1 out) = P(out on turn * blank on river) + P(blank on turn * out on river)
= ((X/47)*((46-(Y-1))/46)) + (((47-X)/47)*(Y/46))

P(hitting either 1 or 2 outs) = P(hitting 2 outs) + P(hitting exactly 1 out)

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Solving for the case of a set where X=7 and Y=10 yields:

P(hitting 2 outs) = 2.91%
P(hitting exactly 1 out) = 30.48%
P(hitting either 1 or 2 outs) = 33.39%

Lost Wages