[BruceZ]

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Anyone know how to calculate the odds of him flopping a straight with AQ when I hold two of the tens?

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2*4*4/C(48,3) =~ 0.185% or 539.5-to-1

That is, 2 possible Ts that can flop, times 4 possible Js, times 4 possible Ks, divided by the total number of possible flops C(48,3) = 48!/(48-3)!/3! = 48*47*46/(3*2*1).

OR if you prefer fractions to combinations:

2/48 * 4/47 * 4/46 * 6 =~ 0.185% or 539.5-to-1

That is, compute the probability of flopping a T (2/48), followed by a J (4/47), followed by a K (4/46). This gives the probability of flopping T-J-K in exactly that order, so we multiply by 6 since there are 3! = 3*2*1 = 6 orders that the T,J,K can flop.