[Buzz]

Suppose you are dealt 88. If we exclude 88X flops (That would be flopping quads!), and also 8XX flops (That would be flopping a full house!), then we're interested in 8XY flops, where neither X nor Y is an eight and where X and Y are not a pair. 2*48*44/2 = 48*44 = the number of these flops.

When we hold 88, there are a total of 50*49*48/6 possible flops.

Dividing the number of flops which could contain exactly one eight and no pair by the total number of possible flops, after cancelling out common terms, we end up with (44*6)/(50*49) = 0.1077551, which is about a ninth. This is the probability of flopping a set when you hold a pair. Note that we are excluding flopped full houses and flopped quads.

When the probability is 0.1077551, the odds against are about 8.3 to 1.

If you wanted to include 88X flops (quads), there are 1*48 =48 of these possible.
If you also want to include 8XX flops (full houses) there are 2*12*6=144 of these possible
Add both of those to 48*44 to get 48*48. Then (48*48)/(50*49*48/6) = (48*6)/(50*49)= 0.117551. This is the probability of flopping a set, a full house, or quads when you hold a pair.

When the probability is 0.117551, the odds against are about 7.5 to 1.

Anonymous - Your answer is basically the same as what I get if I include full houses and quads. (You rounded off 0.0391837 to get 0.39 in an intermediate step). I'm not sure what Bad Beetz wanted, but he asked for the odds of flopping a set, not the odds of flopping a set or better. But I'll agree that the odds of flopping a set or better are probably more meaningful than the odds of flopping a set.

Never a guarantee my math is correct, but I think the odds against flopping a set (not a set or better) are about 8.3 to 1, as shown above.

O.K. I'm vowing to stay out of these Texas hold 'em posts from now on.

Buzz