View Full Version : flopping a set

bad beetz
09-10-2002, 05:46 PM
I thought I read that the odds of flopping a set when you have a pocket pair are 1/11, but it seems to me the odds are 2/50+2/49+2/48=.122483 ~=1/8

this is right, isn't it?

09-10-2002, 06:23 PM
Not sure where you read that. It's incorrect. Your math is correct.

09-10-2002, 07:01 PM
his math is not correct. the expression he's looking for is:
2/50 + (2/49 * 48/50) + (2/48 * (1 - (2/50 + (2/49 * 48/50)))) = 0.04 + 0.039 + 0.0383 = 0.1173, not the 0.122 that he came up with. in this case, it happens to be close, but this type of error is common and can be costly.

09-10-2002, 07:37 PM
Bon Jovi's math isn't correct either.

P(flopping a set)= 1-P(not flopping a set)= 1- 48(47)(46)/[50(49)(48)]=

1- 47(46)/[50(49)]= 0.12.

09-10-2002, 07:40 PM
I meant to say anonymous' math isn't correct either.

I think one should refrain from making comments about 'common errors'
people make when they are not correct in what they are saying.

bad beetz
09-10-2002, 08:18 PM
Is the possibillity of four of a kind being included or excluded. A much more round about way:

Three ways to flop set or better

P(Flopping a set) =
P(X-1-1) = (2/50*48/49*47/48) +
P(1-X-1) = (49/50*2/49*47/48) +
P(1-1-X) = (49/50*48/49*2/48) =

47/1225 +
47/1200 +

There's always gaping holes in my statistics calculations, but it seems right to me. It's been a few years since my statistics classes.

09-10-2002, 08:22 PM
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.