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DoctorWard
08-15-2005, 07:22 AM
Hi All

I want to work out the probability of hitting a set (3) or quads (4) on the flop when you hold a pocket pair in texas hold'em?

I thought it might be:
=(2/50 x 48/49 x 48/48 x 3C1) + (2/50 x 1/49 x 48/48 x 3C2)
=3/25

However I think I might be double counting, can someone please help?

DW

DoctorWard
08-15-2005, 07:35 AM
Well I think I worked it out...

=(2/50 x 48/49 x 47/48 x 3C1) + (2/50 x 1/49 x 48/48 x 3C2)
=(2/50 x 48/49 x 48/48 x 3C1)
=11.755%

Can someone please confirm?

LetYouDown
08-15-2005, 08:39 AM
Set and only a set:

2 * C(48,1) * C(44,1) = 2112/19600 = 10.78%

C(2,2) * 48/19600 = 0.245%

Your math seems right, but a little too complicated and it includes flopping a full house (with the set and not the cases where trips of another rank flop).

DoctorWard
08-15-2005, 09:07 AM
If I said "what is the chance of flopping a set or better when holding two pair"?

=(2/50 x 48/49 x 48/48 x 3C2)
=11.755% or 1 in 8.51 times

Since you can't flop a type of straight or flush when holding a pair then this covers a set, full house and quads.

Correct?

LetYouDown
08-15-2005, 09:39 AM
When holding two pair? I assume you mean a pocket pair. Once again it looks ok, except you're not counting the times when you hold say, 4-4 and the flop comes J-J-J.

hukilai
08-19-2005, 12:18 AM
Your chance to NOT hit your set or better is C(48,3)/C(50,3). So the chance to hit is:
1 - C(48, 3)/C(50,3) = 11.7%