#1
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Pockets hitting set or quads?
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 |
#2
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Re: Pockets hitting set or quads?
Well I think I worked it out...
The answer is: =(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? |
#3
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Re: Pockets hitting set or quads?
Set and only a set:
2 * C(48,1) * C(44,1) = 2112/19600 = 10.78% Quads: 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). |
#4
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Re: Pockets hitting set or quads?
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? |
#5
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Re: Pockets hitting set or quads?
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.
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#6
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Re: Pockets hitting set or quads?
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% |
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