#1
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I should be able to formulate this myself but I can\'t.....
OK, I feel dumb as I can't seem to formulate this properly so, if someone could give me the derivation and answer, I'd greatly appreciate it.
If I start with two suited cards in holdem, what percentage of flops (not boards, just flops) will contain either: 1) 2 or more of my suit 2) quads, full house, trips, or two pair (not counting boards where one of the two pair I make is on the board. That is, if I hold XYs, an XYZ board would count but an XZZ board would not) Assume this is done in a vacuum where nothing at all can be inferred about other players' hands. Thanks in advance. |
#2
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Re: I should be able to formulate this myself but I can\'t.....
Given any 2 starting cards there are C(50,3) = 19,600 possible flops. Assuming that your cards aren't connected so you can't make a straight/straight flush:
Quads By inspection, 2 flops. Full House With cards XY you need flops of XXY or XYY. 2*C(3,2)*3 = 18 flops. Flush C(11,3) = 165 flops. Trips With cards XY you need flops of XXZ or YYZ where Z is not X or Y. 2*C(3,2)*44 = 264 flops. 2 Pair With cards XY you need flops of XYZ where Z is not X or Y. 3*3*44 = 396 flops. Flush Draw Need flop with exactly two of our suit. C(11,2)*39 = 2145 flops. Total = 2990 Probability = 2990/19,600 = 15.3% Lost Wages |
#3
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Re: I should be able to formulate this myself but I can\'t.....
Thanks a lot-- this was exactly what I was looking for. It makes perfect sense when I see it yet when I tried to do it myself I kept screwing it up. [img]/images/graemlins/frown.gif[/img]
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