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View Full Version : I should be able to formulate this myself but I can't.....

Ghazban
12-09-2004, 11:54 AM
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.

Lost Wages
12-09-2004, 01:57 PM
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:

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

Ghazban
12-09-2004, 05:00 PM
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. /images/graemlins/frown.gif