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Festus22
07-08-2003, 01:59 PM
Assuming you have 2 of the same suit, what's the odds of making your flush with the board making EXACTLY a 3-flush. Also, what's the odds of the board making EXACTLY a 4-flush.

How is this calculation performed?

Copernicus
07-08-2003, 03:32 PM
For the first one:

[C(11,1)*C(39,2)/C(50,3)] = P(exactly 1 of suit on flop)

times

[C(10,2)/C(47,2) = runner, runner on the turn and river

Product is 1.73%

Assuming your 2d question is what is the P of a flush if you flop exactly 2 of your suit its:

[C(11,2)*C(39,1)/C(50,3)] = P(exactly 2 of suit on flop)

times

[{C(9,2)+C(9,1)*C(38,1)}/C(47,2) = P(either 2 or 1 on turn and river

Product is 3.83%.

Just to complete it, P(flopping the flush)=

C(11,3)/C(50,3) which is .84%.

The total is 6.4%.

That includes straight flushes if possible from the first two cards.

Festus22
07-08-2003, 03:50 PM
I guess my original question is if I start with 2 cards of the same suit, what's the probability that I'll end up with a flush after seeing all 5 board cards.

Subpart 1: What's the odds that the board will show exactly 3 of my suit?

Subpart 2: What's the odds that the board show exactly 4 of my suit?

Copernicus
07-09-2003, 03:18 PM
The 6.4% is the answer to the big picture..thats your proability of eventually winding up with a flush.

For the entire board to show 3 but not 4 or 5 of your suit (regardless of flop vs turn vs river) its

C(11,3)*C(39,2)/C(50,5) = 5.8% (In english, you need to draw 3 out of the remaining 11 of your suit, AND any two of the 39 non-suited cards, vs all of the ways that 5 cards can be dealt from the remaining 50)