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joeg
02-13-2004, 07:36 AM
Hi, just a quick question, basicaly when you hold Ax suited, what are the odds of making a flush by the river, assuming you never fold, I used the following equation

(12c3 * 47c2)/50c5

i.e. all of the 3 card combinations that make your flush multiplied by the total combinations from all the remaing cards, so your not fussed about the board pairing or 4 flushes etc. This gave me 11%, I just wanted to check if this is correct as I've heard different figures for this.

Thanks in advance

Joe.

Henke
02-13-2004, 10:55 AM
There are a few errors in the equation. First of all, if you hold two suited cards, there are only 11 left (not 12) of that particular suit. Second of all, I think you're "double counting" some boards. The correct way to calculate this is to count the number of boards for the three different types of flushing-boards:
<ul type="square"> Three-flush boards (C(11,3)*C(39,2))
Four flush boards (C(11,4)*39)
Five flush boards (C(11,5))
[/list]
Summing these and dividing by C(52,3) should yield about 5% chance.

toots
02-13-2004, 11:37 AM
[ QUOTE ]

Summing these and dividing by C(52,3) should yield about 5% chance.


[/ QUOTE ]

did you mean C(50,5)?

ThinkQuick
02-24-2004, 05:35 PM
135597/2118760

I get 6.4%

I also did (1- prob(none of the suit) - prob(1 of the suit) - prob(2 of the suit)) and got the same answer

Jonathan