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Horrible Player
10-31-2003, 08:41 PM
can some one tell me if this is right
you have a 0.38 prob on getting 1 of the remaining 9 suited cards from the 47 unseen cards. can someone give me the equation?

thanks

Saborion
11-01-2003, 04:35 AM
I hope I understood your question. And being pretty new myself, I hoep my answer is correct. /images/graemlins/smile.gif

47 cards left, 9 that gives you the flush. 9/47 ~ 0.1915. 19.15 % chans of getting your flush.

In terms of odds. 9 cards of the 47 help you, the remaining 38 doesn`t. 38/9 ~ 4.22.

The chance of you getting the flush is 19.15 %, and the odds of that happening is 4.22.

Bozeman
11-01-2003, 01:32 PM
Actually, you only have 35% probability to hit a flush in 2 cards.

incorrect: 9/47+9/46 since it counts times you hit on the turn and the river twice.

Easiest way: 1-(chance of missing on turn)(chance of missing on river if you miss turn)=1-38/47*37/46=35%

KSU78
11-01-2003, 03:57 PM
p(flush) = [C(9,1)C(47-9,1) + C(9,1)C(8,1)]/C(47,2)

p(flush) = [C(9,1)C(38,1) + C(9,1)C(8,1)]/C(47,2)

p(flush) = (342 + 36)/1081 = 0.3497

Odds against making a flush = (1/0.3497) - 1 = 1.86

Saborion
11-02-2003, 12:29 AM
And of course I was making a mistake. Pay no attention to my post.

BruceZ
11-02-2003, 12:41 AM
Your post is correct for the odds of making the flush on the turn.