View Full Version : odds of no one having an ace at 6max game
Sinnister
07-13-2005, 08:56 PM
hello, lets say everyone is in the hand and i raise with tt, the flop is a 9 x and im in sb position what are the odds no one has an ace, i often dont bet here because i am ignornat
LetYouDown
07-13-2005, 09:44 PM
C(48,12)/(52,12) I believe...or 33.757503%
elitegimp
07-13-2005, 09:58 PM
[ QUOTE ]
C(48,12)/(52,12) I believe...or 33.757503%
[/ QUOTE ]
this is before you see the flop -- after the flop, there are 47 unknown cards (your 2 tens, the 3 on the flop) and 44 of them are not aces:
C(44,10)/C(47/10) [since you know your hand, it's TT)
Unfortunately, my trusty TI is at work. Can someone else turn it into a probability?
Edit: Keep in mind, though, that this is a rough estimate and assumes the odds of your opponent calling your raise is independent of having an A... that's probably not a valid assumption to make!
MickeyHoldem
07-13-2005, 10:05 PM
0.479185939
BruceZ
07-13-2005, 10:06 PM
[ QUOTE ]
C(48,12)/(52,12) I believe...or 33.757503%
[/ QUOTE ]
That's for all 6 players before the flop. He's asking on the flop, with 1 ace on the flop and none in his hand. Assuming random hands, this would be C(44,10)/C(47,10) =~ 47.9% that no one holds an ace. Of course the proper way to do this requires taking into account a realistic set of hands that the players could actually hold.
LetYouDown
07-13-2005, 11:45 PM
Whoops...either I read only the subject or the post was edited. Could have sworn I read it. Oh well! Thanks for the catch.
vBulletin® v3.8.11, Copyright ©2000-2024, vBulletin Solutions Inc.