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 07-16-2012, 05:46 PM #1 centurion     Join Date: Nov 2011 Location: Bromsgrove Posts: 197 Odds of 3 chosen cards on the flop. At a local tournament, me and a friend were having silly £1 bets on flops, all cards over 8, 2 cards the same suit, really trivial stuff like that. I offered a bet at evens, that an 8, 4 or 2 would be on the flop. I know I'm getting the best of it, I was thinking probably 55/45 favourite, but he seemed to think I was 75% favourite-ish. We didn't really talk about it, but it just got me wondering what the actual odds were? Any chance of some help please guys?
 07-16-2012, 06:05 PM #2 adept   Join Date: Aug 2011 Location: Nova Posts: 721 Re: Odds of 3 chosen cards on the flop. That's 12 cards total. You need at least one. Pr(at least 1) = Pr(1) + Pr(2) + Pr(3) = [(12C1)(40C2) + (12C2)*40 + 12C3] / 52C3 = 47/85 = 55.294%
 07-16-2012, 06:14 PM #3 centurion     Join Date: Nov 2011 Location: Bromsgrove Posts: 197 Re: Odds of 3 chosen cards on the flop. I have no idea what any of that means, but you seem to know your stuff. Random internet man, I trust you implicitly. Cheers for the help
07-16-2012, 07:08 PM   #4
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,882
Re: Odds of 3 chosen cards on the flop.

Quote:
 Originally Posted by heehaww That's 12 cards total. You need at least one. Pr(at least 1) = Pr(1) + Pr(2) + Pr(3) = [(12C1)(40C2) + (12C2)*40 + 12C3] / 52C3 = 47/85 = 55.294%
Or

1 - (40/52 * 39/51 * 38/50)

=~ 55.294%

 07-16-2012, 09:22 PM #5 adept   Join Date: Aug 2011 Location: Nova Posts: 721 Re: Odds of 3 chosen cards on the flop. Gahh, I usually would have taken that approach. Something about this forum decreases my brain's efficiency :@
 07-17-2012, 02:15 AM #6 veteran   Join Date: Jun 2011 Posts: 2,260 Re: Odds of 3 chosen cards on the flop. fwiw, heehaw, you answered it in the way that Bruce answers most questions. But that's because Bruce usually has to.
07-17-2012, 05:50 AM   #7
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,882
Re: Odds of 3 chosen cards on the flop.

Quote:
 Originally Posted by tringlomane fwiw, heehaw, you answered it in the way that Bruce answers most questions. But that's because Bruce usually has to.
When the solution with fractions is simple enough, and the solution with combinations is simpler, I usually try to give both since more people understand the solution with fractions. In this case the solution with fractions is much simpler, so I'd just give that one. Using fractions is the same as using permutations instead of combinations.

Then there's the inclusion-exclusion solution:

3*(12/52) -
C(3,2)*(12/52 * 11/51) +
C(3,3)*(12/52 * 11/51 * 10/50)

=~ 55.294%

Last edited by BruceZ; 07-17-2012 at 08:20 AM.

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