View Full Version : AA vs AA with one winning
TheProdigy
11-24-2005, 02:38 AM
I was just involved in a hand where I had AA and another guy had AA, but a four-flush(with diamonds) came out and I ended up winning the hand because I had the ace of diamonds....can anyone tell me the odds of this happening? and maybe a short explanation if possible, I'm trying to figure out how to do odds myself...
EnderW27
11-24-2005, 02:51 AM
Assuming you're heads up and one's all in, one AA will beat the other 2% of the time. So 96% of the time you're coming out with a tie.
So basically what this means is you'll get a four flush of your appropriate suit about 1/50 by the river. Something to keep in mind when you're thinking of odds.
My brain's a bit frazzled tonight so I'll let others explain exactly how the odds are calculated. Hope this helps
BruceZ
11-24-2005, 03:42 AM
[ QUOTE ]
I was just involved in a hand where I had AA and another guy had AA, but a four-flush(with diamonds) came out and I ended up winning the hand because I had the ace of diamonds....can anyone tell me the odds of this happening? and maybe a short explanation if possible, I'm trying to figure out how to do odds myself...
[/ QUOTE ]
Given that you both have AA, the probability that you will win with a flush (incl. str8 flush) is:
2*[C(12,4)*36 + C(12,5)] / C(48,5)
= 45-to-1 =~ 2.17%.
C(48,5) is the total number of boards. C(12,4)*36 is the number of ways to get 4 board cards in one particular suit, and 1 in a different suit. C(12,5) is the number of ways to get all 5 board cards in one particular suit. Then multiply by 2 since we can make a flush in either of 2 suits.
What would be the odds of 2 people both having AA and one winning (the first part being key)?
avisco01
11-26-2005, 03:38 AM
I think whoever has the A/images/graemlins/spade.gif should automatically win, I mean, it is afterall the prettiest card in the deck... For the OP, don't worry about how often this will happen, just be thankful you won! Sorry, I wish I had more to add to this...
BruceZ
11-26-2005, 05:05 AM
[ QUOTE ]
What would be the odds of 2 people both having AA and one winning (the first part being key)?
[/ QUOTE ]
Heads up:
P(both AA) = 1/C(52,4) = 270,724-to-1
P(both AA) * P(1 AA winning) =
1/C(52,4) * 4*[C(12,4)*36 + C(12,5)] / C(48,5)
=~ 6,226,674-to-1
10 players:
P(2 AA) = C(10,2)/C(52,4) = 6015-to-1
P(2 AA) * P(1 AA winning) =
C(10,2)/C(52,4) * 4*[C(12,4)*36 + C(12,5)] / C(48,5)
=~ 138,370-to-1
[ QUOTE ]
I think whoever has the A/images/graemlins/spade.gif should automatically win, I mean, it is afterall the prettiest card in the deck... For the OP, don't worry about how often this will happen, just be thankful you won! Sorry, I wish I had more to add to this...
[/ QUOTE ]
Norm? Is that you? Norman Chad?
/images/graemlins/grin.gif
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