AA vs AA with one winning

[TheProdigy]

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]

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]

[ 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]

I think whoever has the A[img]/images/graemlins/spade.gif[/img] 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]

[ 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[img]/images/graemlins/spade.gif[/img] 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?

[img]/images/graemlins/grin.gif[/img]