Two players dealt AA 4 handed

[Mark1978]

Could someone please check my calcs here? I want to know the odds of two players being dealt AA in a 4 handed game.

The odds of two out of two players being dealt AA is:

52!/48!/4! = 270725

There are 6 ways of dealing AA to two people out of four:

4!/2!/2! = 6

270725 / 6 = 45121

The odds are 1 in 45121.

Is this correct?

Thanks in advance.

[pzhon]

[ QUOTE ]

There are 6 ways of dealing AA to two people out of four:

270725 / 6 = 45121

Is this correct?


[/ QUOTE ]
Yes, this method is correct because the possibilities are mutually exclusive. It's not possible for more than one pair of players to get AA, so the probability is the sum of the probabilities for each pair. You would want to use something like inclusion-exclusion if you wanted the probability that at least one pair of players gets AK.

[RustyBrooks]

This happened to me last week. Lots of action preflop and on the flop, which had a K on it. When the hand was over I turned over my cards and told the guy (a quasi-friend of mine) that if he had KK I was going to stab him in the throat. He laughed and turned over AA also.

[RustyBrooks]

(it was 4 handed)

[Zag]

I don't agree that there are 6 ways for two players to have AA. There are 6 ways for any one player to have AA (leaving only 1 way for another player), and then there are 6 ways that the two AA hands could be distributed around the 4 players.

Also, you are working on the odds of it happening on the very next hand. The odds of it happening are almost guaranteed, if you play enough hands.

Finally, I think that the chances increase slightly due to cards not being as random as you might hope, especially if this is a home game (where insufficient shuffling and cheating are both more prevalent than in a casino).

[Mark1978]

[ QUOTE ]
I don't agree that there are 6 ways for two players to have AA. There are 6 ways for any one player to have AA (leaving only 1 way for another player), and then there are 6 ways that the two AA hands could be distributed around the 4 players.

Also, you are working on the odds of it happening on the very next hand. The odds of it happening are almost guaranteed, if you play enough hands.

Finally, I think that the chances increase slightly due to cards not being as random as you might hope, especially if this is a home game (where insufficient shuffling and cheating are both more prevalent than in a casino).

[/ QUOTE ]

I was shuffling/dealing, so no worries about cheating.

[ QUOTE ]
there are 6 ways that the two AA hands could be distributed around the 4 players.


[/ QUOTE ]

That's exactly what I meant.

[ QUOTE ]

Also, you are working on the odds of it happening on the very next hand. The odds of it happening are almost guaranteed, if you play enough hands.


[/ QUOTE ]

Yes, that's what I wanted to know. I'm trying to settle an argument. I'm not looking for a reason to say "ZOMG!!!! this one hand I played was like 1247138942389124895623478 to 1."

[Zag]

Sorry, I didn't see the /4! in the original expression. However, since you are doing permutations and not combinations in the first expression, I think you have to say that there are 12 ways the two AA hands can be given out to 4 players, not 6. So I think that the answer is 1 in 22560 hands.

So, how many deals would you have to play to have a 50% chance of seeing this happen? Assuming I am correct with the 1/22560 chance in one deal.

(22559/22560) ^h = 0.50

take natural log of both sides

h ln(22559/22560) = ln(0.50)

so h = ln(0.50) / ln(22559/22560) = 15637 hands

If you and your friends have played that many hands, it is about 50% that you've seen this come up at least once.

[Mark1978]

Zag,

Here's another way of putting it to support my first answer. Please tell me if I'm going wrong somewhere.

The total possible ordered ways of dealing 8 cards are:

52!/44! = 3.03423E+13

There are 6*24=144 ways for two aces to be dealt into two of the four hands, because there are 4!=24 ways of ordering the four aces and 6 ways of two different players holding them.

The remaining random cards that fill the other hands are 48!/44!, so:

144*48!/44! = 672468480

672468480 / 52!/44! = 45121