Odds of AA vs. AA - Gambling and Probability

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Odds of AA vs. AA

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01-07-2010 , 04:34 AM

kataruni ochiru

journeyman

Join Date: Aug 2009 Posts: 223

How do you calculate the odds of 2 players getting dealt AA?

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01-07-2010 , 09:43 AM

jay_shark

Pooh-Bah

Join Date: Sep 2006 Posts: 3,655

At an n-handed table, the exact probability that two players get dealt pocket aces is,

n*C(4,2)/C(52,2) - C(n,2)*6/C(52,2)*1/C(50,2)

where C(n,2) is the number of ways that you can select 2 objects from n total objects.

At a 10-handed table table, this is ~ 4.508%

Of course, you don't have to be so nit-picky. You can approximate the solution by computing the probability that one players receives pocket aces and then multiply it by the total number of players. In this case, it is
n*C(4,2)/C(52,2).

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01-07-2010 , 10:15 AM

Ovaltine88

centurion

Join Date: Jul 2008 Posts: 126

What about odds of AA vs. AA then hitting a four-flush?

I actually seen that once.

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01-07-2010 , 10:52 AM

saladspoon

journeyman

Join Date: Jul 2009 Posts: 265

4.508% seems way too high.

The way I would calculate it is to start with the probability that two players both get pocket aces:

(4*3*2*1)/(52*51*50*49)

Then for a 10-handed table, it's C(10,2) for any 2 out of 10 receiving the cards.

C(10,2) * (4*3*2*1)/(52*51*50*49)

Resulting in a 0.0166% chance of it happening at the 10-player table, or about 1 out of every 6016 deals.

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01-07-2010 , 11:02 AM

jay_shark

Pooh-Bah

Join Date: Sep 2006 Posts: 3,655

Quote:

Originally Posted by jay_shark

At an n-handed table, the exact probability that two players get dealt pocket aces is,

C(n,2)*6/C(52,2)*1/C(50,2)

where C(n,2) is the number of ways that you can select 2 objects from n total objects.

At a 10-handed table table, this is ~ 0.0166 %

I realized that I calculated the probability that someone at the table is dealt pocket aces. The second term in my op, which is
C(n,2)*6/C(52,2)*1/C(50,2), is the probability that two players at an n-handed table are dealt pocket aces.

Last edited by jay_shark; 01-07-2010 at 11:14 AM.

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01-11-2010 , 08:47 AM

PokerRon247

Carpal \'Tunnel

Join Date: Jul 2008 Posts: 11,192

Do the above calcluations take into account that there is only one possible combination of AA left after one person has been dealt AA? (ie once one person has been dealt AA, the probability of another person getting dealt it decreases dramatically)

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01-11-2010 , 11:13 AM

jay_shark

Pooh-Bah

Join Date: Sep 2006 Posts: 3,655

Quote:

Originally Posted by PokerRon247

Yes.

It computes the probability that two players at any n-handed table are dealt pocket aces.

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