Number of AA hands in one session -- help me figure this.

[brick]

Sorry if this has been asked before.

I'm curious about how many AA hold'em hands I can expect to get in one 6 hour session that is extremely good. (.5% of sessions)
I figure in one six hour session I should get AA once. (playing 40 hands per hour)
A session that is plus 3 Standard Deviations would happen once every 200 6 hour sessions correct? (3 Standard Deviations accounts for 99%. 1/2 of the other 1% would be 3 SD above normal.)

Do I have this right so far? How many AA hands should I expect in a +3SD outcome?

[aces961]

The standard deviation of the number of AA you will be dealt is equal to sqrt(1/221*220/221*240) which is a little above 1. This would then give you a result a of about 4.2.

However the 3 S.D. estimate is not very accurate in this case since this is a binomial random variable with mean near 1 and a number of trials much higher than that. A few further computations gives you P(5 AA or more)=.005027. P(4 AA exactly)=.0193789, so P(4 AA or more) is above .02. So we se the 3 S.D. estimate isn't extremely accurate in this case.


On a side note if the number of hands was say 2000 and you used the 3 S.D. estimate these would be the results.
mean +3S.D. is 18.05.
P(17 or more AA)=.0114823
P(18 or more AA)=.0055057
P(19 or more AA)=.0025129
So we see the higher N is the better of an estimate the 3 S.D. figure becomes.

[squiffy]

If I recall correctly, Lee Jones states that you can expect to be dealt AA once every 200 hands or so. But he does not show how to calculate this.

You would need to know how many different AA combinations are possible and which ones to throw out.

And you would need to divide by the total number of two card hand combinations that are possible.

1 in 200 deals seems a bit low to me. But there are 4 aces out of the 52 card deck, so about 7.6% of the cards are Aces.

So then you need to figure out how many hands are dealt on average per hour. I have heard some say 100 hands an hour. But that sounds much too high to me, even for online play.

Then you can calculate how many hours you would have to play, on average, to get one AA hand.

If there are four aces AcAdAhAs.

I guess there are 6 unique two card combinations of AA?

AcAd AhAd AsAs
AcAh AhAs
AcAs

I used to remember a simple formula for this. Maybe it is n-1 + n-2 + n-3 . . . . where n is the number of cards, here 4 cases. So 3 + 2 + 1 = 6.

[Copernicus]

This one is much easier to compute directly...the probability of getting any specific pocket pair is 4/52*3/51, which is 1/221.

I wonder what the value of the of knowing the answer to the original question is, though?

[BruceZ]

Everything aces961 said is exactly correct. I'll just elaborate a little. You are asking for a 3 SD result, but your AA are not distributed by a normal distribution (smooth bell curve), but rather by a binomial distribution (discreet distribution that looks like steps approximating the bell curve). The variance of the binomial distribution is 1/221 * 240, and the standard deviation is the square root of this or 1.04. We can sometimes use the normal distribution to approximate the binomial distribution, but when we do this, the variance we use becomes [ 1/221 - (1/221)^2 ] *240, and the standard deviation is the square root of that, which in this case isn't much different from the SD of the binomial distribution since the probability is small. It still comes out to 1.04. This is the same thing aces961 computed in a different form. As he showed, it is not very accurate in this case.

BTW, 3 standard deviations actually includes 99.74% of the sessions, not 99% as you said, so only 0.13% will be above this amount. Using binomdist in Excel, I found that getting AA 5 times or more will occur 0.50%, and getting AA 6 times or more will occur .087%. 0.13% is between 5 and 6, even though 3 SDs of the binomial distribution puts it at 4.2.

So the answer to your question is that it depends on what you mean by 3 SD. If you mean 0.13%, you have to wait for 1/.00087 = 1150 sessions to get 6 or more AA, but that would actually be more than a 3 SD result. If you mean 3 binomial distribution SDs, then you would just need 5 or more AA, and that would again be more than 3 SDs, and it would take only 1/.005 = 200 sessions.

[brick]

Thanks for your help with this problem. Good to hear I that I might get 5 AA hands in one session someday.

[ QUOTE ]
I wonder what the value of the of knowing the answer to the original question is, though?


[/ QUOTE ]

.. I'm trying to learn this game... just curious about what to expect.

[Copernicus]

Thinking about the game and curiosity are good things, but obscure considerations like the 3 SD result for AA wouldnt be at the top of my priority list!

GL. You'll find a lot of good threads on basic poker probability here, and the recommendations of Mike Petriv's Hold Em Odds Book are right on if you want to dig deeper into it.