Math Question.

[RobReality]

Last night we had another one of many private single table tournaments on UB.

In the course of this 2 hour tournament, one of the players, whom we really don't know all that well (friend of a friend situation), got the following hands:

AA x 6
KK x 2
QQ x 1
JJ x 2
TT x 2
99 x 1

And various other pocket pairs and A high suited configurations.

I definitely understand the concept of standard deviation, but this is astounding.

The way I understand it, it is 221 to 1 against getting AA on any given hand. I was going to attempt to calculate the odds of him getting AA 6 times in a two hour span at approximately 75 hands per hour. I realized that I don't necessarily know how to go about figuring that. In my mind it would go something like this:

(221^6)/(75*2)

But that number is simply astronomical. I can't believe that the odds are that terrible.

Can someone show me the proper way of figuring situations such as this, and perhaps a quick explanation of why you are right, and I am wrong?

I would appreciate it immensely.

Thank you very much.

Rob.

[danzasmack]

look at getting AA as a "success", and not as a "failure" - apply the binomial distribution where # of hands = 150, # of successes = 6, and P(success) = 1/221 this gives us:

(150 C 6)*[(1/221)^6]*[(220/221)^144]

this number will be small and when calculated you get:
6.3866E-05

But that is because you are saying EXACTLY 6 times, were you to do at LEAST 6 times it will be a different. It's like the probability of getting aces EXACTLY once in 1k hands is small, simply because you should get it more often than that.

To calculate the probability of getting AA at LEAST 6 times you simply do 1-P(aces 5 or less). Though in this case it will not be very different at all. This method seems right to me.

[pzhon]

[ QUOTE ]
I was going to attempt to calculate the odds of him getting AA 6 times in a two hour span at approximately 75 hands per hour.

[/ QUOTE ]
The probability that a particular 6 hands will all be AA, and the others will not be AA is 1/221^6 * (220/221)^144. The number of ways of choosing 6 hands out of 150 is "150 choose 6" =

150*149*148*147*146*145
---------------------------,
6*5*4*3*2*1

or about 14.3 billion.

The probability of getting AA exactly 6 times in 150 hands is (150 choose 6) * (1/221^6) * (220/221)^144 ~ 1/15657. The probability of getting AA 6 or more times in 150 hands is about 1/14213. However, this underestimates the frequency with which you will see so many AA hands in 2 hours because it will happen much more frequently when there are slightly more hands than average in the 2 hour period. Also, if you can choose the 2 hour period, you might start the period with an AA, or a cluster of AA hands.

[RobReality]

Thank you very much for your reply.

Rob.

[RobReality]

That makes sense in a dizzying sort of way. I appreciate it. Thank you.

Rob.

[EStreet20]

Please tell me this lucky shmuck won the tourney...

[RobReality]

Definitely.