Probability problem - 85K hands, pocket aces 356 times

[michw]

After reviewing my latest Pokertracker database, I realize I'm not getting my fair share of pocket aces and of course I'm irate about this. Now I want to try and figure out if I really have anything to bitch about. I haven't taken a statistics course since college and I'm having a hard time setting this problem up.

Given:
85,592 hands
356 pocket aces

Problem: What is the probability of getting dealt pocket aces 356 times or fewer in 85,572 hands?

Set up:
I'm inclined to set this up as a binomial distribution problem? I either get pocket aces, or I don't.

Probability of success = 1/211 = .004739
Probabilty of failure = 210/211 = .995260

I came across this formula that gives the probability of x successes in n independant trials. It's called the binomial probability function.

F(x) = [n!/(x!(n-x)!)]*(p^x)*(q^(n-x))
Where n = trials
x = # of times dealt aces
p = prob. success
q = prob. failure

Is this the correct approach to solve this problem? If so, I'm having problems with two issues.

1) My calculator can't crunch these large factorial numbers.
2) If I use this equation I would have to calculate 356 different probabilities. I would have to calculate the probability of getting pocket aces one time, and then two times, and so forth, and then add each of the individual probabilities together.

These two problems lead me to believe there may be an easier way to solve this problem. Can any math geeks help me here? I'm going to work on creating a spreadsheet to solve this problem, but am not confident in the set up.

Ok, Excel can't handle these large factorials either.

[numeri]

You are correct. Many calculators have a "cumulative binomial" function, which is the one you want. If your numers were smaller, you could use the BINOMDIST function in Excel.

What you can do instead is use the normal approximation to the binomial. (This works when n is large, as is true in this case.)

The expected number of AA hands is n*p = 85592*(1/221) = 387. The standard deviation is sqrt(np(1-p)) = 19.6.

If you want to use Excel, you can then use the NORMDIST function.

[ QUOTE ]
NORMDIST(x,mean,standard_dev,cumulative)

X is the value for which you want the distribution.

Mean is the arithmetic mean of the distribution.

Standard_dev is the standard deviation of the distribution.

Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, NORMDIST returns the cumulative distribution function; if FALSE, it returns the probability mass function.

[/ QUOTE ]

NORMDIST(356, 387, 19.6, TRUE) = 0.0569

[michw]

Many thanks! I figured there was a simpler way. I'm also glad to find I have a reason to be angry, the math supports it! [img]/images/graemlins/mad.gif[/img]

[numeri]

No problem. It's really not that unusual. Keep in mind that out of 100 85k samples, ~6 will have this result, and another 6 will have the result at the other end of the curve.

On a brighter note, I've played 65k hands of LHE and gotten AA 321 times. The probabililty of getting AA 321 or more times in 65k hands is 0.058. See, it all balances out. [img]/images/graemlins/laugh.gif[/img]