95% confidence in number of AA-JJ in 10,000 hands.

Hi

Hard for me , but easy for the math guys here.

If you get a pair 1 in 16 hands, then that would be 100/1600 hands. Of that hundred, a perfect or mean distribution would be about 23% or 23 hands (4/13) as being AA,KK, QQ, or JJ.
Correct me if I'm wrong.

My question is, if you had 10,000 hands, what would the range of AA -JJ hands you would have with a 95% confidence rate (if rate is the correct word).

I think the mean would be about 144, and the range would be something like XXX to YYY with a 95% confidence rate at 10000 hands? But what would the values of XXX and YYY be?

Also, how does the range gap close at 20,000, 50,000 and 100,000 as your sample grows.

I just want to get a sense of the proportion and range of large pairs in my database. If there is a better way than a 95% confidence rate, please mention that.

PS. I only remember 1/2 of what I took in university stats course...

Thanks

[uDevil]

You will be dealt a pair with probability C(4,2)/C(52,2). That's one time in 221.

So P(AA,KK,QQ,JJ)=4*(1/221)=.0181 or about 1/55.

In N hands, the expected (mean) # of AA, KK, QQ, JJ hands is N*P(AA,KK,QQ,JJ)

If N=10000, mean= 10000*.0181= 181 hands

One standard deviation, SD= sqrt(N*P(AA,KK,QQ,JJ)*(1-P(AA,KK,QQ,JJ))

so SD=sqrt(10000*.0181*(1-.0181))= 13.3

That means that about 68% of the time, in a sample of 10,000 hands, you will get 181 +/- 13.3
hands of AA, KK, QQ, or JJ. 95% of the time, you will get 181 +/- 1.96*13.3, i.e. 181 +/- 26.1 of these hands.

[Phil153]

Best way to get a feel for this is to play around with a binomial calculator.

http://www.stat.sc.edu/~west/applets/binomialdemo.html (java applet with graph)
http://www.swogstat.org/stat/public/...calculator.htm (non java)

N is number of trials, p is probability (for AA, p = 1/221 ~= 0.00452, for AA-JJ, p ~=0.0181). Then enter different values for X or t and see how likely different frequencies are. Use the "at least" or "more than" functions to see how many lie above and below that number.

For example, using the java based calculator above with N=1000, p = 0.0045, X = 0, the probability that you will get zero AA out of N = 1000 hands is 1.1%. The probability of getting more than 7 is 8.6%

thx phil and uudevil