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