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Ok I think I found the solution to my own problem. There was a similar
question asked at the Wizard of Odds website. So I adapted the equation to my purposes.
let p = probability of an event; in this case, the probability of getting dealt at least one 9 in stud
which shown above is ~=0.21738
h = # hands in a given session
Thus hp = expected number of hands dealt in that session with at least one 9
Using the Wizard's formula:
SD = sqrt (hp - (1-p))
= sqrt (200 x 0.21738 - (1-0.21738))
= sqrt (43.5 - .78262)
= 5.83
So in 200 hands within 1 std deviation, you should see at least one Nine in ~37-49 hands.
Not especially useful I guess, especially since this forum is pretty devoid of any 'poker is rigged' posts.
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Did you get your forumla right?
I don't think you did.
I think the appropriate formula is
SD = sqrt(npq) where n = number of hands, p = prob of getting dealt a 9 in a single hand, q = prob of not get dealt a 9 (or 1-p)
E[x] = p*n = 43.5
SD = 1.86
Your results of 55 hands with a 9 was 6 standard devs away from the average.