Someone asked my that I show my work on this.

OK, P(5 KK or AA|2100 hands)

Is mostly P(5) but techncially you want to say "5 KK/AA or _fewer_" so you have to sum P(1)...(2)(3)(4)(5). The are all calcualted the same way, but I will do P(1) (simplest) and P(5) (most germain)

P(1) - what are the odds that you are so pathetic as to get AA or KK exactly ONCE in 2100 hands?

Well, in 1 hand - 1/110 (roughly) (or .00905 exactly, according to excel)

In the first hand, but not the next two: - 1/110 * (1-1/110) = 109/110 * 109/110 = .0088 (a smidge less than 1/110).

Now, in 3 hands - you could get KK on the first, second, or third hand - so there are 3 "combinations" (or permutations... I get them confused), so you have to multiply by 6 = .0267

You can also write this (3 p 1) * P^1 * (1-P)^(2)
or in general
(N p M) * P^M* (1-P)^(N-M)
Where P = the probability of a single event
N = number of trials
M = number of succeses
And (N p M) = N "permute" M, or N!/(M-N)!

Now, some math symbols:
* = "times" but this is the comp sci way of writing it.
X^Y = X to the power Y, i.e, X^2 is X-squared
X! = X "factorial" is X * (X-1) * (X-2)... * 2 * 1

Dang, gotta go. More later... or maybe someone can finish this for me