Easy ? but I have brain-lock

[NajdorfDefense]

How do you calculate the probability of flopping x sets [or fewer] in Y hands.

E.G. I have flopped 3 sets in 77 pocket pairs dealt.

What are the odds against that?

[Masquerade]

It's a binomial distribution (as you either flop a set or you don't). So if the chance of an event happening on a single trial is p and you have N trials total then the probability of p occurring exactly n times is (N choose n) p^n (1-p)^(N-n) where (N choose n) = N!/(n! (N-n)!).

[NajdorfDefense]

so 77! / (3! * 74!)

or 73,150-1 against?

[bigpooch]

With p=1081/1225 as the probability of NOT flopping a set or
better, just add

C(77,0)(p^77)
C(77,1)(p^76)((1-p))
C(77,2)(p^75)((1-p)^2)
C(77,3)(p^74)((1-p)^3)

Altogether, this is about 0.015530244, or about 63.39 to 1
against.

[NajdorfDefense]

I have no idea what C(xxx) means below. Help?

[Tom1975]

C(x,y)= x!/[y!*(x-y)!]