Probability of # of Blackjacks in one hand.

Hey all,

I'm trying to figure out the probability of two different scenarios in blackjack.

The table uses a continuous shuffler so we can assume a fresh 6 decks every hand.

For arguments sake, let's say that there are 7 hands and always 7 hands being played at all times. (Disregarding the dealers hand)

What is the probability that out of the 7 hands that 3 of them are Blackjacks? And then what is the probability that 4 out of the 7 are Blackjacks?

Thanks to anyone that can supply the answer, if the math used to get to the answer could be included with the answer, that would be awesome as well!

Begin with the probability of 7 BJs.
It is a (multi)hypergeometric.
Notation: C(N;a,b,...,c) is the combinatorial N! / a!b!...c! where N=a+b+...+c
C(N;n) = N! / n! (N-n)!

(One ace each) x (One figure each) / (All possible cases) =
C(24;1,...,1,17) x C(72;1,...,1,65) / C(312;2,...,2,288) = 5.7 x 10^-36


Now the probability of 6 or more BJ is equal to

C(7;6) --- number of selections of 6 among 7 players
x C(24;1,...,1,18) x C(72;1,...,1,66) / C(312;2,...,2,290) --- notice there are only six "1" or "2" in the combin
x {1 - C(18;1) x C(66;1) / C(290;2)} --- non-BJ for the seventh player

= 9.5 x 10^-33


Now you can develop for the other numbers. For example,
P(3 BJ) =
C(7;3)
x C(24;1,1,1,21) x C(72;1,1,1,69) / C(312;2,2,2,306)
x {1 - C(21;1,1,1,1,17) x C(69;1,1,1,1,65) / C(306;2,2,2,2,288) }

= 0.0013844728