Three 7's

Using 6 decks, what is the probabilty of a player having three (not suited) 7's and the dealer having three (not suited) 7's? On the same lines, the same 6 decks- same question but the players hand and dealers hand are suited.

Any help ont hsi would be great.
Thanks

Almost frickin' impossible for both. Millions.

Quote: vcowan

Using 6 decks, what is the probabilty of a player having three (not suited) 7's and the dealer having three (not suited) 7's? On the same lines, the same 6 decks- same question but the players hand and dealers hand are suited.

Any help ont hsi would be great.
Thanks

I assumed this is for blackjack and the player didn't split his pair of 7's. The second part of the question is easier; allowing the dealer's suit to be the same or different than the player's, I get a probability of 1.99893E-10 or about 1 in 5,000,000,000. In Excel notation, that's (4*permut(6,6) + 4*permut(6,3)*3*permut(6,3))/permut(312,6).

I am getting the same answer, using
24/312*5/311*4/310 * (18/309 * 5/308 * 4/307 + 3/309*2/308*1/307)= 1.9989E-10

As to the first question, it's easy to answer if suited hands are included:
24/312*23/311*...*19/307 = 1.1027E-7

If you are looking for probability of unsuited hands specifically, it'll be a little bit less, but the difference will be several orders of magnitude smaller than the answer, so with this level of accuracy, the answer won't be different.