What's the probability of getting 3 suited ace-jack blackjacks off the top of a 6 deck shoe?

Question

I don't care what suits they are, nor if they are the same or different suits. And I mean I'm playing three hands, nobody else at the table, and of course the dealer gets a hand.

I'm thinking, to start out with, (48/312)(47/311)(46/310)(6/308)(6/307)(6... 26/billion, or about 1 in 40 million.

Here's my problem. The above only works if we *don't* get, say, an ace of diamonds on hand one, card one, and then a jack of diamonds on hand two, card one. Because if we do that, then the that only leaves 5 jacks of diamonds available to complete hand one. And so forth (i.e., what if you get ace of diamonds hand one card one, and then jacks of diamonds for hand two card one *and* hand three card one)

We also have to account for the probability that the dealer gets one of the 48 cards of interest as his/her first card.

Help.

Answer 1

This a is a very difficult number to find for a lot of reasons.

For ballpark estimates you coudl use the number 1:21 for the frewuency of blackjacks being dealt, and 4 times that for suited blackjacks and rais it to the third power.  *****1:(84^3)****

To be truely accurate you would have to play out every possible combination of cards that could be dealt and look at how many ways it can play out so that you get three suited blackjacks.

It woudl be difficult to do in the way you are suggesting because if your first card is a ten and you work the remaining cards from there, your odds would be different than if your first card was an ace, this woudl effect the first card for each individual blackjack and you would have to compensate for the differences in frewuency of your first card being an ace and it being a ten and average them all togther.

Its a major headache so just use 1:84^the number of hands with suited blackjacks, in your case, raised to the 3rd power.  This isnt totally accurate do to the effect fo removal of cards but its the best way since the true number if incredibly difficult to find.