TITLE: Blackjack Probability
QUESTION [7 upvotes]: Suppose that you are playing blackjack against the dealer. In a freshly shuffled deck (standard $52$ cards), what is the probability that neither of you are dealt a blackjack. Blackjack being $2$ cards adding to $21$ i.e. $Ace + 10,J,Q,or K$ (or vice versa as order does not matter).
The farthest I've really come is that the odds of the first player getting dealt a blackjack is $128\over 2652$. 
First case: Odds of getting an Ace are $4\over52$, odds of the next being 10,J,Q,or K are $16\over51$.
Other case: Odds of getting 10,J,Q,or K are $16\over52$ and Ace $4\over 51$ so ${((4*16)*2)\over (52*51)} == {128\over 2652}$
Not sure where to go from here...

REPLY [3 votes]: It might be a little simpler to deal with three separate probabilities:
Since there are four aces and 16 cards 10-K, the raw probability of a hand being blackjack is:
$$P_\text{blackjack} = \frac {4 \times 16} {\binom {52} {2}}.$$
The probability of both getting blackjack is just the probability of the player getting blackjack and the subsequent probability of the dealer also getting blackjack:
$$P_{\text{both}} = P_\text{blackjack} \times \frac{3 \times 15} {\binom {50} {2}}.$$
By the inclusion/exclusion principle, the probability of at least one getting blackjack is the sum of probabilities of each individual getting blackjack minus the probability of both getting blackjack (removes redundant overlap):
$$P_\text{either or both} = 2 P_\text{blackjack} - P_\text{both}$$
and the final result (neither gets blackjack) is just the inverse probability:
$$1 - P_\text{either or both} = 0.90524.$$