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...

  • $\begingroup$ alright just did that $\endgroup$
    – ADH
    Feb 10, 2015 at 3:29
  • $\begingroup$ Are you assuming two players against the dealer? The use of neither indicates that. Your calculation for the first player is correct. $\endgroup$ Feb 10, 2015 at 4:28

4 Answers 4


This is my second try, but I think is a MUCH better argument than what I had before. So I deleted the more complicated answer I tried to give earlier.

The total # of possible deals to the two players is: $$\binom{52}{2}\binom{50}{2}=1,624,350$$ (i.e., choose two cards for the first player and then two for the second.)

Then, the # of ways to deal a blackjack to both players would be: $$\binom{4}{1}\binom{16}{1}\binom{3}{1}\binom{15}{1}=2880$$ (i.e, choose which ace to give to the first player, then which 10,J,Q,K for the first player, then which ace for the second player, and then which 10,J,Q,K for the second player.)

Also, the # of ways that exactly one player gets a blackjack would be: $$\binom{2}{1}\binom{4}{1}\binom{16}{1}\times(\binom{3}{2}+\binom{3}{1}\binom{32}{1}+\binom{47}{2})=151,040$$ (i.e., choose which player to give the blackjack, then choose which ace to give them, and then which 10,J,Q,K to give them. Then for the other player we either give them two aces from the 3 that are left, give them one ace from the 3 that are left and 1 of the 32 2-9 cards, or give them two non-aces from the 47 non-aces that are left.)

So, the probability of at least one player getting a blackjack is: $$\frac{2880+151,040}{1,624,350}$$

Hence the probability of neither player getting a blackjack is: $$1- \frac{2880+151,040}{1,624,350} \approx 90.5\%$$

  • $\begingroup$ If the player goes blackjack or busts, the dealer doesn't get to draw, so there is some conditioning between the events that may need to be addressed. $\endgroup$
    – Avraham
    Feb 11, 2015 at 4:40
  • $\begingroup$ @Avraham: we are concerned with blackjack before the dealer (or even the players) draws, so this is not an issue. $\endgroup$ Feb 11, 2015 at 4:53
  • $\begingroup$ I had the same interpretation as Ross. $\endgroup$ Feb 12, 2015 at 2:32
  • 1
    $\begingroup$ Very clever Michael Cotton, thank you. I can't help but wonder if there perhaps is a less hideous way of acquiring the same answer though. $\endgroup$
    – David
    Nov 29, 2016 at 13:38

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: $$1 - P_\text{either or both} = 0.90524.$$

  • $\begingroup$ Fantastic explanation! $\endgroup$
    – Beginner
    May 23, 2018 at 22:56

I don't know if this will get you any closer to a closed form solution, but maybe it helps you think about it.

So the odds of not getting blackjack the first hand would be ((52*51)-((4*16)*2))/(52*51) after which point you know the deck is either out one or two face cards, out one or two aces, or neither. This happens every time. (until there are no more other cards, at least)

 hand#       aces,face-cards     total cards
  1            -- * --             52 * 51
             /  / | \  \
  2    2,16 *  *  *  *  * 4,14     51 * 50
            3,16 4,16 4,15 

With a branching factor of five (I know), we have two aces, once ace, two whatevers, one 10, two 10s. The outermost branches are short, but the inside of the tree is massive... Hopefully this at least helps you contextualize the problem. :) Good luck!


The naive assumption is that the chance of each player getting blackjack is independent of the others. As you say, the chance than one player does not get blackjack is $\frac {2524}{2652}$ The chance that none of $n$ players gets blackjack is then $\left(\frac {2524}{2652}\right)^n$. This is not quite correct, as the fact that the first player did not get blackjack enriches the average deck with cards that could make a blackjack for the second player. Because the chance of one player getting blackjack is small, the enrichment is small as well, so this is not far off.


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