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The problem that both you and the dealer are not dealt the blackjack can be solved by finding the probability that at least 1 player is dealt a blackjack and then use the law of total probability to find the probability that no one is dealt a blackjack. This approach is stated here Blackjack Probability).

Now we have 4 players and we want to find the probability that nobody is dealt a blackjack. I wonder if there is a simpler approach than calculating 1 - \sum P(only i player is dealt a blackjack) where i = 1 to 4 since it is more complicated to calculate when i = 3 and 4.

clement
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    Unsure if the following approach is simpler. I would use Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. Using the syntax of the second link, for $~k \in {1,2,3,4},~$ I would let $~S_k~$ denote the subset of all possible deals where Person$-k~$ was dealt a blackjack. – user2661923 Nov 30 '23 at 07:35
  • Re last comment, note that when enumerating (for example) $~S_1 \cap S_2,~$ you have to compute the number of ways that Person$-2~$ could be dealt blackjack, given that Person$-1~$ was dealt blackjack. – user2661923 Nov 30 '23 at 07:38

1 Answers1

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You can solve this using inclusion-exclusion:

$$ 1 - \binom{4}{1} \frac{\frac{4!}{3!} \frac{16!}{15!}} {\binom{52}{2}} + \binom{4}{2} \frac{\frac{4!}{2!} \frac{16!}{14!}} {\binom{52}{2}\binom{50}{2}} - \binom{4}{3} \frac{\frac{4!}{1!} \frac{16!}{13!}} {\binom{52}{2}\binom{50}{2}\binom{48}{2}} + \binom{4}{4} \frac{\frac{4!}{0!} \frac{16!}{12!}} {\binom{52}{2}\binom{50}{2}\binom{48}{2}\binom{46}{2}} $$

where in each term:

  • The leading coefficient is the number of ways to select $i$ out of the 4 players to definitely have blackjack (the non-selected players may or may not have blackjack).
  • The first factor in the numerator is the number of ways in which 1 of the 4 aces can be assigned to each of the $i$ selected players.
  • The second factor in the numerator is the number of ways in which 1 of the 16 ten-value cards can be assigned to each of the $i$ selected players.
  • The denominator is the number of ways to deal 2 cards to each of the $i$ selected players.

You can confirm this using my Icepool Python package, which is not as computationally efficient but requires less manual work:

from icepool import Deck, multiset_function

deck = Deck([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10] * 4)


@multiset_function
def blackjack4(a, b, c, d):
    return a >= [1, 10], b >= [1, 10], c >= [1, 10], d >= [1, 10]


output(blackjack4(deck.deal(2, 2, 2, 2)).map(sum))

This computes the chance of $i$ hands drawing a 2-card blackjack. The chance of $i = 0$ is

$$ \frac{1550115663120}{1896396138000} \approx 81.74\% $$

You can run this in your browser here.

HighDiceRoller
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  • Beat me to it :D! But I'm getting a slightly different answer $\frac{2152938421}{2633883525}$ although our formulas should be the same. I used this Sage-code to calculate: https://www.online-python.com/0LCxrgPR7D – ploosu2 Nov 30 '23 at 08:07
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    It's just a matter of simplifying fractions or not. I chose to keep the raw denominator of the 4-way deal. – HighDiceRoller Nov 30 '23 at 08:23