Probability – Calculate the Probability of No Blackjack in a 4-Player Game

card-gamesprobability

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.

Best Answer

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.

Related Question