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It's Texas Hold'Em with 6 players. What's the probability that at least 1 player is dealt an ace? What is the distribution model for 0..4 aces?

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First calculate the probability that no aces are dealt to 6 players (12 cards):

Prob of no aces 48/52 * 47/51 * ..... * 37/41 = 0.3376 = 33.8%

Therefore the probability of at least one ace being dealt is given by 1 - 0.3376 = 0.6624 = 66.24%


The probability of 1 and only 1 Ace being dealt is given by:

12 * 4/52 * 48/51 * 47/50 * ... * 38/41 = 0.4379 = 43.79%

The probability of 2 and only 2 Aces being dealt is given by:

66 * 4/52 * 3/51 * 48/50 * .... * 39/41 = 0.1902 = 19.02%

Similarly, for 3 Aces, we find the probability to be,

12C3 * (48P9)/(52P12) = 0.0325 = 3.25%

Finally the probability that all 4 aces are dealt is given by,

12C4 * (48P8)/(52P12) = 0.001828 = 0.182%

The accuracy of these number can be verified, by adding them together to give 66.24% - the answer to the first part of the question above.

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