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What are the odds of getting double aces around in a four player game of pinochle?

There are 48 cards in a pinochle deck, eight of which are aces.

Then, all cards are dealt (12 to each player).

Now, "double aces around" means you have double aces

What is dealt to other players is irrelevant. There are:

"48 choose 12" = 69,668,534,468 possible hands (see here for "choose")

"36 choose 4" = 58,905 possible hands that have all eight aces (you only choose the other four cards).

Thus the probability is precisely the fraction:

p = 58,905 / 69,668,534,468 ≈ 8.455x10^(-7) = 0.00008455%

That's about eight in a million (or, even more roughly, one in a hundred thousand).

However, a game of pinochle (played to 1500) can go for several rounds, and many people play pinochle for long periods of time. Although our probability p is very small, if you play 1000 games of pinochle (say, 6 rounds each), the odds of NOT seeing aces all around once is:

(1-p)^6000 ≈ 0.99494

which means there's roughly 0.5% chance of seeing aces all around at least once in 1000 games of pinochle.

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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