(25!/((25-4)!4!))

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(52!/((52-4)!4!))

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(52!/((52-4)!4!))

Solved

Posted on 2007-10-15

In an poker hand (texas Holdem), with 10 players, 25 cards could be dealt if the hand plays to the final community card.

(Each player receives 2 cards and 5 community cards are dealt)

How likely is, or what is a valid approach for calculating the probability of, all 4 aces appearing in those 25 cards, assuming a truly random shuffle.

I think i would look at it in terms of there being no Aces and then taking the complement but I'm not sure that this is the correct logic to use.

My interest is more than purely theoretical. As a player in online games, I see what I perceive to be a larger than expected proportion of hands with high cards - As & Ks. This might be a quite false perception of course - but I'm starting from here and I'm not sure i can get the sums right with my statistics knowledge from 40 years ago.

I know that there are lot of poker stats around but I can't see this one.

(Each player receives 2 cards and 5 community cards are dealt)

How likely is, or what is a valid approach for calculating the probability of, all 4 aces appearing in those 25 cards, assuming a truly random shuffle.

I think i would look at it in terms of there being no Aces and then taking the complement but I'm not sure that this is the correct logic to use.

My interest is more than purely theoretical. As a player in online games, I see what I perceive to be a larger than expected proportion of hands with high cards - As & Ks. This might be a quite false perception of course - but I'm starting from here and I'm not sure i can get the sums right with my statistics knowledge from 40 years ago.

I know that there are lot of poker stats around but I can't see this one.

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