3 Related Poker Probabilities

Consider an ordinary shuffled deck of 52 playing cards and deal out 5 cards (5-card stud comes to mind!) and consider three probability questions:

1. What's the probability that you get at least 2 aces?

2. What's the probability that you get at least 2 aces given that you know that you have at least one ace?

3. What's the probability that you get at least 2 aces given that you know that you have at least the ace of hearts ?

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Note: the only "exotic" Maple command we need here is binomial(n,r) (binomial stands for "binomial coefficient") which is the familiar "n choose r" from class: i.e. binomial(n,r) = .

1. P(at least 2 aces)

This one is easy!

> Total:=binomial(52,5);

=> "total" represents the total number of different 5 card hands.

> NoAce:=binomial(48,5);

=> "NoAce" is the number of hands with no aces : choose 5 of the 48 non-aces.

> ExactlyOneAce:=binomial(4,1)*binomial(48,4);

=> "ExactlyOneAce" is the number of hand with exactly one ace and is arrived at as a two step operation: choose any ace and then choose 4 non-aces.

> AtLeastTwo:=Total-(NoAce+ExactlyOneAce);

=> Take away all the hands with either 0 or 1 ace and you have left the hands with "AtLeastTwo." Our probability of interest is thus P1:

> P1:=AtLeastTwo/Total;

> P1:=evalf(P1,4);

=> You can "expect" to be dealt at least 2 aces only about 4 times out of a hundred. This means that 2 or more aces will win a lot of Poker Hands!

2. P(at least 2 aces given at least one ace)

> AtLeastOne:=Total-NoAce;

These are the number of hands with at least one ace and is the altered Universe that the second probability must be calculated with respect two.

> MoreThanOneGivenOne:=AtLeastOne-ExactlyOneAce;

=> "MoreThanOneGivenOne" is the number in the altered Universe that have more than one ace. Our probability of interest is thus P2:

> P2:=MoreThanOneGivenOne/AtLeastOne;

> P2:=evalf(P2,4);

> P2/P1;

=> Your probability of getting at least 2 knowing that you already have at least one is about triple the unconditioned probability (which is not too surprising).

3. P(at least two aces given at least the ace of hearts )

For this question, we know that we at least have the ace of hearts in particular (why the hell should that make any difference?).

> WithAceHearts:=binomial(51,4);

=> "WithAceHearts" is the number of hands that have the ace of hearts and is the same number as the number of 4-card hands from a deck of 51 (i.e. remove the Ace of Hearts).

> NoOtherAces:=binomial(48,4);

=> All these hands have the Ace of Hearts but no others .

> AtLeastTwoGivenAceHearts:=WithAceHearts-NoOtherAces;

=> These hands have the ace of hearts and at least one other ace. Clearly then, our third probability is P3:

> P3:=AtLeastTwoGivenAceHearts/WithAceHearts;

> evalf(P3,4);

=> It's seems incredible: knowing which ace you happen to hold almost doubles(!) the probability of getting at leats two aces. If anyone can come up with an intuitively pleasing explanation of why this should be I would be most interested in hearing it!