easy probabilty question

[BogusPomp]

You shuffle a deck and deal out 12 cards.

What is the probability there is at least 3 cards of any rank amongst those 12?

[DrVanNostrin]

13*4c3*49c9/52c12

13 ways to pick rank
4c3 ways to pick 3 of that rank
49c9 ways to pick the remaining 9 cards

[BruceZ]

[ QUOTE ]
13*4c3*49c9/52c12

13 ways to pick rank
4c3 ways to pick 3 of that rank
49c9 ways to pick the remaining 9 cards

[/ QUOTE ]

Multiplying by 49c9 quadruple counts each case where there are 4 of a given rank, and multiplying by 13 over counts the cases where there are at least 3 of MORE THAN 1 rank. The probability that there are at least 3 of a GIVEN rank is:

[C(4,3)*C(48,9) + C(4,4)*C(48,8)] / C(52,12)

=~ 3.433%

The second term of the sum C(48,8) is the number of ways to have 4 of the given rank. Note that if we multiply this term by 4 in this expression, it will be equal to your result, since you count each of these 4 times.

Now the probability that there are at least 3 of SOME (unspecified) rank can in theory be computed without over counting by applying the inclusion-exclusion principle to the 13 ranks, where the first term would be 13 times the above expression, but this would be complicated due to the possible combinations of 3 and 4 of each rank. Instead, it is easier to take 1 minus the probability that no rank is repeated more than 2 times. That is:

P(1 or more rank repeated at least 3 times) =

1 - P(no rank repeated more than 2 times) =

1 - [P(0 pairs) + P(1 pair) + P(2 pairs) + ... P(6 pairs)] =

1 - sum{p = 0 to 6}C(13,p)*C(13-p,12-2p)*(6^p)*4^(12-2p) / C(52,12)

=~ 40.124%


Each term in the sum is the probability of exactly p pairs. The first term is for 0 pairs, which is simply 12 different ranks. For each term, there are C(13,p) ways to choose the ranks which will pair, times C(13-p,12-2p) ways to choose the ranks of the 12-2p cards that do not pair out of the remaining 13-p ranks, times 6^p ways to choose one of the 6 possible pairs for each of the p pairs, times 4^(12-2p) ways to choose one of the 4 possible cards for each of the (12-2p) unpaired cards. This is all divided by C(52,12) total ways to choose the 12 cards.

EDIT: Changed "double counts" to "quadruple counts" in first sentence.