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A related, maybe more difficult follow up questions: What is the probability of getting a full house and a flush and a 2 of spades in a 13 card hand? How do I even go about solving this problem? I'm not sure how to divide up the probabilities.
The full house and flush must be separate; no sharing cards.
Added 3+ months ago:
For example, {2,4,5,5,5,6,8,8, 10, J, Q, K, A} is a set with a full house that is not accounted for in the above. Additionally, {2H, 2C, 2S, 3H, 3S, 4H, 5H, 6H, 7D, 8C, 9S, 10S, 10D} where the second letter represents (H=hearts, C=clubs, S=spades, D=diamonds), is a set that would have a house, but shares cards with its flush. Both the flush and the house must exist in the 13 cards without sharing.
Full house -- 3 of a kind + a pair.
The first card is given: chance = 1. After the first card, there are 51 more to deal and 3 of them are of the same kind. Next card, there are 50 more to deal and 2 are the same kind.
Chance = 1 x 3/51 x 2/50
For the pair, there are 49 more to deal and 35 of them will define a pair. (Not 36 -- we can't allow four of a kind.) Now there are 48 more to deal and 3 of them will make a pair.
Chance = 35/49 x 3/48
Total = 1 x 3/51 x 2/50 x 35/49 x 3/48 = 105.042 in a million.
Flush -- 5 of the same suit not in sequence. Do this the same way.
Yawn. At my age I get bored counting cards. I showed you how to do it. You do the rest.
I believe this is the answer to the question: "What is the probability of drawing 3 of a kind followed by 2 of a kind?"
This does not answer my question because it misses other cases in which a full house and a flush can occur. For example, {2,4,5,5,5,6,8,8, 10, J, Q, K, A} is a set with a full house that is not accounted for in the above. Additionally, {2H, 2C, 2S, 3H, 3S, 4H, 5H, 6H, 7D, 8C, 9S, 10S, 10D} where the second letter represents (H=hearts, C=clubs, S=spades, D=diamonds), is a set that would have a house, but shares cards with its flush. Both the flush and the house must exist in the 13 cards without sharing.