Asked by xiny
In a poker game, you get 5 cards from a regular deck of 52 cards.
What is the probability that you get full house (i.e. three of a kind and two of another kind) if the first two cards you've got were two aces?
What is the probability that you get full house (i.e. three of a kind and two of another kind) if the first two cards you've got were two aces?
Answers
Answered by
Tammy
OMG, the bot is right.
But of course it gave no explanation, and both Google and Wikekpedia
give that answer.
3 of a kind = 13*C(4,3) = 52
2 of another kind = 12 * C(4,2) = 72
number of possible "full houses" = 52*72 = 3744
number of combinations of 5 cards = C(52,5) = 2,598,960
prob(full house) = 3744/2,598,960 = appr .0014406
But of course it gave no explanation, and both Google and Wikekpedia
give that answer.
3 of a kind = 13*C(4,3) = 52
2 of another kind = 12 * C(4,2) = 72
number of possible "full houses" = 52*72 = 3744
number of combinations of 5 cards = C(52,5) = 2,598,960
prob(full house) = 3744/2,598,960 = appr .0014406
Answered by
Bot
The probability of getting a full house with two aces is 0.00144057623049219.
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