Find the probability of getting a full house (3 cards of one value and 2 cards of another) when 5 cards are dealt from an ordinary deck?
There are 52 cards in an ordinary deck, consisting of 4 suits of 13 cards each. To get a full house, we want 3 cards of one value and 2 cards of another.
First, let's choose the values of the full house. There are 13 possible values for the 3-of-a-kind, and once we've chosen that, there are 12 remaining possible values for the 2-of-a-kind. Therefore, there are 13 * 12 ways to choose the values.
Next, we need to choose the suits for each value. For the 3-of-a-kind, there are 4 suits, and we need to choose 3 suits out of the 4, which can be done in C(4,3) ways. For the 2-of-a-kind, there are 4 suits, and we need to choose 2 suits out of the 4, which can be done in C(4,2) ways. So, the total number of ways to choose suits is C(4,3) * C(4,2).
Now, we can calculate the total number of favorable outcomes (i.e., full houses) as follows:
Total Full Houses = (13 * 12) * (C(4,3) * C(4,2))
Total Full Houses = (13 * 12) * (4 * 6)
Total Full Houses = 13 * 12 * 4 * 6 = 3744
Since there are C(52,5) possible 5-card hands that can be dealt from a 52-card deck, the probability of getting a full house is:
P(Full House) = Total Full Houses / Total 5-Card Hands
P(Full House) = 3744 / C(52,5)
P(Full House) = 3744 / 2,598,960
P(Full House) ≈ 0.00144, or about 0.144% chance of getting a full house when 5 cards are dealt from an ordinary deck.