Answer: 0.04754
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Step-by-step explanation:
A hand of two pairs is when we have 2 cards of the same value, and another 2 of the same value, then the fifth card is something else entirely.
An example hand is:
- king of hearts, king of clubs ...... 1st pair
- queen of spades, queen of clubs ..... 2nd pair
- 7 of diamonds
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We use the nCr formula since order doesn't matter.
C(n,r) = (n!)/(r!*(n-r)!)
For any suit, there are n = 13 cards. We make r = 2 selections for each of the two pair.
C(n,r) = (n!)/(r!(n-r)!)
C(13,2) = (13!)/(2!*(13-2)!)
C(13,2) = (13!)/(2!*11!)
C(13,2) = (13*12*11!)/(2!*11!)
C(13,2) = (13*12)/(2!)
C(13,2) = (13*12)/(2*1)
C(13,2) = (156)/(2)
C(13,2) = 78
There are 78 ways to select which two ranks will be chosen to represent each of the two pair.
Let's say one of the ranks of the two pair is a king. There are n = 4 kings and we select r = 2 of them. That gives C(4,2) = 6 ways to select those two kings. We'll also have 6 ways to select the other two cards in the other pair.
Therefore, we have 78*6*6 = 2808 ways to pick the first four cards that consist of the two pair (eg: 2 kings and 2 queens).
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So far we've considered just the two pairs. We have one card left to select. There are 13-2 = 11 cards of any particular suit that weren't part of the two pair. There are 4 suits to pick from, which means we have 11*4 = 44 cards to pick from for that fifth and final slot.
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Summary so far:
- 2808 ways to pick the two pair (eg: 2 kings and 2 queens)
- 44 ways to pick the fifth card that differs from the first four cards (anything other than what was chosen for the two pair).
That gives 2808*44 = 123,552 different two pair hands possible.
This is out of C(52,5) = 2,598,960 hands
Divide those two values to get the final answer
(123,552)/(2,598,960) = 0.04753901560624
This then rounds to 0.04754 as the final answer.