### Instruction: 3 cards are chosen at random from a standard 52-card deck.  What is the probability that they form a pair?  (A 3-card hand is a 'pair' if two of the cards match in rank but the third card is different.  For example, 668 is a pair, but 999 is not.).### Response: To solve this problem, I need to count how many 3-card hands are pairs, and how many are possible in total.
Then I can divide the former by the latter to get the probability.
First, I'll count the total number of 3-card hands.
This is just a combination of 52 cards taken 3 at a time, which is denoted by 52 choose 3.
Using the formula for combinations, I get
52 choose 3 = 52! / (3! 49!) = (52 x 51 x 50) / (3 x 2 x 1) = 22,100
So there are 22,100 possible 3-card hands.
Next, I'll count how many of these hands are pairs.
To form a pair, I need to choose a rank (such as 6 or king) for the two matching cards, then choose two suits for those cards, and then choose a different rank and suit for the third card.
There are 13 ranks to choose from, so I have 13 ways to pick the rank of the pair.
Once I have the rank, I need to pick two suits out of four for the pair.
This is a combination of 4 choose 2, which is
4 choose 2 = 4! / (2! 2!) = (4 x 3) / (2 x 1) = 6
So there are 6 ways to pick the suits of the pair.
Now I have the pair, and I need to pick a different rank and suit for the third card.
There are 12 ranks left to choose from, since I can't use the same rank as the pair.
And there are 4 suits to choose from for each rank.
So I have 12 x 4 = 48 ways to pick the third card.
Putting it all together, I have 13 x 6 x 48 = 3,744 ways to form a pair.
Therefore, the probability of getting a pair is