On the first day of Christmas, Calvin gave to me

There are 3 "three's" left, and 51 cards left. Therefore the probability is $\boxed{\frac{3}{51}}$ .

In the deck of 52 cards, there are four 3s.

Since one 3 has been taken without replacement, three 3s are left in a deck of 51 cards.

Hence the probability $\frac{3}{51}$ .

Total cards 52. Already one card draw balance cards 51.Next get a pair of three cards so the probability is 3/51

A standart 52 cards deck has 4 card of each rank (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, King, Queen and Jack), one per suit (club, diamond, heart and spade). You draw two cards, one of them being a 3. Since there are 52 cards in a deck, after you draw one there will remain 51 cards, and between those there are still 3 cards rank 3. Thus, the probability of drawing another 3 is of $\frac{3}{51}$ .

It is given that out of the 2 cards drawn, one is a 3, so we have to find the probability of getting another 3 from the remaining 51 cards. Since, there are four 3's in a deck and one is already taken for the pair, there are three 3's remaining in the deck.

So, probability of getting a pair $= \boxed{\frac{3}{51}}$

As you get that card the total number of cards deduct and its probability would decrease