The classical Ross book has the following problem (25)
Two cards are randomly selected from a deck of 52 playing cards. (a) What is the probability they constitute a pair (that is, that they are of the same denomination)? (b) What is the conditional probability they constitute a pair given that they are of different suits?
The answer for (a) is obviously 3/51. Why do you divide this probability by 39/51 when nominal match is queried under "suits differ" (b) condition? I guess that this is a 4x13 deck and we clearly draw without replacement. That is, if you drawn an ace of spades, there is no chance that you draw another ace of spades. That is why we have 3 in the nominator of (a) -- there are 3 other aces under other suits. If we limit ourselves to the same suit, we have to put 0 there, since we excluding all the possibilities to get another ace. What the hell is happening?
When you do $3/51$ for a, you have three successes out of $51$ possibilities, the remaining cards. For b, you have the same three successes, but only $39$ possibilities, the cards of other suits.