Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33

Whether you pick the four cards simultaneously, or pick them one at a time (without replacement) doesn't actually matter if you are finding a probability; the two situations are mathematically identical. You can see this intuitively by thinking of taking hold of four cards in the deck first. If you take them all out at the same time, or if there is a nanosecond between your removing each, why would the probability that you get a pair be affected? It won't be, so the 'ambiguity' you suggest is present in such questions is no ambiguity at all.

You can see that either perspective will give you the same answer, though it's easier to illustrate with a simpler example. Say you have 3 red marbles and 4 blue marbles in a bag, and you pick two (either simultaneously, or without replacement - it's the same thing), and you want to find the probability of picking two red marbles. If we look at the problem as though we are picking marbles one at a time, we have 3*2 ways of picking two reds, and 7*6 ways of picking two marbles, so the probability would be 3*2/7*6 = 1/7. If we look at the problem as though we're picking two marbles simultaneously, we have 3C2 ways of picking two red marbles and 7C2 ways of picking two marbles, so the probability would be 3C2/7C2 = 1/7. So when you stick your two hands in the bag and grab two marbles, it doesn't matter if you lift your two hands out at the same time, or take them out one at a time; the probability is the same.

Note though that you need to be consistent in the calculation - if you assume order matters when you calculate the numerator, you must also assume order matters when you calculate the denominator.

Let's calculate the opposite probability ans subtract this value from 1.

Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \(\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\).

\(C^4_6\) - # of ways to choose 4 different cards out of 6 different values;
\(2^4\) - as each of 4 cards chosen can be of 2 different suits;
\(C^4_{12}\) - total # of ways to choose 4 cards out of 12.

So \(P=1-\frac{16}{33}=\frac{17}{33}\).

Or another way:

We can choose any card for the first one - \(\frac{12}{12}\);
Next card can be any card but 1 of the value we'v already chosen - \(\frac{10}{11}\) (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left);
Next card can be any card but 2 of the values we'v already chosen - \(\frac{8}{10}\) (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left);
Last card can be any card but 3 of the value we'v already chosen - \(\frac{6}{9}\);

\(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\).

So \(P=1-\frac{16}{33}=\frac{17}{33}\) - the same answer as above.

Answer: C.

Hope it helps.