lunarpower wrote:i think this is one of our questions.

in any case, this is certainly not an abnormally long problem.

if you solve this problem by direct computation, then it can take a long time. however, there's a shortcut.

the key is to notice that "at least one card" is a COMPLEX EVENT, but the OPPOSITE event ("NO matching cards") is a SIMPLE event. therefore, if we can calculate the probability of the simple event, we can just subtract from 1 to find the probability of the complex event.

the probability of the OPPOSITE event:
draw the first card --> 12/12 (any first card will do)
draw the second card --> 10/11 (any card except for the one matching the first card drawn)
draw the third card --> 8/10 (any card except the two that match the ones already drawn)
draw the fourth card --> 6/9 (any card except the three that match the ones already drawn)

therefore, the probability of this opposite event is (12/12)(10/11)(8/10)(6/9), which reduces to (1/11)(8/1)(2/3), or 16/33.

therefore, the probability of the ORIGINAL event is 1 - 16/33, or 17/33.

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there are plenty of official problems that are longer than this one. for instance, see this problem:

https://www.manhattangmat.com/forums/gprep-2-t1737.html