You will *never* be asked this kind of 'poker question' on the GMAT, though they are very popular in undergraduate combinatorics classes. For one thing, these questions are generally quite a bit too complicated for the GMAT, and for another, they rely too heavily on knowledge that not all test takers will share (the composition of a deck of cards). I'll give a solution, but you really don't need to worry about it for the test. If you aren't at all familiar with cards, that's even more reason to skip this question.
In total, there are 52C5 ways to choose a set of 5 cards out of 52. That's the denominator of our probability (where 52C5 = 52*51*50*49*48/5! ).
In the numerator, there are 13 different ways to choose the set of cards which will be our "4 of a kind" (they can be all of the aces, or all of the kings, etc). Once we choose our 4 of a kind, there are 48 cards left from which we can choose our last card. So there are 13*48 hands you can make which contain "4 of a kind". That's the numerator of our probability.
So the answer is 13*48/52C5, which, if you simplify, turns out to be roughly 1/4000.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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