What is the probability of cutting two cards of the same rank from a well-shuffled deck?
Simple answer: You can cut a card from a deck with probability 1, and then cut another card of the same rank, with probability 1/51. There are three such cards, for a final probability of \(1\cdot3\cdot1/51 = 3/51\).
Less simple answer: The sample space is all pairs of cards; the sample-space size is \(C(52, 2) = 52\cdot51\). For a given pair \((c_1, c_2)\), there are 3 cards \(c_2\) that match \(c_1\)’s rank, and there are 52 such cards \(c_1\), for a total of \(52\cdot3\) satisfactory pairs. The probability of an acceptable outcome is \[(52\cdot3)/C(52, 2) = (52\cdot3)/52\cdot51 = 3/51\] Even less simple answer: the previous question made an unwarranted assumption: \((c_1, c_2)\) is a pair different from \((c_2, c_1)\), that is, the pairs are ordered. The problem doesn’t suggest this, and implies otherwise by ignoring card rank, which would make \((c_1, c_2)\) and \((c_2, c_1)\) the same pair. The sample-space size \(C(52, 2)\) is twice as big as it should be. Similarly, the number of satisfactory pairs is twice what it should be.? The adjusted counts give the ratio \(((52\cdot3)/2)/((52\cdot51)/2) = 3/51\). As a wise man once wrote, “what I tell you three times is true.”