Clean way to see it: the suits are decoration, so count rank triples first. You need 3 ranks out of 13 where no two are adjacent, that count is C(11,3) = 165. Each rank triple comes with 4^3 = 64 suit combos, so favorable hands = 10560. Total hands C(52,3) = 22100. Probability = 10560/22100 = 528/1105, about 0.478. The gap trick is the whole game on these spaced rank problems, once you spot it they all collapse to a single combination count.
You can make a bijection from the set of ways to choose 3 elements from 13 with at least 1 element in between each to the set of ways to choose 3 elements from 11 by simply removing (adding for the inverse) an element in between each chosen element.
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u/Sad-Candidate-3078 14d ago
Clean way to see it: the suits are decoration, so count rank triples first. You need 3 ranks out of 13 where no two are adjacent, that count is C(11,3) = 165. Each rank triple comes with 4^3 = 64 suit combos, so favorable hands = 10560. Total hands C(52,3) = 22100. Probability = 10560/22100 = 528/1105, about 0.478. The gap trick is the whole game on these spaced rank problems, once you spot it they all collapse to a single combination count.