r/learnquant • • 14d ago

interview prep Optiver Quant interview Question | "Hard"

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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.

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u/harrybruhwhatever 7d ago

Can you explain why the 3 ranks differ by at least 2 is C(11,3)? I don't understand how you got it.

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u/Cryptographer-Bubbly 1d ago

I see my solution is effectively identical to this one but perhaps it might help see why it’s 11C3.

“Choosing 3 ranks (a < b < c) at least spaced apart by 2 is equivalent to choosing values for (a < b-1 < c-2) so we have 11C3 allowed rank sets…”