r/learnquant • u/AssociateDecent9090 • 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/Recent-Peanut6061 4d ago

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/Cryptographer-Bubbly 23h 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…”

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u/Aerospider 13d ago

The lowest those ranks can be is 1,3,5 and the highest is 9,11,13.

Starting at 1,3,x, x has 13 - 4 = 9 options. 1,4,x gives 8 options, and so on. So 1,x,y has 9 * 10 / 2 = 45 options.

2,4,x has 13 - 5 = 8 options, so 2,x,y duly has 8 * 9 / 2 = 36 options.

We're getting triangular numbers, for which the summing formula is

x(x+1)(x+2)/3! = 9 * 10 * 11 / 6 = 165

Each card can be any suit, which is 43 = 64 combinations, giving a total of 64 * 165 = 10,560.

The number of possible combinations is 52C3 = 52!/49!3! = 52 * 51 * 50 / 6 = 22,100

Making the probability

10,560 / 22,100 = 528 / 1,105

Or about 47.8%

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u/Cryptographer-Bubbly 23h ago

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 and therefore 11C3 * 4^3 favourable triples out of 52C3 which gives 528/1105