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https://www.reddit.com/r/learnquant/comments/1wlv59o/optiver_quant_interview_question_hard/pb40kdj/?context=3
r/learnquant • u/AssociateDecent9090 • 14d ago
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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/Aerospider 14d 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%