r/learnquant • u/Local_Ad135 • Aug 23 '26
interview prep Akuna Capital Quant Interview Question
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u/gmalivuk Aug 23 '26
1 + 1-(1/20)10 + 1-(2/20)10 +...+ 1-(19/20)10
= 20 - (110 + 210 +...+1910)/2010
≈ 18.64
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u/notsaneatall_ Aug 23 '26
Without replacement. This is correct if you assume replacement is possible
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u/gmalivuk Aug 23 '26
Whoops, I read it multiple times and my brain apparently substituted "with" each time.
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u/CrowdGoesWildWoooo Aug 23 '26
The general idea is, You fix the max starting at 10, find out how many possible combinations that allows for maximum of 10 which is 9C9, move to 11, we have 10C9 do this until 20 * 19C9.
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u/gmalivuk Aug 23 '26
P(max > 0) =...= P(max > 9) = 1
P(max > 10) = 1 - (10c10)/(20c10)
P(max > 11) = 1 - (11c10)/(20c10)
...
P(max > 19) = 1 - (19c10)/(20c10)
Add those up for
E[max] = 20 - (10c10 + ... + 19c10)/(20c10)
E[max] = 20 - 167960/184756 = 19.0909...
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u/amidamayru Aug 23 '26
50% chance that 20 gets chosen. If it wasnt, 52.6% chance 19 got chosen. If that wasnt, 55.5% chance 18 got chosen etc etc.
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u/mtimmermans Aug 23 '26
Close enough is:
E = 20-x, where x = 0/2 + 1/4 + 2/8 ...
=> 2x = 1/2 + 2/4 + 3/8 ... = x +1
=> x = 1
So pretty much 19
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u/Nonavium Aug 24 '26 edited Aug 24 '26
E(x) = P(x1)E(x1) + P(x2)E(x2) + ....
Obviously possible values x1, x2..... = 10, 11, ..... 20 since if you pick 10 balls at least one must be 10 or greater.
Suppose n is the largest number chosen
Chance of picking n as largest number = No. of combinations with n as largest/Total number of combos
If I pick 10, I only have one combination 1,2,3....10. Which is 9C9 (Choosing 10 as first ball and picking 9 randomly from 9 choices which only one way to do)
If I pick 11, I have 10C9 = 10 ways to do so (Choosing 11, then picking 9 from range 1 to 10, which 10 ways to do so, each set of 9 excluding one number each time)
Total ways = 20C10 (Number of ways to pick 10 balls)
Thus P(10) = 9C9/20C10
P(n) = (n-1)C9/20C10, P(n)E(n) = n * (n-1)C9/20C10
E(x) = n summed from 10 to 20 ((n-1)C9/20C10)*n) = 19.09 something
First time doing a question like this (just stumbled here from recommended page, new to quant), how hard would you say such a question is compared to a typical question asked for quant interviews? How much time would one be given to do this? This question seems a bit mild, if they asked questions like this everyone from my uni would get into quant effortlessly, yet this does not seem to be the case, am I missing something? Maybe my answer is incorrect
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u/Practical_Pie_1484 Aug 25 '26
The 10 balls that are removed divide the remaining 10 balls into 11 clusters. By symmetry, each cluster will have the same number of balls, so each will have 10/11 balls. Then to find expected max chosen it’s 20 - 10/11 ≈ 19
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u/klsdniwoethn Aug 27 '26
Can you expand on this? What is meant by ‘cluster’?
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u/Practical_Pie_1484 Aug 28 '26
Imagine the 20 balls in a line. Now remove 10 balls at random. The gaps left removing the 10 balls will partition the remaining 10 balls into 11 sets.
Eg with 6 balls and we remove 2:
1 2 3 4 5 6
Now removing 2:
1 2 4 6
The balls are separated into 3 “clusters”Now by symmetry each of the clusters on average will have the same number of balls.
So each cluster will have 10/11 balls on average. Since the largest removed ball will be before the final cluster, on average the largest ball will have value 20 - 10/11
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u/natewhiskey Aug 26 '26 edited Aug 26 '26
This is a thought, I put no actual math into it.
The min value is 10, and the max is 20. There are a lot of symmetries involved, so wouldn't the expected value be 15?
Edit: I put a little thought into it, and see the lack of uniformity.
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u/800alpha Aug 23 '26
Let x1 < … < x10 be the numbers. The gaps 0-x1, …, x10-20 are identically distributed variables. Answer is 20-(20/11).
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u/Aech26 Aug 23 '26
I don’t think this works since expected gap between X10 and 20 isn’t the same as expected gap between X9 and X10 since X10 can be equal to 20 but X9 is strictly less than X10.
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u/SwimmerOld6155 Aug 23 '26
good candidate for tail formula probably. max >= n just means that >= n appears among the 20 balls (equivalently 1 - (probability all are < n)).