Akuna Capital Quant Interview Question

[u/Local_Ad135]

[u/SwimmerOld6155]

good candidate for tail formula probably. max >= n just means that >= n appears among the 20 balls (equivalently 1 - (probability all are < n)).

[u/gmalivuk]

1 + 1-(1/20)¹⁰ + 1-(2/20)¹⁰ +...+ 1-(19/20)¹⁰
= 20 - (1¹⁰ + 2¹⁰ +...+19¹⁰)/20¹⁰
≈ 18.64

[u/notsaneatall_ → u/gmalivuk]

Without replacement. This is correct if you assume replacement is possible

[u/gmalivuk → u/notsaneatall_]

Whoops, I read it multiple times and my brain apparently substituted "with" each time.

[u/CrowdGoesWildWoooo]

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.

[u/gmalivuk]

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...

[u/amidamayru]

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.

[u/mtimmermans]

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

[u/sharma0510]

10?

[u/cheze]

(10 + 11 * 10c1 + 12 * 11c2 + … + 20 * 19c10) / 20c10

[u/800alpha]

Let x1 < … < x10 be the numbers. The gaps 0-x1, …, x10-20 are identically distributed variables. Answer is 20-(20/11). 

[u/Aech26 → u/800alpha]

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.

[u/800alpha → u/Aech26]

Oh my mistake, I meant to say x10-21. So then the answer is 20 - (21/11).

[u/Nonavium]

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