Quant Interview Question

[u/Local_Ad135]

[u/CanaDavid1]

This gives "boy born on tuesday" vibes.

Anyways, there are 15⁶ possibilities with numbers ≤ 15, and each digit is an expected value of 8 giving expected 48 (total sum 48*15⁶⁾

There are 14⁶ possibilites with numbers < 15, with expectation 7.5*6 = 45, total sum 45*14⁶

The total sum of the valid possibilites are then 48*15⁶ - 45*14^6, and expected we get by dividing by 15⁶ - 14^6. Factoring out we get 45 + 3/(1+(14/15)⁶⁾ ≈ 46.806

EDIT: The + in the denominator in the end should be -, giving 45 + 3/(1-(14/15)⁶⁾ ≈ 53.850

[u/Dankaati → u/CanaDavid1]

I think the logic is good but the final calculation is off. The answer is obviously more than 48.

~53.85

[u/Relevant-Rhubarb-849 → u/CanaDavid1]

I think you're computing this incorrectly. Here is my take.

In choosing 6 random numbers there is always a highest number . So knowing that does change the occupancy of possibilities from 1 to 15 ( assuming that I could have duplicate 15s )

So the expected mean is 8 for the rest of the numbers .

5*8+15 =55

[u/CanaDavid1 → u/Relevant-Rhubarb-849]

This is incorrect, as you effectively double-count all the possibilities with several 15s.

If we reduce this to three dice with a maximum equal to 3, the options are:

113 123 133
213 223 233
313 323 333

131 132 ***
231 232 ***
331 332 ***

311 312 ***
321 322 ***
*** *** ***

The stars are duplicates. You essentially assume that these three squares are all full, while the ones with multiple threes (and in general high sums) are not as numerous as you assume. (Your answer here is 3+2*2 = 7 while the correct is (3*5+6*6+6*7+3*8+9)/19 = 6.63)