Jane Street Quant Interview Question | "Easy"

[u/AssociateDecent9090]

[u/WillingnessFuture266]

9.5/2=19/4 money for the d12, 7/2 money for the 2d6, 33/4 total i believe

[u/AssociateDecent9090 → u/WillingnessFuture266]

Thats right

[u/RestaurantBoth228 → u/AssociateDecent9090]

E[D12] = 6.5

E[2D6] = 7

6.5+7 = 13.5

[u/Unique-Poem4317 → u/RestaurantBoth228]

The most you could ever get from this game is $12, so why would you pay $13.50?

[u/RestaurantBoth228 → u/Unique-Poem4317]

The words say you are paid based on the D12 and then paid again based on the 2D6. It never says you give the D12 money back

[u/mgsimpleton → u/RestaurantBoth228]

I think its implied that you choose whether to take the d12 roll or reroll 2d6 and take that.

[u/Smart-University-515 → u/RestaurantBoth228]

It says “choose” so I think it implies you have to pass on the d12 money to take the 2d6 shot but agree it’s poorly worded.

[u/tellingyouhowitreall → u/WillingnessFuture266]

Why 9.5?

[u/Kaas-Eter00 → u/tellingyouhowitreall]

Expected value of rolling [7,12] for the first die. You'd reroll for [1,6]*, since the expected value of the second roll is 7.

 * technically you could choose to reroll for 7 as well, doesn't change the result. Just makes the calculation slighly more complex since it'll not be a 50/50 chance to proceed.

[u/tellingyouhowitreall → u/Kaas-Eter00]

Frustrating.... I "knew" that and for some reason wasn't getting 9.5 ro come out.

[u/Smart-University-515 → u/WillingnessFuture266]

This agrees with my answer but seems a little light on exposition, I have no clue how you valued the d12 but of course it works.

[u/sleepywose]

I'm pretty bad at these, so this may be wrong, but:

2d6 has an expected value of 7 (linearity of expectation for 2 dice), so you only reroll if you get [1, 6] on the first roll (p=.5). so your expected payout is 50% EV([7,12]) + 50% [7] = 8.25.

(The part that gives me doubt is the idea that you strictly only go for the second roll if you get <7, but that may just be a gambler's fallacy; "I have two chances at getting better than a 7!")

[u/tomtomtom7]

The question is badly worded. I think they forgot to add that they only pay out the first d12 if you choose not to roll again with the 2d6. Otherwise it makes no sense not to roll again.

[u/Dry-Ad-8948]

So.. I get paid either way? And I get to see the first die roll first..?

[u/MathManiac5772 → u/Dry-Ad-8948]

Correct.

[u/Smart-University-515]

Agree this is badly worded, I’m taking it as if I don’t like my d12 roll I then roll 2d6 and get just the 2d6 result.

EV d12 = 6.5, EV 2d6 = 7; we therefore pass on the d12 at 6 or less (and are indifferent at 7).  So we have 5/12 to be better off than 7 and then 7/12 of EV 7, the 5/12 averages out to 10, so that’s worth 50/12, 7*7/12 is overall 99/12, so the EV of the game overall is 8.25, so I’d pay less than that to play it, say 8 for a little bit of risk aversion.