Jane Street interview questions

[Andy Nguyen]

Pre-Interview
(Ten minutes)

1) Mental Math: One million minus one hundred eleven.
2) Mental Math: Fifty-four percent of one hundred ten.
3) Game: With one die, suppose in a round, you earn the amount of dollars equal to the value of the upwards face of the die. (eg. you earn $6 if you roll a six.) Now also suppose after your first roll, you are given the opportunity to cancel your first and roll again, taking that value as the final value. What should your strategy be?
4) What's the closest integer to the square root of 1420.
5) You and a roommate are hosting a party. You invite 10 other pairs of roommates. During the party you poll everyone at the party (excluding yourself) and ask how many hands each person shook. Two conditions:
a) Each person did not shake his roommate's hand.
b) Each person shook a different number of hands.
Question: How many hands did you roommate shake?
6) a) You roll a die, and are given an amount in dollar equal to the number on the die. What would you pay to play this game if you played it a lot of times?
b) now say that when you roll the die, you're allowed to either take the money that you'd get with the roll, or roll a second time; if you roll a second time, you're obligated to take the number of dollars that you get with the second roll. Now what is the worth of the game?
c) Same thing as above, except you have an option to play the game a third time.

Interview
(Thirty minutes)

1) Suppose you are given the opportunity to bid for a treasure chest, which you know with 100% confidence to be priced anywhere between $0-$1000. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest. Now, also suppose you have a friend who is willing to buy the treasure chest from you for one and a half times the price of the treasure chest (should you obtain the chest). What should your bid be?

2) In Baseball, the batting average is the number of hits over the number of at bats. Player A has a greater batting average than Player B in the first half of the season and the second half of the season. Is it possible that Player B would have a higher batting average for the entire season?

3) How much calories does a Big Mac have? Would you bet $1 on it? How about $10?

4) How many tons does the ocean weigh?

5) How much would you be willing to bet on it being within 25% of that at even odds?

6) A company has a value V which is uniformly distributed between 0 and 1. you are planning to place a bid B for the company. If B is smaller than V, then your bid loses and you get nothing; if B is larger than V, you get to purchase the company at price B, and the company will end up being worth 1.5 * V. What price B should you bid to maximize your profit?

7) On a sheet of paper, you have 100 statements written down. the first says, "at most 0 of these 100 statements are true." the second says, "at most 1 of these 100 statements are true." ... the nth says, "at most (n-1) of these 100 statements are true. ... the 100th says, "at most 99 of these statements are true." how many of the statements are true?

8) You have two decks of cards: one has 13 reds and 13 blacks, and the other has 26 reds and 26 blacks. We play a game in which you select one of the two decks, and pick two cards from it; you win the game if you select two black cards. Which deck should you select to maximize your chances of winning? Try to do this problem in your head, without writing any calculations down.

9) You have a deck of 52 cards, and you keep taking pairs of cards out of the deck. if a pair of cards are both red, then you win that pair; if a pair of cards are both black, then I win that pair; if a pair of cards has one red and one black, then it's discarded. If, after going through the whole deck, you have more pairs than I do, then you win $1, and if I have more pairs than you do, I win $1. What is the value of this game in the long run?

 

[Prasad]

For the mental math questions, are we expected to just rattle off the answer, or talk out loud through the steps?

For example,
-> Whats 54% of 110?
->(wait couple seconds to calculate) 59.4

or

-> Whats 54% of 110?
-> Well that's 54 * 11 /10 = 594/10 = 59.4

For the square root of 1420, is there an efficient way of doing this? We know that 35^2 is 1225, and 40^2 is 1600, but what about the numbers in between?

So, what's 38^2? It'll take me 10 seconds to answer that.

 

[Andy Nguyen]

I don't think the purpose for these calculations is to get the answer but see how you would skip the brute force route and come up with some other method.

For example, (38=2*19) so (38^2=4*19^2)
Anyone who is good with number can do mental calculation for things like square of numbers up to a certain x
So, automatically, you will know that (19^2=361) which leads to the result (38^2=1444)

This can be done in few seconds (less than 5) compared to 10 seconds it took you.

Now, (\sqrt{1420}=\sqrt{355*4}=2\sqrt{355})
As in the example above, we know 19^2=361 so (\sqrt{355}) is very close to 19. I would then guesstimate my answer at ~37.9 give or take.

Now you see the relationship between these two questions.

 

[silkentouch]

Andy,
There is a much more intuitive way of doing this problem:
As Prasad pointed out, 35^2 = 1225 and 40^2 = 1600. 1420 is almost exactly between these two. We know that distance between two consecutive squares increases as numbers grow. Therefore, 1420 is closer to square of 38 instead of square of 37.

---------- Post added at 09:38 PM ---------- Previous post was at 09:28 PM ----------

And just to confirm that answer mentally before speaking out aloud in the interview:
39^2 would be 79 less than 40^2 (lets say roughly 80 less). 38^2 would be say another 80 less. So 38^2 would be around 1440, which confirms our answer.

 

[Prasad]

I don't think the purpose for these calculations is to get the answer but see how you would skip the brute force route and come up with some other method.

For example, (38=2*19) so (38^2=4*19^2)
Anyone who is good with number can do mental calculation for things like square of numbers up to a certain x
So, automatically, you will know that (19^2=361) which leads to the result (38^2=1444)

This can be done in few seconds (less than 5) compared to 10 seconds it took you.

Now, (\sqrt{1420}=\sqrt{355*4}=2\sqrt{355})
As in the example above, we know 19^2=361 so (\sqrt{355}) is very close to 19. I would then guesstimate my answer at ~37.9 give or take.

Now you see the relationship between these two questions.

Thanks for your response! When I did it, I pretty much just figured out it was between 35 and 40, and then started squaring them one by one.

What about my other question? Would you outline the method out loud as you were doing it, or just be silent for 5 seconds until you've figured out the answer?

 

[Andy Nguyen]

Of course they want you to explain as you think. Even on phone interview, they would ask you to think out loud.

 

[Prasad]

Andy,
There is a much more intuitive way of doing this problem:
As Prasad pointed out, 35^2 = 1225 and 40^2 = 1600. 1420 is almost exactly between these two. We know that distance between two consecutive squares increases as numbers grow. Therefore, 1420 is closer to square of 38 instead of square of 37.

---------- Post added at 09:38 PM ---------- Previous post was at 09:28 PM ----------

And just to confirm that answer mentally before speaking out aloud in the interview:
39^2 would be 79 less than 40^2 (lets say roughly 80 less). 38^2 would be say another 80 less. So 38^2 would be around 1440, which confirms our answer.

Ah, thank you! I like your approach, and it'll reduce the number of numbers you'll have to square.

38^2 = 40*36 + 4 = 1440+4 = 1444
37^2 = 40*34 + 9 = 1360+9 = 1369
36^2 = (35 + 1) ^2 = 1225 + 70 + 1 = 1296

I can get those relatively quickly using the above method, the problem was that there were too many numbers to check. You're right though, since it's halfway, I only need to start at 38.

Also, would you say outloud those intermediate steps, or just do it?

 

[silkentouch]

Depends on the overall feel of the interview. I generally outline the steps, but sometimes, if I can get an answer really really quickly...I don't outline the steps. I have had an interviewer stop mid-way and ask me, how did I do it so quickly.

Part of the magic is not telling up front how the trick is done.

 

[Prasad]

Of course they want you to explain as you think. Even on phone interview, they would ask you to think out loud.

Haha, thank you! It may be totally obvious to you, but it wasn't to me, which is why I asked. Sorry for wasting your time.

 

[silkentouch]

Andy, did you get the first question right?

Interview
(Thirty minutes)

1) Suppose you are given the opportunity to bid for a treasure chest, which you know with 100% confidence to be priced anywhere between $0-$1000. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest. Now, also suppose you have a friend who is willing to buy the treasure chest from you for one and a half times the price of the treasure chest (should you obtain the chest). What should your bid be?

How can I place a bid if I don't understand the distribution of the values? e.g. If I bid say 600, I make money if the chest is valued over 400 and less than 600, and lose money if the chest if valued under 400.
But as per your question, I do not know what is the probability that the chest is valued under 400.

 

[Ryan]

I got the same one. The correct answer is 0 which is break even. Any bid above 0 will be a lost in profit.

 

[silkentouch]

Ryan,
That is true if the distribution is uniform.
For non-uniform distribution, a positive bid may make sense.
e.g. assume the distribution from [0,1000] to be a uniform 'near exhaustive' distribution over [750,1000]. Then a bid of 1000 would be the best bid....

 

[Joe Solly Val]

This is gonna sound cheeky, but any chance you can explain your asnwer to this? I dont particually want a short-n-sweet answer, but more of 1 that i can then expand and explore myself

Joe

 

[silkentouch]

Hi Joe, Here is a dirty answer:
For a uniform distribution, say you bid 300. If the cost is over 300, you make or lose nothing.
If the cost is less than 200 (probability 2/10), you lose an amount equally distributed over 0-300.
If the cost is between 200-300 (probability 1/10), you make an amount equally distributed between 0-150.
Do you make or lose money, on average if you bid 300? Why?

 

[NeedOPT]

The mathematical way to do this is to write down the expected profit P from a bid B, given a probability density function p(V) of the initial "true value" of the firm V

(E(P) = \int_{0}^{B} (1.5V - B)p(V) dV)

Viewing this expectation as a function of B, we can see that the maximum must be reached either with B = 0 (boundary of available bids) or, by taking the derivative, at a solution to the equation:

(Bp(B) = 2\int_{0}^{B} p(V) dV)

or, if we rewrite P(V) as the cumulative distribution function of the true value of the firm:

(Bp(B) = 2P(B))

 

[Joe Solly Val]

Thank you for ur reply.
I followed it until u formed the first (Bp(B)). I understood the method to create u expected profit, its the natural/instinctive idea. Then it is maximal at a boundary or a max-turning point. My issue is then how to differentiate the equation. Using FTC I got:

(d(E(P))/dB = 2p(B))

where (p(B)) is the probability V=B. As we are talking about a continuous probability then that is zero, thus (d(E(P))/dB = 0) for all B. And all higher deriratives are 0.

Eeek!

For uniform distribution the intuitive thing, for me, if to just calculate the integral. (Areas under two triangles). But this would become a little harder given a different distribution.



Interview Q9 = 0?

The mathematical way to do this is to write down the expected profit P from a bid B, given a probability density function p(V) of the initial "true value" of the firm V

(E(P) = \int_{0}^{B} (1.5V - B)p(V) dV)

Viewing this expectation as a function of B, we can see that the maximum must be reached either with B = 0 (boundary of available bids) or, by taking the derivative, at a solution to the equation:

(Bp(B) = 2\int_{0}^{B} p(V) dV)

or, if we rewrite P(V) as the cumulative distribution function of the true value of the firm:

(Bp(B) = 2P(B))

 

[NeedOPT]

Differentiation under the integral sign - Wikipedia, the free encyclopedia

Using their notation, x = B, t = V, f(B,V) = (1.5V-B)p(V), a(B) = 0, b(B) = B

Applying the chain rule formula + FTC (they give the answer in the above link)

(d(E(P))/dB = (1.5B-B)p(B) - \int_{0}^{B}p(V)dV)

I suggest you brush up on your calculus.

Also, I suggest that you brush up on your ideas about probability. p(B) is not the probability that the random variable is B. p(B) is the probability DENSITY. It is a finite-valued function. Its only measurable meaning is in integral form.

Thank you for ur reply.
I followed it until u formed the first (Bp(B)). I understood the method to create u expected profit, its the natural/instinctive idea. Then it is maximal at a boundary or a max-turning point. My issue is then how to differentiate the equation. Using FTC I got:

(d(E(P))/dB = 2p(B))

where (p(B)) is the probability V=B. As we are talking about a continuous probability then that is zero, thus (d(E(P))/dB = 0) for all B. And all higher deriratives are 0.

Eeek!

For uniform distribution the intuitive thing, for me, if to just calculate the integral. (Areas under two triangles). But this would become a little harder given a different distribution.



Interview Q9 = 0?

 

[Wannabe Quant]

Slight Confusion

For the treasure chest question, what is the expected value of your net profit? Is it
(
\frac{1}{1000} \int_0^b (1.5v - b) dv = - \frac{b^2}{4000}
)

or is it
(
\frac{1}{b} \int_0^b (1.5v - b) dv = - \frac{b}{4}
)


I think it's the first one but a friend of mine insists it's the second without a good explanation.

The second one looks like the conditional expectation (E[1.5v-b|b \geq v])

But surely what we want is (E[(1.5v-b)1_{[b \geq v]}]) where (1_{[b \geq v]}) is an indicator function.

I hope this all makes sense.

 

[goodstudent]

For the mental math questions, are we expected to just rattle off the answer, or talk out loud through the steps?

For example,
-> Whats 54% of 110?
->(wait couple seconds to calculate) 59.4

or

-> Whats 54% of 110?
-> Well that's 54 * 11 /10 = 594/10 = 59.4

For the square root of 1420, is there an efficient way of doing this? We know that 35^2 is 1225, and 40^2 is 1600, but what about the numbers in between?

So, what's 38^2? It'll take me 10 seconds to answer that.

For 38*38. There is a really neat trick for multiplication for any 2 two-digit numbers:
If you are given a problem of solving x^2, do 100*(x-25)+(50-x)^2. It might seem tedious but after some practice you can do it in your head easily.

So for example, 38^2 = 100(38-25)+(50-38)^2 = 1300+144 = 1444.

---

For sqrt(1420), people have given many ways, but another approach is to see it as sqrt(100*14.2) = 10*sqrt(14.2). Which is near 10*3.7 or 10*3.8-ish

---------- Post added at 12:17 PM ---------- Previous post was at 11:12 AM ----------

For the treasure chest question, what is the expected value of your net profit? Is it
(
\frac{1}{1000} \int_0^b (1.5v - b) dv = - \frac{b^2}{4000}
)

or is it
(
\frac{1}{b} \int_0^b (1.5v - b) dv = - \frac{b}{4}
)


I think it's the first one but a friend of mine insists it's the second without a good explanation.

The second one looks like the conditional expectation (E[1.5v-b|b \geq v])

But surely what we want is (E[(1.5v-b)1_{[b \geq v]}]) where (1_{[b \geq v]}) is an indicator function.

I hope this all makes sense.

The second one makes sense to me. But I am not sure.

 

[Wannabe Quant]

I don't agree.

The second one says "because you're bidding b, you now think the distribution of the real value is uniform on [0,b]" which does not make sense. This is just a conditional expectation and no where in the question does it say that your assumption changes with your bid.

The actual payout is (1.5V-B) if (B \geq V) and 0 otherwise.

This is equivalent to ((1.5V-B)\mathbb{I}_{[B \geq V]}).

Hence the first integral is the correct one.