7.7 Jane Street Phone Interview

[postelrich]

I have a phone interview with Jane Street in a few days for a trader assistant position. Not sure how much it would differ for a more quantatitve role, as the questions posted here, http://www.quantnet.com/forum/threads/jane-street-interview-questions.3039/, are mainly riddles. If someone has something to add on top of those, that would be great since the last other thread is a little old.

 

[SYau]

Good luck Rich!

 

[goldski]

I have a phone interview with Jane Street in a few days for a trader assistant position. Not sure how much it would differ for a more quantatitve role, as the questions posted here, http://www.quantnet.com/forum/threads/jane-street-interview-questions.3039/, are mainly riddles. If someone has something to add on top of those, that would be great since the last other thread is a little old.

Jane street asks a lot of brainteasers. Tons of it. So go through one of the interview books and learn all the brainteasers. They ask you to do mental math too. It's a really great place.

 

[Paul Mithouard]

I have a phone interview with Jane Street in a few days for a trader assistant position. Not sure how much it would differ for a more quantatitve role, as the questions posted here, http://www.quantnet.com/forum/threads/jane-street-interview-questions.3039/, are mainly riddles. If someone has something to add on top of those, that would be great since the last other thread is a little old.

Without being indiscrete, could you pleaase share your profile with us ?
Thanks a lot in advance Postelrich
Paul

 

[postelrich]

MA Econ with no solid finance experience yet.

discrete enough!?

 

[postelrich]

Got asked two math questions, can't remember the first one for the life of me but I know I answered it right. The second one was a probability ques. If you flip four quarters and get $1 for each head, what is the expected value? I said $2 because the flips are independent, easy enough. Then he asked, now assume after you flip the four quarters you can flip one of them again(if it was tails), now what is the expected value? I said, $2.50 figuring in another 50cents for the extra flip. He says no, gives the hint that would you flip again in every outcome...oooh I got his hint but got flustered and couldn't figure it out. Looking back in retrospect now, its quite easy though I'm not sure it's right because he said it would have to be less than $2.50. Now, you take the four states, 4H, 3H, 2H, 1H. 4H gives you $4 x 1/4 . 3H gives you $3 x 1/4 + 50cents for the extra flip. And so on till you get 11.50/4 or $2.75

Any thoughts on what I did wrong here?

 

[Tom Maloney]

It's kind of like flipping 5 coins, except you can't get 5 heads. If you could, the expected value would equal $2.50.

For instance if the outcome is HHTH, you can clip again to either get HHTHH or HHTHT, with respective payoffs $4 or $3, with respective probabilities both (1/2)^5. So you can count the number of ways to get each outcome. I get that you can win $4 six ways (I think you have to assign the event HHHHHH to give $4), $3 5 choose 2 ways, $2 5 choose 3 ways, and $1 5 choose 4 ways. So E = $ (4*6 + 10*3 + 10*2 + 5*1)/2^5. It's late, so I welcome any corrections.

 

[AI_DD]

It's like flipping 5 coins, but different. The expected value of flipping 5 coins is 2.5. This take into the situations that HHHHH and HHHHT happen, which actually can't happen when u flip 4 coin and pick a tail to flip it again. Besides the above, it is the same with flipping 5 coins. So the expected value you want is 2.5 minus the expected value of the scenario of HHHHH and HHHHT, that is 2.5-4*1/(2^5)-5*1/(2^5)=2.5-9*1/(2^5)=2.21875

 

[ashwani]

As I see, there are four coins, and the expected payoff in first round is $2 as all said. You get to flip a second time only one coin only if all four didn't show up H in the first. And the probability of getting a chance for a second flip is 1-(1/2)^4=15/16 i.e. a little less than 1.
Irrespective of which coin you chose for second time, the payoff would be (15/16) * 0.5 making a total of

2 + [ (15/16)*0.5] = $2.46874

 

[Tom Maloney]

As I see, there are four coins, and the expected payoff in first round is $2 as all said. You get to flip a second time only one coin only if all four didn't show up H in the first. And the probability of getting a chance for a second flip is 1-(1/2)^4=15/16 i.e. a little less than 1.
Irrespective of which coin you chose for second time, the payoff would be (15/16) * 0.5 making a total of

2 + [ (15/16)*0.5] = $2.46874

I like it. Note this result agrees with my answer above.

 

[postelrich]

Hmmm both solutions sounds good but since its order doesn't matter aren't you over counting.

It's like flipping 5 coins, but different. The expected value of flipping 5 coins is 2.5. This take into the situations that HHHHH and HHHHT happen, which actually can't happen when u flip 4 coin and pick a tail to flip it again. Besides the above, it is the same with flipping 5 coins. So the expected value you want is 2.5 minus the expected value of the scenario of HHHHH and HHHHT, that is 2.5-4*1/(2^5)-5*1/(2^5)=2.5-9*1/(2^5)=2.21875

So for this don't you have to also subtract THHHH, HTHHH, HHTHH, HHHTH as well?

As I see, there are four coins, and the expected payoff in first round is $2 as all said. You get to flip a second time only one coin only if all four didn't show up H in the first. And the probability of getting a chance for a second flip is 1-(1/2)^4=15/16 i.e. a little less than 1.
Irrespective of which coin you chose for second time, the payoff would be (15/16) * 0.5 making a total of
2 + [ (15/16)*0.5] = $2.46874

This makes pretty good sense and I'm inclined to trust Tom's approval since he just took the probability refresher.

 

[Tom Maloney]

The order matters in terms of calculating the probability. So for instance, let's say the problem was this: flip two fair coins, what's the probability that you get 1 head? Well you can get HT (first coin is head, second is tail) or TH. There are four equally probable events TT, TH, HT, HH. So the probability is 2/4 = 1/2. I'd check out some of the links listed at http://www.google.com/search?q=probability+order+doesn't+matter.

 

[postelrich]

Ok thanks! My brain is still hardwired for linear algebra.

 

[AI_DD]

Hmmm both solutions sounds good but since its order doesn't matter aren't you over counting.

So for this don't you have to also subtract THHHH, HTHHH, HHTHH, HHHTH as well?

No, you don't. THHHH is like you have 1 tail and 3 heads in the first round, then you pick the tail and reflip it and get a head. HTHHH, HHTHH, HHHTH are similar.

 

[postelrich]

I wish we had someone to give a definitive answer, I couldn't find anything searching online. Both solutions sound like logical answers but they differ.

 

[Tom Maloney]

I wish we had someone to give a definitive answer, I couldn't find anything searching online. Both solutions sound like logical answers but they differ.

I disagree with AI_DD's answer: it doesn't make sense to subtract $5*1/(2^5) because you can't win $5.

 

[AI_DD]

I disagree with AI_DD's answer: it doesn't make sense to subtract $5*1/(2^5) because you can't win $5.

Yes, you can't win $5. so I subtract this scenario from 2.5$, which is the expect value of flipping 5 coins that could generate 5$.

 

[Tuan Pham]

How about this:
- For the original problem, there are 5 outcomes (0, 1, 2, 3, 4) depending on the number of heads. So the expected value is
(1/2^4)*1*$0 + (1/2^4)*4*$1 + (1/2^4)*6*$2 + (1/2^4)*4*$3 + (1/2^4)*1*$6
=(1/2^4) *( 1*$0 + 4*$1 + 6*$2 + 4*$3 + 1*$4 ) = $2.

- For the new problem, the number of outcomes are the same, except that the value of each outcome is different because we are allowed to flip a coin that is T. This means the value for each outcome will increase by .5 (except the last outcome). So the new expected value is
(1/2^4)*(1*$.5 + 4*$1.5 + 6*$2.5 + 4*$3.25 + 1*$4) = $2.46875

 

[postelrich]

ah a new way to think about that arrives to the same solution, thank Tuan.

 

[PrancingPeach]

I don't think anybody so far has taken into account all the variables in the problem. The best way to do this without drowning in complexity is to work backwards. There are three questions you must ask: (1) if you do re-flip, what is your expected gain from doing so? (2) under what circumstances should you re-flip, and how much profit do you expect to make in such a situation before re-flipping? (3) what is your expected profit in all situations where you do not decide to re-flip?

In this case, the analysis breaks down easily:

You would only re-flip a coin if you didn't get four heads. If you do get four heads, which happens with frequency 1/16, your payoff is $4. In the case that you don't get four heads, which happens with frequency 15/16, you expect to gain $.50 from the re-flip plus $1.50 from the other three coins. Hence your expected payoff overall using this strategy is (1/16) * $4 + (15/16) * ($1.50 + %.50) = $2.125. The hardest part of this, for me anyway, is actually doing that arithmetic in my head.