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Jane Street Quant Interview Question

I interviewed in New York, NY and was asked:
"Unfair coin with P(H) = 1/3 and P(T) = 2/3. a) How to make an event with 50% probability? b) Expected number of flips until a realization occurs? c) Can you create a strategy to reduce the number of flips necessary? d) Can you create a strategy to reduce the number of flips necessary for an unfair coin with any bias?"
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Answer

Part of a Quant Interview Review - one of 418 Jane Street Interview Reviews

Answers & Comments

4
of 7
votes
a) Event 1 = {T,H}, Event 2 = {H,T}, if any other outcome, then re-roll
b) The probability of Event 1 or Event 2 occurring is 1/9+1/9=2/9. The expected number of 2-roll "tries" is 9/2. And each "try" consists of two rolls so 9 expected rolls for a realization.
c) Event 1 = {T,T}, Event 2 = any other combination. Probability of either event is 4/9.
d) Many solutions. Trick is to not discard any rolls. Use strategy from part (a), but if you roll {T,T}, then continue and put {T,T,H,H} in Event 1 and {H,H,T,T} in Event 2.
- Interview Candidate on Apr 1, 2012
0
of 0
votes
the c) is wrong. probability of event 2 is 5/9 not 4/9
- mokhlos on Jul 1, 2012
1
of 1
vote
mokhlos is right. Discard HH.
- Interview Candidate on Jul 1, 2012
1
of 1
vote
Interview Candidate,

P(H) = 1/3 and P(T) = 2/3 so a H-T combo will have a 2/9 prob.
For a), the probability of Event 1 or Event 2 occurring is 2/9 + 2/9 = 4/9.
Therefore, the expected number of 2-roll "tries" is 9/4.
Each try consists of two rolls - so an expected 9/2 rolls until a realization.
- Neil on Jan 3, 2013
4
of 4
votes
best answer:

TT is counted as case 1 -> p = 4/9
TH or HT is case 2 -> p = 2/9 + 2/9 = 4/9
only discard HH

Expected # of flips E = 8/9*2 + 1/9*(2 + E) -> E = 9/4 or 2.25 flips
- Anonymous on Feb 20, 2013

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