Jane Street Interview Question: There are 3 coins. One coin h... | Glassdoor

Interview Question

Quantitative Trader Interview Hong Kong (Hong Kong)

There are 3 coins. One coin has heads on both sides, one

  coin has tails on both sides, the third one has head on one side and tail on the other side. Now I pick up one coin and toss. I get head. What is the chance that the coin I picked has heads on both sides?
Answer

Interview Answer

13 Answers

6

2/3

Interview Candidate on Oct 17, 2010
0

i got this question too, but i am still not sure how it is calculated

still dun quite understand on Oct 19, 2010
0

why not 1/2?

Anonymous on Oct 19, 2010
6

Because you have 1/3 chance to get double head coin and you will surely get head, 1/3 chance to get single head coin and then 1/2 chance to get head. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. Then, given you get head after tossing, then chance that you chose double head coin is (1/3)/(1/3+1/6) = 2/3

Anonymous on Oct 19, 2010
0

That can't be right. We know that the chosen coin landed heads up, so we can disregard the third (Tails/Tails) coin. Of the remaining 2 coins, there's a 1/2 chance that the chosen one is the double-headed coin. The answer should be 1/2, no?

Are you sure? on Nov 15, 2010
1

2/3 is correct... review http://en.wikipedia.org/wiki/Bayesian_inference

Yellow on Nov 17, 2010
0

Let HH, TT, HT be the events defined by picking the head-head, tail-tail, and head-tail coins, respectively. Given that we pick one of the coins randomly, we obtain
P(HH)=P(TT)=P(HT)=1/3. Let H and T be the head and tail events respectively. The probability of event H given HH is clearly 1. That is, P(H|HH)=1. Furthermore,
P(H|HT)=1/2 and P(H|TT)=0. The probability of a heads is:
P(H)=P(H int HT)+ P(H int HH)+P(H int TT)=
P(H|HT)P(HT)+ P(H|HH)P(HH)+P(H|TT)P(TT)=(1/2)(1/3)+ (1)(1/3)+(0)(1/3)=
(1/3)(3/2).
Note that P(H int HH)=P(H|HH)P(HH)= 1/3.
Therefore, the probability of event HH given H is:
P(HH|H)=P(H int HH)/P(HH)= (1/3)/ ((1/3)(3/2))=2/3.

Interview on Jan 2, 2011
2

2 heads on double headed coin, 1 head on the other, P(head is coming from double headed) = 2/3

jkramer on Jan 28, 2011
0

conditional probability?
probability the double headed given a head = probability (head unoin probability double headed coin)/ probability head is thrown.

= 1/3 / 1/2 = 2/3

anom on Jan 29, 2011
0

most of you are missing a fact that can't be ignored. Read the question - is says if you draw heads, meaning only 2 coins can have heads, what are the odds that you drew the 2 headed coin is 50%!

prm on Jun 21, 2011
2

C1 := 2 tail coin
C2 := 1 head 1 tail coin
C3 := 2 head coin

P(C3|H)=P(H|C3).P(C3)/P(H) = 1.(1/3)/(1/2) = 2/3
P(C2|H)= P(H|C2).P(C2)/P(H) = (1/2).(1/3)/(1/2) = 1/3

Anonymous on Nov 21, 2011

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0

2/3 : this is an application of Bayes theorem

Roland on Oct 14, 2015

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