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# Jane StreetQuantitative Trader Interview Question

I interviewed in Hong Kong (Hong Kong) and was asked:
"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?"

Part of a Quantitative Trader Interview Review - one of 418 Jane Street Interview Reviews

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2/3
- Interview Candidate on Oct 17, 2010
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i got this question too, but i am still not sure how it is calculated
- still dun quite understand on Oct 19, 2010
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why not 1/2?
- Anonymous on Oct 19, 2010
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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
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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
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2/3 is correct... review http://en.wikipedia.org/wiki/Bayesian_inference
- Yellow on Nov 17, 2010
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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
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- jkramer on Jan 28, 2011
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conditional probability?

= 1/3 / 1/2 = 2/3
- anom on Jan 29, 2011
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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
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C1 := 2 tail coin
C2 := 1 head 1 tail 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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This is kind of like the Monty Hall Problem so check that out too.

Easiest way to see that it is definitely not 1/2 is to think of the same problem with more possible out comes than just heads or tails:

Imagine there are 101 boxes each with 100 balls in The first box has all blue balls, the next all red... etc with different colours in each box until the 100th which is all yellow say. The 101st has 1 ball of each colour. Suppose you randomly pick one box and pick one ball out at random.

The ball is blue.

Which is more likely?

That you happened to pick the 1 multi coloured box out of the 101 boxes and picked the one ball in that box that was blue.

Or

That you picked one of the 100 boxes that were all one colour and that colour happened to be blue.

Hopefully you can see that these are not equally likely events.
- Not 1/2 on Mar 17, 2012