Interview Question

Quantitative Trader Interview

-Hong Kong

Jane Street

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?

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14 Answers

11

2/3

Anonymous on

7

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

2

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

jkramer on

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

0

2/3 : this is an application of Bayes theorem

Roland on

0

So your coins are H/H H/T T/T You see an H so you can't be holding the T/T coin. You are either holding the H/H coin or the H/T coin Now lets mark the H/H coin as H1/H2 and the H/T coin as H3/T If you're looking at the H1 then the other side is H2 If you're looking at H2 then the other side is H1 If you're looking at H3 then the other side is T So 2 out of the three times the other side would be heads. Fun problem pretty easy though.

Shiang Chen on

0

Another way to look at is between the H/H and H/T coin there's 3 heads and 1 tails. You're already looking at a head so there's 2 more heads and 1 tail left. so 2/3 of the time the other side will be heads.

Shiang Chen on

1

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

Yellow on

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

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

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

0

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

still dun quite understand on

0

why not 1/2?

Anonymous on

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

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