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## Interview Question

Interview Los Angeles, CA

# We play a game. You flip two coins, if you get both heads I

give you \$1. If you get both tails you flip again. If you get one of each, I pay you nothing. What's the expected value of this game?
Tags:

1

Dismiss TT since it is not part of the sample space (ie there is no outcome from it) You are twice as likely to get one of each (H-T or T-H) versus getting both heads answer: (2/3)(\$1) + (1/3)(0) = 66.7 cents

Interview Candidate on Aug 17, 2011
0

E=0.25x1+0.5x0+0.25xE 0.75xE=0.25 E=0.33

kerzane on Oct 18, 2011
0

interviewer told me .667 is correct

Kerzane is wrong on Oct 24, 2011
0

HT and TH add up to 1/2 of the probability with zero payout. HH is 1/4 probability with \$1 payout. TT results in a "reset" so we can disregard. After we normalize the probabilities, we get: E[payout]= (1/3)*\$1 + (2/3)*\$0 = \$0.33

Kerzane is right on Oct 28, 2011
0

You can also see it a geometric series. By writing E=0.25x1+0.5x0+0.25xE, Kerzane could as well have written E?0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(....))) This is quite simply the geometric sum: (n-> infinity) E=0.25 + 0.25^2 + 0.25^3....0.25^n which is equal to 0.25/(1-0.25) = 0.25/0.75 = 1/3 or 0.333

Kerzane is really right on Dec 7, 2011