Interview Question

Quantitative Researcher InterviewNew York, NY

A tosses n+1 coins. B tosses n coins. B wins if he has at least as many heads as A. What is the probability that B wins?

4

Use symmetry. If both A and B had N coins, we can say that prob(A>B) = a, prob(A

Candidate on
2

Question rephrased: what's the prob that A has more heads? First n throws, they have equal number on the average. So, A gets a chance to have more heads on the last throw, 50% chance. -> P(A wins) = 1 - P(B wins) = 50% -> P(B wins) = 50%

0

^ is a great way of thinking about it.

OPERATOR on
1

1/2 or 0.5

sys on
0

I don't think 50% can be right, that's already the chance for when they have an exactly equal amount.

Anonymous on
0

Let A, B be two binomial r.v.’s with means (n+1)/2 and n/2 respectively and variances (n+1)/4 and n/4 respectively. Let Z = B-A then we can derive E[Z] = -1/2 and Var[Z] is about n/2. By Chebyshev’s inequality, Pr[B wins] = Pr[Z>=0] = sigma(Z) sqrt (1/2n)]<= 1/(2n).

Joseph Lin on
2

See the other guy's solution

Anonymous on