Question

Alice and Bob play a game in which they flip a coin repeatedly. Each time the coin is heads, Alice wins $1 (and Bob loses $1). Each time the coin is tails, Bob wins (and Alice loses) $2. They continue playing until Alice has won three flips. Prove that the expected value of Bob’s winnings is $3. (Hint: Use linearity of expected value to consider the expected value of each flip separately, with flips being worth $0 if they do not occur.)

Answer #1

Let X be the number of flips done until Alice has won three flips. Then X will follow Negative Binomial distribution with parameters r = 3 and p (probability of heads) = 1/2

X ~ NegBinomial(r = 3, p = 1/2)

E(X) = r/p = 3/(1/2) = 6

Let X1, X2 be the number of flips won by Alice and Bob respectively. Then, for a given value of X, X1, X2 will follow Binomial distribution with parameters n = X and p (probability of tails) = 1/2

That is, X1 | X ~ Binomial(n = X, p = 1/2) and X2 | X ~ Binomial(n = X, p = 1/2)

and E(X1 | X) = E(X2 | X) = np = X/2

By law of iterated expectation, E(X1) = E[E(X1 | X)] = E[X/2] = E[X]/2 = 6/2 = 3

Similarly, E(X2) = 3

Bob’s winnings = -X1 + 2X2

Expected winnings for Bob's =E[-X1 + 2X2]

= -E[X1] + 2E[X2] (Using linearity of expected value)

= -3 + 2 * 3

= $3

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