Question

In: Statistics and Probability

:) Alice and Bob play the following game: in each round, Alice first rolls a single...

:)

Alice and Bob play the following game: in each round, Alice first rolls a single standard fair die. Bob then rolls a single standard fair die. If the difference between Bob’s roll and Alice's roll is at most one, Bob wins the round. Otherwise, Alice wins the round.

(a) (5 points) What is the probability that Bob wins a single round?

(b) (7 points) Alice and Bob play until one of them wins three rounds. The first player to three wins is declared the winner of the series. What is the probability that Bob wins the series?

(c) (7 points) In a single series, what is the expected number of wins for Bob?

(d) (6 points) In a single series, how many more games is Alice expected to win than Bob? That is, what is the expected value of the number of wins for Alice minus the number of wins for Bob?

(e) (5 points) In a single series, what is the variance of the expected number of wins for Bob?

Expert Solution

Answer:

Alice and Bob play the following game: in each round, Alice first rolls a single standard fair die. Bob then rolls a single standard fair die. If the difference between Bob’s roll and Alice's roll is at most one, Bob wins the round. Otherwise, Alice wins the round.

When they both roll a single standard fair die, the total number of possible outcomes = 62 = 36

Since, Bob wins the round if the difference between Bob’s roll and Alice's roll is at most one, therefore the cases where Bob wins are:

(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (4, 5), (5, 4), (5, 5), (5, 6), (6, 5), (6, 6)

So, the number of cases where Bob wins the round = 16

Hence, the number of cases where Alice wins the round = 36 - 16 = 20

(a) Therefore, the probability that Bob wins a single round = (Number of cases where Bob wins the round)/(Number of total outcomes) = 16/36 = 4/9

Answer: The probability that Bob wins a single round is 4/9.

(b) The probability that Alice wins a single round = (Number of cases where Alice wins the round)/(Number of total outcomes) = 20/36 = 5/9

The first player to three wins is declared the winner of the series. The probability that Bob wins 3 consecutive games = (4/9)3 = 64/729

Hence, the probability that Bob does not win 3 consecutive games = 1 - 64/729 = 665/729

Let X be the random variable denoting the number of rounds Bob loses before the wining 3 consecutive games.

Then X ~ Geo (p = 64/729)

The p.m.f. of X is given by,

P(X = k) = (1 - 64/729)k - 1(64/729), k = 1, 2, 3, ....

In a single series i.e. when k = 1,

P(X = 1) = 64/729

Answer: The probability that Bob wins the series is 64/729.

(c) In a single series, the expected number of wins for Bob

= E(X) = 1/p = 729/64

Answer: In a single series, the expected number of wins for Bob is 729/64.

(d) Let, Y be the random variable denoting the number of rounds Alice loses before the wining 3 consecutive games.

Then X ~ Geo (q = 665/729)

The p.m.f. of Y is given by,

P(Y = k) = (1 - 665/729)k - 1(665/729), k = 1, 2, 3, ....

In a single series i.e. when k = 1,

P(Y = 1) = 665/729

In a single series, the expected number of wins for Alice

= E(Y) = 1/q = 729/665

Therefore, the expected value of the number of wins for Alice minus the number of wins for Bob

= E(Y) - E(X) = 729/665 - 729/64 = -10.29

Answer: In a single series, the expected value of the number of wins for Alice minus the number of wins for Bob is - 10.29.

(e) In a single series, the variance of the number of wins for Bob

= Var (X) = (1 - p)/p2 = (1 - 64/729)/(64/729)2 = 484785/4096

Answer: In a single series, the variance of the number of wins for Bob is 484785/4096.

orchestra answered 2 years ago

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