U4 Alice Bob and Confucius are bored during recess so they d

U4. Alice, Bob, and Confucius are bored during recess, so they decide to play a new game. Each of them puts a dollar in the pot, and each tosses a quarter.
Alice wins if the coins land all heads or all tails. Bob wins if two heads and one tail land, and Confucius wins if one head and two tails land. The quarters are fair, and the winner receives a net payment of $2 ($3 - $1 = $2), and the
losers lose their $1.
(a) What is the probability that Alice will win and the probability that she will lose?
(b) What is Alice\'s expected payoff?
(c) What is the probability that Confucius will win and the probability that
he will lose?
(d) What is Confucius\' expected payoff?
(e) Is this a zero-sum game? Please explain your answer.

Solution

(a)

As it is given that each quarters are fair, so the probability of \'Heads\' and \'Tails\' will be equal on each quarter.

There are 3 coins being tossed. Alice win if all the coins land either heads or all of them land in tails. The probability that all the three coins land as heads is equal to the probability that all coins land in tails because the coin is unbiased.

And the probability of all heads/tails is given by - P = (0.5) (0.5) (0.5) = 0.125.

This is because each coin has a probability of 0.5 to fall as heads/tails. And all the three tosses are independent from each other. So, P(HHH) = 0.125 = P(TTT).

So, P(HHH) + P(TTT) = 0.125+0.125 = 0.25

Hence, the probability that Alice will win is = 0.25 and the probability that she will lose = 1-0.25 = 0.75.

-------------------------------------------------------------------------------------------------------

(b)

Alice wins with a probability of 0.25 and loses with a probability of 0.75.

So, she earns $2 with a probability of 0.25 and loses $1 with a probability of 0.75.

Hence, the expected playoff of Alice will be = 0.25($2)+0.75(-$1) = -$0.25.

------------------------------------------------------------------------------------------------------------

(c)

Confucius wins if one head and two tails occur. Now, the probability of winning of Confucius can be written as -

P(1H+2T) = P(HTT)+P(THT)+P(TTH)

So, P(1H+2T) = (0.5)(0.5)(0.5)+(0.5)(0.5)(0.5)+(0.5)(0.5)(0.5) = 3 x 0.125 = 0.375

Again the above equation holds because of the independency of outcomes of coins and the unbiasness of the quarter coin.

So, probability that Confucius will win = 0.375 and the probability that he will lose = 1- 0.375 = 0.625.

______________________________________________________________________________

(d) Confucius wins with a probability of 0.375 and loses with a probability of 0.625.

So, he wins $2 with a probability of 0.375 and loses $1 with a probability of 0.625.

Hence, the expected playoff = 0.375($2)+0.625(-$1) = $0.75 - $0.625 = $0.125.

_____________________________________________________________________

(e)

The expected playoff of Bob would be same as the Confucius because of the symmetry. The only difference is that the number of heads and tails get exchanged for Bob. But the probability remains same as the coins are unbiased. So, P(H) = P(T) = 0.5.

So, expected playoff of Bob = $0.125.

Total Expected Playoff = Expected playoff of Alice + Expected playoff of Confucius + Expected playoff of Bob

So, Total Playoff = -$0.25+$0.125+$0.125 = 0

As the net playoff is zero for the game, so we can say that it is a zero-sum game.