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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. What is the probability that Alice will win and the probability that she will lose? What is Alice's expected payoff? What is the probability that Confucius will win and the probability that he will lose? What is Confucius's expected payoff? Is this a zero-sum game? Explain.

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Answer:

1/4 Alice wins and 3/34 she loses; Alice's expected payoff -$0.25; 3/8 Confucius wins and 5/8 he loses; Confucius's expected payoff $0.125; yes it is a zero sum game.

Step-by-step explanation:

First we create the sample space for flipping three coins:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

There are a total of 8 outcomes.

Alice wins if all coins land either all heads or all tails; this is either HHH or TTT, giving her a 2/8 = 1/4 chance of winning.  This means she has a 6/8 = 3/4 chance of losing.

Alice's expected payoff is found by multiplying her chances of winning, 1/4, by the amount she wins, $2, and her chances of losing, 3/4, by the amount she loses, $1:

1/4(2) + 3/4(-1) = 2/4 + -3/4 = -1/4 = -$0.25.

Confucius wins if there is one head and two tails; this is HTT, THT, or TTH.  This gives him a 3/8 chance of winning, which means he has a 5/8 chance of losing.

Confucius's expected payoff is found by multiplying his chances of winning, 3/8, by the amount he wins, $2, and his chances of losing, 5/8, by the amount he loses, $1:

3/8(2) + 5/8(-1) = 6/8 + -5/8 = 1/8 = $0.125.

Bob wins if two heads and one tail land; this is either HHT, HTH, or THH.  This gives him a 3/8 chance of winning, which means he has a 5/8 chance of losing.

Bob's expected payoff is found by multiplying his chances of winning, 3/8, by the amount he wins, $2, and his chances of losing, 5/8, by the amount he loses, $1:

3/8(2) + 5/8(-1) = 6/8 + -5/8 = 1/8 = $0.125.

This makes the sum of the expected values

-$0.25+$0.125+$0.125 = 0; this makes it a zero sum game.

MrRoyal

Probabilities are used to determine the chances of events

  • Alice's probability of winning is 1/4
  • Alice's probability of losing is 3/4
  • Alice's expected payoff is -$0.25
  • Confucius's probability of winning is 3/8
  • Confucius's probability of losing is 5/8
  • Confucius's expected payoff is $0.125
  • The game is a zero-sum game

The given parameters are:

[tex]n =3[/tex] --- number of people

[tex]r = 2[/tex] --- sides of a coin

So, the number of possible outcome of the game is:

[tex]Toss = r^n[/tex]

[tex]Toss = 2^3[/tex]

[tex]Toss = 8[/tex]

The sample space (S) is then listed as:

[tex]S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}[/tex]

(a) Probability that Alice wins and lose

Alice wins in the following outcomes

[tex]Win = \{HHH,TTT\}[/tex] -- all outcomes are the same.

So, the probability of winning is:

[tex]P(Win) =\frac{n(Win)}{Tosses}[/tex]

[tex]P(Win) =\frac{2}{8}[/tex]

[tex]P(Win) =\frac{1}{4}[/tex]

Using the complement rule, the probability of losing is

[tex]P(Lose) = 1 - P(Win)[/tex]

[tex]P(Lose) = 1 - \frac 14[/tex]

[tex]P(Lose) = \frac 34[/tex]

She gets $2 if she wins, and she loses $1, if she lost.

So, her expected payoff of the game is:

[tex]Expected = \sum x \times P(x)[/tex]

This gives

[tex]Expected = 2 \times \frac 14 - 1 \times \frac 34[/tex]

[tex]Expected = \frac 24 - \frac 34[/tex]

Simplify

[tex]Expected = - \frac 14[/tex]

[tex]Expected = - 0.25[/tex]

Hence, her expected payoff is -$0.25

(b) Probability that Confucius wins and lose

Confucius wins in the following outcomes

[tex]Win = \{HTT, THT, TTH\}[/tex] -- one head, two tails.

So, the probability of winning is:

[tex]P(Win) =\frac{n(Win)}{Tosses}[/tex]

[tex]P(Win) =\frac{3}{8}[/tex]

Using the complement rule, the probability of losing is

[tex]P(Lose) = 1 - P(Win)[/tex]

[tex]P(Lose) = 1 - \frac 38[/tex]

[tex]P(Lose) = \frac 58[/tex]

He gets $2 if she wins, and she loses $1, if she lost.

So, his expected payoff of the game is:

[tex]Expected = \sum x \times P(x)[/tex]

This gives

[tex]Expected = 2 \times \frac 38 - 1 \times \frac 58[/tex]

[tex]Expected = \frac 68 - \frac 58[/tex]

Simplify

[tex]Expected = \frac 18[/tex]

[tex]Expected = 0.125[/tex]

Hence, his expected payoff is $0.125

(c) Is it a zero-sum game?

To do this, we start by calculating the probabilities of Bob winning and losing

Bob wins in the following outcomes

[tex]Win = \{HHT, HTH, THH\}[/tex] -- two heads, one tail.

So, the probability of winning is:

[tex]P(Win) =\frac{n(Win)}{Tosses}[/tex]

[tex]P(Win) =\frac{3}{8}[/tex]

Since Bob and Confucius have the same probabilities of winning, then they would have the same probabilities of losing and the same expected payoff, as follows:

[tex]P(Lose) = \frac 58[/tex]

[tex]Expected = 0.125[/tex]

Calculate the sum of the expected values of the three participants

[tex]Total = -0.25 +0.125 + 0.125[/tex]

[tex]Total = 0[/tex]

Since the total expected value is 0, then the game is a zero-sum game

Read more about probabilities and expected values at:

https://brainly.com/question/13934271

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