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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.
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
