Question

Consider a pay-to-play game which involves flipping a coin three (3) times. The payout for the...

Consider a pay-to-play game which involves flipping a coin three (3) times. The payout for the game depends on the number of heads obtained in the three coin flips. Let the discrete random variable X represent the number of heads.

a) What is the probability P{X = k} associated with each value k of the random variable?

b) Suppose that the game has a payout of X^2 dollars. What is the minimum amount that should be charged for admittance (player gets one game of three coin flips) so that the person running the game won’t lose money on average? Hint, determine E[X^2].

c) Suppose that the game has a new payout of 2X + 3X^2 dollars. Now what is the minimum amount that should be charged for admittance?

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Answer #1

Number of times a coin is flipped : n=3

Probability of obtaining a head in a single flip : p = 1/2 =0.5

q=1-p=1-0.5 =0.5

X : Number of heads

X follows a binomial distribution with n=3 and p=0.5;

Probability mass function of X :

Probability of getting 'k' heads in 3 flips of a coin = P(X=k)

x P(x) P(x)
0 0.1250
1 0.3750
2 0.3750
3 0.1250

b)

Payout for this game = X2

E(X2) is the amount the player expected to win when the player plays once :

Therefore, minimum amount that should be charged for admittance (player gets one game of three coin flips) so that the person running the game won’t lose money on average = E(X2)

Minimum amount that should be charged for admittance (player gets one game of three coin flips) so that the person running the game won’t lose money on average = 3

c) Suppose that the game has a new payout of 2X + 3X2 dollars. Now what is the minimum amount that should be charged for admittance

Therefore, minimum amount that should be charged for admittance (player gets one game of three coin flips) so that the person running the game won’t lose money on average = E(2X+3X2)

E(2X+3X2) = 2 E(X) + 3E(X2)

For Binomial distribution , E(X) = np = 3 x 0.5 =1.5

From (b) E(X2) = 3

E(2X+3X2) = 2 E(X) + 3E(X2) = 2 x 1.5 + 3 x 3 = 3 + 9 =12

E(2X+3X2) = 12

Suppose that the game has a new payout of 2X + 3X2 dollars, the minimum amount that should be charged for admittance = 12

Alternatively

E(2X+3X2) = 12

Suppose that the game has a new payout of 2X + 3X2 dollars, the minimum amount that should be charged for admittance = 12

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