1. What is the fair price that must be paid to enter a game in which...

1. What is the fair price that must be paid to enter a game in which you can win $ 950 with 0.1 of probability, $ 2.5 with 0.4 probability and lose $ 14 with 0.5 probability?

2. A person throws 2 dice and receives the sum of the results obtained in pesos, how much must he pay per game to have a fair game?

Solutions

Expert Solution

Probability of winning $950 = 0.1
Probability of winning $2.5 = 0.4
Probability of losing $14 = 0.5
So expected winning from the game = $9500.1 + $2.50.4 - $140.5 = $89
So any amount less than $89 is the fair price that must be paid to enter the game.

Let result on first die be X and result on second die be Y.

Clearly, X and Y are independent.

P(X+Y=2) = P(X=1,Y=1) = P(X=1)P(Y=1) = (1/6)(1/6) = 1/36

P(X+Y=3) = P(X=1,Y=2) + P(X=2,Y=1) = P(X=1)P(Y=2) + P(X=2)P(Y=1) = (1/6)(1/6) + (1/6)(1/6) = 2/36

P(X+Y=4) = P(X=1,Y=3) + P(X=2,Y=2) + P(X=3,Y=1) = P(X=1)P(Y=3) + P(X=2)P(Y=2) + P(X=1)P(Y=3)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 3/36

P(X+Y=5) = P(X=1,Y=4) + P(X=2,Y=3) + P(X=3,Y=2) + P(X=4,Y=1)
= P(X=1)P(Y=4) + P(X=2)P(Y=3) + P(X=3)P(Y=2) + P(X=4)P(Y=1)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 4/36

P(X+Y=6) = P(X=1,Y=5) + P(X=2,Y=4) + P(X=3,Y=3) + P(X=4,Y=2) + P(X=5,Y=1)
= P(X=1)P(Y=5) + P(X=2)P(Y=4) + P(X=3)P(Y=3) + P(X=4)P(Y=2) + P(X=5)P(Y=1)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 5/36

P(X+Y=7) = P(X=1,Y=6) + P(X=2,Y=5) + P(X=3,Y=4) + P(X=4,Y=3) + P(X=5,Y=2) + P(X=6,Y=1)
= P(X=1)P(Y=6) + P(X=2)P(Y=5) + P(X=3)P(Y=4) + P(X=4)P(Y=3) + P(X=5)P(Y=2) + P(X=6)P(Y=1)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 6/36

P(X+Y=8) = P(X=2,Y=6) + P(X=3,Y=5) + P(X=4,Y=4) + P(X=5,Y=3) + P(X=2,Y=6)
= P(X=2)P(Y=6) + P(X=3)P(Y=5) + P(X=4)P(Y=4) + P(X=5)P(Y=3) + P(X=6)P(Y=2)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 5/36

P(X+Y=9) = P(X=3,Y=6) + P(X=4,Y=5) + P(X=5,Y=4) + P(X=6,Y=3)
= P(X=3)P(Y=6) + P(X=4)P(Y=5) + P(X=5)P(Y=4) + P(X=6)P(Y=3)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 4/36

P(X+Y=10) = P(X=4,Y=6) + P(X=5,Y=5) + P(X=6,Y=4) = P(X=4)P(Y=6) + P(X=5)P(Y=5) + P(X=6)P(Y=4)
= (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/6) = 3/36

P(X+Y=12) = P(X=6,Y=6) = P(X=6)P(Y=6) = (1/6)(1/6) = 1/36

Expected amount of money won in pesos
= E(X+Y)

= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36) = 7

So the person must pay less than 7 pesos per game to have a fair game.

orchestra answered 1 year ago

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