Question

There are four sealed boxes. There are 100 pounds in one box and the others are empty. A player can pay X to open a box and take the constants as many as they like. Assuming this's a fair game.

a) What's the value of X

b) Instead of maintaining the constant price, if you choose incorrect boxes after 2 times the price changes to Z. What's the fair value of X and Z?

Answer 1

p(100 pound) = 1/4

p(empty = 0 pound) = 3/4

a.

value(x) = E(winning)

E(winning) = 100*1/4 + 0*3/4

= 25 pounds

value (X) = 25 pounds

b.

fair value of X = 25 pounds { as calculated above }

fair value of Z

after 2 incorrect boxes there remain only 1 correct and 1 incorrect box

p(correct) = 1/2

p(incorrect) = 1/2

E(winning) = 100*1/2 + 0*1/2

= 50 pounds

fair value of Z = 50 pounds

P.S. (please upvote if you find the answer satisfactory)