Question

Conditional Expectation

You are offered the following game:

-A 6-faced fair die is rolled. Call the result of this roll J

-A coin is flipped, if it lands heads you win 2^J dollars, if it lands tails you win J.

a) What is the probability you win more than $25 at this game?

b) Compute the conditional expectation conditional on J and then use the Law of Iterated expectations to compute the expected value of playing the game.

NOTE: this problem is from an Econometrics class

Answer 1

We first compute the various outcomes of the given game here as:

The probability of each outcome is computed here as:
= (1/6)*0.5 = 1/12

P(X = 1) = P(J = 1, Tails) = 1/12
P(X = 2) = P(J = 2, Tails) + P(J = 1, Heads) = 2/12 = 1/6
P(X = 3) = P(J = 3, Tails) = 1/12
P(X = 4) = P(J = 4, Tails) + P(J = 2, Heads) = 2/12 = 1/6
P(X = 5) = P(J = 5, Tails) = 1/12
P(X = 6) = P(J = 6, Tails) = 1/12
P(X = 8) = P(J = 3, Heads) = 1/12
P(X = 16) = P(J = 4, Heads) = 1/12
P(X = 32) = P(J = 5, Heads) = 1/12
P(X = 64) = P(J = 6, Heads) = 1/12

a) Therefore the probability that:
P(X > 25) = P(X = 32) + P(X = 64) = 2/12 = 1/6

Therefore 1/6 is the required probability here.

b) The conditional expectation conditional on J is computed here as:
E(X | J = 1) = (1 + 2) / 2 = 1.5
E(X | J = 2) = (2 + 2²) /2 = 3
E(X | J = 3) = (3 + 2³)/2 = 5.5
E(X | J = 4) = (4 + 2⁴)/2 = 10
E(X | J = 5) = (5 + 2⁵)/2 = 18.5
E(X | J = 6) = (6 + 2⁶)/2 = 35

Therefore the conditional expectation of total amount won now using the tower law of expectation is computed here as:

Therefore 12.25 is the required expected value here.