Question

In a dice game, Alice wins 5 (tokens) if she guesses the roll in advance; otherwise she loses 1. The dice shows six equally likely values as usual. Verify that Alice's expected win G1 is 0.

In a second experiment, Bob has two different dice in his sleeve, of which he selects one with equal probability and rolls them. One shows only even numbers, the other only odd numbers. The expected profit G2 is still 0.

In the third case, Alice learns from an oracle which dice Bob will use, before she makes her choice. She adjusts her strategy and uses the additional information. How high will her expected profit G3 be?

Answer 1

In G1, probability of guessing correctly = 1/6

So the probability of guessing wrong = 1-(1/6) = 5/6

So, the Expected winnings = (5*1/6)+(-1*5/6) = (5/6)-(5/6) = 0

In G2, the probability of choosing the dice which shows only even numbers = 1/2

And the probability of showing a particular even number out of the three even numbers = 1/3

So, the probability of showing any even number = (1/2)*(1/3) = 1/6

In the same way, the probability of choosing the dice which shows only odd numbers = 1/2

And the probability of showing a particular odd number out of three odd numbers = 1/3

So the probability of showing any odd number = (1/2)*(1/3) = 1/6

So, the probability of any number remains the same as 1/6.

So the expected winnings = (5*1/6) + (-1*5/6) = (5/6)-(5/6) = 0

In game G3, as Alice will already learn from an oracle which dice Bob will use , before making the choice, so she make choice accordingly as Alice will choose a even number , if he knows that Bob will choose the dice that shows only odd number and vice versa. In odd dice is chosen then the number of possible events will be 3 and in the case of even number dice, the number of possible events is also 3.

When Alice knows which dice Bob will choose, she will have to choose a number out of the possible 3 numbers.

So the probability of Alice having a correct prediction = 1/3 and the probability of Alice having a wrong prediction = 2/3.

So the expected winnings = (5*1/3) + (-1*2/3) = (5/3)-(2/3) = 3/3 = 1