Question

Suppose it costs $8 to roll a pair of dice. you get paid the sum of the numbers in dollars that appear on the dice.

a. what is the expected value of the game?
b. is it a fair game?

Answer 1

Hello!

there is no specific expected value, but I predict that is is going to be in the negatives.

a player is able to gain a total of 12 dollars, but has a HIGH probability of making less than they spent. If they had to pay $6, they would have a 50/50 chance of making their money back.
so no it is not a fair game either.

I hope this helps, and have a nice day!

Answer 2

a. The expected value ( or mean ) is -1.

b. It is not a fair game.

Explanation:

a.

First we look into all possible values for a random variable X, that is defined as:

X = The sum of the number on both dice.

Next there is a table, that shows us the frequency of a certain sum in all possible combinations:

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

You can see that the sum 2 appears only one time, as opposed to the sum 7, that appears 6 times.

Now, let's create a random variable called Y, so that

Y = The sum of the number on both dice minus 8.

So if we count the frequency of all possible sums minus 8, we get the following:

`\left[\begin{array}{ccc}Y&P(Y=y)\\-6&1/36 \\-5&2/36\\-4&3/36\\-3&4/36\\-2&5/36\\-1&6/36\\0&5/36\\1&4/36\\2&3/36\\3&2/36\\4&1/36\end{array}\right]`

So from now on, we just need to calculate the expected value:

`E(Y) = (-6)(1)/(36) + (-5)(2)/(36) \cdots + 4(1)/(36) = \sum_(i=-6) ^(4)iP(Y=i) = -1`

b.

From this, as the expected value isn't equal to zero, that means it's a unfair game or a biased game, it's so that one side always triumph over the other.