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

Answer:

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

b. It is not a fair game.

Step-by-step 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:

$E(Y) = (-6)\frac{1}{36} + (-5)\frac{2}{36} \cdots + 4\frac{1}{36} = \sum_{i=-6} ^{4}iP(Y=i) = -1$

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.