**Example Question:**

Suppose you are offered 10 to 4 odds that you cannot roll two even numbers with the roll of **two** fair six-sided dice. In other words, you will win $10 if you succeed (and roll two even values) and you will lose (pay) $4 if you fail to roll two even values. **What is the expected value of this game? What is the expected value if you play 400 times?**

**Solution and Explanation:**

Here you are rolling two fair dice (6 sided) and you are looking for **two even numbers**.

Here are all the Possibilities:

2, 2

2, 4,

2, 6

4, 2

4, 4

4, 6

6, 2

6, 4

6, 6

So there are **9 possible ways** for this to happen.

When rolling two fair dice, you can get any of 6 outcomes on die1 and any of 6 outcomes on die 2. So the **sample space** (all possibilities) has **36 values**.

This means your chance of rolling two even values is 9/36 Therefore, the **probability of winning** is 9/36 = .25

**Probability of losing** is 1 – .25 = .75

**The expected value is the prob of winning * the value you get when you win + prob of losing* value you lose (which is negative as it is a loss).**

**Expected Value = .25 * $10 + .75*(-$4) ==> **Notice that the $4 is negative because it is a loss

= .25*10 – .75*4 = 2.50 – 3 = – .50

The expected value in this example in negative which tells us that over time (as you play) you are expected, on average, to be at a loss at the end. The more you play, the more you are likely to lose.

**Overall if you play 400 times your expected win/loss is: 400* (-.50) = – $200**

This is a loss of 200 dollars.

**NOTE: in some cases, the result will be positive and will allow an overall gain over time. Try this problem again but using $20 if you win and -$3 if you lose. In that case, you expected value will be positive and therefore the more you play, the more you expect to win.**