Expected Value Probability and Odds

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?

Red and Blue Dice

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.