Question 281388: Suppose that the following game is played. A woman rolls a fair die. If she rolls a 1 or 6, she loses $1. If she rolls a 2 or a 5, she loses $2. If she rolls a 3 or 4, she wins $9. What is the expected value of the game?
Answer by Mathematicians(84)   (Show Source):
You can put this solution on YOUR website! I would first set up a table where P(x) is the probability for event x and the outcome if x happens (or Y). For example, your question would start like this. Some basic information is that a die has 6 sides, therefore a fair die has a probability of landing 1/6 on each side.
X -- P(x) -- Y
1 -- 1/6 -- -$1
2 -- 1/6 -- -$2
3 -- 1/6 -- +$9
4 -- 1/6 -- +$9
5 -- 1/6 -- -$2
6 -- 1/6 -- -$1
When you have a table like this, you want to multiply each P(x) * y and add them up.
So in your case we would get:
1/6 * -$1
+1/6 * -$2
+1/6 * +$9
+1/6 * +$9
+1/6 * -$2
+1/6 * -$1
= $(-1/6 - 2/6 + 9/6 + 9/6 - 2/6 - 1/6) (It is helpful if you keep the same denominator)
= + $12/6 = $2
$2 is the expected value. What this represents is that if you somehow played infinite games, you would win an average of $2 per game.
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