SOLUTION: The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four are blue, two are yellow and one is green. A player wins $4 i

Question 951522: The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four are blue, two are yellow and one is green. A player wins $4 if the spinner stops on red and $2 if it stops on green. The payer loses $2 if it stop on blue and loses $3 of it stop on yellow. What is the expected value of the game? Please help. I'm confused.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
expected value = sum (p(x) * v(x)) where sum (p(x)) = 1.
p(x) is the probability of an event occurring.
v(x) is the value you get when that event occurs.

p(red) = .3 and v(x) = 4
p(blue) = .4 and v(x) = -2
p(yellow) = .2 and v(x) = -3
p(green) = .1 and v(x) = 2

sum of p(x) is equal to 1.

expected value = sum of p(x) * v(x) = (.3 * 4) - (.4 * 2) - (.2 * 3) + (.1 * 2) = 0

your expected value is equal to 0.

this means that, if you played the game many times, you would expect to gain an average of 0 dollars each game.

consider you played 1000 games and the probability of each occurrence held up.

.3 * 1000 = 300 * 4 = + 1200
.4 * 1000 = 400 * -2 = - 800
.2 * 1000 = 200 * -3 = - 600
.1 * 1000 = 100 * -2 = + 200

your net gain would be equal to + 1200 - 800 - 600 + 200 which is equal to 0.

you neither won or lost any money.

your average net gain per play would be 0 / 1000 = 0.

it's easier to see how this works if the expected value isn't 0.

assume that you lost 2 dollars rather than won 2 dollars when you hit green.

the probabilities would then be:

p(red) = .3 and v(x) = 4
p(blue) = .4 and v(x) = -2
p(yellow) = .2 and v(x) = -3
p(green) = .1 and v(x) = -2 *****

sum of p(x) is still equal to 1.

expected value = sum of p(x) * v(x) = (.3 * 4) - (.4 * 2) - (.2 * 3) - (.1 * 2) = -.4

when you lost 2 on hitting green rather than winning 2 when hitting on green, your expected value becomes -.4 dollars per game.

this means that, if you played the game many times, you would expect to lose an average of 40 cents per game.

consider you played 1000 games and the probability of each occurrence held up.

.3 * 1000 = 300 * 4 = + 1200
.4 * 1000 = 400 * -2 = - 800
.2 * 1000 = 200 * -3 = - 600
.1 * 1000 = 100 * -2 = - 200

your total net gain would be + 1200 - 800 - 600 - 200 which would be equal to - 400.

that's your total net gain.

divide that by 1000 times that you played and the average becomes 400 / 1000 = .4 dollars per game.

you would lose an average of .4 dollars per game which is the same as losing 40 cents per game.

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