A spinner is divided into 4 equal spaces. A player bets $2.00 on a given space. If the spinner lands on that space, he is paid $8.00 and gets his $2.00 back. If the spinner lands on on any space not chosen, the player loses the $2.00. (Assume that landing on a line is voided and spinner is respun) What is the expected value of the game?
Please show your work....I'm trying to get the logic down.
Pi R Squared
Favorite Answer
Hi,
¼ of the time he will make a profit of $8 and ¾ of the time he will loose $2.
The expected value is ¼*8 + ¾*-2 = 2 - 1.50 = .50
A player should average winning $.50 per hand. That's unusual for an "odds" game where the house wants to make money!!
I hope that helps!! :-)
Merlyn
for any discrete random variable the expectation is:
E(X) = μ = âx * P(X = x)
Let X be the amount of money you win
E(X) = 8 * 1/4 + -2 * 3/4 = 0.50
asldas a
so when wins he gets 8
and he loses he loses 2 or wins -2
so probability of winning is 1/4
and probability of losing is 3/4
so expected value is (1/4)*8+(3/4)*(-2)= 2-1.5=0.5
so expected value is $0.5
hope this helps