Probability Question?

Question

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 · Accepted 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 · Answer

for any discrete random variable the expectation is:

E(X) = Î¼ = âx * P(X = x)

Let X be the amount of money you win

E(X) = 8 * 1/4 + -2 * 3/4 = 0.50

asldas a · Answer

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

Probability Question?

3 Answers