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The Taxpayer
Lv 7
The Taxpayer asked in Science & MathematicsMathematics · 1 decade ago

Probability 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.

3 Answers

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  • Pi R Squared
    Lv 7
    1 decade ago
    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
    Lv 7
    1 decade ago

    for any discrete random variable the expectation is:

    E(X) = μ = ∑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
    1 decade ago

    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

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