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Math Help - Discrete Probability Distribution Problem

  1. #1
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    Discrete Probability Distribution Problem

    Hi everyone, I hope you can help me with the following problem, especially the part 2 of the problem.

    Three dice are thrown in a game. If 1 or 6 turns up, you will be paid 1p. If neither 1 nor 6 turn up, you pay 5p. How much gain or loss is expected in 9 games?

    Let say you are given opportunity to change the rule of the game if 1 or 6 appears. To make the game worthwhile, what is the minimum amount in everyday currency that you would suggest for both the payments?

    Thanks in advance.
    Last edited by ose90; June 29th 2009 at 02:33 PM. Reason: Sorry for the typo error
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  2. #2
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    Outcome | Probability | Payoff
    1 or 6 | 1/3 | 1
    2-5 | 2/3 | -5

    I apologize for the appearance of my table. The '|' lines are supposed to represent column separators Unfortunately, I cannot get the spacing correctly.

    To find the expected return, multiply the probabilities with the payoffs and add.

    (1/3)*1 + (2/3)*-5 = 1/3 -10/3 = -9/3 = -3

    That would be for 1 game. For 9 games, multiply it by 9: -3*9 = -27

    The changing the game question has many possible answers. Choose the expected return you want, then change the payoff from either or both of the outcomes to match that return.
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  3. #3
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    Hello, ose90!

    Three dice are thrown in a game.
    If 1 or 6 turns up, you will be paid $1.
    If neither 1 nor 6 turn up, you pay $5.
    How much gain or loss is expected in 9 games?

    You will lose $5 if no 1 or 6 appear on any of the three dice.
    The probability is: . \tfrac{4}{6}\cdot\tfrac{4}{6}\cdot\tfrac{4}{6} \:=\:\tfrac{8}{27}

    Otherwise, you win $1.
    The probability is: . 1 - \tfrac{8}{27} \:=\:\tfrac{19}{27}

    We have: . \begin{array}{|c|c|c|} \hline<br />
\text{Outcome} & \text{Prob.} & \text{Payoff} \\ \hline \hline \\[-4mm]<br />
\text{1 or 6} & \frac{19}{27} & +1 \\ \\[-4mm] \text{Other} & \frac{8}{27} & \text{-}5 \\ \hline \end{array}

    EV \;=\;\left(\frac{19}{27}\right)(+1) + \left(\frac{8}{27}\right)(\text{-}5) \;=\;-\frac{7}{9}

    In 9 games, your expectation is: . (9)\left(\text{-}\frac{7}{9}\right) \;=\;-7

    You can expect to lose $7.




    Say you are given opportunity to change the rule of the game if 1 or 6 appears.
    To make the game worthwhile, what is the minimum amount for both the payments?
    A sloppy question! . . . There is no one right answer.
    Besides, what does "worthwhile" mean?



    Let W = amount we win, L = amount we lose.

    Then: . EV \:=\:\frac{19}{27}W + \frac{8}{27}(-L) \;=\;\frac{19W - 8L}{27}


    If the game is to be "fair", the EV should be zero:

    . . \frac{19W - 8L}{27} \:=\:0 \quad\Rightarrow\quad W \:=\:\frac{8}{19}L

    The amount won should be about 42.1% of the amount lost.
    . . (In the original game, it is only 20%.)

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