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Math Help - Find expectated payoff

  1. #1
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    Find expectated payoff

    Hi

    There is a bag containing numbers from 1 to n. A number is drawn at random from the bag and put back.(let the number be "X") A second number is drawn after that. (let the number be "Y")

    THe person gets 10$ if X>y, he has to pay 5$ is x<y and gets 0$ is x=y.

    What is the expected payoff of the game?

    (PS: I am not sure how to calculate the probabilities of x>y,x<y). The prob of x=y should be 1/n^2 most probably but not sure.. Any help is appreciated)

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by champrock View Post
    Hi

    There is a bag containing numbers from 1 to n. A number is drawn at random from the bag and put back.(let the number be "X") A second number is drawn after that. (let the number be "Y")

    THe person gets 10$ if X>y, he has to pay 5$ is x<y and gets 0$ is x=y.

    What is the expected payoff of the game?

    (PS: I am not sure how to calculate the probabilities of x>y,x<y). The prob of x=y should be 1/n^2 most probably but not sure.. Any help is appreciated)

    Thanks
    By symmetry p(x>y)=p(x<y), p(x=y)=1/n and

    p(x>y)+p(x<y)+p(x=y)=1

    which will allow you to find the three probabilities, then the expected payoff is:

    E=10p(x>Y)-5p(x<y)

    CB
    Last edited by CaptainBlack; July 6th 2009 at 06:24 AM.
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  3. #3
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    Hello, champrock!

    There is a bag containing numbers from 1\text{ to }n.
    A number is drawn at random from the bag and put back. (Let the number be x.)
    A second number is drawn after that. (Let the number be y.)

    The player: . \begin{array}{cc}\text{wins \$10} & \text{ if }x > y \\ \text{pays \$5} & \text{if }x < y \\ \text{gets \$0}& \text{if }x = y \end{array}

    What is the expected payoff of the game?
    There are: . n^2 possible draws.


    In n cases, the numbers are equal.
    . . P(x = y) \:=\:\frac{n}{n^2} \:=\:\frac{1}{n}


    Among the other n^2-n cases, half of them have x greater than y.
    That is, in \frac{n^2-n}{2} cases, x > y
    . . Hence: . P(x > y) \:=\:\frac{\frac{n^2-n}{2}}{n^2} \:=\:\frac{n-1}{2n}

    Similarly: . P(x < y) \:=\:\frac{n-1}{2n}


    Therefore: . EV \;=\;\left(\frac{1}{n}\right)(0) + \left(\frac{n-1}{2n}\right)(10) + \left(\frac{n-1}{2n}\right)(-5) \;=\;\boxed{\frac{5(n-1)}{2n}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Translation: . EV \:=\:\frac{5n-5}{2n} \:=\:\frac{5n}{2n} - \frac{5}{2n} \;=\;\frac{5}{2}-\frac{5}{2n}

    On the average, you can expect to win slightly less than $2.50 per game.

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