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Math Help - Expected value Problem

  1. #1
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    Expected value Problem

    For a game played using the dice in Problem #13, you make a 2$ bet on the outcome of a roll of the dice. If the roll is 5 or lower you win 3$. If the roll is 11 or 12, you win 10$. What is the expected value of this game?(note the winning amounts are net amounts.)

    Here is the dice probability distribution for problem 13



    Here is what i did



    Is this correct?
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  2. #2
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    Quote Originally Posted by solace View Post
    For a game played using the dice in Problem #13, you make a 2$ bet on the outcome of a roll of the dice. If the roll is 5 or lower you win 3$. If the roll is 11 or 12, you win 10$. What is the expected value of this game?(note the winning amounts are net amounts.)

    Here is the dice probability distribution for problem 13



    Here is what i did



    Is this correct?
    I can't read this last table, but the probability of rolling a 5 or less is 0.3225
    and the prob of rolling a 11 or 12 is 0.0175, and the probability of any other
    result is 1-0.3225-0.0175 = 0.66.

    Now the value of the game is:

    0.3225*5 + 0.0175*10 - 0.66*2 = +0.4675

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    I can't read this last table, but the probability of rolling a 5 or less is 0.3225
    and the prob of rolling a 11 or 12 is 0.0175, and the probability of any other
    result is 1-0.3225-0.0175 = 0.66.

    Now the value of the game is:

    0.3225*5 + 0.0175*10 - 0.66*2 = +0.4675

    RonL
    Why .3225*5? Should it be *3 since thats how much money you get if you roll a 5 or lower?

    Also why .3325?

    .09+.09+.1425+.15= .4725 right?

    THen 1-.4725+.0175 = .5100

    so: .4725*3+.0175*10-.51*2= .5725

    Also, doesnt the net value mean that you lose the 2$ you bet and any winnings are -2? So the 3$ win is actually 1$ net profit, and the 10$ win is actually 8$ net profit?

    So should i times the probabilities by 1 and 8 instead?
    Last edited by solace; May 20th 2007 at 11:30 PM.
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  4. #4
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    Quote Originally Posted by solace View Post
    Why .3225*5? Should it be *3 since thats how much money you get if you roll a 5 or lower?
    If that is what you mean net amount. It is an odd way of expressing it but
    if you have it defined that way OK.


    Also why .3325?

    .09+.09+.1425+.15= .4725 right?

    THen 1-.4725+.0175 = .5100

    so: .4725*3+.0175*10-.51*2= .5725
    yes, slip of the calculator

    Also, doesnt the net value mean that you lose the 2$ you bet and any winnings are -2? So the 3$ win is actually 1$ net profit, and the 10$ win is actually 8$ net profit?

    So should i times the probabilities by 1 and 8 instead?
    If that is what you mean by net winnings (I would interpret net winnings
    to be how much better off you are after compared to before the bet), then:

    0.4725*1 + 0.0175*8 - 0.51*2 = +0.4675

    My experience is that when you bet your returned stake is not counted
    part of your winnings. Just think of the fun the tax man would have if it did.

    RonL
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  5. #5
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    Hello, solace!

    You make a $2 bet on the outcome of a roll of the dice.
    If the roll is 5 or lower you win $3.
    If the roll is 11 or 12, you win $10.
    What is the expected value of this game?
    Rolling a pair of dice, there are 36 possible outcomes.

    The totals and their respective probabilities are:

    . . \begin{array}{cc}2 & \frac{1}{36} \\ 3 & \frac{2}{36} \\ 4 & \frac{3}{36} \\ 5 & \frac{4}{36} \\ 6 & \frac{5}{36} \\ 7 & \frac{6}{36}\end{array}
    . . \begin{array}{cc}8 & \frac{5}{36} \\ 9 & \frac{4}{36} \\ 10 & \frac{3}{36} \\ 11 & \frac{2}{36} \\ 12 & \frac{1}{36}<br />
\end{array}

    P(\text{5 or lower: win \$3}) \;=\;<br />
\frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}

    P(\text{11 or 12: win \$10}) \;=\;\frac{2}{36} + \frac{1}{36}\;=\;\frac{3}{36}

    P(\text{lose \$2}) \;=\;\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} \;=\;\frac{23}{36}


    Hence: . E \;=\;\frac{10}{36}(3) + \frac{3}{36}(10) + \frac{23}{36}(-2) \;=\;\frac{14}{36} \;=\;0.30555...


    Therefore, you can expect to win an average of about 30 per game.

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, solace!

    Rolling a pair of dice, there are 36 possible outcomes.

    The totals and their respective probabilities are:

    . . \begin{array}{cc}2 & \frac{1}{36} \\ 3 & \frac{2}{36} \\ 4 & \frac{3}{36} \\ 5 & \frac{4}{36} \\ 6 & \frac{5}{36} \\ 7 & \frac{6}{36}\end{array}
    . . \begin{array}{cc}8 & \frac{5}{36} \\ 9 & \frac{4}{36} \\ 10 & \frac{3}{36} \\ 11 & \frac{2}{36} \\ 12 & \frac{1}{36} \end{array}" alt="
    \end{array}" />

    P(\text{5 or lower: win \$3}) \;=\; \frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}" alt="
    \frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}" />

    P(\text{11 or 12: win \$10}) \;=\;\frac{2}{36} + \frac{1}{36}\;=\;\frac{3}{36}

    P(\text{lose \$2}) \;=\;\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} \;=\;\frac{23}{36}


    Hence: . E \;=\;\frac{10}{36}(3) + \frac{3}{36}(10) + \frac{23}{36}(-2) \;=\;\frac{14}{36} \;=\;0.30555...


    Therefore, you can expect to win an average of about 30 per game.
    They are not fair dice.

    RonL
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