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Math Help - Elena' s Dice

  1. #1
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    Smile Elena' s Dice

    Hello,
    this is the question:

    Elena plays a game where she tosses two dice.
    If the sum is 6, she wins 3 point. (The probability of the sum being 6 is 5/36)
    If the sum is greater than 6, she wins 1 point (The probability of the sum being greater than 6 is 21/36).
    If the sum is less than 6, she loses k points. (The probability of the sum being less than 6 is 10/36).

    Find the value of k for which Elena's expected number of points is zero.

    Thanks a lot!
    Amine.
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  2. #2
    Super Member

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    Hello, Amine!

    Do you know the Expected Value formula?
    They have already done the hard work for you.


    Elena plays a game where she tosses two dice.
    If the sum is 6, she wins 3 points. (The probability of the sum being 6 is 5/36)
    If the sum is greater than 6, she wins 1 point (The probability of the sum being greater than 6 is 21/36).
    If the sum is less than 6, she loses k points. (The probability of the sum being less than 6 is 10/36).

    Find the value of k for which Elena's expected number of points is zero.

    . . \begin{array}{|c|c|c|}<br />
\text{Event} & \text{Prob.} & \text{Payoff} \\ \hline \\[-4mm]<br />
= 6 & \frac{5}{36} & +3 \\ \\[-4mm]<br />
> 6 & \frac{21}{36} & +1 \\ \\[-4mm]<br />
< 6 & \frac{10}{36} & -k \end{array}

    We have: . \left(\frac{5}{36}\right)(+3) + \left(\frac{21}{36}\right)(+1) + \left(\frac{10}{36}\right)(-k) \;=\;0

    . . . . \frac{15}{36} + \frac{21}{36} - \frac{10k}{36} \;=\;0 \quad\Rightarrow\quad 1 - \frac{5k}{18} \:=\:0 \quad\Rightarrow\quad \frac{5k}{18} \:=\:1 \quad\Rightarrow\quad k \:=\:\frac{18}{5}


    Therefore: . k \:=\:3.6

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