Elena' s Dice

**Aminekhadir** · March 6th 2010, 12:56 AM

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.

**Soroban** · March 6th 2010, 05:42 AM

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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