Expected value for a sum of two special dice

Question

I am trying to find the expected value for the sum of two dices, given that if both dices shows an identical number, then the sum is doubled.

Basically if we get $1$ for a dice and $1$ for the other one, this will count as: $4$, similarly: $2+2=8$, $3+3=12$, $4+4=16$, $5+5=20$, $6+6=24$. If any other combination appears then it is counted as for a normal dice.

I know how to find the expected value for two normal dices, since I just find it for one, then double it. $$E(x)={\frac16} (1+2+3+4+5+6)=\frac16\frac{6\cdot 7}{2}=\frac72$$ And doubling it gives the expected value for two dices to be $7$. I would expected that in the question case the expected value to be a little higher, but I don't know how to start calculating it.

I don't need a full solution, only some good hints that will help me to calculate it later.

Edit. I think I have gotten an idea to draw a matrix.

$$\begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{ x } & 1 &2 &3 &4 &5 &6 \\ \hline \hline \hline 1 &4 &3 &4 &5 &6 &7 \\ \hline 2& 3 & 8 &5 &6 &7&8 \\ \hline 3& 4 &5 &12 &7 &8&9\\ \hline 4 &5 &6 &7&16&9&10 \\ \hline 5 &6 &7&8&9&20&11 \\ \hline 6&7&8&9&10&11&24 \\ \hline \end{array}$$ And now the expected value in our case is: $$E=\frac1{36}\left(l1+l2+l3+l4+l5+l6\right)$$ Where $l_k$ is the sum of the numbers in $l_k$. This gives: $$E=\frac1{36}\left(29+37+45+53+61+69\right)=\frac{294}{36}=8.1(6)$$ Is this fine?

Answer 1

The probability of throwing two $1$s, two $2$s, two $3$s, two $4$s, two $5$s or two $6$s all equal $\frac{1}{36}$ each. In these special cases, we must add $2$, $4$, $6$, $8$, $10$ and $12$ to the sum respectively. We thus find:

$$E(X) = 7 + \frac{2 + 4 + 6 + 8 + 10 + 12}{36} \approx 8.17$$

Answer 2

Linearity of expectation works. Each of the die is worth $1$ with probability $\frac 16 \times \frac 56$, worth $2$ with probability $\frac 16 \times \frac56+\left(\frac 16\right)^2$ and so on Keep in mind that it might have value $8,10,12$ as well as the usual.

Answer 3

Given that there are two dice with 6 sidesand we are considering all possible pairs of outcomes, there will be a total of 36 possible sum outcomes ranging from 2 to 12. You can find the average of these outcomes by $$(2 + 3 + 3 +4 + 4 + 4 +...+12)/36$$ To get the proper average with the new weighted outcome you could simply multiply by 36 (to just get the sum of the outcomes) add the sum of the like pairs in (doubling their original weight), then divide by 36 (since the number of actual outcomes is still 36) again. However, we can do the same thing in less steps by using the linearity of expectation.