Magic dice question.

Question

A pair of dice are magic. When they are rolled together, then half the time, they behave like a normal pair of dice, but the other half of the time they miraculously roll a double ( equally likely to be any of the six possible doubles). What is the expected value of the product of the two scores when the pair of dice are rolled?

Answer 1

HINT

There are 2 cases, on average with equal weight:

1. Magic happens: then there are 6 outcomes with equal probability, can you compute the expected value for this case?
2. No magic happens: then you have regular dice, can you compute the expected value for this case?
3. Average the expectations for (1) and (2).

Answer 2

Think of the sample space as $(M,n,m)$ where $M=0$ if the dice behave normally and $M=1$ otherwise. $n,m \in \{1,...,6\}$. $M, n,m $ are independent and uniformly distributed as appropriate. Hence $P[(M,n,m) ] = {1 \over 2 \cdot 6^2}$.

Define $f(M,n,m) = \begin{cases} nm,& M=0 \\ n^2, & M=1\end{cases}$.

Then $E[f] = {1 \over 2 \cdot 6^2} (\sum_n \sum_m nm + 6 \sum_n n^2) = {1 \over 2 \cdot 6^2} ((\sum_n n)^2 + 6 \sum_n n^2) = {329 \over 24} \approx 13.708$.

Answer 3

The easiest way to calculate the 36 cross sums (1/2) the problem is to note that

$$(1 + 2 + 3 + 4 + 5 + 6) \times (1 + 2 + 3 + 4 + 5 + 6) = 441.$$

Therefore, 1/2 of the time, the expected product is

$$\frac{441}{36}.$$

$$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 91.$$

Therefore, 1/2 of the time, the expected product is

$$\frac{91}{6} = \frac{546}{36}.$$

$$\left(\frac{1}{2}\right) \times \left(\frac{441}{36} + \frac{546}{36}\right) ~=~ \frac{987}{72} ~\approx~ 13.708.$$