1. ## Rolling 5 Die

Let a fair die be rolled 5 times. Find the math expectation of the product of the outcomes.

2. Originally Posted by Aryth
Let a fair die be rolled 5 times. Find the math expectation of the product of the outcomes.
If the outcomes are 1 1 1 1 1 then the product is 1 with a probability of $\frac{1}{6^5}$.

A product of 2 could happen in 5 different ways, so the probability is $\frac{5}{6^5}$.

Same probability applies to an outcome of 3 and 5.

The largest out come is five 6s, which has a probability of $\frac{1}{6^5}$

Hope this helps.

Outcome Ways
1 1
2 5
3 5
4 5 + 10
5 5
6 5 + 10 + 10
...

3. Originally Posted by Aryth
Let a fair die be rolled 5 times. Find the math expectation of the product of the outcomes.
Hi Aryth,

Let's see if we can solve the corresponding problem for 2 dice. With luck, maybe this will show us the way to a more general solution.

To that end, let's say x and y are the numbers on the two dice. We assume each pair (x,y) has equal probability, i.e. 1/36. By definition, the expected value of their product is

$E(XY) = \sum_{x,y} xy \; p(x,y)$
$= \sum_{x=1}^6 \sum_{y=1}^6 \frac{xy}{36}$
$= \frac{1}{36} \left( \sum_{x=1}^6 x \right) \left( \sum_{y=1}^6 y \right)$
$= \frac{1}{36} \left( \frac{6 \cdot 7}{2} \right) ^2$

Maybe you can take it from there...

4. Looks like... for k dice...

$\frac{1}{6^k} \left( \frac{6 \cdot 7}{2} \right) ^k$

which also works when k=1.