What would be the average result given limited reroll and unique die faces?

Question

What would be the average result when rolling these 5 dice?

Each die has 6 sides and has the following distribution: 2, 1, 1, 0, 0, 0

Up to two dice may be re-rolled. To be re-rolled a die must have rolled a 0 on the initial roll.

Answer 1

The expected contribution of a die to the sum is

$$\frac{2+1+1+0+0+0}6=\frac23\;.$$

The expected contribution of a die to the sum, given that the die doesn't show $0$, is

$$ \frac{2+1+1}3=\frac43\;. $$

The probability to roll a $0$ is $\frac12$, so in your initial roll of $5$ dice you have probability

$$ \frac{\binom5n}{2^5} $$

of rolling $n$ zeros and $5-n$ non-zeros. Now you can reroll up to $2$ zeros. For $n\le2$, you end up with $n$ arbitrary dice contributing $\frac23$ each and $5-n$ non-zero dice contributing $\frac43$ each. For $n\gt2$ you end up with $2$ arbitrary dice, $n-2$ zero dice and $5-n$ non-zero dice. Thus the expected sum is

\begin{eqnarray*} &&\frac1{2^5}\left(\sum_{n=0}^2\binom5n\left(n\cdot\frac23+(5-n)\cdot\frac43\right)+\sum_{n=3}^5\binom5n\left(2\cdot\frac23+(5-n)\cdot\frac43\right)\right) \\ &=& \frac1{3\cdot2^5}(5\cdot4+5(1\cdot2+4\cdot4)+10(2\cdot2+3\cdot4)+10(2\cdot2+2\cdot4)+5(2\cdot2+1\cdot4)+2\cdot2) \\ &=& \frac{20+90+160+120+40+4}{96} \\ &=& \frac{217}{48} \\ &\approx& 4.52\;. \end{eqnarray*}

Answer 2

Here's a hint.

One of three (disjoint) cases happens after the first roll:

• None of the dice show zero.
• Exactly one die shows zero.
• Two or more dice show zero.

Some pieces of what you'll need to solve the problem:

• The probability of each case above occurring.
• The expected value showing on one die, given that it's non-zero.
• The expected value showing on one die, given that any face can show.

Can you take it from here?