Roll a cube with a tetradron?

If he rolls a $1$ he is done. So, cubed numbers containing a $1$ (other than $1$ itself) won't effect the expected value. So, other than the number $1$ , we only need to consider cubed numbers that contain only the digits $2$ , $3$ , and $4$ , that are less than one million.

It turns out, only one number fits the bill, $343 = 7^3$ .

So, with only two numbers to consider, $1$ and $343$ , we have the following states to consider:

• $0 \rightarrow$ Starting state
• $3 \rightarrow$ Last number rolled was a $3$
• $4 \rightarrow$ Last two numbers were a $3$ followed by a $4$

And, let $E_n =$ Expected number of rolls to get a cube from state $n$ .

So, we get the following set of linear equations:

• $E_0 = 1 + (1/4)*E_3 + (1/2)*E_0$
• $E_1 = 0$
• $E_2 = E_0$
• $E_3 = 1 + (1/4)*E_0 +(1/4)*E_3 + (1/4)*E_4$
• $E_4 = 1 + (1/2)*E_0$

The above were derived, by considering the probability of going to each state from where they are. For example, from state $3$ ( $E_3$ ) you use up one move (the next one) for sure, plus you have a $1/4$ probability of rolling a $1$ , at which point you are done, so the expectation from there is zero, plus a $1/4$ probability of rolling a $2$ , in which case you effectively end up where you started, state $0$ ( $E_0$ ), plus a $1/4$ probability of rolling a $3$ , putting you in state $3$ ( $E_3$ ), plus a $1/4$ probability of rolling a $4$ , putting you in state $4$ ( $E_4$ ).

Solving, $E_0 = \frac{34}{9}$

$34+9 = \boxed{43}$