Expected number of rolls until lcm is greater than $2000$? 
You continually roll a fair $10$ sided dice. What is the expected number of rolls until the lowest common multiple of all numbers that have appeared is greater than $2000$?

The primes in the numbers $1$ to $10$ are $2,3,5,7$. The lowest common multiple of these numbers  is $210$. However, different numbers could come in different frequencies (one $9$ is worth two $3$s). How can you deal with this?
I am happy for solutions with approximate answers. The true value by simluation is around $18.8$.
 A: (sketch of the method)
Best to proceed by states.
We remark that $$\text{lcm}(2,3,4,5,6,7,8,9,10)=2^3\times 3^2\times 5\times 7=2520$$
is so close to $2000$ that no divisor of it will suffice.  It follows that you need to see the $8$, the $9$, the $7$ and either a $5$ or a $10$.
Introduce states $(a,b,c,d)$ according to whether or not the relevant prime power, $(2^3,3^2,5,7)$, has been fully seen.  Here $a,b,c,d\in \{0,1\}$ and, $(1,1,0,0)$, say, means that you have seen $8,9$ but none of $5,7,10$.  We remark that there are $16$ states which we can number according to the binary representations of the numbers $0,\cdots, 15$ in an obvious way.  Since $14 =1110_2$  we'd have state $14$ be $(1,1,1,0)$, meaning that you have only to see the $7$.  Similarly, the state $8=(1,0,0,0)$ would mean that you had seen the $8$ roll but sill needed the $9$, the $7$ and either the $5$ or the $10$.
Let $E_i$ be the expected number of tosses it will take given that you are starting in state $i$.  Of course, the answer to the question is $E_0$.
We see that $E_{15}=0$ since state $15$ means you have already seen enough tosses.
$E_{14}=10$ since you expect it to take $10$ tosses to see the $7$.
$E_{13}=5$ since you expect it to take $5$ tosses to see either the $5$ or the $10$.
For $E_{12}=E_{(1,1,0,0)}$ we note that $$E_{12}=\frac 7{10}\times (E_{12}+1)+\frac 2{10}\times (E_{14}+1)+\frac 1{10}\times (E_{13}+1)\implies$$ $$\implies E_{12}=\frac {35}3$$
Continue in this manner.  As you see, the computations are tedious but not difficult.  They are also a bit error prone, so I suggest simulation to confirm the result.
