Lets assume we have a weighted dice roll with sides $ \lbrace 1,2,3,4,5,6 \rbrace$ given $p(i)= p_i $ and $\sum_{i=1}^6 p_i=1$ for some $p_i \in \left(0,1\right)$. Denote by $X_i^{(t)}$ the random variables that are $1$ if event $i$ occured on the $t$-th roll and $0$ if not, for some $t \in \mathbb{N}$. Let further $S_i^{(t)} = \sum_{k=1}^t X_i^{(k)}$ be the number of times that outcome $i$ occured after $t$ rolls.
Now i want to roll that dice until $1,2,3$ appear a certain number of times without regards to the order, so for example the outcome that $1$ occured $100$ times, $2$ occured $200$ times and $3$ occured $150$ times, more precisely \begin{equation} Y := \min\lbrace t > 0 \space | \space S_{1}^{(t)} \geq 100, S_{2}^{(t)} \geq 200 , S_{3}^{(t)} \geq 150 \rbrace . \end{equation}
I am interested in the expected number of rolls for that to happen so $E[Y]$. I would like to have a theoretical expression at the end but I also want to calculate it directly or at least approximate it to a certain degree.
My attempt was to look at the $P(Y = t)$, which I broke down recursively and I ended up with a mess of a formula, which theoretically gives me a way to calculate it iteratively for each $t$ but I ended up realizing that especially for large $t$ calculating the faculties $t!$, which appeared in my formula over and over would be unfeasible.
I know that $X_i^{t}$ are Bernoulli distributed, so i tried to consider an elegant way to maybe boil down the expectation of $Y$ to some formula depending on the expectations of $X_i^{t}$, which I can calculate but I was not really successful yet. I feel like I am missing something obvious that brings about an elegant solution.
Thanks for your help in advance!
Edit: I have now thought of maybe modeling the problem as a markov chain with absorbance state $(0,100,200,150)$, the involving matrices however would be very tedious to calculate, so it does not really solve the problem for now.