Imagine a game where there are $3$ buckets ($1$, $2$, and $3$) full of an unlimited amount of prizes. You choose a bucket at random and pull out a prize - the prize you pull out has a chance of having "choose another" written on it as well, in which case you once again choose a bucket at random and pull out a prize. Any given prize has a chance of giving you the opportunity of choosing another. If you pull out a prize that doesn't say "choose another" then the game ends.
I want to know what the expected total winnings from this game is.
Say bucket $1$ gives an average prize of $x_1$, bucket $2$ an average prize of $x_2$ and bucket $3$ an average prize of $x_3$, and the corresponding probabilities of giving you another go are $p_1$, $p_2$ and $p_3$.
I thought about setting up an absorbing Markov chain with transition matrix: $$\begin{pmatrix}\frac{p_1}{3} & \frac{p_1}{3} & \frac{p_1}{3} & 1-p_1 & 0 & 0 \\ \frac{p_2}{3} & \frac{p_2}{3} & \frac{p_2}{3} & 0 & 1-p_2 & 0\\ \frac{p_3}{3} & \frac{p_3}{3} & \frac{p_3}{3} & 0 & 0 & 1-p_3 \\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix}$$
Where the states in order across the top are you pick from bucket $1$, you pick from bucket $2$, you pick from bucket $3$, you finish the game on bucket $1$, you finish the game on bucket $2$ and you finish the game on bucket $3$.
I then tried to do some stuff from the wikipedia page https://en.wikipedia.org/wiki/Absorbing_Markov_chain, but then got stuck.
The matrix $N$ there can give me the probability that I end at a particular bucket, and also the average number of steps at each bucket, but this didn't seem to help me when I tried this experiment in some Python code with actual numbers.
Any help would be appreciated.
