# Expected winnings in a game with buckets of prizes

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.

Let $$W$$ be the expected total winnings
$$W = \frac{1}{3}(x_1 + p_1W) + \frac{1}{3}(x_2 + p_2W)+ \frac{1}{3}(x_3 + p_3 W)$$
Rearranging for $$W$$ it follows that $$W = \frac{x_1 + x_2 + x_3}{3 - p_1 - p_2 - p_3}$$