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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.

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Assuming that buckets are chosen uniformly at random, then we can say:

Let $W$ be the expected total winnings

By the law of total expectation, conditioning on the first step of your process, we have

$W = \frac{1}{3}(x_1 + p_1W) + \frac{1}{3}(x_2 + p_2W)+ \frac{1}{3}(x_3 + p_3 W)$

Rearranging for $W$ it follows that $W = \frac{x_1 + x_2 + x_3}{3 - p_1 - p_2 - p_3}$

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  • $\begingroup$ Beautiful. Thank you. $\endgroup$ Feb 3, 2021 at 12:18

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