Is it possible to extend Polya's Urn problem to balls with $3$ different colors instead of just $2$?
ie. An urn contains $1$ red, $1$ blue, and $1$ green ball. At each turn you draw one ball and put it back along with another ball of the same color. Let $X_R$ be the number of red balls in the urn after turn $n$. What is the distribution for $X_R$?
From this post, in a regular Polya's urn with two colors, the number of red balls in the urn after $n$ draws is uniform over $\{1, ..., n+1\}$ ie: $$P(X_R = k) = \frac{1}{n+2} \text{ for } 1 \le k \le n+1$$
How would I need to change this for my scenario in which there are three colors? Any help would be great!
After $n$ turns the probability distribution for $1,2,3,...,n+1$ red balls is $$\frac{2(n+1)}{(n+1)(n+2)},...,\frac{6}{(n+1)(n+2)},\frac{4}{(n+1)(n+2)},\frac{2}{(n+1)(n+2)}.$$ In other words, the probability of there being $r$ red balls is proportional to $n+2-r$.
This is straightforward to prove by induction. For the inductive step, consider the probability of having $r+1$ balls after $n+1$ turns.
This is the probability of having $r$ balls after $n$ turns multiplied by the probability a red ball is then chosen, $\frac{r}{n+3}$, plus the probability of having $r+1$ balls after $n$ turns multiplied by the probability a red ball is not then chosen, $\frac{n+2-r}{n+3}$.
The probability is therefore proportional to $$r(n+2-r)+(n+2-r)(n+1-r)=(n+2-r)(n+1)$$ and is thus proportional to $$n+2-r=(n+1)+2-(r+1).$$