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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!

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    What I would do is pretend the blue and green balls are the same color. You are then working in a two color urn but starting with one red and two bleen balls. Find the distribution of red balls following the two color approaches. By symmetry the distribution is the same for all three colors. In particular the mean of the distribution is $\frac {n+3}3$ after $n$ draws. – Ross Millikan Jan 28 '21 at 19:51

2 Answers2

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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 $$n+2-r=(n+1)+2-(r+1).$$

  • Could please you explain how you get the $(n+1)(n+2)$ on the bottom? I'm a bit confused there. –  Jan 29 '21 at 15:32
  • Sure. The probabilities are proportional to $n+1,n, ..., 1$. So, since the actual probabilities add up to $1$, I've just divided by $\sum_{i=1}^{n+1}i$. I hope that's clear now. –  Jan 29 '21 at 15:36
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The existing answer shows by calculation that the probability for $r$ red balls after $n$ draws is proportional to $n+2-r$, one less than the number of balls that are not red. Since this is the number of different states (indexed by the counts for all three colours) that are accessible after $n$ draws and contain $r$ red balls, this suggests that the urn has the same property as the $2$-colour urn, that all accessible states are equiprobable.

This is indeed the case for any number of colours. Consider an urn with $1$ ball of each of $c$ colours, with a ball of the colour drawn added after each draw. It’s straightforward to prove by induction that all states accessible after $n$ draws are equiprobable.

This is clearly the case for $n=0$. So assume that it’s true for some $n$. Consider some state $(n_1,\ldots,n_c)$ that’s accessible after $n+1$ draws, i.e. $\sum_in_i=n+c+1$ and $n_i\ge1$. It can arise from $c$ different states, which by the induction hypothesis all have the same probability $p$, and the probability for this to occur is

$$ \sum_ip\cdot\frac{n_i-1}{n+c}=p\cdot\frac{\sum_in_i-c}{n+c}=p\cdot\frac{n+1}{n+c}\;, $$

independent of the state. Thus all $\binom{n+c-1}{c-1}$ accessible states have the same probability $\binom{n+c-1}{c-1}^{-1}$.

The marginal probability for there to be $r$ balls of a particular colour, which allows for $\binom{n+c-r-1}{c-2}$ different states, is

$$ \frac{\binom{n+c-r-1}{c-2}}{\binom{n+c-1}{c-1}}\;. $$

For $c=3$, this is

$$ \frac{n+2-r}{\binom{n+2}2}\;, $$

the result derived in the existing answer.

joriki
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