# Symmetry argument for Pólya's urn scheme

In case of Pólya's urn model, an urn initially contains $r$ red balls and $b$ blue balls. At each stage a ball is randomly selected from the urn and is then returned along with $m$ other balls of the same color. How to prove that at stage $n$ a red ball is selected with probability $\frac{r}{b+r}$ using symmetry argument. (Note that it is same as the probability at stage 1).

• any solution hint on it ? – RIchard Williams Apr 26 '13 at 1:25

Initial case: after the first draw, either you have $r+m$ red balls and $b$ blue balls with probability $\frac{r}{r+b}$,or you have $r$ red balls and $b+m$ blue balls with probability $\frac{b}{r+b}$, so in the second draw the probability of drawing a red ball is $\frac{r}{r+b}\frac{r+m}{r+b+m} + \frac{b}{r+b}\frac{r}{r+b+m} = \frac{r}{r+b}$.
Inductive case: suppose after $k$ draws you have pattern indexed by $K_j$ with $r_{k_j}$ red balls and $b_{k_j}$ blue balls having probability $p_{k_j}$, such that $r_{k_j}+b_{k_j}=r+b+km$ and $\displaystyle \sum_j p_{k_j} \frac{r_{k_j}}{r_{k_j}+b_{k_j}} = \frac{r}{r+b}$, i.e. the probability of drawing a red ball in the $k+1^\text{th}$ draw is $\frac{r}{r+b}$, so $\sum_j p_{k_j} {r_{k_j}} = (r+b+km)\frac{r}{r+b}$ and since $\sum_j p_{k_j} =1$ you have $\sum_j p_{k_j} m=m$.
Then the probability of drawing a red ball in the $k+2^\text{th}$ draw is $$\displaystyle \frac{r}{r+b}\sum_j p_{k_j} \frac{r_{k_j}+m}{r_{k_j}+b_{k_j}+m} + \frac{b}{r+b}\sum_j p_{k_j} \frac{r_{k_j}}{r_{k_j}+b_{k_j}+m}$$ $$=\frac{r(r+b+km)\frac{r}{r+b} + rm + b(r+b+km)\frac{r}{r+b}}{(r+b)(r+b+km+m)}$$ $$=\frac{r}{r+b}.$$