In an urn with $b$ blue and $r$ red balls, each time (call it a trial) a ball is chosen at random and then put again in the urn along with $c$ extra balls of the same color. What is the "intuitive" (possibly symmetry) argument to prove that the number of red balls in the first $n$ trials follow an uniform distribution between $0$ and $n$ when $b=r=c$.
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1The second part of this answer also contains an explanation of this phenomenon: The Polya urn model is equivalent to uniformly picking a probability $p\in[0,1]$ and then drawing red balls with probability $p$. – joriki May 05 '13 at 17:23
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When $b=r=c$ the balls always exist in multiples of $c$, so for purposes of the probabilities we can divide through by $c$ and consider $b=r=c=1$, i.e. a Polya urn starting out with one ball of each colour.
For a symmetry argument that the number of red balls in this case follows a uniform distribution, imagine the balls in a line, with an additional white ball marking the boundary between red and blue balls. A draw corresponds to uniformly choosing one of the spots between any two adjacent balls (possibly including the white ball) at which to add another ball, with the ball's colour determined by which side of the white ball it's on. Thus we fill the space between the initial red and blue ball with $n+1$ more balls, starting with the white ball and then choosing a spot uniformly for each of the remaining $n$ balls. It's clear that this chooses a uniformly random permutation of the $n+1$ balls (this is straightforward to prove by induction), so the white ball is equally likely to end up in any of the $n+1$ positions, corresponding to the $n+1$ possible numbers of red balls.
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