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Math Puzzle: Rich Get Richer?

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Today I'm going to talk about a math problem whose answer is simultaneously simple and unintuitive. Here is the problem statement:

You have a urn that has $1$ red ball and $1$ blue ball. You repeatedly remove one ball from the urn randomly, then put two balls of the same color back into the urn. How does this system behave over time? Specifically, what is the probability that at some point the urn will contain exactly $r$ red balls and $b$ blue balls?

If you want to see my solution, go ahead and expand the solution tab. Otherwise, you can try to figure it out for yourself. Also, if you have other interesting questions you want to ask about this system, feel free to post them in the comments below.

Intuitively, this seems like a "rich get richer" problem", in that the long term behavior of the system depends on how "lucky" we are in the early stages (i.e., how many red and blue balls we pick). For example, if we pick mostly red balls, then we'd expect the long term behavior to be overwhelmed by red balls, and the same goes for the blue balls. However, if the proportion starts off nearly equal for each color, we'd expect it to remain nearly equal long term. That was my initial intuition, but maybe yours is different. Anyway, I will now show that the initial intuition contradicts the mathematics.

First notice that one way to get to $r$ red balls and $b$ blue balls is to pick a red ball $r-1$ times then a blue ball $b-1$ times. Picking $r-1$ red's in a row occurs with probability:
$$ \frac{1}{2} \cdot \frac{2}{3} \dots \frac{r-1}{r} $$
and following that up with $b-1$ blues in a row occurs with probability: $$ \frac{1}{r+1} \cdot \frac{2}{r+2} \dots \frac{b-1}{r+b-1} $$
so the overall probability is the product of these two probabilities. Now notice that no matter what sequence of balls we pick, the unreduced denominator will always be the same: $ (r+b-1)! $. Further, if we end up with the same number of red and blue balls, the numerator will be the same (since multiplication is a commutative operation) -- only the order will be different. This means that every way to end up with $r$ red balls and $b$ blue balls is equally likely, and each path has probability:
$$ \frac{(r-1)! (b-1)!}{(r+b-1)!} $$
There are $ \binom{r+b-2}{r-1} = \frac{(r+b-2)!}{(r-1)! (b-1)!} $ such paths, so the marginalized probability of ending up with $r$ red balls and $b$ blue balls is
$$ \frac{1}{r+b-1} $$
This is the answer to the original question, and it means is that every urn with $n$ balls is equally likely.

The answer to this problem turned out to be disappointingly simple. I typically like to solve problems with simple/elegant closed form solutions (e.g., involving e), but this is a little bit too simple I think. I think it was worth blogging about because the math contradicts the intuition (at least for me). Note that you could also solve this problem using induction on the number of balls in the urn, but that would require you to "assume" that every urn will $n$ balls is equally likely, and that would be an unintuitive assumption to make without working out some examples or already knowing the answer.

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