This its very similar to another result: at the beginning, before seeing any draws, you believe that at the end of $n$ draws, every possible number of green draws is equally likely, i.e, $\int_0^1 \binom{n}{k}x^{n-k}(1-x)^{k} dx = \frac{1}{n+1}$. The proof: if you draw $n+1$ IID uniform $[0,1]$ random variables, on the one hand, the first one its equally likely to have any particular rank, so the probability it has rank $k+1$ is $\frac{1}{n+1}$. On the other hand, the probability it has rank $k+1$ is exactly the probability that $k$ of the remaining $n$ uniform random variables take a value greater than it, and $n-k$ of the remaining $n$ take a value lesser than it, which equals the integral.
This its very similar to another result: at the beginning, before seeing any draws, you believe that at the end of $n$ draws, every possible number of green draws is equally likely, i.e, $\int_0^1 \binom{n}{k}x^{n-k}(1-x)^{k} dx = \frac{1}{n+1}$. The proof: if you draw $n+1$ IID uniform $[0,1]$ random variables, on the one hand, the first one its equally likely to have any particular rank, so the probability it has rank $k+1$ is $\frac{1}{n+1}$. On the other hand, the probability it has rank $k+1$ is exactly the probability that $k$ of the remaining $n$ uniform random variables take a value greater than it, and $n-k$ of the remaining $n$ take a value lesser than it, which equals the integral.