I've got a question on a martingale that i've been stuck on for a good while.
An urn contains $n$ white and $n$ black balls. We draw them one by one without replacement. We pay £1 for any black ball drawn but receive £1 for any white one. Denote by $ X_i $ our money after the $i$-th draw $X_0 = 0$. Let
$$ Y_i = \frac{X_i}{2n - i} \quad (1 \leq i \leq 2n-1), $$
We want to show that $Y_i$ is martingale.
Now I've been able to prove so far that $X_i$ is a martingale via the property when we take the separate events of probabilities of picking a black ball $ \frac{n - B_i}{2n - i} $ and white, but it looks like I'm left with $\frac{X_i}{2n - (i + 1)}$ when all is inserted into $Y_i$. I'm not sure if I'm missing anything, but I don't know where to go with this or if there's some kind of logic i'm missing??