We have an urn that has 5 yellow balls and 3 black balls. After we choose a ball we replace it with 1 yellow and 1 black together. Let $Y_m$ be the amount of yellow balls after m draws.
Is $$X_m = \frac{Y_m}{2m+8}$$ a martingale?
So I know that at the end we will have $2m + 8$ balls. Then, $Y_m = n + 5$, which means that $$P(Y_{m+1}|Y_m=k) = \frac{n+5}{2n + 8}(k+1) + k \left(1-\frac{ n+5}{2n+8} \right)$$.
And that's where I stop and need guidance.
Define the event $E_m$ as $$E_m =\{\text{ the $m$th ball is yellow }\}.$$
Then, you can compute the conditional expectation using $$\mathbf E[X_{m+1}|X_m] = \mathbf E[X_{m+1},E_m|X_m] + \mathbf E[X_{m+1},\neg E_m|X_m].$$
Now note that $$\mathbf E[X_{m+1},E_m|X_m]=\mathbf E[X_{m+1}|E_m,X_m]P(E_m|X_m),$$ with a similar equation for $\mathbf E[X_{m+1},\neg E_m|X_m],$ where $P(E_m|X_m) = X_m$ is the probability of drawing a yellow ball at the $m$th draw.