There is a basket with balls. There are $M$ white balls and $N-M$ black balls. We take out all balls one after another. Find probability of $k$th ball to be white.
Okay, let $A_k$ - be event of $k$th ball is white. I've started: $$P(A_1)=\frac{M}{N}\\ P(A_2)=(1-\frac{M}{N})\frac{M}{N-1}+\frac{M}{N}\frac{M-1}{N-1}=\frac{M}{N}$$ Wow, I think $P(A_k)=\frac{M}{N}$ for any $k$. That is the problem. I've tried to use induction: $$P(A_k)=(1-\frac{M}{N})^{k-1}\frac{M}{N-k+1}+(1-\frac{M}{N})^{k-2}\frac{M}{N}\frac{M-1}{N-k+1}+...+(\frac{M}{N})^{k-1}\frac{M-k+1}{N-k+1}$$
I see there something like $((N-M)+M)^{k-1}$, but there are no binomial coefficents.
So, what should I do?
You are correct in noticing that the probability $P(A_k)$ will be $\frac{M}{N}$ for all $k$. Instead of proving this via induction, let us instead prove combinatorially:
Instead of picking out balls and placing them into locations labeled "first", "second", "third", ... in the same order of "first", "second", "third",... and asking about the $k^{th}$ location housing a white ball...
we instead pick out the balls and place them into the the locations in the order "$k^{th}$", "first", "second", ..., "$(k-1)^{th}$", "$(k+1)^{th}$",...
What we have done is create a bijection between the two scenarios. As such, the probability that the $k^{th}$ location has a white ball is the same as the probability that the first space has a white ball, which is as already noted equal to $\frac{M}{N}$.
A more direct proof where $M>1$ is arbitrary becomes rather infeasible to attempt. Thankfully, in combinatorics, proofs like these where we appeal to the intuitive nature of the problem are incredibly common and well received. Read more at wiki - Combinatorial Proofs