I'm having trouble with a question. Say I have two urns, one with r amount of broken pieces and s "normal" pieces, and another urn with s broken pieces and r normal pieces. At every stage, I draw a piece from a urn and then return it to the same urn. If the piece is broken, the next piece I draw will come from the same urn, but if not, the next piece I draw will come from the other urn. I start by drawing a piece from the first urn (r broken pieces and s normal ones). What is the probability that the piece extracted in the nth stage is broken?
So, I made a tree diagram to understand the beginning of the sequence. In the first draw, the probability of drawing a broken piece is r/N (where N is r + s). In the second stage, the probability is (r/N)(r/N) + (s/N)(s/N). In the third, (r/N)^3 + (r/N)(s/N)^2 + (s/N)^3 + (s/N)(r/N)^2, and so on...
For the n position, I know that, regardless of the urn, I will extract a piece from a set of N pieces. Could it be that the possibility of drawing a broken piece at the nth position is just a weighted average of the possibilities of extracting a broken piece from each urn? like [r (from urn 1) + s (from urn 2)]/2*N?
Thanks!
Instead of trying to compute the probability of drawing a broken piece directly, let's try to compute the probability that we are drawing from urn $1$ or urn $2$. Let $p_n$ be the probability that we draw from urn $1$ on turn $n$, and let $q_n$ be the probability that we draw from urn $2$. Obviously, $p_n+q_n=1$. We can compute $p_{n+1}$ in terms of $p_n$ and $q_n$ by using the law of total probability. If we draw from urn $1$, then we will make the next draw from urn $1$ with probability $\frac r{r+s}$ and if we draw from urn $2$, the probability that we make the next draw from urn $1$ is $\frac r{r+s}$. Therefore,$$p_{n+1} = p_n\frac r{r+s}+q_n\frac r{r+s}=\frac r{r+s}(p_n+q_n)=\frac r{r+s},$$ and so $$q_{n+1}=\frac s{r+s}.$$
Now you should have no trouble working out the probability that we draw a broken piece.
I haven't checked your tree diagram, but you must have made a mistake somewhere, since the probability is constant.