Probability of winning 2nd trial if 1st trial is different? My company is running a contest right now where a winner is chosen from two pools: Pool A might have $n$ entries and Pool B might have $m$ entries. One person is chosen from Pool A and one person is chosen from Pool B, and then one is chosen from either of those two people.
For sake of example, say $n = 20$, $m = 5$. How do I determine the final probability of winning the entire contest? It's not just a $1/25$ chance, is it? Pool A winner has a $1/20$ chance, Pool B winner has a $1/5$ chance, and then the winner between the pool winners is $1/2$, how do I find the probability of the winner between the two pools?
Is it perhaps if you were in Pool A, you have a $(1/20 * 1/2)$ or $(1/5 * 1/2)$ for Pool B?
 A: For someone in group A to win, they have to be drawn, which has chance $\frac 1n$, then be chosen, which has chance $\frac 12$. The chance of both events is the product. In your example, people in group B have a much higher chance than people in group A.
A: I interpret the question to be:  Given that you select a person uniformly at random from the pool of $n+m$ candidates, what is the probability that your selected person will win.
The eventual winner has to win two contests:  the one for their pool, and then the over all contest.  Specifically, there are two independent paths to victory:
Path I:  Be in group $A$ (probability = $\frac n{n+m}$). Win group $A$ (probability $\frac 1n$).  Win the final contest (probability =$\frac 12$).  Thus $$\frac n{n+m}\times \frac 1n\times \frac 12=\frac 1{2(n+m)}$$
Path II:  Be in group $B$ (probability = $\frac m{n+m}$). Win group $B$ (probability $\frac 1m$).  Win the final contest (probability =$\frac 12$).  Thus $$\frac m{n+m}\times \frac 1m\times \frac 12=\frac 1{2(n+m)}$$
Thus, while it is certainly true that a person in the smaller group has a higher chance of winning, a randomly selected person has a much lower probability of being from the smaller group...and these two effects exactly cancel out.  In the end, the answer really is just $\boxed {\frac 1{n+m}}$
A: If your question is "how do I find the probability of the winner between the two pools?", the answer is simply $\frac12$, as the winner is decided between $2$ people, one each from pool $A$ and pool $B.$
If your question, say, is what is P(a particular person from pool $A$ wins), $Pr=\frac1{20}\cdot\frac12$ for the example with $n=20, m=5$
and P(a particular person from pool $B$ wins) $=\frac15\cdot\frac12$,
as you have computed at the end of your question. 
