$\textbf{Animal Kingdom}$
In an animal kingdom, there are $n$ carnivores and $m$ herbivores. When two herbivores meet, nothing will happen. When two carnivores meet, both will die. If one herbivore meets one carnivore, the herbivore will die. All such meets can only happen between two animals. All living animals will meet another one sooner or later. If a new animal, either a carnivore or a herbivore, enters this kingdom, what is its probability of survival?
$\textbf{Answer}$
The new comer's chance of survival is the same regardless of its kind. If $n$ is odd, then there is no chance for it to survive. Otherwise, if $n$ is even, then it has a survival probability of $\frac{1}{n+1}$.
$\textbf{Analysis}$
Firstly, the existence and the number of herbivores do not matter because meeting herbivores does not kill the new comer. So let's ignore them, i.e. we assume there are only $n$ carnivores in the kingdom. Secondly, when the new comer meets a carnival, it will always die. Therefore the new comer's kind does not matter, either.
This also means that the new comer will survive only if it never meets a carnival. This is impossible when $n$ is odd because in this case there is always at least one carnival left to meet this new comer. When $n$ is even, we have totally $(n+1)$ animals. Let's imagine that we line them up randomly and pair them up. Meeting starts from the front. Apparently, only the last one can survive. This means that the new comer must occupy the last position in order to survive. Hence the probability is $\frac{1}{n+1}$.