A weird problem of probability! Consider three events $A,B,C$ such that $P(A)>0$, $P(B)>0$, and $P(C)>0$. The events are linked to each other through the constraints $P(A\cup B\cup C)=1$ and $P(A)=P(\overline{B})$.
We suppose $P(A)>P(B)$, strictly. Therefore, we expect the event $A$ to occur before the event $B$, which is more unlikely. 
However, if we require $P(A)=P(\overline{B})$, we are asking that the time we need to wait for getting a success of $\overline{B}$ (i.e. to not observe the occurrence of the event $B$) is the same we expect to wait to see a success of $A$. 

But this means that the event $B$ must have been already occurred, and this is in contradiction with the fact that $P(A)>P(B)$.

I am probably doing some conceptual mistake, but I cannot see it. Moreover, what is the role of the event $C$ in all of this, since  $P(A\cup B\cup C)=1$?
This post refers to this one A problem of conditional probability

In this picture (sorry for raw format) $t_A,t_B$ represent the time we expected to wait to see the event $A$ and the event $B$, respectively (the dashed lines represent the time in which we expect the events not to happen). How does the constrain $P(A)=P(\overline{B})$ look like in this scheme?
Thanks for your suggestions!
 A: In any kind of settings where the temporal language that you are using makes sense, what you forget here (when you say "this means that the event B must have been already occurred") is that $\bar B$ could occur first, before anything else happens. Especially before $B$ happens.
Therefore there is no contradiction.
Edit: an example to clarify things. Say you roll a die with $100$ equiprobable faces again and again, and "time" corresponds to the increasing number of tries.
Say $A=$ "get a number larger than $1$" and $B=$ "get $100$", and $C$ is whatever you like. Clearly $P(A)=P(\overline B)=\frac{99}{100}$ and $P(A\cup B\cup C)=1$. 
As you roll the die (as time goes) you see that more likely than not the event $\overline B$ will occur immediately at the start of the experiment. It is wrong to think that to negate $B$ you first need to have $B$. 
A: Your story is about a Venn diagram, and nothing more. Not about events happening sooner or later, or more often, etc. Just happening or not happening is the question. It could belong to the following setup: 
Yesterday the cookie jar was full, and now half of the cookies are gone. There are three children who could have eaten from the cookies. Denote by $A$ the event that Alice has eaten some cookies, and similarly for $B$ and $C$.
The corresponding Venn diagram contains $7$ subsets with probability entries $x_k\geq 0$, whereby $\sum_{k=1}^7 x_k=1$. Furthermore there are two easily satisfiable linear conditions among the $x_k$. No big deal.
