The Abel-and-Cain Urn Problem An urn contains three distinguishable kinds of balls, say $A,B,C$.
Abel bets to get, in $t$ trials with replacement, at least one ball of kind $A$ and at least one ball of kind $B$. 
Cain bets to get, in $t$ trials with replacement, exactly $t$ balls of kind $C$. 
We want Abel and Cain to have the same chance to win.
My solution is: No matter the number of balls of each kind in the urn, if Abel and Cain have the same chance to win at the end of the game, then it must be $t=2$. 
My reasoning is: Abel can win at any trial between $2$ and $t$, whereas Cain can possibly win only at the end of the game. Since we asked that at the end of the game Abel and Cain must have the same chance to win, then the last trial must represent the only possible success also for Abel, and this implies $t=2$.

Is this reasoning correct?

A further question, which might be a bit naive (or even silly), so please apologize me in that case:

How do we take into account (e.g. in terms of conditional probability) the fact that Cain already knows that Abel cannot win at the first trial and that Abel already knows that Cain cannot win at any trial a part the last one?

EDIT: I attach this scheme to explain the reasoning (see also the comments for further details).

Here we interpret each trial as a shot. And the probability to get a success for Abel in each trial $k$ as a target of a certain area (green targets, top scheme). The area of the $Ab_k$ targets increases as $k$ increases, and the area of the target in correspondence of $t$ is $Ab_t=p$. For Cain, there is only one target (blue target, bottom scheme), the last one, since he cannot win in the middle of the game. The area of his last target is $Ca_t=q$. 
The request is that $p=q$, in correspondence of the last trial. Now, Abel can hit a target (and therefore win the game) at any trial (a part the first one). So if the last one has the same area for Abel and Cain, there must be only one target, otherwise Abel has more chance to win.
 A: I do not understand your reasoning, and think that is not correct.
If I am wrong in this then it seems that you found a nice way to prove Fermat's last theorem.
That would be wonderful of course, but I have not much hope.

Let $a,b,c$ denote number of balls of kind $A$, $B$, $C$ respectively.
Let $A$ denote the event that after $t$ trials at least one of the selected balls will be of kind $A$.
Let $B$ denote the event that after $t$ trials at least one of the selected balls will be of kind $B$.
Let $Ab$ denote the event that Abel wins.
Let $Ca$ denote the event that Cain wins.
Then: $$P(Ab)=1-P(A^{\complement}\cup B^{\complement})=1-P(A^{\complement})-P(B^{\complement})+P(A^{\complement}\cap B^{\complement})=$$$$1-\left(\frac{b+c}{a+b+c}\right)^t-\left(\frac{a+c}{a+b+c}\right)^t+P(Ca)$$
So 
$$P(Ab)=P(Ca)\iff(b+c)^t+(a+c)^t=(a+b+c)^t$$
A: If you replace the urn with a magical process that returns a ball $A$ or a ball $B$ with probability $a=b=1 - 2^{-1/3}$ each and a ball $C$ with probability $c=-1 + 2^{2/3}$ then you can copy paste your reasoning without any change. At no point does your reasoning make use of the fact that $a,b,c$ were supposed to be rationals.
However, with $t=3$ the probability that Abel wins is $1-(b+c)^3-(a+c)^3+c^3 = 1 - 1/2 - 1/2+c^3 = c^3$, which is also the probability that Cain wins.
Since your reasoning proves something false, it is invalid.
