A problem of points An urn contains $c$ elements of 3 different types: There are $\alpha>0$ elements of type $A$, $\beta>0$ elements of type $B$ and $\gamma>0$ elements of type $G$, and $c=\alpha+\beta+\gamma$.
Alex bet $S$ dollars to get, in $n>0$ independent trials (i.e. with replacement), at least one element of kind $A$. Therefore, he can win the game with probability $P(Alex)=1-\left(\frac{c-\alpha}{c}\right)^n$.
Bart bet $S$ dollars not to get, in $n>0$ independent trials (i.e. with replacement), any element of kind $B$. Therefore, he can win the game with probability $P(Bart)=\left(\frac{c-\beta}{c}\right)^n$.
For the sake of simplicity, we assume that $\alpha,\beta,\gamma,n$ are such that $P(Bart)=P(Alex)$. Thus, since Alex and Bart have the same chance to get the $S+S$ dollars at stakes, if they both win or if they both lose, they get back their money, i.e. $S$ dollars.

But if the game is interrupted at the trial $k<n$, how should Alex and Bart  correctly divide the stakes?

The solution should be given both in case the trials are performed from the same urn, and in case the trials are performed separately from two identical urns (one for Alex, one for Bart).
 A: HINTS:
Alex gets this much money: $M_A = S \times \mathbf{1}[Alex\ wins] + S\times\mathbf{1}[Bart\ loses]$
Bart gets this much money: $M_B = S \times \mathbf{1}[Alex \ loses] + S \times \mathbf{1}[Bart \ wins]$
Here, the notation $\mathbf{1}[event]$ is the indicator random variable for the event, i.e., its value is $1$ if the event occurs, and $0$ if not.
Taking advantage of linearity of expectation: $E[M_A] = S \times P(Alex\ wins) + S \times P(Bart \ loses)$, and similarly for $E[M_B]$.  The $2S$ total should be split according to $E[M_A]$ and $E[M_B]$.
The answer now splits into 4 cases, depending on whether Alex has won already, and, whether Bart has lost already.
After $k<n$ trials, Alex could have won already, i.e. $P(Alex \ wins) = 1$, or his outcome could still be undecided in which case he still has $n-k$ trials left, so $P(Alex \ wins) = 1-\left(\frac{c-\alpha}{c}\right)^{n-k}$.
Similarly, Bart could have lost already, or he still has to survive $n-k$ trials to not lose.
In each of the 4 cases it should now be easy to calculate the answers.
Re: the version of the question where they play from the same urn - this makes their winning/losing dependent, but expectations are still linear even for dependent variables, so the answer does not change at all.
A: [1] Alex has already won, i.e. $P(Alex \ wins) = 1$, and Bart's outcome is undecided. He will lose only if, in the remaining $n-k$ trials, at least one element of kind $B$ is extracted. This can occur with probability $P(Bart \ loses)=1-\left(\frac{c-\beta}{c}\right)^{n-k}$. Therefore,
$$
E[M_A]=S\times 1+S\times \left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right],\,\,\,\,
E[M_B]=S\times \left(\frac{c-\beta}{c}\right)^{n-k}+S\times 0.
$$
[2] Alex has already won, i.e. $P(Alex \ wins) = 1$, and  Bart has already lost, i.e. $P(Bart \ loses) = 1$. Therefore,
$$
E[M_A]=S\times 1+S\times 1,\,\,\,\,
E[M_B]=S\times 0+S\times 0.
$$
[3] Alex's outcome is undecided, i.e. $P(Alex \ wins)=1-\left(\frac{c-\alpha}{c}\right)^{n-k}$. Bart's outcome is also undecided, i.e. $P(Bart \ loses)=1-\left(\frac{c-\beta}{c}\right)^{n-k}$. Therefore,
$$
E[M_A]=S\times \left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times \left[1-\left(\frac{c-\beta}{c}\right)^{n-k}\right],
$$
$$
E[M_B]=S\times \left(\frac{c-\beta}{c}\right)^{n-k}+S\times \left[\left(\frac{c-\alpha}{c}\right)^{n-k}\right].
$$
[4] Alex's outcome is undecided, i.e. $P(Alex \ wins)=1-\left(\frac{c-\alpha}{c}\right)^{n-k}$. At the same time, Bart has already lost, and $P(Bart \ loses) = 1$.  Therefore,
$$
E[M_A]=S\times \left[1-\left(\frac{c-\alpha}{c}\right)^{n-k}\right]+S\times 1,\,\,\,\,
E[M_B]=S\times 0+S\times \left(\frac{c-\alpha}{c}\right)^{n-k}.
$$
