Betting on the outcome of a semidecidable computation If my neighbor Colin insists that

Bruce Willis will be the winner of the 2024 U.S. presidential election

we can bet on it, and (barring some very strange outcomes) there is some well-defined future time, no later than 20 January 2025, at which the bet will have matured, the outcome will be known, and the loser will be obliged to pay the winner.
But suppose Colin insists instead that

No woman will ever be elected president of the United States.

Here if a woman is elected, I have clearly won the bet, but Colin can never win, because I can always claim that a woman might be elected at some future time.  This is clearly unfair to Colin because I never have to pay him anything.
My question is:

Is there a fair and unbiased way to administer a bet like this, one in which the victory condition can be decided in only one direction?

My idea is that one can naturally think about decomposing the bet into multiple sub-bets: instead of making a single bet of value $x$ on “never”, we might make a series of bets on the 2024 election ($x_1$), the 2028 election ($x_2$), and so on, whose values sum to $x$.
Then in January 2025, if a woman has won the election, Colin immediately pays me the full value $x$.  Otherwise I pay Colin $x_1$ and we meet again four years later to resolve the next bet: if a woman has won, Colin refunds the  $x_1$  I erroneously gave him in 2025, and pays me an additional amount of $x$.  But if not, I pay Colin an additional $x_2$ and we part ways for another four years.
Or perhaps the right way to do it is that if a woman wins in 2024 Colin pays me only $x_1$, and his future payments to me of $x_2, x_3, \ldots$ become foregone conclusions, but are still not collectible by me until the relevant elections have occurred?
I am curious about how one might compute an equitable sequence $x_1, x_2,\dots$, taking into account such matters as future value discounting.
Note that the answer to the question could be useful even for bets that are guaranteed to end, if they do so in the sufficiently far future.  Suppose Colin and I want to bet whether a woman will be elected President of the United States in his lifetime.  We might want to adopt a protocol of this sort, rather than waiting until Colin's death to settle the bet all at once.
To avoid edge cases and the complications of the real world, let's consider this abstract example:  Colin and I agree on a specific Turing machine $\mathcal M$.  I assert that $\mathcal M$ will halt when started with an empty tape; Colin asserts that it never halts.  We want to bet €1 on the outcome.
Can this bet be made fairly?
Consider the following protocol:   Run $\mathcal M$ one step at a time.  If it has not halted after step $n$, I pay Colin $€2^{-n}$.  If it does halt, Colin refunds my payments so far and pays over an additional €1.
This seems unfair to me because I will have very shortly paid Colin almost the whole €1, even if $\mathcal M$ halts relatively quickly, say after only 1000 steps.  Can this be repaired by requiring Colin to pay me back with interest if $\mathcal M$ halts?
Mathematically this may be a somewhat nebulous question, but I guess that in the domain of financial instruments it is well-studied.
 A: You could pay Colin \$1 right now, and agree on an interest rate (e.g. 2% or the inflation rate). If a woman is then elected, Colin pays you \$2 times the appropriate amount of interest.
A: Let the amount Colin owes you if a woman wins be $a$. (I'm conveniently skipping the dollar sign throughout the answer).  I will first express your strategy and its pros and cons.
You need to pay Colin a certain amount every election a man wins. Now, mathematically, say a woman wins an election at infinity. This implies Colin will have to pay you $a$ at infinity. You would have also paid him a certain amount by that time. The amounts should be equal, since Colin was wrong, yet no woman won the election in a finite time. Call the amount you pay Colin after the $i$th election (if a man wins) $x_i$. We have:
$$\sum_{i = 1}^{\infty} x_i = a$$ Set $x_i = ap_i$:
$$\sum_{i = 1}^{\infty} ap_i = a$$
We get a nice expression:
$$\sum_{i = 1}^{\infty} p_i = 1$$ Now, we can model the infinite series so that the sum is 1. An immediate thought would be to set $$p_i = {1\over 2^i}$$ But this seems wrong, this is a decreasing sequence. As more elections pass without a woman winning, Colin's words feel truer. So, we need to find an increasing sequence that converges, which is not possible (at least from whatever I know). I am compelled to leave this hindrance for later. 
Suppose a woman wins the $n$th election. You have already paid all the partial sums till the $n-1$th election, so you want that back along with $a$. So, Colin will pay you
$$\left(\sum_{i = 1}^{n-1} ap_i\right) + a$$ which is just
$$a\left(\frac{2^{n+1}-1}{2^n}\right)$$ I think this part of the situation works out right. As far as I understand, you have discussed this in your post.
Here come our problems - the bet is extremely unfair for Colin since the amount is decreasing (and moreover never gets to $a$ in a finite amount of time, but he has to pay you the full amount $a$ as soon as a woman wins). But if a woman winning an election wouldn't be so unlikely in the US, Colin would have had an even unfair bet (The probability that no woman has won by the $n$th election is just $P = (\frac12)^n$, where $P \to 0$ as $n \to \infty$). So, we need another strategy.
You and Colin are both going to die someday (I refuted this fact without stating in the above part). So, you and Colin agree on a certain number of elections (say 9). You agree to give a certain $x_i$ such that the $x_i$'s sum to $a$ in the decided (9 elections) amount of time. The $x_i$'s can be increasing. If a woman wins before this duration is over, Colin returns everything you had paid him till now along with $a$. Again, this situation seems fair. Without loss of generality (I'm joking here!), let us assume Colin is honest, i.e, he tells you if he is going to invest the $a$, at a certain rate. Then, he has to give you the proceeds too, if a woman wins within the stipulated duration (since I'm assuming you would also have invested the money had you been not deprived of it). However, if Colin's words remain true throughout the stipulated time period, we tweak the conditions. If a woman wins after the stipulated time period, Colin gets to keep the proceeds (I think it's fair this way, you can nullify this condition if you want). After that time period, we do a different process(which continues only till no woman wins). Let us say that today, the results of the last election under the time period were declared, and you give Colin the due amount, that make his total winnings $a$. For the next election (if a man wins), you give Colin $$a \left(\frac{\frac12+\frac23}{2}\right) = \frac{7a}{12}$$ money. Colin only returns the amount after the next (meaning after the 4-year term), keeping the interest. In general, for every $n$th election after the time period, if $n$ is odd, and a man has won that year, you give Colin
$$\frac a2\left(\frac{n}{n+1}+\frac{n+1}{n+2}\right) \space\space\space... \space\space(*)$$ which Colin returns after the $n+1$th election (if a man has won), keeping the proceeds. However, at any time if a woman wins, Colin returns all the money he's taken from you (in regards to this bet), except the interest received in the stipulated number of elections (9 elections in our example).
This is my strategy. Not that robust, but it at least guarantees that Colin will feel that the bet was fair. You can tweak the $(*)$ a little bit, and adjust the number of (9 was only an example) elections (based on $a$, the expected profit Colin gets by investing, your expected life etc.) such that you both agree.
I agree that this does not seem to be a perfect answer, yet I doubt any purely mathematical answer would be fair for Colin. Please feel free to add suggestions.
$\color{grey}{\text{(This has become my longest answer on MathSE yet).}}$
