The problem there, and the problem with Pascal's Mugging in general, is that outcomes with a tiny amount of probability dominate the decisions. A could be massively worse than B 99.99999% of the time, and still naive utility maximization says to pick B.

One way to fix it is to bound utility. But that has its own problems.

The problem with your solution is that it's not complete in the formal sense: you can only say some things are better than other things if they strictly dominate them, but if neither strictly dominates the other you can't say anything.

I would also claim that your solution doesn't satisfy framing invariants that all decision theories should arguably follow. For example, what about changing the order of the terms? Let us reframe utility as after probabilities, so we can move stuff around without changing numbers. E.g. if I say utility 5, p:.01, that really means you're getting utility 500 in that scenario, so it adds 5 total in expectation. Now, consider the following utilities:

1<2 p:.5

2<3 p:.5^2

3<4 p:.5^3

n<n+1 p:.5^n

...

etc. So if you're faced with choosing between something that gives you the left side or the right side, choose the right side.

But clearly re-arranging terms doesn't change the expected utility, since that's just the sum of all terms. So the above is equivalent to:

1>0 p:.5

2>0 p:.5^2

3>2 p:.5^3

4>3 p:.5^4

n>n-1 p:.5^n

So your solution is inconsistent if it satisfies the invariant of "moving around expected utility between outcomes doesn't change the best choice".