If k is tiny, this is only a minute chance of free money. I agree that it seems absurd to turn down that deal, but if the only cost of solving Pascal's mugger is that we avoid advantageous lotteries with such minute payoffs, it seems a cost worth paying.

But recall - k is not a constant, it is a function of how often the "situation" is repeated. In this context, "repeated situation" means another lottery with larger standard deviation. I'd guess I've faced over a million implicit lotteries with SD higher than k = 0.1 in my life so far.

We can even get more subtle about the counting. For any SD we have faced that is n times greater than the SD of this lottery, we add n to 1/k.

In that setup, it may be impossible for you to actually propose that free money deal to me (I'll have to check the maths - it certainly is impossible if we add n^3 to 1/k). Basically, the problem is that k depends on the SD, and the SD depends on k. As you diminish the SD to catch up with k, you further decrease k, and hence p, and hence the SD, and hence k, etc...

Interesting example, though; and I'll try and actually formalise an example of a sensible "SD adjusted EU" so we can have proper debates about it.