If many of your input variables only have two digits of accuracy the end result shouldn't have four digits of accuracy.

Almost-inaudibly, whispering in a small corner of the room while scribbling in a notebook that the teacher is totally stupid while said teacher says something similar to the quote above:

under the assumption that all variables have equivalent ratios of weight to the final result and that the probability distribution of the randomness is evenly distributed across sub-digits of inaccuracy, along with a few other invisible assumptions about the nature of the data and calculations

Yep, that's me in high school.

In your example, the cited specific case only means that the final accuracy to be calculated is +- 0.01 individual ship relevance, which means that at the worst this one instance, by the standard half-the-last-significant-digit rule of thumb (which is not by any means an inherent property of uncertainties) means that there's +- 0.5% * 1 ship variance over the 542 : 2 000 000 ratio for this particular error margin.

Note also that "24% weight of the relevance of 1 ship in the odds" translates very poorly in digit-accuracies to "3745 : 1", because 3745:1 is also 0.026695141484249866524292578750667% chance, which is a shitton of digits of accuracy, and is also 111010100001 : 1, which is 12 digits of accuracy, and is also (...) *

As you can see, the "digits of accuracy" heuristic fails extremely hard when you convert between different ways to represent data. Which is exactly what happened several times in the steel-vulcan's calculations.

Moral of the story: Don't work with "digits of accuracy", just memorize your probability distribution functions over uncertainty and maximal variances and integrate all your variables with uncertainty margins and weights during renormalization, like a real Vulcan would.

Edit: * (Oh, and it's also 320 in base-35, so that's exactly two significant digits. Problem solved, move along.)