Followup: By choosing any 25 primes (not necessarily the 25 smallest), can you achieve 11 or even 12 matching sums?

Unless I'm missing something, it's not 80, which can be formed as 2*5*8. Anyway, you have a finite number of choices at each position, so in theory you could just exhaustively calculate all 10*4*10*4*10 = 16,000 combinations (or more likely write a computer program to do so), then sort the distinct results and look for a gap.

Is the function the same for each couple, with only the two inputs differing?

As Jaap noted on a recent duplicate: If there was only one bridging domino crossing the border between such a pair, then that domino plus the rest of that border would split the rest of the board into areas of (6a+5) and (6b+5) squares (for some integers a and b), neither of which can be tiled by dominoes.

Being difficult doesn't make it a good puzzle. What's difficult isn't even realizing that "the third player plays the game" is true; it's realizing that, despite how obvious and boring that is, it is nevertheless the intended answer. Contrast (for instance): "Three men agree to take turns shooting at each other, in this order: Alan hits his target 1/3 of the time, Bob hits his target 2/3 of the time, Charlie hits his target every time. Who should Alan shoot at on his first turn?" Now that takes some actual cleverness (if you don't already know the answer from hearing it before).

If you actually had a serious answer in mind, then I suggest you just present it openly at this point. Then we can help you improve the question so that it does a better job of hinting at that answer and ruling out others.

J S-R M: Consider the first example again, except that Bob (Insane) claims Alice is guilty. In that scenario, the cops can't deduce which of Alice/Dave/Eve is the Goon (Alice can rule out herself but not Dave, and Dave can rule out himself but not Alice).

Are rot13(gur pbybef bs gur obeqref) significant?

Yeah, one of the comments on the source translates to rot13("Sebag rzcgl obk = 1 / Onpx rzcgl obk = 4"), which alas I was still unable to catch on to the pattern.