IronAxe
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My point is that if I'm not playing Heroes of Might and Magic II, and you're not playing Heroes of Might and Magic III, then I don't care how our strength ratings compare. We don't have a way to verify this in equal conditions (on the same map). Would a chess player be interested in the rating of checkers players if they don't play checkers themselves? No, they wouldn't...
Therefore, we can tentatively consider my described absurd situation to be fair :) If you want to prove that players in another version of the game are weaker, come to a tournament there and crush them, completing the next map in 30 days, while their leader completes it in 1000 days :)
Thereby, you'll increase your overall current rating and lower theirs :)
Added 1 minute ago
I checked it in my head... Nonsense! You can't reduce the second parameter proportionally to the first... it's wrong... there's no direct relationship because the first depends only on the leader's result, and the second depends on the number of participants...
Added 5 minutes ago
I agree, but I'm just afraid that you're leading to a comparison of players from different versions...
Added 1 minute ago
There's no problem, though - these players already have a rating, and at the moment it's the same for everyone, so this doesn't interfere with calculating the final formula... I'm just pointing out that this parameter will be dimensionless, i.e., fractions of a whole - for example, 0.75 or 1.3... And it won't be related to the nominal rating of the leaders (which could be 150 or 1500)...
Let me explain: the parameter that determines the strength (qualification) of the tournament will be calculated, for example, like this:
the ratio of the average sum of the ratings of the five strongest participants in the tournament (regardless of their placement) to the average sum of the ratings of all participants in this tournament. The higher the coefficient, the higher the tournament rating.