this post was submitted on 01 Dec 2024
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[–] Sas@beehaw.org 2 points 3 weeks ago (1 children)

I feel they might've left something out. If you're at base value still an additive 100% increase (1+1=2) is better than a multiplicative 25% (1×1.25=1.25) increase but in games where bonuses stack another additive 100% increase would raise the effective value by 50% instead (1+1+1=3) whereas another multiplicative 25% would still raise the total by that much (1×1.25×1.25=1.56) so if you're stacking a lot of bonuses, eventually the multplicative ones are more effective. As for how many steps it would take to be equal in our example... 1+1×X=1×1.25^X I'm not gonna do this in my bed on my phone but that equation should already tell you that the right side grows faster when X -> infinity

[–] sukhmel@programming.dev 2 points 3 weeks ago

It'll become greater after 12 applications:

  1. For 11 times 1.25¹¹ ≈ 11.64 < 12 = 1+ 1×11
  2. For 12 times 1.25¹² ≈ 14.55 > 13 = 1 + 1×12

There's no need for a precise solution since it's integers anyway.