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If we are talking about opposed 1d100 rolls then:I actually always wondered what the advanced math behind -/+ 5 in a roll off is.
I know that it's higher bonus/disadvantage then is obvious. That's why a lot of systems use it as a base.
but want is the actual probability advantage of say... 20 vs 25 +d100? (Min: 21 vs 26) (Max: 120 vs 125)
at equal skill levels you have 49.5% to roll strictly better than opponent (with 49.5% chance of losing and 1% tie)
with +5 advantage chances are 54.4%|0.95%|44.65%
with +10 advantage: 59.05%|0.9%|40.05%
with +15 advantage: 63.45%|0.85%|35.65%
with +20 advantage: 67.6%|0.8%|31.6%
with +25 advantage: 71.5%|0.75%|27.75%
with +30 advantage: 75.15%|0.7%|24.15%
Larger skill gaps are unlikely to appear in the quest. Reverse numbers if we are on the losing side.
General formula is:
with +X advantage: (49.5+1.005X-X^2/200)%|(1-X/100)%|(49.5-0.995X+X^2/200) for win|tie|lose
The mechanical advantage isn't as big as can be expected (+30, which is a master-level swordsman against an untrained peasant, only results in a 75% chance of victory). The reason for that is a relatively large dispersion of a uniform distribution. Using something that generates bell-like curves (sum of several dice is the easiest example) makes modifiers much more important (compare, for example, D&D (1d20) and GURPS (3d6) systems).
On the other hand this variance is mitigated in the quest by two factors:
1) A fair share of encounters are multiple-roll affairs. If you have 75/25 chances on any given roll, you are about 90/10 to win at least 3 out of 5.
2) Oftentimes an extreme difference removes the need for a roll altogether or changes its context (killing mooks narratively or rolling against whole enemy squad instead of a single soldier)
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