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Topic | Hey math people - Reduce fail rate, or reroll failures? |
Lokarin 09/29/20 4:10:33 AM #1: | Scenario: You're playing a 2d6 game where under default conditions a 7+ is a success. Your AGI bonus adds to your roll, enemy stats reduce that roll, but basically under neutral conditions you just need a 7+ (which is about 58.3%) ... Ok, you have an option of 2 items: One cuts your enemy's bonus in half, so if the enemy was getting a +2 they only get a +1. The other just rerolls if you fail once. Or, to be more precise, your chance to hit is "2d6+Offensive Modifier >= 7+Defensive Modifier" But you are in a bad position: Making a ranged attack at 8AGI (+1) against a unit that is 4 tiles horizontally/5 tiles vertically (16+25)=41, sqrt(41)=6.4, 6.4/2 = 3.2, floored = 3. ... would have a bonus of +3. Your 2d6+1 would have to equal or beat 10. If you take the first item which cuts the enemy bonus in half it would get reduced to 1 since rounds are floored (1.5=1), or a reroll.So, which is better: 2d6+1 >= 10 with reroll, or 2d6+1 >= 8 Or ONE attempt at 41.7%, or TWO attempts at 16.7% (EDIT: I was off by 1 looking at the chart, it should be 58.3% and 27.8%) ... Under this scenario the first item is better since 41.7% is better than 30.6111% ............ --- So what's the question? When is rerolls ever better than reducing penalty? Is there an easy way to output all possible outcomes? At a range of 1.999 or less then rerolls will probably be better since the enemy bonus would always be 0 and therefor the first item would do nothing... --- "Salt cures Everything!" My YouTube: https://www.youtube.com/user/Nirakolov/videos ... Copied to Clipboard! |
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