Showing my work[]
This is for xbony2 ❣️:
Let for the chance a point of durability will be restored in a tick.
Therefore,
For a second there are 20 ticks, each of which are independent events. We are going to be focusing on 1 second, or . For all the calculations determining whether a point is regenerated in those 20 ticks, there will be twenty . To calculate the probability of the outcomes of events happening, we'd have to do Event A AND Event B AND Event C... or mathematically:
If we are looking for the probability that happens exactly once in a second (and all the other times doesn't happen exactly once):
In other words, the probability that a point is regenerated exactly once in a second (and all other times it doesn't happen, including the times where it happens more than once) is a little under 1%.
If we are looking for the probability that happens at any point in one second (and multiple times are allowed), it would be the complement of a second where no point is regenerated. So we first take the probability that all 20 ticks will not regenerate a durability point, or ; there is a 77.76% chance that no durability will be restored in a second. The events where that doesn't happen would be if any of the ticks regenerate a durability point; this can happen once, twice, thrice, as many times so long as at least one tick regenerates a point. This would be mathematically expressed as
In other words, the probability that regeneration occurs in any given second (and can happen more than once) is roughly 22.25%. --SirMoogle (talk) 22:29, 9 October 2017 (UTC)