For those of you into STEM. Here is a proof:
I am attempting to do statistics. In a game based around percentages I want to find the optimal one. Just like poker, or the three door and 2 goats and one car doors problem problem everyone has encountered in school.

Probabilities:
- Probability of Elite Rerolling to Vanity (P(E→V)):
If you get an Elite item, the probability that it rerolls to Vanity is simply L. Thus, P(E→V)=L

- Probability of Vanity Rerolling to Mythical (P(V→M)):
If you get a Vanity item, the probability that it rerolls to Mythical is also L. Thus, P(V→M)=L

Objectives:
- Maximize: P(E→V) which is the probability of Elite becoming Vanity.
- Minimize: P(V→M) which is the probability of Vanity becoming Mythical.
- 0≤L≤1 (Luck percentage can't be negative & does not matter if > 100%).

Goal:
Find the value of L that maximizes P(E→V) while minimizing the value of P(V→M).

Thus:
- Equal Probabilities: Since both probabilities are equal then P(E→V)=P(V→M)=L. This means that increasing L improves your chances of getting to reroll a Vanity from Elite drops but also increases the risk of losing a Vanity to a reroll to a Mythical drop.

Solution Approach:
- Utility Function: Define a utility function that weighs the benefit of getting Vanity items against the cost of losing them to Mythical rerolls. For example, if you value getting a Vanity item twice as much as you dislike losing one to Mythical, you could use a utility function like U(L)=2⋅P(E→V)−P(V→M)U(L)=2⋅P(E→V)−P(V→M)

Optimization: Find the value of L that maximizes this utility function.
For example, If a vanity is worth 100m and a mythical 10m gold then I can comfortably write a utility function that will calculate the optimal L value. Finding that L value is still impossible without the base drop rate percentages (%).


The Saving Grace
A luck percentage in the mid-range of 60%-80% is as optimal as possible and a reasonable heuristic given the equal probability of rerolling up from either Elite or Vanity

Explanation:
(Disclaimer, this is calculated without the base drop rates of each level of the loot table for the game or whatever map is being ran. It is as optimal as the available math will allow me to calculate)

A mid-range luck percentage from 60%-80% is a heuristic that attempts to balance the competing priorities between P(E→V), the probability of an Elite re-rolling to a Vanity & P(V→M), the probability of a Vanity re-rolling to a Mythical .

It isn't derived from the optimal mathematical formula due to the lack of base drop rates and also a lack of map specific loot tables. Instead, it's a practical approach that acknowledges the equal probability of re-rolling up and down while trying to maximize the benefits and minimize the risks within a reasonable range.

100% Luck is NOT OPTIMAL for re-rolling to and acquiring Vanity rarity loot.

P.S. You can't make a guide about "understanding the loot table" without some sort of proof. If nobody knows the base drop rates of the different rarity of items then nobody besides the code of the game has ever truly "understood" the loot table. STS Please release the loot table base drop percentages. Also if people are to keep spending real money on platinum to open crates I will be requesting the drop rates of those crates!