Arch-Glacor/Hard mode unique drop rate equation

An equation can be estimated based on the drop rates provided from Elder God Wars Drop Rates.

First, we assume that the equation that governs drop rates is linear and independent in regards to the Streak (s) and the Enrage (e).

• $P(s,e)=A\times s+B\times e+C$ Then, we can take advantage of the examples where the Streak is constant and only the Enrage value varies:

• Streak: 1, Enrage: 100 - 1/494 chance of a unique drop
• Streak: 1, Enrage: 500 - 1/166 chance of a unique drop
• Streak: 1, Enrage: 999 - 1/91 chance of a unique drop

We can solve for A (the slope) by using:

• $Slope={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$ • {\begin{alignedat}{3}Slope(e_{500},e_{100})&={\frac {{\frac {1}{166}}-{\frac {1}{494}}}{500-100}}\\&={\frac {.004}{400}}\\&=.00001\\\end{alignedat}} We can then validate the slope, by comparing the result with another example:

• {\begin{alignedat}{3}Slope(e_{999},e_{500})&={\frac {{\frac {1}{91}}-{\frac {1}{166}}}{999-500}}\\&={\frac {.005}{499}}\\&=.00001\\\end{alignedat}} With B solved for, we can then move onto solving for C. Here, we assume that the contribution from the Streak value is 0, and use the previous data to determine a rough value for C:

• $P(0,e)=0.00001\times e+C$ • Streak: 1, Enrage: 100 - 1/494 chance of a unique drop
• Streak: 1, Enrage: 500 - 1/166 chance of a unique drop
• Streak: 1, Enrage: 999 - 1/91 chance of a unique drop
• ${\frac {1}{494}}=.002024=0.00001\times 100+C$ • ${\frac {1}{166}}=.006024=0.00001\times 500+C$ • ${\frac {1}{91}}=.010989=0.00001\times 999+C$ Solving for C in each, we get C = .001 (approximately).

Now we solve for A by introducing another example with Streak and Enrage:

• Streak: 5, Enrage: 200 - 1/320 chance of a unique drop

We start by using the above equation and replacing what information is known:

• ${\begin{array}{lcl}P(s,e)&=&A\times s+0.00001\times e+.001\\{\frac {1}{320}}&=&A\times 5+0.00001\times 200+.001\\0.003125&=&A\times 5+0.003\\\end{array}}$ Solving for A, we get: A = .000025.

Thus, the equation ends up being:

• $P(s,e)=0.000025\times s+0.00001\times e+0.001$ Or with a denominator:

• $P(s,e)={\frac {25\times s+10\times e+1000}{1000000}}$ 