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  1. Hey guys, penguin here, I have a degree in math and decided to solve the question, what is the optimal way to refine? so i wrote two programs, outlined below, to simulate 1 million refines and what the averages of 2 different methods are the results are expected First method, use BSB/old elu on +7 to +8, use HD oridecon otherwise (solveRates), second method refine BSB -> old ori all the way through (solveRates2) solveRates() ; n counter,ln,refine,success,iteration s refine=7 f ln=1:1:1000 q:refine=10 d . i refine=7 s success=$$roll(50) . i refine=8 s success=$$roll(30) . i refine=9 s success=$$roll(19) . i 'success,refine>7 s refine=refine-1 . e i success s refine=refine+1 . i refine=7 s counter=counter+1 . s iteration=iteration+1 q counter_"|"_iteration solveRates2() ; n counter,ln,refine,success,iteration s refine=7 f ln=1:1:1000 q:refine=10 d . i refine=7 s success=$$roll(50) . i refine=8 s success=$$roll(50) . i refine=9 s success=$$roll(30) . i refine=7 s counter=counter+1 . i refine=8 s counter=counter+2 . i refine=9 s counter=counter+4 . i success s refine=refine+1 . s iteration=iteration+1 q counter_"|"_iteration ; roll(value) ; n roll s roll=$r(100) i roll<value q 1 q 0 i put these guys in a wrapper to iterate 1 million, these are the averages BsB used from solveRates method: 25.593615 BsB used from solveRates2 method: 19.328085 total iterations solveRates method: 49.350849 total iterations solveRates2 method: 7.328922 I think its right because BSB all the way through is solvable, you have (2)(1) + 2(2) + 3.3333(4) = 19.333 which is what my averages are converging to BUT WAIT, what if we used safe refines instead of BSB and old ori ALL THE WAY THROUGH??? So use HD oridecon to go higher than +7, if we drop to +6 safe refine back to +7 solveRates3() ; n counter,ln,refine,success,iteration s refine=7 f ln=1:1:1000 q:refine=10 d . i refine=7 s success=$$roll(30) . i refine=8 s success=$$roll(30) . i refine=9 s success=$$roll(19) . i 'success,refine>6 s refine=refine-1 . e i success s refine=refine+1 . i refine=6 s counter=counter+1,refine=7 . s iteration=iteration+1 q counter_"|"_iteration iterating, we get Safe Refine +7 needed: 30.947449 HD Ori needed: 67.090516 assume BsB = 10,000,000 HD Ori = 1,500,000 Safe Refine = 4,000,000 Old Ori = 4,000,000 Refine to +10 using BsB/old ori = 222,720,000 zanels to refine Refine to +10 using BsB/old ori/HD ori = trash, infinity zanel don't do it Refine to +10 using Safe Ticket/HD ori = 224,395,000 zanels to refine Takeaways: 1. BsB + old ori is better than BsB at 7 and HD ori onward, even in terms of BsB used 2. Safe Refine +7 + HD ori is actually a reasonably cost effective method to refine something to +10 3. HD ori should never be used outside of refine event 4. Optimal cost efficiency during refine event is safe to 7 to preserve +7, use HD ori to get to +9, use BsB/old to get from 9 to 10
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