I was stuck in tunnel vision. Due to using a closed bernouilli in RL all the time I didn't account for the fact a ws can fail even after 10 tries. You can expect , like you said in your probability calculator, that 1 passed by the x scrolls (%) 65.13, You have over 50% chance of having 1 ws passed after 7 slots but as @silv pointed out - and which i was stupid enough to not understand at the time - this is a mean and not an average. Since it is possible that a 1st ws is going to pass only after the 20th slot, 25th, 30th, ... So we can conclude that we expect to have to use 10 scrolls to pass 1 10% but the mean for x amount of tests would be 7. Basicly out of 100 people, 10 would pass on the 1st slot, 9 on 2nd, 8 on 3rd, 7 on 4th, 7 on 5th, 6 on 6th, 5 on 7th (= 51 people) . It is expected tho that someone is only gonna pass his 1st scroll on his 29th slot. Kinda insane what a new look on the thread gave me for a rather simple calculation. Sorry again for the misunderstanding.
Could someone help explain the visualization graph to me? In the explanations, how I'm understanding what it says is: If you are perfecting a 1b item, if its worth <1603m (1.6b??) perfected, then use only 30%s on it. What I don't understand is since that graph simulation is based on a 1b item, how is the end-perfected result costing between 1603m (1.6b) and 2850m (2.85b)? Another part I don't understand is why the line for 10%+ws a flat line? How I understand it, this mean that the price of one success using only 10%+ws is always 5000m (5b?) I'm not the most mathematically proficient person so maybe I am misunderstanding what is put here. Thank you for the great write up though and I hope to see it in its completion!
Hey, The worth of the item is 1b before scrolling it. Usually unscrolled. So if your clean item is worth less than 1.6b you should scroll with 30% for the first slot. The risk of destroying it will not cost as much as the WS. For the 10%. It is a flat line because on average it will take you 10 scrolls and it will never destroy your equip. To explain the opposite: The 30%'s are not flat because they destroy the item, which makes the price of getting a success based on the price of the item. Hope this clarified
Neat! I'm honestly fascinated by all the mathematical analysis a lot of you put in these guides; scrolling probability, magic/damage calculators, hp washing 101. I guess it comes as we got older and actually buckled down on look at this game in a different lense. Not like when we were just kids and teenagers LOL. I remember just trying my luck at 10%s and blasting away at a failed attempt to make a 15 wg. Just poppin in, maybe I'll relook this guide when I'm even at that point of the game where I can afford to gamble with ws.
Hello everyone, I'd like to jump in a little bit late on this, partly because your debate got me a little confused, partly because I see the problem differently. The way I see it, it doesn't matter whether you are trying to pass the 1st, 2nd, 3rd,...or last slot. To me, the question comes down to "Should I (10% + WS), or (30% + WS)?". Now, one could also add "30% alone" as part of the question, but the option of 30% alone (w/o WS) is only valid if the value of the item (before passing the additional slot) is not too high compared to the value of WS. It's also valid if you're a real gambler who has some serious balls. Assumptions: 10% price - negligible 30% price - let's say 10m WS price - 0.5b, for simplicity Also, please note that we are working in billions, (so if you see 0.51, it translates to 510m). The answer isn't that complicated when you think of it this way. The expected cost of passing an additional slot, using (10% + WS), is 5B. As someone mentioned in the comments above, the number of trials before the first success follows a Geometric distribution, with p = 0.1, and the expected number of trials = 1/p = 10. Thus, expect 10 trials (so 10 WS, each 500m). This cost remains fixed. The expected cost of passing an additional slot, using (30% + WS), varies based on the value of the item before passing that additional slot. Let C be the random variable representing that cost. Let a be the the value of the item before passing that additional slot. Now, I'd like to point out that you could do otherwise if you want to gamble, and your potential gain from it may be very tempting. That being said, probabilities don't lie. Peace
Guide is updated as of today, it only took three years to bother. Bumping in hopes of feedback and announcing that it is hopefully not a WIP anymore.
