UPDATED AS OF 01.07.2023 Hello, I started this project three years ago and didn't finish it, I decided that I wanted to finish it once and (hopefully) for all. So you can now know which scroll-combination you should use for your perfect equips. I wanted to make a guide that tells you what scrolls you should use to optimize your chances of getting a perfect item and minimizing the cost. This guide will tell you you how many 30% scrolls you should use before you 30%+WS and how many times you 30%+WS before you 10%+WS your perfect item. Spoiler: TLDR Don't 10% your item without WS, this is only applicable for low-worth items, or where 30%'s are worth way more than 10%'s. If an item is worth less than 2.857WS you should 30% it. If an item is worth between 2.857WS and 5.714WS you should 30%+WS it. If an item is worth more than 5.714WS you should 10%+WS To start with such an experiment you need to know what it will on average cost to make a scroll work on an item. When you have the cost of making one scroll work on the item, you can simply update the price of the item and apply the same calculation again. Note that this is will be the average cost, the average cost is not to be confused with the median number of tries you need to succeed a scroll. The median, means that you will have a 50% chance of succeess preceeding a specific attempt, and 50% chance of succeess on the succeeding attempts. For a 10% the median is 7, the average is still 10. Below is a list of the cost of acquiring an item scrolled once, this also serves as the updated value of that item. What will it cost to 10% an item? This will on average, cost you 10 items, and 10 scrolls. What will it cost to 10%+WS an item? This will on average cost you 1 item, 10 scrolls, and 10 WS. What will it cost to 30% an item? This will on average cost you 3.33 items, and 3.33 scrolls. What will it cost to 30%+WS an item? This will on average cost you 2.167 items, 3.33 scrolls and 3.33 WS. Spoiler: WHY 2.167? Someone can surely explain this better. Expected number of items used (E): Theres a 30% chance of passing, 35% chance of spending an item, and 35% chance of having to use another WS. So first of all there will at least be 0.35 additional items spent on average. Then there is a 70% chance of spending the expected amount of items E, due to it failing, and us having another go. We simply multiply 0.7 *E So the expected number of items used is: E = 0.35+0.7*E 0.3E = 0.35 Expected number of items = 0.35/0.3 =1.167 Then we add 1 more, because we have to pay for the initial item aswell. =2.167 I have also simulated this, and it is correct, but I'm open for better explanations. So what can we deduce from this knowledge? We can figure out when to use which scroll to have the statistics in your back, and in the long run would lead to you making the most out of your scrolling. In this ideal scenario I will neglect the price of the 10% and 30% scrolls, they are of course not irrelevant and sometimes their price will matter, but comparing them to the price of white scrolls that are 400m (as of June 2023) they are negligible. I will also not calculate prices in mesos, rather in WS as they are the leading factor of the calculations. The order in which to scroll I believe all people can easily figure out from intuition, the order is: Step 1. 30% Step 2. 30% + WS Step 3. 10% + WS Spoiler: Hidden 0’th step This step is actually to 10% without WS, but is only applicable if the the price of the item is ridiculously low. What do we want to know here? Well it is the point at which to go from step 1 to 2, and step 2 to 3. This transition is dependent on the value of the item. So with a clean item you can easily calculate its worth, it is whatever you bought it for. But what is a +1, +2 or +3 item worth? I would argue that the worth of a +x item is worth the same as the cheapest average way of obtaining it. Step 1 to 2: The cost of step 1 is 3.33*[item value] The cost of step 2 is 3.33*[WS value] + 2.167*[item value]. Let’s equal them and find the item value. (1/0.3) / (1/0.3 - 60/35) = 2.857 WS Is the value of the item below this -> step 1. Is the value of the item over this -> step 2. Step 2 to 3. The cost of step 2 is 3.33*[WS value] + 2.167*[item value]. The cost of step 3 is 1[item value] + 10 [WS value] Let’s again equal them and find the item value. (10-(1/0.3) )/ (35/30) = 5.714 WS Is the value of the item below this -> step 2. Is the value of the item over this -> step 3. Spoiler: Graph Illustrating When To Go To Next Step So this is basically it, but you also need to know the value of your item as you scroll it, because you want to 30% or 30%+WS it 1,2,or 3 times before you start doing 10%'s. I am going to show this with a simple example, which is the Red Craven. Today (30.6.2023) I found a clean RC in FM for 555m, which is the same as 555/400 = 1.39 WS. So now we have three options, are we using step 1, step 2 or step 3? What would it cost me to: 30% a clean RC into +1: 3.33 RC = 3.33 * 1.39 = 4.63 WS. 30%+WS a clean RC into +1: 2.17 RC + 3.33 WS = 2.17 * 1.39 + 3.33 = 6.34 WS 10%+WS a clean RC into +1: 1 RC + 10 WS = 1.39 + 10 = 11.39 WS So the obvious choice here is to 30% the first slot, so we are going to use step 1. What about the second slot? To figure out what to do now, you have to know the value of the +1 RC, I didnt find any in FM for reference, but as I stated earlier the value of a +1 item, should be the cheapest method of obtaining it. So the value of a +1 RC (given that the value of a clean RC is 555/400 WS) should be 4.63 WS as shown in the calculation above. Im going to do the calculation for the second slot as well: What would it cost me to: 30% a +1 RC into +2: 3.33 RC = 3.33 * 4.63= 15.42 WS 30%+WS a +1 RC into +2: 2.17 RC + 3.33 WS = 2.17 * 4.63 + 3.33 = 13.33 WS 10%+WS a +1 RC into +2: 1 RC + 10 WS = 4.63+ 10 = 14.63 WS So this is actually a close race, on average it will be the cheapest to 30%+WS it for the second slot using step 2. Since it is such a close race I would say that all of the above methods are viable, do a 30% if you feel lucky, a 30%+WS if you scroll alot and want to win in the end, and do a 10% if you want to make your own perfect weapon without the hassle of buying more clean RC's. If 30% scroll prices are much higher than 10%, you could also consider doing 10%. For the 3rd slot and onwards you should always use 10% +WS, step 3. You can of course get lucky with 30%, but in the end you will lose if you do it. I did also simulate this in a python-script, so here is my python script telling me which method to use: Keep in mind this is a simulation, some numbers may vary a tiny bit. Interestingly I have also estimated the value of a perfect RC, given that a clean RC is worth 1.39 WS. A perfect RC should be worth 63.37 WS, which given a WS price of 400m equals 25.35b. This is a fairly good estimate, knowing that perfect RC usually go for around that price, usually a bit lower. Browsing the forums today(30.06.2023) I see that there is a perfect RC up for sale at for 24b. I just saw a smega for someone selling a +2 RC for c/o 4.7b and a/w 5.5b. My estimate is 13.32ws, ws go for around 400m now, so that would be 5.3b, mentioning as a reassuring observation. Spoiler: Python Code For The People That Care Beware!! This is in no way optimized, nor spent a lot of time on. Code: import math from random import randint success = False avg = 0 number_of_tries = 50000; ws = 0 scroll = 0 item = 0 ws_price = 400 scroll_price = 0 price_30_ws = 0 price_30 = 0 price_10_ws = 0 price_10 = 0 item_price = 600 print("item_price", "\t10% \t\t 10%+ws \t 30% \t\t 30%+ws") for item_price in range(555,600,50): for i in range(0,7): price = item_price for i in range(0,number_of_tries): success = 0 price += item_price while success ==0: scroll = randint(0,9) price += ws_price price += scroll_price if scroll >=7: success +=1 if scroll <7: if randint(0,1) == 1: price += item_price price_30_ws=price/number_of_tries price = item_price for i in range(0,number_of_tries): success = 0 price += item_price while success ==0: scroll = randint(0,9) price += scroll_price if scroll >=7: success +=1 if scroll <7: price += item_price price_30=price/number_of_tries price = item_price for i in range(0,number_of_tries): success = 0 price += item_price while success ==0: scroll = randint(0,9) price += scroll_price if scroll ==9: success +=1 else: price += item_price price_10=price/number_of_tries price = item_price for i in range(0,number_of_tries): success = 0 price += item_price while success ==0: scroll = randint(0,9) price += ws_price price += scroll_price if scroll ==9: success +=1 price_10_ws=price/number_of_tries round_int = 2 print(round(item_price/ws_price,round_int),"\t\t",round(price_10/ws_price,round_int),"\t",round(price_10_ws/ws_price,round_int), "\t" , round(price_30/ws_price,round_int) , "\t" , round(price_30_ws/ws_price,round_int)) item_price = min(price_10, price_10_ws, price_30, price_30_ws) if item_price == price_10: print("you should: 10%") if item_price == price_10_ws: print("you should: 10%+ws") if item_price == price_30: print("you should: 30%") if item_price == price_30_ws: print("you should: 30%+ws") print("final item price in ws: ", round(item_price/ws_price,round_int)) Thanks for reading the guide, the discussion has already been somewhat heated. I will gladly take further feedback if there is anything I have missed, or there are misunderstandings.
beep boop statistics bot incoming. Interesting idea and I will read the full text once you're ready but just wanted to point out that the expected average success on your first 10% scroll happens at the 7th try, not on the 10th. Just like a 30% scroll has an expected average success on the 2nd try, not on the 3rd. If you want a more in-depth explanation feel free to ask.
Perhaps the median success happens on the 7th try, but the expected number (mean) of tries is 10. The median doesn't matter in this type of cost analysis.
This is a binominal distribution where we are solely looking at the first pass of the ws, The expected number is thus 7 slots for your first 10% scroll to pass. What you are saying is that it is a given that you are using 10 scrolls, the expected amount of passes is indeed 10% which results in 1 but it is expected to happen on the 7th slot on average.
This is incorrect. The random variable here is "number of tries until first pass," whose distribution is not binomial at all. All I was saying is the the median of this distribution is around 6-7, but the mean is still 10, and one shouldn't confuse the median (i.e. average "scenario") with the mean (i.e. average cost) when doing this type of scrolling cost analysis. Spoiler It's not difficult to explicitly calculate the expectation as the following sum 0.1 x 1 + 0.9 x 0.1 x 2 + 0.9 x 0.9 x 0.1 x 3 ... = 0.1 / (1 - 0.9)^2 = 10 If this were not true, you should try making some serious profits with WS + 10%.
So what you can calculate is after how many tries you are more likely to succeed than fail. This would be done by doing 0.9^x =0.5. Here we get that x= 6.58. Which means as @Coryn stated earlier you are more likely to succeed than fail after 7 attempts. I am still not sure what to do with this. I will do some simulations later to check what happens. But I believe what @silv is saying is correct. We are looking at the average cost of getting 1 successful scroll. Not when we expect to get our first success.
I'm not sure what you are trying to say here as this is a clear case of a binominal distribution / Bernouilli experiment (Success/Fail). Are you just making up the number 7-8 on the spot ? Insert pro calculation: We are talking about 1 success only, not about 2,3,4,... It might make it clear though if we are trying to take 2 slots with a 10% scroll + WS The expected slot to land your 2nd 10% scroll is on the 17th slot. with p = 0.1 n = 17 k = 1 The probability of P(X>k) is 51,82% with Contemplating whether to post as this is getting kinda cringy for a mapleroyals forum but here goes..
The 7-8 number was just a rough heuristic estimate, which turns out to be off. The correct median is between 6-7 as your “pro calculation” (or simply 0.9^x) shows. If you believe that’s the right number to look at, I’d suggest you start WS + 10% for exactly one slot only and resell them for 3 WS of profit. The k such that Pr(X > k) = 0.5 defines the median of the distribution. It is a very different quantity from the expectation Ex(X) = 10 in this case, which is what you should care about by the law of large numbers.
Should've used /s, agreed Going to leave it here though because I simply can't explain it well enough in English. In all honesty, good luck with the project.
The math is black and white here. All I was trying to say this whole time is that you are confusing the mean with the median.
One more comment since you're misunderstanding my logic By your interpretation of my logic you think I'm saying 10s will work on 7th,14th,21th slot while I'm saying it's 7,17,27,...97...
Leaving this here as someone else has attempted to build a similar calculator in the past. It definitely does not look to include the level of statistical rigor discussed above: https://royals.ms/forum/threads/scrolling-calculator-for-perfect-items.127383/
So you’re saying that everyone should make a new character, 10% + WS exactly 1 slot, and sell for 3 WS profit, because somehow the very first slot only takes 7 WS on average? And note that what I’m saying is not that if you use 10 scrolls, you are expected to pass 1 (you seem to think this is what I’m saying). What I’m saying is exactly the fact that if you keep scrolling until the first pass, the expected number of scrolls used is 10 and not 7. If you don’t trust my math, feel free to run a simulation. I can promise you that the mean outcome for any slot (including the first slot) is 10 white scrolls.
@ok50 very interesting, I hope you will finish this wip -- I still do not get it, guys, why the 7th 10%-scroll should work and not the 10th. Doesnt 10% mean, in 10 independent events, it is likely that 1 passes? 10%? If it is more like to happen at the 7th instead of the 10th trial, wouldnt that mean that its more like a 15% scroll? Edit: I mean it is more like in e.g. 100 trials - they are all independet of each other - so you will get 10 - 90 in average after a lot of experiments. But it could happen at the first slot or at the last. Isnt thinking that it gets more likely, called "gamblers fallacy"?
I previously made a calculator based on the negative binomial distribution, just for your reference If you are looking at the first pass of the ws, then the random variable will be the number of trials until the first success (or the number of failures until the first success), which satisfies the assumption of the geometric distribution. The binomial distribution is more like "if we use 20 ws, what is the probability of passing exactly 5 times?". These two are very similar but they are not the same. I still don't understand how you draw the conclusion where 7th, 17th, 27th trials will work. Do you use the median in your conclusion? In my understanding, each trial is fundamentally an independent Bernoulli trial and therefore the number of trials shouldn't affect the probability of other trials. it might be something that I don't know yet and I am really interested in your explanation! How did you come up with 2.33 items? wouldn't it be 3.33 items per success? Even it is a success, the cost would still be 3.33 items. ___________________________________________________________________________________________________________________________________________ To help other players to understand this more easily, imagine we are going to scroll a number of 333 equips with 30% alone, we are going to buy 333 30% scroll and 333 equips. In the end, 100 equips (333*30%) will pass, 117 equips (333*35%) will lose a slot but not be destroyed, and 117 equips (333*35%) will be destroyed. Hence, the cost for 30% alone will be 333*(30% scroll price+item price). And for the 30%+ws, 100 equips (333*30%) will pass, 117 equips (333*35%) will be saved by the ws, and 117 equips (333*35%) will be destroyed. In this case, the saved equips by the ws will be reused in the scrolling process and therefore you only need to buy 217 equips instead of 333 equips, the cost for 30% + ws will be (333*30%scroll price+ws price)+(217*item price). Same concepts apply to the rest
Hey, thanks for the reply. I hope the 10%-scroll argument is over soon. As for my calculations of getting 2.23, and 1.17, they are the same numbers you got, just 1 lower. This is because I calculated the average cost of getting a success on your item AFTER already having the item. I assumed that you have the item, and want to scroll it, after that it will cost you an additional 1,17 and 2,23 on average. I could be clearer on that in the post, of course.
