---------- In Progress ---------- 0. Dummy Scrolling One sentence says all: dummy scrolling doesn't work. Ask yourself, if dummy scrolling works, why are you not using the dummy scrolling pattern to scroll 10%+ws at the cost of 2b or lower each slot and making 20b a week? Spoiler: the Haida and Tlingit tribes airplane summoning dance In the early 20th century, after witnessing airplanes fly overhead, some community members of the Pacific Northwest, specifically the Haida and Tlingit tribes began to perform dances and rituals in hopes of summoning these planes, which they viewed with a mix of awe and curiosity. The dancing itself did not actually summon airplanes; it was more of a symbolic act reflecting the community's fascination and hope. 1. 30% (to be updated. Mostly said in this thread: https://royals.ms/forum/threads/tec...lculation-and-market-price-estimation.125996/) 2. 70% 70% succeed, 15% fail, 15% boom. Shield ATT, bottom DEX, top LUK, etc.: 70% gives 2 stat, and 30% gives 3 stat. Most of the time, it is not worthwhile using 30% (rare cases: scrolling perfect equipment, being desperate, etc.). Spoiler: example According to @Sylafia 's price guide, 3 LUK 16 att and 17 att Khanjars are 40m and 70m. To get 19 att / 6 slot, the cost of scrolling each of the Khanjars: { [16 att Khanjar] + [30% shield ATT] - [ 16/6 khanjar] × 35% } / 30% ≈ 150m or { [17 att Khanjar] + [70% shield ATT] - [ 17/6 khanjar] × 15% } / 70% ≈ 100m You will better be scrolling 17/7 with 70% instead of scrolling 16/7 with 30% if your goal is a 31 att Khanjar. In the above example, I used [19 att 6 slot Khanjar] = { [16 att Khanjar] + [30% shield ATT] - [ 16/6 khanjar] × 35% } / 30%. It comes from: Spoiler: expression [19 att 6 slot Khanjar] × 30% + [0] × 35% + [ 16/6 khanjar] × 35% = [16 att Khanjar] + [30% shield ATT] Also, [19 att 6 slot Khanjar] × 70% + [0] × 15% + [ 17/6 khanjar] × 15% = [17 att Khanjar] + [70% shield ATT] Again, we take Khanjar for example. What's the cost of scrolling a 3 LUK 31 att Khanjar? Lets now consider using 70% scrolls only. Spoiler: Probability of getting 31 - 25 Khanjar Getting 31 att Khanjar: p1 = 0.7^7 × 0.15^0 × 1 Getting 29 att Khanjar: p2 = 0.7^6 × 0.15^1 × 7 Getting 27 att Khanjar: p3 = 0.7^5 × 0.15^2 × 7 × 6 / 2 Getting 25 att Khanjar: p4 = 0.7^4 × 0.15^3 × 7 × 6 × 5 / (3 × 2) Spoiler: combination number 1, 7, 7 × 6 / 2, 7 × 6 × 5 / (3 × 2) are combination numbers C(7,0), C(7,1), C(7,2), and C(7,3). p1 = p2 / 1.5 p1 = p3 / 0.964 p1 = p4 / 0.344 Lower attack is pretty much worthless, so I ignore them here. (3 LUK Khanjars: ) We need 1/p1 = 12.14 [17 att Khanjar]'s, and about 37 × 70% scrolls to get one [31 att Khanjar]. 12.14 × [17 att Khanjar] + 37 × [70% shield ATT] ≈ [31 att Khanjar] + 1.5 × [29 att Khanjar] +0.964 × [27 att Khanjar] + 0.344 × [25 att Khanjar] (Price of Khanjars retrived from Sylafia's price guide.) Thus [31 att Khanjar] = 520m. According to the price guide, 3 LUK 31 att Khanjar is 810m. So there is 290m profit. Spoiler: ez 3. 60% 4. White Scroll if the scrolling fails, the equipment will NOT lose a slot, and the WS will be consumed; if the scrolling succeeds or booms (with dark scroll), the WS will be consumed. When should one use a WS? Whether or not to use a WS is solely determined by the cost of losing a slot on the equipment. When one chaos (60%) an equipment, if the equipment loses more than WS/40% value when it loses a slot, a WS should be used. When one 30% an equipment, if the equipment loses more than WS/35% value when it loses a slot, a WS should be used. Spoiler: boom? This 35% is the probability of failing. The 35% of booming doesn't affect whether or not to use a WS. 70%: WS/15%. Again 15% is the prob of failing. 60%: WS/40%. Choice of 10% or 30%, example: Spoiler: 30%+ws OR 10%+ws Weapons: 10% or 30% + WS? 110 Kanzir: just 10%+ws. How about 109 Kanzir (2.1b)? 30%: Spoiler: 30%+ws With WS 30% × [114/6 Kanzir] + 35% × 0 + 35% × [109/7 Kanzir]= [109/7 Kanzir] + [dagger 30%] + WS The cost to make a [114/6 Kanzir] with 30%+ws is: [114/6 Kanzir] = { 65% × [109/7 Kanzir] + [dagger 30%] + WS } / 30% = 6240m 10%: Spoiler: 10%+ws The cost to make a [114/6 Kanzir] with 10%+ws is: [114/6 Kanzir] = [109/7 Kanzir] + { [dagger 10%] + WS } / 10% = 7100m The cost to make a [114/6 Kanzir] with 30%+ws is 6240m, and the cost to make a [114/6 Kanzir] with 10%+ws is 7100m. Thus one should 30%+WS a 109 Kanzir (current price 109 Kanzir 2.1b, WS 500m). 5. Clean Slate Scroll (CSS) CSS can only use on equipment that has lost at least ONE slot (total clean slots - slots landed - slots remaining >= 1). CSS 20%: Spoiler: CSS 20% 20% the equipment recovers 1 slot; 40% the equipment booms; 40% the equipment remains intact. CSS 5%: Spoiler: CSS 5% 5% the equipment recovers 1 slot; 9.5% the equipment booms; 85.5% the equipment remains intact. CSS 3%: Spoiler: CSS 3% 3% the equipment recovers 1 slot; 5.82% the equipment booms; 91.18% the equipment remains intact. CSS 1%: Spoiler: CSS 1% 1% the equipment recovers 1 slot; 1,98% the equipment booms; 97.02% the equipment remain intact. Most of the time, tradable equipments are not worth CSS'ing. Example: 20 att CSS'able SCG (https://royals.ms/forum/threads/s-20-scg-css-able-sold.218924/). 20 SCG 18.5b CSS 20% 280m, 5% 60m, 3% 35m, 1% 20m (as of Oct 2024). 23 gloves 36b, and [20 att 1 slot gloves] = 31b Each time you use a CSS on the SCG, the average gain is: Spoiler: gain If you choose to CSS 20 CSS'able gloves, the average to make one 20/1 SCG/BWG is: Spoiler: cost to be updated 6. Chaos Scroll: check out SCRL 127 (about attack gloves, to be written) and SCRL 217 (capes and shoes) 7. Advanced Scrolling: Risk Estimation and Minimal Funding (SCRL 475) (to be written). If you know how to do Monte Carlo simulation, then excellent, you prob can easily do your own simulation and draw your own conclusion with a few basic rules. If not, a rough estimation is also realizable based on Excel/Numbers.
Dummy scrolling works fine for me. At the end, scrolling depends on probabilities + how is you luck. ps: everytime I hit scroll day when I'm high, somehow I end up booming all my stuff . Has anyone else experienced this?
My 1+1 = 2 mind couldn't comprehend all of your posts but I'm damn sure that almighty Matt himself should give you the Royals PhD medal or sumn like that.
Hi! As a statistician I found this guide a really interesting read. I would like to urge caution when discussing the outcomes and estimating costs with white scrolls, as their very nature can involve more than a single attempt to reach the desired outcome. However, your calculations (particularly with 30%ing a Kanzir) included probabilities reflecting a single attempt only. If we wish to estimate the costs of making a Kanzir whose first slot succeeded with a 30% and white scroll, we must consider the probability of the second attempt, third attempt, and so on as there is no 100% guarantee that it will ever succeed/boom. Therefore, the 30%/35%/35% split cannot apply when estimating the costs of making the 114/6 Kanzir in your example. To begin, I'll outline the actual probabilities which exclude the chance of a 30% dagger failing on the first slot without a boom (because white scrolls prevent this): Chance of success: 46.154% Chance of boom : 53.846% Spoiler: why these numbers? I have two separate arguments that could convince you. First, consider the original ratio defining the probability of an event A: P(A) = n(A) / n(S) where S represents the entire realm of possible outcomes and n(A) or n(S) is some counting mechanism of the events. (Warning: I am going to refer to this loosely without actual counts but the theory and formula still remain true.) Bear in mind that if we are using white scrolls there is no scenario in which we have a +0 Kanzir with 6 slots. Therefore we can eliminate this from the set of possible outcomes in the denominator, and so we have: Chance of success: 30% / (30% + 35%) = 46.154% Chance of boom: 35% / (30% + 35%) = 53.846% Okay, so I admit there was some hand waving between counts of an actual trial and the raw probability percent. If you're still unconvinced consider the following! For my second explanation, we start by breaking this down by each attempt. The pattern would look like this: First scroll: Success: 30% Boom: 35% Fail, on to second attempt: 35% Second scroll: (Note we are assuming this is within the realm of the 35% chance it failed on the first attempt, so EVERYTHING gets multiplied by 35%) Success: 30% × 35% = 10.5% Boom: 35% × 35% = 12.25% Fail, on to third attempt: 35% × 35% = 12.25% Third scroll: Success: 30% × 35% × 35% = 3.675% Boom: 35% × 35% × 35% = 4.2875% Fail, on to 4th attempt: 35% × 35% × 35% = 4.2875% You probably see the pattern by now right? To know our actual chance of success or boom at some arbitrary attempt we would have: P(Success on attempt n) = 0.3×0.35^(n-1) P(Boom on attempt n) = 0.35×0.35^(n-1) Remember how I mentioned there is no guarantee any individual scroll would pass/boom? That was a hint that this could go on for an infinite number of attempts. Keep this in mind as we calculate the combined probability for each case: P(Success) = 0.3×0.35 + 0.3×0.35^2 + 0.3×0.35^3 + ... P(boom) = 0.35×0.35 + 0.35×0.35^2 + 0.35×0.35^3 + ... As this goes on infinitely you may be concerned about how we're supposed to do anything to tame an infinite sum. Lucky for us, the sum does converge to a finite number if you remember the formula from calc 2 As our common ratio 0.35 is indeed positive and less than 1, we can use the infinite geometric sum formula to arrive at: P(success) = 0.3 / (1 - 0.35) P(boom) = 0.35 / (1 - 0.35) I'll let you do the final arithmetic here to see that the numbers agree Because these numbers differ from a single attempt breakdown of 30/35/35 I think the calculation of the true cost of using 30%s for the 114/6 Kanzir would arrive at a different number. In my next post I plan to break down my calculation methodology and give the actual cost estimate for a 30% passing on the first slot when white scrolls are involved. Please look forward to it!
Hi, thank you for reading and discussing the details of the OP. The calculations in your reply was exactly something I did when I double-checked the math: And in the OP: which is equivalent to: You probably would get the same result if you take 30% scroll and WS into account.
Hi! Thanks for responding to me I'm a little confused by your reply as my point was never to say your arithmetic was wrong, but rather that the cost estimation methodology for use of a white scroll was a bit oversimplified. Since you're saying you double checked the math, I must admit that I'm definitely not following the reasoning of your quoted calculation for expected costs. You're multiplying the combined chance of booming and succeeding (65%) by the cost of the dagger, then adding the cost of the scrolls separately, and finally dividing this sum by the chance of the scroll succeeding. Could you please elaborate on why you computed it this way? It simply doesn't match any form of a statistical Expected Value I could imagine is applicable here
Hi, sure. Scrolling a 30% with WS on 109/7 kanzir, possible results are: 1. successfully landed, output 114/6 kanzir (30%) 2. boomed, output 0 (35%) 3. failed, output 109/7 kanzir (35%) Thus LHS: 30% × [114/6 Kanzir] + 35% × 0 + 35% × [109/7 Kanzir] Cost of scrolling: 109/7 kanzir, 30%, and a WS. Thus RHS: [109/7 Kanzir] + [dagger 30%] + WS I put this formula in a spoiler of a spoiler, sorry to confuse you. I will update the thread to make it more readable.
