Difference between revisions of "User talk:Jdavis/UVs"

Revision as of 13:24, 21 March 2020

Data sets

Here are some data sets on single-UV rates. Maybe I'll integrate this information into a proper wiki page somewhere.

• roflmao35 has 108 craftings resulting in single UVs. 71 +1s, 23 +2s, 8 +3s, 6 +4s. Also some non-UVs, although total sample size for that is not stated precisely.

• Lancer Knightz (Guild)/Unique Variants has 135 craftings resulting in single UVs. 90 +1s, 37 +2s, 6 +3s, 2 +4s. Also some double UVs and non-UVs.

• Lightningfireninja has 15 1-UV tickets. 15 +1s, 0 +2s, 0 +3s, 0 +4s. Also gives full detail of UV type and item type. But seems to have decided to post results after seeing the data, possibly introducing an observational bias.

• Spiral-Spy has interesting results with not enough detail. Perhaps by reverse-engineering, claims that UV type is uniformly random and UV strength is ~73% +1, ~20% +2, ~5% +3, ~2% +4. Not clear why percentages are approximate.

• Blue-Panther has 268 1-UV tickets. 186 +1s, 64 +2s, 11 +3s, 7 +4s. Also breaks down UVs by type on swords, finding not much deviation from 1/8 chances.

• Donkeyhaute has unpublished data on 207 1-UV tickets. 158 +1s, 38 +2s, 8 +3s, 3 +4s. Also promises full detail.

If we aggregate the results from roflmao35, Lancer Knightz, Blue-Panther, and Donkeyhaute, then we have 108 + 135 + 268 + 207 = 718 UVs, among them

71 + 90 + 186 + 158 = 505 +1s,

23 + 37 + 64 + 38 = 162 +2s,

8 + 6 + 11 + 8 = 33 +3s, and

6 + 2 + 7 + 3 = 18 +4s.

So the estimated rates are 505 / 718 = 70.3%, 162 / 718 = 22.6%, 33 / 718 = 4.6%, and 18 / 718 = 2.5%. If we use only Blue-Panther and Donkeyhaute (because the others are from crafting), then we get similar rates. Jdavis (talk) 14:50, 14 August 2019 (UTC)

I edited the post above. Jdavis (talk) 20:08, 14 August 2019 (UTC)

I edited the post above again. Jdavis (talk) 12:05, 15 August 2019 (UTC)

2-UV strategies

In your discussion about 2-UV strategies (where you already have 1 UV on the item and want a specific second UV), your conclusion says the expected costs for "locking 1 rolling 1" and "locking 1 rolling 2" would even out if the cost for a 2UV ticket = the cost for a 3UV ticket divided by 2.

How is this possible? When I try to take a logical approach, this is what I find: I will use shields as an example because the numbers are smaller. Assume Grey owlite shield with Normal max on it, and the goal is to add Elemental max as a 2nd UV.

Locking the normal max, rolling with a 2UV ticket, I get: 1/3 chance for it to be elemental. Locking the normal max, rolling with a 3UV ticket, I get: 1/3 chance for it to be elemental on the 2nd slot; if no success, my 3rd slot has 1/2 chance to be elemental.

Assume C2 = C3 / 2, then every roll with C2 has 1/3 chance to be elemental, while every roll with C3 / 2 has either 1/3 or 1/2 chance. Is this because the probability to get to the 1/2 is 2/3, (2/3 * 1/2 = 1/3) causing it to even out as well? I think I may have answered my own question. [Waterbeat]

Hi. Thanks for writing. I need to think about what you wrote. Jdavis (talk) 13:24, 21 March 2020 (UTC)