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

Latest revision as of 21:48, 26 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• Donkeyhaute has unpublished data on 207 1-UV tickets. 158 +1s, 38 +2s, 8 +3s, 3 +4s.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)

I think this is a case of conditional probability vs intuition where it'll always feel a little bit weird at first glance. I guess you can rationalize it a little by saying that if you get elemental on the first of the 2 UVs (with the triple ticket), the probability of getting elemental on the second UV is zero. Intuitively we tend to ignore that because we're happy we got the UV we were looking for. The chance of that happening is 1/3. The chance of getting elemental on the second UV is 2/3 * 1/2 = 1/3. The chance of getting elemental on neither roll is 2/3 * 1/2 as well, which is our remaining 1/3 to add up to 1. --Waterbeat (talk) 18:03, 21 March 2020 (UTC)

Hi again. Sorry that I forgot to follow up.

I have not found an error in your thinking. :)

My intuition is roughly like this: A 2-UV ticket gives you one chance to get the second UV you wanted, while a 3-UV ticket gives you two chances. So they break even when the cost of the 3-UV ticket is twice the cost of the 2-UV ticket.

But with probability theory, I tend not to trust my intuition. I'm always worried that I'm overlooking or oversimplifying. I trust only algebra carried out from precisely stated assumptions. So please do let me know if you find an error in that. Cheers. Jdavis (talk) 21:48, 26 March 2020 (UTC)