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Getting the best UVs

10 replies [Last post]
Tue, 07/29/2014 - 14:35
Skyshroud

I am in the process of trying to build the best gear possible for farming specific areas. After getting the best gear, the only way to make it better is to get the right unique variants on it. So, if I am trying to make sure my weapon has IAS, damage vs specific monster types, or charge speed: VH or max, would it be a better option to keep using 1* UV tickets until I get one, or keep using 3* tickets.

The way I currently see it, is a 3* ticket is 11x more expensive than a 1*, but a 3* ticket guarantees that it will have 3 independent stats. In other words, for a 1* if I get damage vs fiend low, I would have to roll it again, but with a 3* if I have damage vs fiend low, the other two slots cannot be damage vs fiend so it increases my odds of getting at least 1 of the stats I want significantly. However, with the 1* they are cheap enough that I can just keep rolling them until I get 1 stat I want then roll a 3* holding that slot.

Any information or input is appreciated.

Tue, 07/29/2014 - 16:05
#1
Bopp's picture
Bopp
asked occasionally

This question is asked occasionally. You might try a Google search for something like "UV ticket site:forums.spiralknights.com", without the quotation marks. But the right answer changed at some point, when Three Rings added the ability to keep some of your old UVs while trying for new ones.

The short answer is: Buy 1-UV tickets until you get a UV that you really like. Then buy 2-UV tickets, holding the one good UV that you already have, until you get a second UV that you really like. Then buy 3-UV tickets, holding the two good UVs, until you get a third UV that you really like.

The details of why this is the best strategy are not especially hard to understand, but they take a while to lay out --- especially because the specifics depend on which item you're using (helmet, sword, bomb, etc.), how many UVs you're going for (1, 2, or 3), and how drastic they are (+1, +2, +3, or +4). Basically, it comes down facts like this: A 3-UV ticket costs much more than three times as much as a 1-UV ticket.

For example, suppose that you want a CTR+3 or CTR+4 UV on a Brandish. A sword can have eight types of UVs, so suppose that you have 1/8 chance of getting CTR. Now, our data from crafting suggests that, when UVs happen, they are +3 or +4 about 1/8 of the time. And it's reasonable to assume that these chances are independent of each other. So your chance of getting CTR+3 or CTR+4 is 1/8 * 1/8 = 1/64, your expected number of 1-UV tickets is 64, and your expected cost is 64 * 20,000 = 1,280,000 crowns. (This "expectation" is a concept from probability theory. Think of it as the average. You might get lucky and require fewer than 64 tries, or you might get unlucky and require more than 64 tries, but 64 is "average luck".)

You might want to try buying that first UV from another player, though the Auction House or some other avenue. Players tend to undervalue their UVs, relative to the cost of getting them from Punch. For example, a CTR+3 Brandish might cost only 500,000 from the Auction House (I haven't checked in a while).

If you do end up using Punch to get some UVs, please record each try in this thread: which item, which ticket, which old UVs kept, and the result. We could always use more data on how Punch works, to better answer questions from people like yourself.

Mon, 08/04/2014 - 14:14
#2
Skyshroud
More math

So... I actually went to college for mathematics and given the math you provided I went a little more in depth.
Since on a weapon there are 8 possibilities
6 monster types
1 increase attack speed
1 increase charge speed
For sake of demonstration, let us assume you want exactly two items off the list
For each item you want High or better (+3, +4) and the chance of getting any one of these is 1/8
Math time:
A 1-UV ticket is 20,000
The chance of getting one of the two desired stats is (1/8)*2 and the chance of getting high or better is (1/8)
So the cost of getting one high quality item is approximately 20000/(((1/8)*2)*(1/8)) = 20000/(1/32) = 640,000
The result is one item with one high+ UV of desire on it.

A 3-UV ticket is 225,000
The chance of getting at LEAST one of the two desired stats is 1-((6/8)*(5/7)*(4/6)) = 1-(120-336) = 1-(5/14) = 9/14 and the chance of exactly that one being high+ is 1/8
So the cost of getting one choice item with high+ quality is approximately 225000/((9/14)*(1/8)) = 225000/(9/112) = 2800000
The result is one item with 3 UV's one of which is desired and high+

Conclusion by using the 1 UV method you will achieve your 1 UV item of choice for 640,000cr and using the 3UV method is approximately 4.5x more expensive. However, buy using the 3UV tickets you will always have 3 items on it and the other "non-choice" items have a chance of 1 being other good stats, and 2 being of any quality low, med, high, or max.

Best method based on these statistics - Buy 1 UV's until you get the stat you want, then buy 3UV's. This will save significantly on cost but you lose the option of having what you want show up as a bonus item (about 20% chance of that happening) and it discounts the fact that you will have 2 other potentially good UVs.

Mon, 08/04/2014 - 16:08
#3
Bopp's picture
Bopp
two desired stats?

In your example, are you saying that you'd be happy with EITHER of the two desired UVs? Or are you saying that you want BOTH of them? The latter makes sense to me. The former seems like an untypical player desire.

Tue, 08/05/2014 - 09:20
#4
Skyshroud
Eventually you'd want both,

Eventually you'd want both, but to get the first of the two that would be the cost.

Tue, 08/05/2014 - 10:01
#5
Bopp's picture
Bopp
to get two UVs +3 or +4 on a sword

Here's the cost to get both. It's more interesting, because it exemplifies the extremely player-friendly feature of holding one UV while rolling for more. Let's say that you're making a sword or handgun (not a bomb), so there are 8 types of UVs. And you want ASI and CTR, each at +3 or +4. I evaluate three strategies.

A. 1-UV tickets until you get one of the UVs you want; then 2-UV tickets, holding the first UV, until you get the other UV: Each 1-ticket has 1/4 chance of getting the right type, and then a 1/8 chance of getting +3 or +4. So you have 1/32 chance, and expect to buy 32 1-tickets. Then each 2-ticket has a 1/7 of getting the right type and 1/8 chance of getting +3 or +4. So you expect to buy 56 2-tickets. The total cost is 32 * 20,000 + 56 * 75,000 = 4,840,000 crowns.

B. 2-UV tickets until you get both UVs that you want: There are "8 choose 2" = 28 pairs of UVs, so each ticket has a 1/28 chance of getting ASI and CTR. Then it has a 1/8 * 1/8 chance of getting +3 or +4 on both ASI and CTR at the same time. So the chance of hitting the desired UVs is 1/28 * 1/8 * 1/8 = 1/1792. You expect to buy 1,792 2-tickets, for a total cost of 134,400,000.

C. 3-UV tickets until you get the two UVs that you want: There are "8 choose 3" = 56 triples of UVs, 6 of which have ASI and CTR (and some damage bonus, that we don't care about). So each ticket has a 6/56 = 3/28 chance of getting ASI and CTR. Then it has a 1/8 * 1/8 chance of getting +3 or +4 on both ASI and CTR at the same time. So the chance of hitting the desired UVs is 3/28 * 1/8 * 1/8 = 3/1,792. You expect to buy 1,792/3 = 597.333... 3-tickets, for a total cost of 134,400,000 again.

So strategy A costs about 3.6% as much as strategies B and C. As an aside, strategy C is better than strategy B, because you end up with some third UV (although that was not our stated goal).

Tue, 08/05/2014 - 18:25
#6
Plancker's picture
Plancker
well

Strategy B and C can also get you 1 of the 2 desired UVs. Its normal that it costs so much more if you are using the lock UV feature in strategy A, but not for B or C. Right now thats not being factored in and only the very rare situation (getting both of the desired UVs with the same ticket) is counted.

I think you meant strategy B to be like this:

2UV tickets until you get both UVs that you want OR until you get one of the UVs you want; then 2UV tickets, holding the first UV, until you get the other UV.

and strategy C:

3UV tickets until you get both UVs that you want OR until you get one of the UVs you want; then 3UV tickets, holding the first UV, until you get the other UV.

Wed, 08/06/2014 - 04:52
#7
Bopp's picture
Bopp
good idea

I didn't mean that, although you're right that that would be a better strategy. So...

D. Buy 2-tickets until you get at least one of the UVs you want; then buy 2-tickets, keeping that UV, until you get the other, if you don't already have it. For each ticket, there are "8 choose 2" = 28 combinations of UV types, divided into four disjoint events.

  1. Getting both CTR and ASI has a 1/28 chance, and within that a 1/8 * 1/8 chance of getting the +3s or +4s that you want, for a total 1/1792 chance of getting both desired UVs (as above).
  2. Getting CTR but not ASI has a 6/28 chance, and within that a 1/8 chance of getting +3 or +4, for a total 6 / (28 * 8) = 3/112 chance of getting the desired CTR UV but not the desired ASI UV.
  3. Similarly, getting the desired ASI but not CTR has a 3/112 chance.
  4. Getting neither desired UV is the rest of the probabillity: 1 - (1/1792 + 3/112 + 3/112) = 1695/1792.

We also need this fact: If you buy a 2-ticket and keep one desired UV, then the chance of getting the other desired UV is 1/7 * 1/8 = 1/56, and the chance of not getting that desired UV is 55/56. Now if you buy 2-tickets repeatedly, there are three disjoint kinds of events that can happen.

  1. You keep getting garbage for K tickets, until you suddenly get both desired UVs in the (K+1)th ticket. The chance of this happening is (1695/1792)^K * (1/1792), and the cost (in thousands) is 75 * (K + 1).
  2. You keep getting garbage for K tickets, until you get the desired CTR UV on the (K+1)th ticket. Keeping that UV, you get garbage for another L tickets, until you get your desired ASI UV on the (L+1)th ticket. The chance is (1695/1792)^K * (97/1792) * (55/56)^L * (1/56), and the cost is 75 * (K + L + 2).
  3. The same thing happens as in the previous case, but with ASI and CTR switched.

The expected cost is the sum of the chances weighted by their costs:

SUM_K (1695/1792)^K * (1/1792) * 75 * (K + 1)
+
SUM_K SUM_L 2 * (1695/1792)^K * (97/1792) * (55/56)^L * (1/56) * 75 * (K + L + 2).

The first term is an arithmetico-geometric series (see Wikipedia) with R = 1695/1792, A = 75/1792, D = 75/1792, for a total of A/(1 - R) + (R D)/(1 - R)^2 = 134400/9409 = 14.2842. (This seems low, but I guess the probability is really small.) The second term can be rewritten

2 * (97/1792) * (75/56) * SUM_K [(1695/1792)^K * SUM_L (55/56)^L (L + (2 + K))].

The inner sum over L is an arithmetico-geometric series with R = 55/56, A = 2 + K, D = 1, for a total of 3192 + 56 * K. Then the sum over K is an arithmetico-geometric series with R = 1695/1792, A = 3192, D = 56, for a total of 724942848/9409 = 77047.8. The entire second term is therefore 1083600/97 = 11171.1. The sum of the two terms is 14.2842 + 11171.1 = 11185.4. Therefore the expected cost of this strategy is 11,185,400 crowns.

I won't bother to do strategy E. This is all pretty laborious.

Wed, 08/06/2014 - 05:21
#8
Plancker's picture
Plancker
One more thing

Another thing about UV tickets: is there a possibility, or have other people actually tested this, that a 2UV or 3UV ticket has a better chance of getting a higher value UV? Its just a feeling and I have never tested or kept data records on it, but every time I have rolled a 2nd UV or 3rd UV on an item so far, I have gotten a high/max one in a fairly low amount of rolls, while I have rolled so many lows with 1UV tickets. Then again, I have rolled much more with 1UV tickets than with the other ones, so it might just be luck on my part, but I just want to know if there are people who have tested this.

Wed, 08/06/2014 - 05:46
#9
Bopp's picture
Bopp
no data

I have never seen data on that (and I keep an eye out for such data). I have also never heard of that possible effect anecdotally until now.

If you do such tickets in the future, please keep careful data, and maybe we can tease out whether that effect exists.

Thu, 08/07/2014 - 16:29
#10
Bopp's picture
Bopp
wiki guide

I have automated a lot of these calculations, and I am making a guide on the wiki to explain strategies for obtaining various combinations of UVs:

http://wiki.spiralknights.com/User:Jdavis/UVs

By the way, the guide disagrees with my post #7 above, so I'm checking the algebra. In either case, strategy A still beats strategy D.

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