**RESOLVED ISSUE WITH SIDE QUEST KEYS**
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More information and timeline here.
*This includes currently unclaimable keys as well*
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As previously announced, the team will be distributing an additional point toward milestones to anyone who completed the Absorbing Man fight in the first step of the Winter of Woe.
This point will be distributed at a later time as it requires the team to pull and analyze data.
The timeline has not been set, but work has started.

Couples Nexus Math (yeah yeah they're gone now sue me)

So I actually didn't know until not long ago that the Couples Nexus cavs that were in the unit offers were actually not normal nexus cavs, but a new kind of cavalier nexus crystal that displayed five options and allowed the player to pick two of them. So unfortunately everything I'm about to say about them is a bit dated. But I figure these things could come around again and this might be useful or interesting information for the future.

So, first of all for those who don't know, what's the deal with the Couples Nexus Cavalier crystals? Well, normal Nexus Cavalier crystals display three different options (they are always different) and the player can choose one of those. The Couples Nexus Cavalier crystal displays five different options and the player selects two of them and gets both of them. So the crystal drops two champions, not one, and you have more options to choose from. An interesting question is: how good is this crystal?

Well, the standard Cav crystal has a 3% chance to drop a 6* champion, a 16% chance to drop a 5* champion, a 31% chance to drop a 4* champion, and a 50% chance to drop a 3* champion. A standard Nexus Cav has he same chances to display those rarities per display slot, but because the Nexus shows the player three options and the player can pick the best one, assuming the best one is always the highest rarity champ the odds of a Nexus producing those rarities is different. It is approximately 8.73% for 6*, 38.12% for 5*, 40.64% for 4*, and 12.5% for 3*. I'll explain how to calculate those numbers at the end, for those that care to know.

We can't easily compare the Couples Nexus to the standard Nexus Cav, because the couples crystal drops two champs. We can ask what the odds of dropping a 6* champ are and what the odds of dropping a 5* are, but we will run into a problem. Because we get two champs, the odds of dropping a 6* and the odds of dropping a 5* are not exclusive. You could get both. Instead, we have to consider all possible outcomes of the crystal and ask what the odds of those are. Those possibilities are: 66, 65, 64, 63, 55, 54, 53, 44, 43, 33. Ten total possibilities. We could compare those to the odds of those same possibilities occurring for two comparable situations: opening two Cavs, and opening two Nexus Cavs. In both cases, we have the same ten possible outcomes in total. And when we do, we get:



There's no question that the couples nexus is better than a normal cav or a normal nexus, obviously (you get two champs). But is it better than two Cavs or two Nexus Cavs? It depends on how you judge. The odds of a total strikeout are actually higher for the couples nexus than two normal nexus crystals. That's because the odds of being forced to choose a 3* twice are independent for two nexus cavs, but slightly correlated for a single couples nexus. On the other hand, the odds of getting all 5s or 6s is slightly higher for the couples nexus: 24.24% vs 21.95% (just add up the percentages for the three possibilities 66, 65, and 55). To figure out whether the couples nexus generates better results than two standard nexus crystals, you would simply look at all ten possible options, pick the ones you find favorable, and add them up for both crystals.

I'm not saying Kabam is likely to actually present this option for players to choose, although they might. Rather, the question is how much should players value these kinds of couples nexus crystals. Is it more than a Cav? Absolutely. Is it more than a nexus? Of course. But how much more? 10% more? 50% more? Twice as much? Intuition can often be misleading when it comes to odds, so knowing what the actual numbers look like can help people make a more informed judgment. It still comes down to preference. Do you want better odds of this or lower odds of that. But at least the numbers tell you where your preferences stand.


Everything below here is math. If you aren't interested in the process of calculating these numbers, here's where you check out.


How do you figure out what the odds of this sort of crystal dropping certain things? There are some complex Probability and Statistics 201 and 301 techniques, but I'm only going to show the straight forward answer which resorts to first principles, with the catch that it is often the case that people screw this up by not accounting for things properly.

First step is to determine what the actual possible outcomes are, as I did above: 66, 65, 64, 63, 55, 54, 53, 44, 43, 33. Let's take the first one. How do we calculate the odds of at least two 6s showing up? Well, the odds of one spot showing a 6* is 3%. The odds of two of them showing 6s is 0.03 x 0.03 = 0.0009 or 0.09%. But that's the odds of two 6s showing up in exactly two slots. How do we calculate the odds of at least two showing up? Well, we recognize that there are ten ways for two 6s to show up in five slots. They look like this:

6 6 X X X
6 X 6 X X
6 X X 6 X
6 X X X 6
X 6 6 X X
X 6 X 6 X
X 6 X X 6
X X 6 6 X
X X 6 X 6
X X X 6 6

Some combinatorial math can get here directly: there are ten ways to arrange two in five: 5!/(3!2!) = 5*4/2 = 10. Now, for each of those possibilities, the odds of that one specific possibility happening is just the odds of each individual option showing up in that spot. So the odds of a 6* showing up is 3%. The odds of a 6* NOT showing up (all the Xs) is 97%. So we get each option having the odds 0.03 x 0.03 x 0.97 x 0.97 x 0.97 ~= 0.08%. There are ten of them and the odds of each are the same (we're just rearranging the order of multiplying terms, so the value will be the same for all of them), so the odds of any one of them happening is 0.03 x 0.03 x 0.97 x 0.97 x 0.97 x 10 ~= 0.82%.

Let's try 65. You might think its the same: we have the same ten possible options as before. But that's actually not true, because in the case above both options are identical. If we try the same thing with 65, replacing one of the 6s with a 5 we would get:

6 5 X X X
6 X 5 X X
6 X X 5 X
6 X X X 5
X 6 5 X X
X 6 X 5 X
X 6 X X 5
X X 6 5 X
X X 6 X 5
X X X 6 5

But that isn't all the options. That's all the options where the 6* shows up before the 5*. It misses half the options where the 6 and the 5 are swapped. So there's twenty options, not ten. Right?

Actually, this is still wrong for a very subtle reason. In fact, I made this exact error initially, then caught it during double checking. Let's say we go with this. What are the odds of the first possibility: 6 5 X X X. Well, the odds of the 6 is 3%, and the odds of the 5 is 16%. What are the odds of the Xs? Well, what are the Xs? They cannot be 6s. If any one of them was a 6, then this Nexus drops two 6s, not a 65. So no X can be a 6. Can the X be a 5? Well, it can, because 6 5 5 3 3 and 6 5 4 3 3 are both 6 5 X X X and both ultimately drop 65. So if we calculate using the same math as in the 66 case we get:

0.03 x 0.16 x 0.97 x 0.97 x 0.97 x 20 = 0.0876 = 8.76%.

If we do this for every possibility and then add up all the probabilities, we should get 100%. Instead we get something like 129%, which is impossible. The total odds for all possible outcomes of the couples nexus must be 100%, because that's the odds of just anything happening. What's the problem?

Consider these two options:

6 5 A B C
6 A 5 B C

We are counting those as two different possibilities above. But what if ABC in both options are identical, and also the As in both options happens to be a 5? In that case, they could be identical options. We would be double counting that option, and that's why the total odds add up to more than 100%. This is a very common way for probability calculations to be double checked. When you calculate the odds of a bunch of stuff happening, and all the different options collectively are supposed to represent all possible options and they are supposed to be distinct and separate from each other, they must add up to 100%, or you've counted wrong.

6 5 X X X and 6 X 5 X X are only different possibilities if none of the Xs are 5s. But we can't just do this:

0.03 x 0.16 x 0.81 x 0.81 x 0.81

Because now we're *failing* to count possibilities: the possibilities where there is more than one 5. So what we *really* have to do is this:

6 5 X X X where all Xs are 4* or lower
6 5 5 X X where all Xs are 4* or lower
6 5 5 5 X where X is 4* or lower
6 5 5 5 5

and then multiply by all possible rearrangements of each one of those. It looks something like this:

0.03 x 0.16 x 0.81 x 0.81 x 0.81 x 20 (one 5)
+ 0.03 x 0.16 x 0.16 x 0.81 x 0.81 x 30 (two 5s)
+ 0.03 x 0.16 x 0.16 x 0.16 x 0.81 x 20 (three 5s)
+ 0.03 x 0.16 x 0.16 x 0.16 x 0.16 x 5 (four 5s)
= 0.0682 ~= 6.82%

Notice this is smaller than the 8.76% above, because we are now correctly not double counting.

Do this correctly for each possibility, breaking it down into its component possibilities, and you get the table above.

There's a lot of complex math that can do this directly, but I prefer to explain it by resorting to first principles. In statistics, you're always ultimately counting and accounting for possibilities. If you can construct a list of all possibilities, such that none of them overlap and all of them collectively represent everything possible, then it generally becomes straight forward to calculate the odds of anything happening.

What are the odds of getting at least one 6*? That would be the sum of the odds of getting 66 and the odds of getting 65, which is 0.85% + 6.82% ~= 7.67%. What are the odds of not getting any 3* champs at all? Just add the odds of getting all possibilities that do not include a 3, and you get 81.25 (the only way to be forced to take a 3* champ is if at least four 3* champs show up, which is why the odds are so low). Once you know the odds of all the individual possibilities, it becomes straight forward to calculate the odds of more complex sets of possibilities.

As promised, we can apply the same techniques towards the standard Nexus crystal. We calculate the odds of 6XX, 5XX, 4XX, and 3XX with the same caution as above, breaking up 6XX as 6XX, 66X, 666, where X is 5* or lower, then 5XX, 55X, 555 where X is 4* or lower, and so on. We end up with:

6* odds = 0.03 x 0.97 x 0.97 x 3 + 0.03 x 0.03 x 0.97 x 3 + 0.03 x 0.03 x 0.03 = 0.0873 = 8.73%
5* odds = 0.16 x 0.81 x 0.81 x 3 + 0.16 x 0.16 x 0.81 x 3 + 0.16 x 0.16 x 0.16 = 0.3812 = 38.12%
4* odds = 0.31 xx 0.5 x 0.5 x 3 + 0.31 x 0.31 x 0.5 x 3 + 0 .31 x 0.31 x 0.31 = 0.4064 = 40.64%
3* odds = 0.5 x 0.5 x 0.5 = 0.125 = 12.5%

In English, the first calculation boils down to (Odds of 6*) times (Odds of 5* or lower) times (Odds of 5* or lower) times (3 ways to arrange that) plus ... etc. You get the idea.

Comments

  • willrun4adonutwillrun4adonut Posts: 2,849 ★★★★★
    edited February 2023
    DNA3000 said:

    So I actually didn't know until not long ago that the Couples Nexus cavs that were in the unit offers were actually not normal nexus cavs, but a new kind of cavalier nexus crystal that displayed five options and allowed the player to pick two of them. So unfortunately everything I'm about to say about them is a bit dated. But I figure these things could come around again and this might be useful or interesting information for the future.

    I might have bought some had I known this. Oh well. And this more than made up for your other recent post with no words.
  • ButtehrsButtehrs Posts: 4,496 ★★★★★
    My man really pushing that word count limit on this one
  • spigwenderspigwender Posts: 473 ★★★
    DNA3000 try not to make every post an entire essay challenge

    Like the posts are informative but dang you don’t have to do this lmao
  • Fit_Fun9329Fit_Fun9329 Posts: 1,775 ★★★★★
    TLDR where
  • DNA3000 try not to make every post an entire essay challenge

    Like the posts are informative but dang you don’t have to do this lmao

    They aren't all like that.
  • TLDR where


  • FrostGiantLordFrostGiantLord Posts: 1,500 ★★★
    I can't wait for a TED talk by @DNA3000 about MCOC statistics. Keep up the good work!
  • BuggyDClownBuggyDClown Posts: 2,032 ★★★★★
    edited February 2023
    I read half of it and got some of the idea. Could grasp rest .
    But it was really good effort. Kudos mate. Keep it up
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