More dragon egg issues

1

Ok, this is getting ridiculous… Hatched 2 more eggs…19 or 20 eggs BOUGHT…Still no emerald or garnet dragons…there is NO WAY the dragons have the same drop rate…the devs are lying about something… there’s no doubt in my mind it’s a scam to get you to spend more gems…what are the odds of 19 to 20 eggs…dropping the same 4 dragons over and over…until they fix this…I will not financially support this game…VIP 9…and they’re still not fixing a F****** thing…they need to fix the drop rate…meaning they need yo stop giving multiple copies after 4…or sticking to the PROMISE of crafting targeted dragons…they have copy protection in deals…ang guild key exclusive cards…so…they aren’t getting a penny more from me until they stop lying and giving us what they promised…and stop screwing Us with multiple copies…
@Kafka , @Jeto , @OminousGMan

3 Likes
2

Assuming an equal drop rate, missing two dragons after 20 eggs is … an approximately 1/3300 incidence, or about halfway (exponentially speaking) between pulling a Mythic from a Glory Chest, vs. a Gem Chest.

So take a number, get in line, join the club (we’ve got jackets). EVERYONE has voiced complaints about these feels-bad streaks, even Kakfa, but please don’t go making claims you can’t prove/investigate.

Here's some cherry-picked data from other users
7 Likes
3

With this amount of unnecessary tags, I wouldnt be surprised if the devs have already muted OP

4

Why do people keep picking up the poopy dollar? The devs are around the corner laughing

Quit buying dragonite and opening eggs until we get the vault event. Its that simple

2 Likes
5

Why does the OP keep making a new thread about this every week?

2 Likes
6

Slight variations of a similar post… Trying to get the devs to do as promised and… maybe… Just maybe listen to the community.

8

Right after Thanksgiving, I opened 11 dragon eggs. I got 5 of 6 dragons and the rest were dups. I am missing Rubirath. A few weeks ago, I opened 5 dragon eggs at the same time and got 1 copy of all the dragons. No Ruby. I like to own 2 copies of each mythic, so I wasn’t upset, nor frustrated. No hurry to get the other 2 dragons, since they are premium content. My focus is to get my kingdoms to 20 and I don’t need those last 2 dragons. Maybe 1 year from now or so, I may try again. If I didn’t spend the gems on dragonite and instead bought deeds, I would have 2 or 3 kingdoms to 20 right now. All of them are at 16 and I have 18 to 20 gold troops per kingdom.

9

I think the math is wrong for 20 eggs.
There are 6^20 combinations.
There are 6* 5^20 combinations where you don’t get all 6.
There are (6* 5/2) * 4^20 where you only get 4 out of 6.
6* 5^20/6^20 ~ 0.1565 = 15.65 % or about 1 in 6.4
15*4^20/6^20 ~ 0.0045 = 0.45 % or about 1 in 222

It’s bad luck, but still relatively likely within a player base this size.

2 Likes
10

Combinatorial analysis is the type of math where it’s super easy to get something subtle wrong and not be able to figure out what until much later…

If the dragons are equally distributed for each roll, then the chance for any specific Dragon is obviously 1/6, with the chance of getting a duplicate equal to (# unique dragons owned)/6.

So if each roll was always one of 4 specific Dragons, that’s ostensibly (4/6) ^ 20 – BUT this doesn’t take into account (among other things) the intuitive fact that the first roll is never a duplicate.

Hmm.

12

This doesn’t take into account (among other things) the intuitive fact that the first roll is never a duplicate.

That’s what the 6*5/2 is for. There are 15 different combinations of 4 out of 6 dragons. => ABCDEF, excluding, AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF.

13

Yeah, I want to work out a general formula that I can build into my spreadsheet but I’m having problems getting it to work consistently across different cases.

For example, if you craft 2 eggs then the breakdown is intuitively:

  • 36 total possible outcomes
  • 6 are the same color dragon each time
  • 30 are differing color dragons (in specific order, i.e. AB vs. BA are counted separately)
  • All outcomes accounted for

And for the “3 egg” problem the breakdown is:

  • 216 total possible outcomes
  • 6 outcomes are 1 color 2 duplicates
  • 90 are 2 colors 1 duplicate (e.g. 6 * 5 * 3)
  • 120 are 3 colors no dupes (e.g. 6 * 5 * 4)
  • All outcomes accounted for

But as soon as I expand to the “4 egg” problem, either my breakdown doesn’t add up to the total or I can’t identify a consistent logic (i.e. formula) that yielded the result:

  • 1296 possible outcomes (6^4)
  • result: AAAA (1 color 3 dupes) = (6) outcomes
  • result: AAAB (2 colors 2 dupes) = (6 * 5) * 4 = 120 outcomes
  • result: AABB (2 colors 2 dupes) = (6 * 5) * 3 = 90 outcomes
  • result: AABC (3 colors 1 dupe) = (6 * 5 * 4) * 6 = 720 outcomes
  • result: ABCD (4 colors no dupes) = (654*3) = 360 outcomes
  • All outcomes account for

Intuitively this should be (# of color permutations) * (# orders in which each color was received – AABB vs. ABAB vs. ABBA; or AABC vs. ABAC vs. ABCA vs. BAAC vs. BACA vs. BCAA) but also intuitively, it should be possible to calculate a probability based solely on # colors across # eggs, without regard for the exact distribution of any dupes.

14

Number of eggs = N
Ways of picking X types * combinations of those X types / all outcomes
XC6 * X^N / 6^N

where nCr = n!/(r!(n−r)!)

15

Funny thing, I’ve used that formula before with no problems but it doesn’t seem to work here, because of a few obvious cases where the result is anywhere from obviously incorrect to patently absurd – for example, what is the probability of getting all 6 colors in just 6 eggs?

  • Assume: X = N = 6
  • nCr(6,X) = nCr(6,6) = 1 (intuitive)
  • X^N / 6^N = 6^6/6^6 = 1 (intuitive)
  • nCr(6,X) * X^N / 6^N = “100%” probability

lol yeah we WISH that were true right? But the correct answer is actually:

  • Given X=N=6
  • result = 6! / 6^6 = about 2% probability
The full breakdown of up to 6 eggs

gems.dragons.wtf
gems.dragons.wtf783×278 17.9 KB

(Apologies for swapping “X” and “N” here, but the rest is the same)

16

I see my mistake. I forgot to subtract the cases where there isn’t at least one of each of the selected, but then you have to re-add each of those double subtracted and so forth. I think there is a nicer formula for it, but I have forgotten how to formulate it.

For 1 out of 6 it is 6C1 * 1^6 / 6^6 = 1/6^5 = 1/7776 or 6/46656
For 2 out of 6 it is (6C2 * (2^6 - 2C1 * 1^6) / 6^6) = 930/46656
For 3 out of 6 it is (6C3 * (3^6 - 3C2 * 2^6 + 3C1 * 1^6) = 10800/46656
For 4 out of 6 it is (6C4 * (4^6 - 4C3 * 3^6 + 4C2 * 2^6 - 4C1 * 1^6) = 23400/46656
For 5 out of 6 it is (6C5 * (5^6 - 5C4 * 4^6 + 5C3 * 3^6 - 5C2 * 2^6 + 5C1 * 1^6) = 10800/46656
For 6 out of 6 it is (6C6 * (6^6 - 6C4 * 5^6 + 6C4 * 4^6 - 6C3 * 3^6 + 6C2 * 2^6 - 6C1 * 1^6) = 720/46656
Summing up to double check, 6 + 930 + 10800 + 23400 + 10800 + 720 does indeed sum to 46656.

17

Turns out the combinatorial coefficient (in this case, total unique distributions of N dragons across X eggs) is already known as a Stirling number of the second kind. Which wasn’t too difficult of a thing to find once I remembered that we’re dealing with a textbook “coupon collector’s problem” here.

So I plotted out the Stirling table at 6 columns wide and extended it out to a count of 30 eggs. A bit of minor formula debugging here and there (this is my first time using the VLOOKUP() function ) and … it’s all done!

gems.dragonlist.3

So, yeah, the “worst luck” I have recorded so far is only about a 1/750 incidence, or slightly better than pulling a base Mythic from a Gem Chest.

Thanks for all the help!

3 Likes
18

After pulling yet another dupe today, I realized that I was overcomplicating this problem. If I have X out of 6 dragons after Y pulls, the probability of me getting a new one is (6-X)/6. This is more easily written as a recursive function. Let P(x,y) be the probability of having x dragons after y pulls. Obviously P(1,1) = 1, as a base.
Let’s say I have 4 dragons after 7 pulls. After 8, I have a 2/6 chance of getting 5, 4/6 of remaining at 4.
(x,y) has a (x/6) chance to become (x,y+1), (6-x)/6 chance to become (x+1,y+1). Reversing, we get:

P(x,y) = P(x,y-1) (x/6) + P(x-1,y-1)((6-(x-1))/6)
This is much easier to write as an excel formula.

6 digit probability up to 50 eggs
1 2 3 4 5 6
1 1.000000
2 0.166667 0.833333 0.000000 0.000000 0.000000 0.000000
3 0.027778 0.416667 0.555556 0.000000 0.000000 0.000000
4 0.004630 0.162037 0.555556 0.277778 0.000000 0.000000
5 0.000772 0.057870 0.385802 0.462963 0.092593 0.000000
6 0.000129 0.019933 0.231481 0.501543 0.231481 0.015432
7 0.000021 0.006752 0.129029 0.450103 0.360082 0.054012
8 0.000004 0.002268 0.069016 0.364583 0.450103 0.114026
9 0.000001 0.000759 0.036020 0.277563 0.496614 0.189043
10 0.000000 0.000254 0.018516 0.203052 0.506366 0.271812
11 0.000000 0.000085 0.009427 0.144626 0.489656 0.356206
12 0.000000 0.000028 0.004770 0.101131 0.456255 0.437816
13 0.000000 0.000009 0.002404 0.069806 0.413923 0.513858
14 0.000000 0.000003 0.001208 0.047739 0.368204 0.582845
15 0.000000 0.000001 0.000606 0.032430 0.322750 0.644213
16 0.000000 0.000000 0.000304 0.021923 0.279768 0.698004
17 0.000000 0.000000 0.000152 0.014767 0.240448 0.744632
18 0.000000 0.000000 0.000076 0.009921 0.205296 0.784707
19 0.000000 0.000000 0.000038 0.006652 0.174387 0.818923
20 0.000000 0.000000 0.000019 0.004454 0.147540 0.847988
21 0.000000 0.000000 0.000010 0.002979 0.124434 0.872577
22 0.000000 0.000000 0.000005 0.001991 0.104688 0.893317
23 0.000000 0.000000 0.000002 0.001329 0.087904 0.910765
24 0.000000 0.000000 0.000001 0.000887 0.073696 0.925415
25 0.000000 0.000000 0.000001 0.000592 0.061709 0.937698
26 0.000000 0.000000 0.000000 0.000395 0.051622 0.947983
27 0.000000 0.000000 0.000000 0.000264 0.043150 0.956586
28 0.000000 0.000000 0.000000 0.000176 0.036046 0.963778
29 0.000000 0.000000 0.000000 0.000117 0.030097 0.969786
30 0.000000 0.000000 0.000000 0.000078 0.025120 0.974802
31 0.000000 0.000000 0.000000 0.000052 0.020959 0.978989
32 0.000000 0.000000 0.000000 0.000035 0.017483 0.982482
33 0.000000 0.000000 0.000000 0.000023 0.014581 0.985396
34 0.000000 0.000000 0.000000 0.000015 0.012159 0.987826
35 0.000000 0.000000 0.000000 0.000010 0.010137 0.989852
36 0.000000 0.000000 0.000000 0.000007 0.008451 0.991542
37 0.000000 0.000000 0.000000 0.000005 0.007045 0.992950
38 0.000000 0.000000 0.000000 0.000003 0.005872 0.994125
39 0.000000 0.000000 0.000000 0.000002 0.004895 0.995103
40 0.000000 0.000000 0.000000 0.000001 0.004080 0.995919
41 0.000000 0.000000 0.000000 0.000001 0.003400 0.996599
42 0.000000 0.000000 0.000000 0.000001 0.002834 0.997166
43 0.000000 0.000000 0.000000 0.000000 0.002362 0.997638
44 0.000000 0.000000 0.000000 0.000000 0.001968 0.998032
45 0.000000 0.000000 0.000000 0.000000 0.001640 0.998360
46 0.000000 0.000000 0.000000 0.000000 0.001367 0.998633
47 0.000000 0.000000 0.000000 0.000000 0.001139 0.998861
48 0.000000 0.000000 0.000000 0.000000 0.000949 0.999051
49 0.000000 0.000000 0.000000 0.000000 0.000791 0.999209
50 0.000000 0.000000 0.000000 0.000000 0.000659 0.999341

To answer the OP, after 20 eggs, missing 1 dragon probability is 0.147540, about 1 in 6.778, missing 2 is 0.004454, 1 in 224.5.

3 Likes
19

a typical reminder that 0.01% chance is still bigger than 0, and the event may even happen twice in a row. Humans are really bad at ‘natural’ understanding of probability theory.

1 Like
20

28 eggs now… Still missing 2… This may be more than typical RNG… there may be a bug… the first 4 I opened were all new dragons… one of each of the 4 I do have, then the duplicates started.

21

With this Rng issue, yes the odds are low for one to get so many duplicates, but indeed not zero. That’s not the issue as i, and many others getting excessive dupes see it.

One could theoretically open say 28 eggs, and all get Rubirath. It’s not a zero percent chance outcome, but it feels awful. Especially when you consider that if they set this up such as with the guardian troops, that same pull drop would have been done at most, by the 21st egg.

The extra dragonite would have obtained the weapons and last dragon.

The RNG crafting is annoying for the effort needed for dragonite, but it’s the lack of a failsafe that really makes this an issue in my opinion.

The guardians were also far more impactful in the meta requiring a massive change to gameplay and giving chills when hearing the Justice League.

2 Likes
22

24 eggs still missing 1