I just can’t believe it. How unlucky do you want to be? yes
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Are you not having fun? 
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You know what to do when you next have 550 dragonite 
Maybe you don’t want an actual answer, but I think it’s interesting to try to provide one.
If your question is: how unlucky is it to open 34 eggs and to not have Lunarelleon, this is (5/6)^34 = 0,203%, or about a chance of 1 in 492 people.
If your question is: how unlucky is it to open 34 eggs and to still miss at least one dragon, I think this is (5/6)^34x6-(4/6)^34x15+(3/6)^34x20-(2/6)^34x15+(1/6)^34x6 = 1,217%, or about 1 in 82 people.
So definitely unlucky, but also not very uncommon given the size of the playerbase.
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welcome to the club… took me 33 eggs for the gem dragons
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Look how much fun you’re having now!
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Wait, shouldn’t those two calculations yield the same value as each other?
Your first example counts all streaks to not drop dragon “X” (X could actually be any arbitrary color of the six, as long as it stays consistent throughout the run), regardless of how many types actually dropped throughout the streak, thus it is the probability of missing at least one dragon. While your second example is … explicitly described as the probability to “still miss at least one dragon” ?
It is somewhat simpler to calculate these long probabilities recursively – for example, after any number of unique dragons (D) from a specific run of eggs (N), your chance of crafting a dupe is (D/6) while your chance of crafting a new dragon is (6-D)/6. Thus, any specific probability(D,N) is the sum of: getting a dupe ( = probability(D,N-1) * (D/6) ) or getting a new color ( = probability(D-1,N-1) * (6-D)/6 ).
Me, I stockpiled Dragonite until I could craft a batch of 10 eggs, but it still took 8 crafts to land that last (Gem) Dragon. Which is unlucky, yes, but “only” garden-variety unlucky.
This is the scary bit, hey
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its 1/82 for EACH of the 6 dragons to be the one you specifically did not get, so yea, if you add up all the odds of the 6 dragons being the specific one missing, you get the same odds as just missing any 1 of them.
I thought my dupe count was high i got 16 extras each dragon round but that is so sad
Cap em at 4 like seal cards
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Almost. The 1 in 82 includes people who miss more than one dragon after opening 34 eggs. That’s why the formula gives a chance of 1 if you replace 34 by 1,2,3,4 or 5.
Recursion can probably also work. I used the inclusion-exclusion principle to find a direct formula and personally prefer working with that.
I see, you’re referring to permutations then (not combinations). Got it.
However, something about your calculations doesn’t add up … I’m trying to spreadsheet this using your method but I’m getting some patently absurd results so far – e.g:
-
Probability of exactly 2 colors after 2 eggs:
= combin(6,2) * (2/6)^2 - combin(6,1) * (1/6)^2
= 15 * (4 / 36) - 6 * (1 / 36)
= (60 / 36) - (6 / 36)
= (54 / 36)
= 150% (correct answer = 5 / 6) -
probability of exactly 5 colors after 5 eggs:
= (5/6)^5 x6 - (4/6)^5 x15 + (3/6)^5 x20 - (2/6)^5 x15 + (1/6)^5 x6
= 6 x (3125/7776) - 15 x (1024/7776) + 20 x (243/7776) - 15 x (32/7776) + 6 x (1/7776)
= (18750 / 7776) - (15360 / 7776) + (4860 / 7776) - (480 / 7776) + (6 / 7776)
= (3390 / 7776) + (4860 / 7776) - (480 / 7776) + (6 / 7776)
= (3390 / 7776) + (4380 / 7776) + (6 / 7776)
= (7776 / 7776)
= 100% (correct answer = permut(6,5) / 6^5 or ~9.2%)
…Help? I think the problem could lie in specifying the correct color combinations per argument (and this is why I hate combinatorics)…
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I didn’t expect someone to go that deep into my math, but hopefully this clarifies things:
My claim is that if you open n > 0 eggs, then the chance of still missing at least one dragon, which is the same as the chance of not having all six dragons is equal to:
(5/6)^n x6-(4/6)^n x15+(3/6)^n x20-(2/6)^n x15+(1/6)^n x6
So your calculation for 5 eggs is correct, because you can’t have all 6 dragons after opening 5 eggs. And for 2 it becomes:
(5/6)^2 x 6 -(4/6)^2 x 15 + (3/6)^2 x 20 - (2/6)^2 x 15+ (1/6) ^2 * 6 =
1/36 x (25 x 6 - 16 x 15 + 9 x 20 - 4 x 15 + 6) =
1/36 x (150 - 240 + 180 - 60 + 6) = 1
Posting this again:
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 |
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I guess the actual reason I’m persisting here is because the particular method I’m currently using in my spreadsheet (which matches the results in James’s post above) suddenly breaks down after 30 eggs:
(graphed: probability of exactly 5 colors after N eggs)
(the error presumably occurs during intermediate steps of the calculation)
…so of course I want to fix it.
I like inclusion-exclulsion as a method since it doesn’t involve recursion all the way down to 1 egg, but the formula calculation (as I have it currently written) is … well, just look at it:
(The formula used is consistent throughout the table)
Obviously, something in my formula is not working for the simple cases, implying I cannot trust it for the more complex cases – though it DOES somehow yield the correct probability for acquiring all 6 colors after any N>5 eggs. Any idea what’s going wrong here?
Your formula for 1 is correct.
Your formula for 6 is correct because of cancellation?
For 2 2, (Cell R5)
So if you pick 2 colors, the odds of only getting those 2 is indeed(2/6)^2 = 1/9.
Let’s count the combinations
12 13 14 15 16
23 24 25 26
34 35 36
45 46
56
That’s 15 combinations.
These are fulfilled 15 * 1/9 times, = 15/9
Removing doubles:
11 also fulfills only getting 12, 13, 14, 15, 16
22 fulfills getting 12, 23, 24, 25, 26.
So both 11 and 22 fulfill only getting 12.
Each double is actually counted 5X2 times.
The chance of each combination counted in the previous cell is = Q5 / COMBIN(6/Q$2) = 1/6 /6 = 1/36
There are actually 2 * 15 = 30 overlaps.
15/9 - 30 * 1/36 = (60-30)/36 = 30/36 = 5/6
The formula then becomes (R$2/6)^$A5*COMBIN(6,R$2) - (Q5/COMBIN(6,Q$2)*COMBIN(R$2,Q$2)*COMBIN(6,R$2))
Note, COMBIN(R$2,Q$2) = R$2
=> (R$2/6)^$A5 * COMBIN(6,R$2) - (Q5/COMBIN(6,Q$2) * R2 * COMBIN(6,R$2))
You needed to cancel the first cell 30/6 =5 times, not once.
The combination overlap is consistent as you increase eggs. I haven’t yet figured out how to pick 3 out of X with this method. Is your formula for the last column the same, subtracting the cell before it?
For 3, we have 20 combinations out of 6. 3 ways to remove a dragon from the selection, 3 ways to remove 2. 20 * 3 /15 = 4
20 * 3/6 = 10
So we remove the previous cell 4 times, the one before that 10 times.
For 4, we have 15 combinations out of 6. 4 ways to remove 1 dragon, 6, to remove 2, 4 to remove 3.
15 * 4/20 = 3
15 * 6/15 = 6
15 * 4/6 =10
For 5 we have 6 combinations out of 6. 5 ways to remove 1, 10 for 2, 10 for 3, 5 for 4.
6 * 5/15 = 2
6 * 10/20=3
6 * 10/15=4
6 * 5/6=5
For 6 dragons, it’s 6,15,20,15,6. For completion’s sake
1 * 6/6 = 1
1 * 15/15=1
1 * 20/20 = 1
1 * 15/15 = 1
1 * 6/6 = 1
Thank you for getting me to revisit this.
For dragons 1-6 for each row, let’s call the combinations without subtraction A to F,
your cells A’ - F’,
and the actual values A" - F"
A" = A
B" = B - 5A"
C" = C - 4B" - 10A"
D" = D - 3C" - 6B" - 10A"
E" = E - 2D" - 3C" - 4B" - 5A"
F" = F - E" - D" - C" - B" - A"
A’ = A"
B’ = B - A’ = B" + 4A"
C’ = C - B’ = C - (B"+4A) = C - B" - 4A"
compare this to C" = C - 4B" - 10A", C’ = C" + 3B" + 6A
D’ = D - C’ = D - (C" + 3B" + 6A)
compare this to D" = D - 3C" - 6B" - 10A", D’ = D" + 2C" + 3B" + 4A"
E’ = E - D’ = E - (D" + 2C" + 3B" + 4A")
compare this to E" = E - 2D" - 3C" - 4B" - 5A", E’ = E" + D" + C" + B" + A"
F’ = F - E’ = F - (E" + D" + C" + B" + A") = F" !!!
Amazingly, all the errors cancel out!
(edit, * becomes italics)
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Thanks! It’s not pretty, but the table is nonetheless working now without recursion into the previous row (not that it makes a difference for practical cases, but still).
Those are … binomial coefficients, aren’t they?