That might be the first time I got the whole set in one drop
Only 1 out of 5 was usable though 😎👍
Commentaires
Fuchsfell02026-01-10 01:04
That's pretty impressive.
I was curious now: that's about a 1 in 35.000.
Fuchsfell02026-01-10 01:45
If anyone's curious how I got that:
Assuming the wikis data is correct.
6.5% (0.065) for two Artifacts.
0.5 for each set.
0.2 for each piece.
Getting 5 pieces can be model via binomial distribution with n=3, p=0.065. Since one artifact is guaranteed, we get a minimum of 3 so 2 successes are needed to get to 5, which comes out to 0.01185 (~1 in 84).
Getting the same set all five times, assuming that it doesn't matter which set it is, comes out to 2*0.5^5 = 0.0625 (1 in 16).
The first artifact can be any one piece. The second one has to be any of the other four and so forth. This comes out to 1*4/5*...1/5 or (5!)/(5^5 ) = 0.0384 (~1 in 26).
Now multiply all that up and you get 0.00002844 = 0.002844% or ~1 in 35.000.
That was fun to write, maybe someone cares
Edit: Reddit messed up some calculations
CriticalDetective4312026-01-10 01:51
I care! Thank you so much for writing that up! You are the kind of person that makes the internet a better place.
acidrainfalls2026-01-10 02:30
I understand very little of that, (I do not have math brain) but it is very fascinating nonetheless!! :D
Own_Construction_9652026-01-10 05:54
r/theydidthemath
Cocoayashi2026-01-10 06:16
That was fun! Thank you!
ellopon2026-01-10 10:26
don't listen to the heater, math is fun some times
Flaky-Significance692026-01-10 19:50
That’s so cool. I took stats class back in school but no way I can understand that now without doing some serious studying.
I just know that math can be really beautiful sometimes.
Key_Judgment_41282026-01-10 23:09
Since the post is about just getting a full set you could calc as a little challenge the chance if you also include drops with 6 artifacts. Would look way less neet tho in the screenshot.
Fuchsfell02026-01-11 00:35
Done.
Something interesting happens here.
One could expect that it becomes more likely, since there is now one more artifact that helps to get all five pieces. But because its so much more unlikely to get 5 artifacts, the odds drop pretty hard.
Assuming, again, that it doesn't matter which set get all five pieces, there will be two cases.
Case A: One artifact doesn't match the other 5.
Case B: All artifacts are the same set.
For both cases, getting 6 artifacts, has the odds 0.000275 (\~1 in 3636)
Case A: 2 Possible Sets to choose from, then 6 possible drops to choose which is off-set. Every combination appears with odds (0.5)\^6 which gives 3/16 chance of case A happening. Then from the five artifacts all must be different pieces so its the same odds as before 0.0384. For case A we get 3/16 \* 0.0384 = 0.0072.
Case B: 2 \* (0.5)\^6 = 0.03125 (1 in 31) odds of getting all 6 pieces from the same set (times 2, because, it doesn't matter what set it is). Now we gotta count all different outcomes. Each artifact can be of five pieces, so we get 5\^6 = 15.625 possible outcomes. Since we have 6 artifacts and 5 pieces, there has to be at least one dupe (The Pigeonhole Principle, fun one, you should look it up).
Which artifact is duped gives us 5 options, with 6 over 2 = 15 options for which two contain the dupe. The remaining 4 can be arranged in the other 4 slots in any order, so we get 4! = 24 options. This leaves us with 5 \* 15 \* 24 = 1800 possible outcomes that have all five different pieces, so 1800/15.625 = 0.1152 odds of getting this. Which leaves 0.03125 \* 0.1152 = 0.0036 odds of getting Case B and 5 different pieces.
Combining both results gives us the combined odds of 0.000275 \* (0.0072 + 0.0036) = 0.00000297 (\~1 in 336.700, or about ten times less likely then the five piece case).
This was a test, Case A was easy, but Case B was quite difficult. Thanks for the task :D
Key_Judgment_41282026-01-11 01:11
Nice work. Can confirm the result, tried it myself after I commented earlier. Ended up with the same ideas as you but for case 2. I drew a probability tree since it was small enough. Had a neet shortened formula to clac the probability which is probability related to your thoughts.
Fuchsfell02026-01-11 01:39
So I didn't accidentally mess something up, that's good lol
It was good practice aswell for statistics exam \^-\^
SoC1752026-01-10 01:35
Whatever the stats are, you now have to use it in exactly that combination
dragosslash2026-01-10 01:57
Witnessed. Impressive!
Yuukiko_2026-01-10 04:55
Pretty much impossible(I think?) before the Luna patches though, you could only get 4 max
RemarkableRing27762026-01-11 21:13
Not impossible, as youd have to spend a fragile resin but thst makes it fsrrrr less likely to see
MightySamMcClain2026-01-10 05:33
There was a cave with a whole set in it, i think was by starsand shoals iirc but that's pretty wild from a domain
Ok_Might_46912026-01-10 05:54
Wow! Crazy! What are the odds?
Titania-sama2026-01-10 12:24
Apparently 1 in 35 000 😂
Ok_Might_46912026-01-10 13:01
That's less than a triple 5* pull
Flaky-Significance692026-01-10 19:50
Holy
TheToppatAmongUs2026-01-10 08:19
Now do it again but with the artifact types in order (Flower Feather Sands Goblet Circlet)
Fuchsfell02026-01-10 23:06
That would be crazy, when the order gets specified like that the odds of it happening jumps from 1 in 35.000 to about 1 in 4.2 Million or half that if you specifically want one set lol
Commentaires
I was curious now: that's about a 1 in 35.000.
Assuming the wikis data is correct.
6.5% (0.065) for two Artifacts.
0.5 for each set.
0.2 for each piece.
Getting 5 pieces can be model via binomial distribution with n=3, p=0.065. Since one artifact is guaranteed, we get a minimum of 3 so 2 successes are needed to get to 5, which comes out to 0.01185 (~1 in 84).
Getting the same set all five times, assuming that it doesn't matter which set it is, comes out to 2*0.5^5 = 0.0625 (1 in 16).
The first artifact can be any one piece. The second one has to be any of the other four and so forth. This comes out to 1*4/5*...1/5 or (5!)/(5^5 ) = 0.0384 (~1 in 26).
Now multiply all that up and you get 0.00002844 = 0.002844% or ~1 in 35.000.
That was fun to write, maybe someone cares
Edit: Reddit messed up some calculations
I just know that math can be really beautiful sometimes.
Something interesting happens here.
One could expect that it becomes more likely, since there is now one more artifact that helps to get all five pieces. But because its so much more unlikely to get 5 artifacts, the odds drop pretty hard.
Assuming, again, that it doesn't matter which set get all five pieces, there will be two cases.
Case A: One artifact doesn't match the other 5.
Case B: All artifacts are the same set.
For both cases, getting 6 artifacts, has the odds 0.000275 (\~1 in 3636)
Case A: 2 Possible Sets to choose from, then 6 possible drops to choose which is off-set. Every combination appears with odds (0.5)\^6 which gives 3/16 chance of case A happening. Then from the five artifacts all must be different pieces so its the same odds as before 0.0384. For case A we get 3/16 \* 0.0384 = 0.0072.
Case B: 2 \* (0.5)\^6 = 0.03125 (1 in 31) odds of getting all 6 pieces from the same set (times 2, because, it doesn't matter what set it is). Now we gotta count all different outcomes. Each artifact can be of five pieces, so we get 5\^6 = 15.625 possible outcomes. Since we have 6 artifacts and 5 pieces, there has to be at least one dupe (The Pigeonhole Principle, fun one, you should look it up).
Which artifact is duped gives us 5 options, with 6 over 2 = 15 options for which two contain the dupe. The remaining 4 can be arranged in the other 4 slots in any order, so we get 4! = 24 options. This leaves us with 5 \* 15 \* 24 = 1800 possible outcomes that have all five different pieces, so 1800/15.625 = 0.1152 odds of getting this. Which leaves 0.03125 \* 0.1152 = 0.0036 odds of getting Case B and 5 different pieces.
Combining both results gives us the combined odds of 0.000275 \* (0.0072 + 0.0036) = 0.00000297 (\~1 in 336.700, or about ten times less likely then the five piece case).
This was a test, Case A was easy, but Case B was quite difficult. Thanks for the task :D
It was good practice aswell for statistics exam \^-\^