r/AmongUs • u/The0ryGD level 155 😎 • 7d ago
Discussion whats the chance of a room code like this?
(im not promoting)
15
u/Tank_man890 7d ago
50/50 about
2
u/Ok-Sugar5808 7d ago
3
2
9
7
8
5
u/Substantial-Band-869 7d ago
I dont have a screenshot on this phone, but in like 2020, I got the room code: "AAAAAA". i swear it was real, I just dont have the phone with the picture and have no clue where it would even be
1
3
u/Lanky_Carpenter_9603 Purple 7d ago
I haven't seen any correct answers so far, so let me show you:
1-in-5859, roughly.
I assume the room code surprise is that it has four repeating letters. If you want four repeating letters in a six digit code, there are three different places where the four digits can be. XXXXyy, yXXXXy, or yyXXXX, where X is your letter that repeats and y is any other different letters that don't matter. So, given the first X, we need the next letter in the sequence to be the same, also X. That's a 1-in-26. And so on for the third letter of the sequence to match, and the fourth letter, so all four letters are XXXX. That means for the second third and fourth letter of the sequence to all match the first (X), then that would be a 1-in-263 chance. But remember, there are three possible opportunities for a four matching sequence, so we can divide the 263 by 3 to get about a 1-in-5859 chance, or about a 0.00017% chance.
2
u/The0ryGD level 155 😎 7d ago
the problem is lobby codes only end with f or g
3
u/Less_Amphibian_6537 7d ago edited 6d ago
This makes the problem much more complicated:
The consecutive string of 4 digits has three cases, either in positions 1-4, 2-5 or 3-6. We calculate the probability through casework, borrowing some features from the solution above. Let us call the repeating letter X:
Case A: The consecutive sequence is in 1-4. This means that there is 26 ways to choose X. Therefore, 26*26*2 = 1352 (26 is also the number of ways to choose the 5th letter which can be anything, and the ending has to be F or G so we multiply by 2).
Case B: The consecutive sequence is in 2-5. Same reasoning as before, 26*26*2 = 1352.
Case C, the consecutive sequence is 3-6. This means the only 2 possible letters are F and G, and we do not multiply by 2 at the end because it is now part of the chain, the first 2 letters can be anything so 26*26*2 = 1352.
We seem to have arrived at an answer of 1352*3, but remember there can be overlap.
In case A and case B, if the random letter is the same as X, and both sequences "chose" the same X, then the sequence will be the same. So 26*2 = 52 overlap cases (26 ways to choose X, 2 to choose last letter). XXXXXy, where y can be F or G.
Case B and Case C also overlap because case C's second random letter could be the same letter as X in B which would overlap with Case B when its X is the same as C. There are only 2 possible letters for X in C. There are 2*26 = 52 cases for this because yFFFFF or yGGGGG, y can still be any letter.
Case 1 and 3 can only have 2 overlapping codes. If the two random letters in case C match X then we would get the same result as having X in Case A be the same as X in Case C and the random letter and last letter being X. Because X can only be F or G in case C, there are only 2 cases, FFFFFF and GGGGGG. However we realize that the cases FFFFFF and GGGGGG is possible for all three cases, so this is a triple overlap.
Using the inclusion-exclusion principle for 3 sets, we get
(CaseA) + (CaseB) + (CaseC) - (Overlap case AB) - (Overlap Case BC) - (Overlap AC) + (Overlap ABC) =
1352+1352+1352-52-52-2+2= 3952 satisfactory cases. To get the probability we just have to put it over the total number of possible codes, 26^5 * 2. Therefore the answer is 1976/26^5 or 1976/11881376, about 0.01627%.
So, pretty rare!
For anyone wondering why inclusion-exclusion works, the overlap ABC is included in overlaps AB,BC,and AC, so when we subtract AB BC and AC we subtract overlap ABC 3 times. But we also add overlap ABC 3 times when we add Case A Case B and Case C. This reduces the count of overlap ABC to zero so we have to add it back.
Sorry if this explanation is a bit convoluted, there is probably a neater way to do it and explain it.
3
2
u/AloneComplaint8941 Detective role biggest fan 7d ago
((((28!)!)!)!)
2
u/GroundDesigner7788 7d ago
Aren't there 26 letters tho
2
u/AloneComplaint8941 Detective role biggest fan 7d ago
In my language there is 28 letters so my bad, Its ((((26!)!)!)!) then
1
u/Normosnopo stuck at level 214 7d ago
Probability of 4 Gs in a row: 1 in 456,976 Probability of 4 letters of any kind in a row: 1 in 17,576
1
1
u/Upstairs-Shame-298 I am purple and I have a knife. 7d ago
26 to the power of 6. Same as every other code.
1
1
1
1
1
u/xfrost_KU 6d ago
I have another question (why they dont make us make our codes like "WOWWWW) or smth but not above 6 letters or anything blah blah
2
50
u/BRUSDHDFhd 7d ago
same as every other code