What are the chances...

[Leiana]

Now, I'm not especially good at maths, but I think that I calculated this correctly.

I have 2 towns. In my first town I have

Rodney
Apollo
Renee
Tabby
Walker
Antonio
Frobert
Celia
Simon

and in my second town I have

Rodney
Apollo
Renee
Tabby
Chevre
Beau
Jay
Broffina
Gabi

With the exception of Antonio in the first town (he replaced Baarbara who I lost in a TT accident) all the villagers are original from when I started the towns, yet I have 4 that are in both towns. I've counted there are at least 343 possible villagers in ACNL so having 2 the same would be 1/117,649. 3 the same 1/40,353,607. And 4 the same 1/13,841,287,201 (multiplying the previous chance by 343 because there are 343 villagers)

I'm either getting the maths COMPLETELY wrong here, or there's some other mechanism that decides what villagers you'll get when you start a town, because I just don't believe that having 4 villagers the same would be a 1 in almost 14 billion chance.

What do you think?

 

[Louis]

( 4) x (339)
(343) (343)
------------- x 2 x 100% = ?
(343)
(343)

This is the correct mathematical response. I don't have a calculator at the moment, so I don't know the last answer

 

[Yui Z]

I think I just found out I suck at math... ._.

 

[Yen Quest]

There are words going around about how certain villagers are rarer than the others because it can't be a part of the starter 5 (such as the octopus and certain type of personality). This should cut the probability down some.

 

[momayo]

The math for having a specific villager is actually a bit more complicated than that.

The probability of getting those 4 villagers is:

(1/no. of peppy) * (1/no. of cranky) * (1/no. of uchi) * (1/no.of smug) * 24
Case: neither one of the four of them is the ninth villager. That is, there are no repeating peppies, crankies, uchi or smugs.

PLUS

(1/no. of peppy) * (1/no. of uchi) * (1/no. of smug) * ((no. of cranky - 1)/no. of cranky) * 120 * (1/335)
Case: The ninth villager is a cranky, meaning there are 2 crankies in the village. The ninth cranky is the cranky you are looking for. There are 120 different ways to order 5 villagers.
335 = 343 - 8 (all villagers available minus the 8 that are already in the village)

PLUS

all the cases where the peppy, uchi, smug is the ninth villager.

Assume that we don't care whether or not the peppy or the cranky came from your starting roster, also that we don't care about the 6 other villagers.

No. of uchi = 21
No. of cranky in New Leaf = around 45
No. of peppy in New Leaf = around 44
No. of smug = 31

Summing it all up:

(1/no. of peppy) * (1/no. of cranky) * (1/no. of uchi) * (1/no.of smug) * 24 = .001862%
2 crankies in your village, one of which is Apollo = .0012230%
2 peppies in your village, one of which is Tabby = .0011950%
2 uchi in your village, one of which is Renee = .0005580%
2 smug in your village, one of which is Rodney = 0.0008337%

Total is 0.005672%, or 5 out of 100,000.

I'll stop, I keep waffling between different approaches D:

Fairly sure that the answer is most likely between 1-5 out of 100,000, though.

 

[Leiana]

Thank you for that explanation! I can't say I really understand it though XD

 

[momayo]

Thank you for that explanation! I can't say I really understand it though XD

I changed it, because I was dumb, haha! But, bottom line is, you can't just multiple 1/343 because your first 8 villagers are constrained by their personality type!

 

[Siren137]

Plus each time you get a villager the total number of villagers left you can select is one less.

First villager is picked from a total of 343, you can't get them again so you minus one from the total so next villager is 1 in 342 etc etc.

That's not including the different villager types of course!

 

[kasane]

Oh wow dem Maths o.0