Still neck division

[.Memories.]

I'm getting .012%, but still very unlikely.

[onealexleft]

(01-15-2014 02:31 PM).Memories. Wrote:  I'm getting .012%, but still very unlikely.

doh! I was calculating the odds of pulling the same names out twice in a row (dependent events) and not independently. I've always sucked at probabilities/combinatorics. Yes, hitting any particular name is 0.012%, therefore hitting 'still neck' again is the same 0.012%.

[TheGreatErenan]

90 adjectives * 89 nouns = 8010 possible division names. So each unique name has a likelihood of appearing that's 1/8010. That's about 0.0125%.

But then the likelihood of getting the same name twice depends upon how many times you generate names. Say you generate a name, and we get "Still Neck." Then the number of subsequent generations you plan on making affects your big picture likelihood of getting that same name again. If you generate one additional division name, then your likelihood of it being "Still Neck" again is [(1/8010) * 1] = ~0.0125%. If you generate two additional division names, then the expected likelihood would be [(1/8010) * 2] = ~0.025%. So the chance of getting any name to appear twice is actually [(1/8010) * (N - 1)] where N is the number of divisions.

EDIT: Wait, actually hang on...

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Yeah, I think I'm wrong, but it's still definitely higher than 0.012%.

[HuskyPete]

I believe the likelihood is the same no matter how many divisions you create. Each event is separate. That's why Vegas is kind enough to put up a list of numbers at the roulette table. "12 red in a row! I bet it all on black!" We all know how that ends up. ;-).

[TheGreatErenan]

I want to say it's this:

1 - [(8010!/(8010 - N)!) / (8010^N)]

8010^N is the number of possible outcomes after the total series of N division name generations.

(8010!/(8010 - N)!) is the number of possible ways we could generate N division names without any repeats.

Therefore (8010!/(8010 - N)!) / (8010^N) is the probability of getting no repeats at all. Subtract this value from 1 and we get the probability of getting at least one repeat.

All of this, of course, assumes that we have 8010 divisions or fewer. If we had more than 8010, then a repeat would be guaranteed.

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(01-16-2014 04:09 AM)HuskyPete Wrote:  I believe the likelihood is the same no matter how many divisions you create. Each event is separate. That's why Vegas is kind enough to put up a list of numbers at the roulette table. "12 red in a row! I bet it all on black!" We all know how that ends up. ;-).

This is only true if you're evaluating the likelihood of getting a particular name on any particular generation. If you're talking about generating 1000 names and calculating the likelihood that two of those names are the same, it's a different story.

To illustrate, if you flip a fair coin, it's 50% that you'll get heads. But if you flip it 1000 times, it's not 50% that one of those flips was heads. You're practically guaranteed that one of them will be heads, since while you have 2^1000 possible outcomes, only one of those outcomes is flipping tails 1000 times. So you've got 1 - 1/(2^1000), which is over 99%.

[Chemoeum]

Math.

[amoffett11]

This is the same as the classic Birthday problem: given 25 people in a room, what are the odds that at least 2 of them have the same birthday. People are usually quite surprised to find that the odds are slightly higher than 50% that 2 people will be born on the same day.

Given the number of division there are, and the number of seasons now that divisions have been generated (4), the odds are very high that we'd get at least one collision. I'd bet that there's been some before and it's just gone unnoticed.

[TheGreatErenan]

(01-16-2014 04:23 AM)amoffett11 Wrote:  Given the number of division there are...

I wonder how many there are... And I'm assuming we're counting 1v1 and 2v2 (do 2v2 arranged and 2v2 random have separated divisions?).

[.Memories.]

(01-16-2014 04:34 AM)TheGreatErenan Wrote:  (do 2v2 arranged and 2v2 random have separated divisions?).

Yes.

[TheGreatErenan]

I'm doing some calculations based on my above post, and unless my program is buggy or my formula is wrong, it appears that with 105 divisions, you'd have about a 50% chance of getting at least one repeat name.

Looks like you pass 1% probability at around 13 divisions.

Furthermore, if you had 500 divisions and wanted to have 1% probability of repeat names, then you'd need about 12.5 million possible names. That means that if nouns and adjectives were still about equal, you'd need about 3536 of each.

Assuming my math and programming are right, obviously.