Still neck division
01162014, 04:12 AM
(This post was last modified: 01162014 04:18 AM by TheGreatErenan.)
Post: #25




RE: Still neck division
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. (01162014 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%. 

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