Rising in the Rankings

[Flarp55]

(06-01-2014 03:50 PM)TheGreatErenan Wrote:  

(06-01-2014 03:31 PM)Bbobb555 Wrote:  Ok, so let x be the number of P1 games for any given player. Then the probability that he/she will have no P1 games on some map is about 12*(1-(11/12)^x). So if a player has played 50 games as P1, then there is actually still about a 12% chance that there are no games for some map. Note that only P1 season 5 games should be counted.

Um, wait. Assuming each map is equally likely to appear, the number of possibilities for map distribution over x matches played is 11^x (eleven possible maps for each of x matches). The number of ways you can play x matches without playing a single match on some particular map is (11-1)^x or 10^x. So I would think the actual probability of this happening is (10^x)/(11^x). If you play 50 matches, that's (10^50)/(11^50), which is less than 1% (about 0.8518551%)

Note that you can't simply multiply this number by 11 to get the probability of not getting every map at least once during your match history, as 10^x already includes a lot of scenarios that exclude other maps. For instance, if the map excluded is, say Glitch, that 10^x includes the scenario of getting SFI x times, which excludes every map except SFI.

Actually, you can simply multiply it by 11, the reason being that the probability that two maps have no matches is extremely small. It only makes less than a 2.5% (9^50/10^50 * 10/2 if you are interested) change in the overall result. (Note: I am ignoring the chance that 3 maps have no matchups, etc. because these have almost no effect and furthermore, they actually increase the probability) When the probability is extremely small, terms further in the PIE expansion tend to have next to no effect at all. I was aware of this and that is why I said "about."

Aside from that, I did make 2 stupid mistakes. First, you were right, there should be no subtraction from 1, and that was a silly mistake. Also, I thought there were 12 maps instead of 11.

Revised:
The probability is, as you said, 11*(10/11)^50, which is still about 9.37%. If we include the probability of two maps being the same, it is (11*10^50-55*9^50)/11^50 which is about 9.13%. As you can see, there is a very small difference in the two, simply because the probability of two maps both having no matchups is negligible.

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(06-01-2014 08:40 PM)Demon Wrote:  Don't over think it. Chance of getting SFI is 1/11, roughly 9%. Chance of not getting SFI is 10/11, roughly 91%. Chance of not getting SFI 50 times in a row is (10/11)^50, so about 0.852%. Very unlikely. Note that while Erenan used different logic, he arrived at the same calculation and is certainly correct

SFI is just an example, of course, this calculation will work for all maps.

Not different logic, it's the exact same logic. But you have to consider that you can get 0's not only on SFI, but on other maps, too, and the probability that some map has no matchups is over 9%.

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The full expression for the probability is (11*10^50-55*9^50+165*8^50-330*7^50+462*6^50-462*5^50+330*4^50-165*3^50+55*2^50-11*1^50)/11^50 = 9.130942597%

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Also note that this probability varies a lot as x varies. For example:
p(10) is 1.
p(20) is about 88.72%. I included all the terms in this.
p(30) is about 50.80%. I included 5 terms in this calculation since it is a large probability.
p(40) is about 22.56%.
p(50) is about 9.13%.
p(60) is about 3.58%.
p(70) is about 1.39%.
p(80) is about 0.536%.
p(100) is about 0.0798%.