foosas Posted February 19, 2020 Share Posted February 19, 2020 (edited) Hey there, Being roasted by a lot of my own team and would love to just have a GM or Dev step in to help settle a debate. First, this post is not a "hot take" on how fair shiny rate is. It's 1/30K. This is to discuss if you can apply a binomial distribution to the approximation of outcomes over a fixed number of encounters (trials) Based on this statement we know that the odds of getting a shiny in any encounter is INDEPENDENT from any other encounter. We also know that the odds do not improve from encounter to encounter for a majority of players (anti-cheat measures excluded here) Thus we can use a binomial distribution to tell someone their statistical likelihood of seeing at least 1 shiny in X encounters. This does not guarantee anyone a shiny on their next encounter or mean that their ODDS in any one encounter are improving, but that their overall chances are improving. Then we can use a calculator to express the P(shiny) = 1/30,000 and make up a number of trials = encounters as 50,000. The odds of not seeing at least one shiny over 50,000 encounters is 81.1%. This does not mean you have an 81.1% chance of getting a shiny in 50,000 encounters. This does not mean that you have to wait 50,000 encounters to see a shiny, or that you can't get a shiny on encounter 1, 2, 3, 4, 5 One way to interpret this is that if 1,000,000 players did 50,000 encounters each, each one would have ~81% chance to see at least 1 shiny in their 50K encounters. I would love to hear from some other stats enthusiasts or a GM to help settle it! https://stattrek.com/online-calculator/binomial.aspx Edited February 19, 2020 by foosas Link to comment
GodofKawaii Posted February 19, 2020 Share Posted February 19, 2020 and you needed to make 2 pointless topics even tho...Shiny Rate is Fair! Koushi 1 Link to comment
Archiver Posted February 19, 2020 Share Posted February 19, 2020 (edited) And what's your point? Quote One way to interpret this is that if 1,000,000 players did 50,000 encounters each, ~81% of them would see at least 1 shiny in those encounters. Also that is wrong. If 1million players did 50k encounters then 100% of them would have an 81% chance of seeing a shiny in those 50k encounters. Edited February 19, 2020 by Archiver FighterChamp and Rittz 2 Link to comment
pachima Posted February 19, 2020 Share Posted February 19, 2020 There are Maths, Bad Maths, Good Maths, useless Maths, and easy Maths. The easy Math is to get a calculator, insert 1-[(29999/30000)]^n, where n is the number of encounters and voilá. No need for further Bad and useless maths. Enjoy. OrangeManiac, Lin, Archiver and 1 other 4 Link to comment
foosas Posted February 19, 2020 Author Share Posted February 19, 2020 3 hours ago, Archiver said: And what's your point? Also that is wrong. If 1million players did 50k encounters then 100% of them would have an 81% chance of seeing a shiny in those 50k encounters. Thank you for that correction. The point was that at any given number of encounters you have an idea of how probable or improbably you might be on your journey to the magic encounter. The odds of how against you or for you the odds can be. Does a binomial fit here? Link to comment
Archiver Posted February 19, 2020 Share Posted February 19, 2020 38 minutes ago, foosas said: Thank you for that correction. The point was that at any given number of encounters you have an idea of how probable or improbably you might be on your journey to the magic encounter. The odds of how against you or for you the odds can be. Does a binomial fit here? I'm pretty sure people already knew this. And what you're doing is the Bernoulli process, not binomial. Link to comment
foosas Posted February 20, 2020 Author Share Posted February 20, 2020 6 hours ago, Archiver said: I'm pretty sure people already knew this. And what you're doing is the Bernoulli process, not binomial. Bernoulli process is a list of independent variables that meets some conditions. It leads to a distribution where if n>1 then its a binomial distribution. In this case we are taking N>1 independent events with a yes (shiny) and no (not shiny) outcome with p(yes) as 1/30000. This (and independence of the events) make for a binomial distribution. When n = 1 a bernoulli distribution is a special case of binomial distribution Link to comment
Kole Posted February 26, 2020 Share Posted February 26, 2020 (edited) To be clear we had a different idea of what independent odds mean as it applies to pokemmo shinies and were relaying information given by the developers over several posts, covering several years. I don't believe "roasted" applies here, at least certainly not intentionally. I do welcome the discussion and any input from developers, however. Edited February 26, 2020 by Kole Link to comment
TohnR Posted February 27, 2020 Share Posted February 27, 2020 On 2/19/2020 at 6:10 PM, pachima said: No need for further Bad and useless maths. Enjoy Thanks, imma use this when @pachima annoys me with his irrelevant calcs ^^ Revz 1 Link to comment
OrangeManiac Posted February 27, 2020 Share Posted February 27, 2020 (edited) The exact opposite. It's 81,1% chance of getting at least one shiny at 50k encounters. 29999/30000 = not finding a shiny (29999/30000)^n = not finding a shiny at n amount of encounters. So the number you get from it is the chance of not finding a shiny. Other than that this post basically just recapped everything that Desu wrote. Edit: Ah later in the post it was corrected Edited February 27, 2020 by OrangeManiac Link to comment
Riesz Posted February 28, 2020 Share Posted February 28, 2020 No, shiny odds is 50%, the pokemon you encounter is either shiny or non shiny imphilxd 1 Link to comment
kevola Posted February 28, 2020 Share Posted February 28, 2020 Yeah, you can use Binomial distribution. If you're trying to find the probability of finding exactly one shiny after n encounters, it would work, however the way Orange stated is a lot easier, and gives you the same answer. Having said that, you can do more cool things with Binomial distribution, i.e if you want to find the probability of exactly 3 shiny encounters, or at least 4 shiny encounters etc after n encounters in total. Does that answer your question? Link to comment
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now