Bytefixer Posted January 9 Share Posted January 9 Hi, I'm fairly new in the game and I'm looking for some clarifications regarding shinny rates and roll. I understand that each roll is an individual roll, and shinny rate does not stack upon each count. (correct me if im wrong) If this is the case and I don't mind not having all OT shinnies under 1 name, can someone please confirm below: Account 1: 10,000 encounter (magikarp) Account 2: 5,000 encoutner (magikarp) Account 3: 10,000 encoutner (magikarp) Condition: All with donator status and shiny charm popped during shunting. From my understanding that the chances of shinny from the above situation would be the same as to Account 4 with 25,000 encounter of magikarps both donator and shiny charm popped. Am I correct? Bottomline is I just wanna reduce the time to reach x number of encounter. What I understand is the more I roll, the higher the chance is. Can someone confirm this? Really appreciate it. Link to comment

CaptnBaklava Posted January 9 Share Posted January 9 1 hour ago, Bytefixer said: Hi, I'm fairly new in the game and I'm looking for some clarifications regarding shinny rates and roll. I understand that each roll is an individual roll, and shinny rate does not stack upon each count. (correct me if im wrong) If this is the case and I don't mind not having all OT shinnies under 1 name, can someone please confirm below: Account 1: 10,000 encounter (magikarp) Account 2: 5,000 encoutner (magikarp) Account 3: 10,000 encoutner (magikarp) Condition: All with donator status and shiny charm popped during shunting. From my understanding that the chances of shinny from the above situation would be the same as to Account 4 with 25,000 encounter of magikarps both donator and shiny charm popped. Am I correct? Bottomline is I just wanna reduce the time to reach x number of encounter. What I understand is the more I roll, the higher the chance is. Can someone confirm this? Really appreciate it. Every encounter is a roll from 1 to 30.000, they do not stack nor does having more encounters increase your odds. True is that if you do more encounter you are more likely to get a shiny but thats just the simple art of probability. To your example: if every account uses the same mulitpliers they all have the same odds of encountering a shiny. Seth did a nice guide on shiny hunting consider reading it: Link to comment

Doctor Posted January 9 Share Posted January 9 1 hour ago, Bytefixer said: What I understand is the more I roll, the higher the chance is. Can someone confirm this? Really appreciate it. Yes and no, it doesn't increase the individual chance for a Pokémon to be shiny, it will still be 1/30000. But the way compound probability works is, the more you roll the dice, the more likely you are in the next dice roll to have a specific result. This doesn't mean you're guaranteed to have a shiny at any point, but similarly to flipping a coin multiple times, the more you flip the coin, the more likely you are to flip heads. Baklava also explained it very nicely, and @Seth's guide on the matter is also really simple and easy to follow. Probability is sometimes hard to understand, that's why it seems, following the logic I explained above, that the more encounters you do, the higher your odds, but it's a slightly different concept. If you wanna learn more about this specific type of probability, you can google "compound probability", which refers to this very situation: what are the odds of a certain result from an isolated event happening (getting a Pokémon encounter doesn't depend on previous encounters, so it's an independent event; the opposite would be drawing a card from a card deck and trying to get a 5: that event DOES care about previous events because you're removing cards from the deck) if I repeat that event multiple times. The coin example is very illustrative because it's basically the same as trying to roll a shiny, in the sense that the coin doesn't care about previous attempts, and its own probability will stay 1/2 in every single throw. But of course, the more you throw it, the more likely you are to get heads or tails, and the probability is calc'd by multiplying the odds by themselves: Probability of getting heads on 1 throw: 1/2 Probability of getting two heads in a row: 1/2 * 1/2 = 1/4 Probability of getting three heads in a row: 1/2 * 1/2 * 1/2 = 1/8 Four heads is 1/16, five heads is 1/32, etc As you can see, the more rolls you do, the less likely it is for you to NOT see a tails coming up in the next throw. Same thing happens with shiny hunting, the more encounters you do, the less likely it is for you to NOT see a shiny (again, Seth has a really good explanation on that on his thread). This doesn't mean that your overall odds increase, the shiny rate stays 1/30000 during every roll you make; and this also doesn't mean that you're ever gonna be guaranteed to get a shiny beyond certain point: you can theoretically throw an infinite number of coins and never see a tails; it's just very, very unlikely. It's also worth to mention that, following everything I've said up to this point, hordes don't have an increased chance of giving you a shiny by themselves, but rather they speed up the process by letting you throw 5 dies at the same time. It might seem like it's the same thing, but it's not, you just go through the entire dice roll thing five times faster than doing it individually, but you're gonna have to do the same amount of encounters before you encounter a shiny. You just do 5 of them at a time. Forgive me if I muddied the question further, I just woke up, english is not my mothertongue and I'm a graphic designer so explaining these things isn't exactly my strong point. Like the wise @Teddiursa once said to me, may RNG be in your favor and shiny rate always stay fair for you. Link to comment

Jgaw Posted January 9 Share Posted January 9 And here i am the pessimist shiny adviser Don't shunt when your account is old If u re new, shunt and repeat this process with different IP adresses Do not try to understand how it works Do not waste time Have fun wih sthg else Live long happily Seth and Quakkz 1 1 Link to comment

Damian Posted January 9 Share Posted January 9 (edited) Hi, The probability of succeeding a random event with probability p exactly x times given n trials is given by the following formula: In your case: - n=25000 - p=1/24,300 - x=1 This gives: P(x) = 25000C1 * (1/24300)^1 * (1-1/24300)^(25000-1) Simplifying: P(x) = 25000 * (1/24300) * (0.9999588)^24999 P(x) = (25000/24300) * 0.357 P(x) = 0.3677 Or approximately 37% chance to have found a shiny given the number of trials and conditions you specified (note that this is the chance for exactly one, the chance for at least one is a bit higher than this). I believe your original question was: 16 hours ago, Bytefixer said: What I understand is the more I roll, the higher the chance is. Can someone confirm this? Really appreciate it. The chance to succeed on any given trial (p above) does not change if you change the number of trials (n above). You can clearly see that they have different values in the formula, so changing one doesn't influence the other. If however you change the value of n, the value of P(x) will change. That is, as you do more trials, the chance that you succeed once will increase. This is the part most people get confused about - they mix up the chance to succeed on a particular trial (p) with the chance to succeed after some number of trials (P(x)). The incorrect belief that increasing the number of trials influences the chance of success on a particular trial where the events are statistically independent is known as The Gambler's Fallacy. I hope this clarifies. Edited January 9 by Damian typo in the equation TohnR, Doctor, Seth and 1 other 3 1 Link to comment

RysPicz Posted January 9 Share Posted January 9 7 minutes ago, Damian said: Hi, The probability of succeeding a random event with probability p exactly x times given n trials is given by the following formula: In your case: - n=25000 - p=1/24,300 - x=1 This gives: P(x) = 25000C1 * (1/24300)^1 * (1-1/24300)^(25000-1) Simplifying: P(x) = 25000 * (1/24300) * (0.9999588)^25999 P(x) = (25000/24300) * 0.357 P(x) = 0.3677 Or approximately 37% chance to have found a shiny given the number of trials and conditions you specified (note that this is the chance for exactly one, the chance for at least one is a bit higher than this). I believe your original question was: The chance to succeed on any given trial (p above) does not change if you change the number of trials (n above). You can clearly see that they have different values in the formula, so changing one doesn't influence the other. If however you change the value of n, the value of P(x) will change. That is, as you do more trials, the chance that you succeed once will increase. This is the part most people get confused about - they mix up the chance to succeed on a particular trial (p) with the chance to succeed after some number of trials (P(x)). The incorrect belief that increasing the number of trials influences the chance of success on a particular trial where the events are statistically independent is known as The Gambler's Fallacy. I hope this clarifies. I can sum this entire post in just 4 words Shiny rate is fair Doctor, Seth, Shadow and 3 others 5 1 Link to comment

Jgaw Posted January 9 Share Posted January 9 My account's equation 0(0) = 0 +0*0/(0*0x ± 0) * {0%-00*0,000} violationirons 1 Link to comment

Sourcerer Posted January 9 Share Posted January 9 17 hours ago, Bytefixer said: Bottomline is I just wanna reduce the time to reach x number of encounter. All you need is 6,000 accounts to have just 1 5x horde encounter. That's how you can reduce the time to reach 30k encounters to a second. That is what you asked right? Not sure if its worth it though... Link to comment

Bytefixer Posted January 10 Author Share Posted January 10 14 hours ago, Damian said: Hi, The probability of succeeding a random event with probability p exactly x times given n trials is given by the following formula: In your case: - n=25000 - p=1/24,300 - x=1 This gives: P(x) = 25000C1 * (1/24300)^1 * (1-1/24300)^(25000-1) Simplifying: P(x) = 25000 * (1/24300) * (0.9999588)^24999 P(x) = (25000/24300) * 0.357 P(x) = 0.3677 Or approximately 37% chance to have found a shiny given the number of trials and conditions you specified (note that this is the chance for exactly one, the chance for at least one is a bit higher than this). I believe your original question was: The chance to succeed on any given trial (p above) does not change if you change the number of trials (n above). You can clearly see that they have different values in the formula, so changing one doesn't influence the other. If however you change the value of n, the value of P(x) will change. That is, as you do more trials, the chance that you succeed once will increase. This is the part most people get confused about - they mix up the chance to succeed on a particular trial (p) with the chance to succeed after some number of trials (P(x)). The incorrect belief that increasing the number of trials influences the chance of success on a particular trial where the events are statistically independent is known as The Gambler's Fallacy. I hope this clarifies. Thank you for a math lesson sensei. Damian 1 Link to comment

violationirons Posted January 10 Share Posted January 10 On 1/9/2024 at 3:03 PM, Jgaw said: My account's equation 0(0) = 0 +0*0/(0*0x ± 0) * {0%-00*0,000} I have about the same 😀 Link to comment

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