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blankf4ce

DON'T GIVE UP/ shiny rate is actually fair

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Posted (edited)

I'm making this post for all the people who complain about not having caught a shiny. The probability of flipping a coin and getting heads twice in a row is 1/4, so if you flip a coin and get heads, your chances of getting tails on the next flip are going to be 3/4. I'm just going to prove to you that every-time you fail at finding a shiny, the probability of finding one increases. I'm am going to do this under the assumption that everything the developers have said is true, and the shiny rate truly is independent odds. I am also going to assume you do not have donator's, to show that anyone can find a shiny if they are determined enough. The rate of finding a shiny is 1/30,000. That would mean that the rate of failing at finding a shiny is 29,9999/30000, or 0.99996666666. To find the probability of two events happening you multiply the probabilities of each event. In this case we want to know the probability of failing at finding a shiny twice in a row, so 29,999/30,0000*29,999/30,000. This will give you 0.9993333444. Which is less than 0.99996666666. You can continue doing this and the number will become smaller and smaller each time. Keep in mind that this is the probability of failure, so the probability of success will be the 1 minus said number, as you will either get one or the other 100% of the time, or 1/1. Meaning that as your probability of failure continues to decrease, your probability of success will continue to increase. You can save yourself a lot of time by simply putting 0.99996666666 to the power of the total number of encounters. For example, when finding the probability of finding a shiny within 30,000 encounters, you would put 0.99996666666^30,000, get 0.36787323618, 1-.036787323618=0.63212676381, so your chances of encountering a shiny within 30,000 encounters is ~63.2%, much higher than it was on your first encounter. This will continue to increase the more encounters you do. If you don't believe me you can go do it yourself. If you don't believe what I'm saying is true, and think that the probability of finding a shiny is 1/30k regardless of how many encounters you do, then I suggest you educate yourself on independent odds and how they actually work rather than making assumption based on the use of the word independent, here is a good place to start : https://learn.problemgambling.ca/probability-odds-random-chance . Luckily, it is a proven fact that isolated sets mimic trends from larger sets of data, so you only have to worry about yourself not every player in the entire game. If we both start flipping a coin, we will both get heads roughly half the time, both together and separate. So, if you are one of those players that has not found a shiny yet, I envy you. The probability of you getting a shiny soon are much higher than mine or anyone else who has recently found a shiny/has had good luck finding shiny pokemon, but not if you give up and don't keep looking for one. You should probably go hatch some eggs of your favorite poke, and see what happens. This also applies to real life, DON'T GIVE UP. Unless its the lottery, you'll never be able to increase your chances enough to give yourself a good enough chance of winning, and if you do, you'll have spent more than the jackpot. Now go get that shiny. 

Don't actually believe this nonsense^^^ the gig is up. 

Edited by blankf4ce

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Posted (edited)
2 hours ago, fredrichnietze said:

title says shiny rate is fair but post only goes into whether it is possible. i *could* win the lottery but does that make it wise or fair? 

Well since the probability of getting a shiny within the 30,000 encounters is higher than the probability of not finding one, I consider that fair. You would never be able to guarantee yourself a shiny, but you can get extremely close simply by doing more encounters. You also have to pay for lottery tickets, if they were free then you could give yourself a good chance of winning(>50%) without paying more than you would win, which is what makes it unwise. They're priced the way they are for a reason, buying another lottery ticket increases your chances of winning, but not enough to be worth it, the real winners are the people selling lottery tickets. That's also the reason I suggested going for a shiny rather than trying to win the lottery...

Edited by blankf4ce

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Posted (edited)
1 hour ago, blankf4ce said:

he probability of flipping a coin and getting heads twice in a row is 1/4, so if you flip a coin and get heads, your chances of getting tails on the next flip are going to be 3/4.

No. Your odds of getting tails on the next flip is 1/2.

All possible 2 flip combinations:

TT

TH

HT

HH

 

So if getting HH is a 1/4 chance and getting HT is a 3/4 chance like you claimed, then in your scenario, it is impossible to get TT or TH. This is a contradiction, so your statement is false.

 

1 hour ago, blankf4ce said:

I'm just going to prove to you that every-time you fail at finding a shiny, the probability of finding one increases. I'm am going to do this under the assumption that everything the developers have said is true, and the shiny rate truly is independent odds. I am also going to assume you do not have donator's, to show that anyone can find a shiny if they are determined enough. The rate of finding a shiny is 1/30,000. That would mean that the rate of failing at finding a shiny is 29,9999/30000, or 0.99996666666. To find the probability of two events happening you multiply the probabilities of each event. In this case we want to know the probability of failing at finding a shiny twice in a row, so 29,999/30,0000*29,999/30,000. This will give you 0.9993333444. Which is less than 0.99996666666. You can continue doing this and the number will become smaller and smaller each time. Keep in mind that this is the probability of failure, so the probability of success will be the 1 minus said number, as you will either get one or the other 100% of the time, or 1/1. Meaning that as your probability of failure continues to decrease, your probability of success will continue to increase. You can save yourself a lot of time by simply putting 0.99996666666 to the power of the total number of encounters. For example, when finding the probability of finding a shiny within 30,000 encounters, you would put 0.99996666666^30,000, get 0.36787323618, 1-.036787323618=0.63212676381, so your chances of encountering a shiny within 30,000 encounters is ~63.2%, much higher than it was on your first encounter. This will continue to increase the more encounters you do.

While your math is correct, your explanation is not. Your odds do not increase as you go along. The next encounter does not care about any previous encounters, and the odds you find a shiny are still 1/30,000. What you should be saying is that the summarized observation of your sample size shows that there was a 63.2% chance that you found at least 1 shiny out of 30,000 encounters.

 

I'll break it down into a reverse explanation of what you did, to explain my point about your odds not increasing. Let's start at the sample size of 30,000; the odds that we find at least 1 shiny in 30,000 encounters is 63.2%. Let's actually encounter a pokemon in that sample size and assume it wasn't shiny. That means that we can eliminate 1 pokemon from our original sample size. So, now we only have 29,999 pokemon left. The odds that there is at least 1 shiny in those 29,999 pokemon is slightly less than 63.2%. If we continue to iterate through this process, we see that the % returns to 1/30,000. At no point did our odds get any better.

 

At each stage, we only ever had a 1/30,000 chance to find a shiny in any given encounter. And, it is quite literally a fallacy to think that next time the odds are more in your favor. The 36.8% (100%-63.2%) removes the notion of "process of elimination" and is why the odds do not increase as you go along. The only time that your argument works, is if it was a 100% chance that a shiny existed within the population of 30,000; because, then you know one exists, and the equation simply turns into: p(x) = 1/(30,000-x) which clearly gives you better odds as you go along, by simple process of elimination (aka, the next encounter does care about all previous encounters).

 

 

tl;dr if this was a meme, then yes I took the bait cause bad probability keeps me up at night.

Edited by Gilan

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2 minutes ago, Gilan said:

No. Your odds of getting tails on the next flip is 1/2.

 

While your math is correct, your explanation is not. Your odds do not increase as you go along. The next encounter does not care about any previous encounters, and the odds you find a shiny are still 1/30,000. What you should be saying is that the summarized observation of your sample size shows that there was a 63.2% chance that you found at least 1 shiny out of 30,000 encounters.

 

I'll break it down into a reverse explanation of what you did, to explain my point about your odds not increasing. Let's start at the sample size of 30,000; the odds that we find at least 1 shiny in 30,000 encounters is 63.2%. Let's actually encounter a pokemon in that sample size and assume it wasn't shiny. That means that we can eliminate 1 pokemon from our original sample size. So, now we only have 29,999 pokemon left. The odds that there is at least 1 shiny in those 29,999 pokemon is slightly less than 63.2%. If we continue to iterate through this process, we see that the % returns to 1/30,000. At no point did our odds get any better.

but it does, for example, let's assume you have already done 30,000 encounters and haven't found a shiny yet. You do 10,000 more encounters. With your logic the odds of finding a shiny within those 10,000 encounters would be ~33%, however you did not do 10,000 encounters, you did 40,000, and the probability of finding a shiny within 40,000 encounters would be ~80%. This would apply to the entire set of data, and not just cut off at the present moment. This is a trick caused by the concept of time making you think your past attempts didn't matter. The data found in the past still belongs to the data set. The past present and future results all belong to the same set of data. You only know one of the three however. 

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18 minutes ago, Gilan said:

No. Your odds of getting tails on the next flip is 1/2.

All possible 2 flip combinations:

TT

TH

HT

HH

 

So if getting HH is a 1/4 chance and getting HT is a 3/4 chance like you claimed, then in your scenario, it is impossible to get TT or TH. This is a contradiction, so your statement is false.

 

While your math is correct, your explanation is not. Your odds do not increase as you go along. The next encounter does not care about any previous encounters, and the odds you find a shiny are still 1/30,000. What you should be saying is that the summarized observation of your sample size shows that there was a 63.2% chance that you found at least 1 shiny out of 30,000 encounters.

 

I'll break it down into a reverse explanation of what you did, to explain my point about your odds not increasing. Let's start at the sample size of 30,000; the odds that we find at least 1 shiny in 30,000 encounters is 63.2%. Let's actually encounter a pokemon in that sample size and assume it wasn't shiny. That means that we can eliminate 1 pokemon from our original sample size. So, now we only have 29,999 pokemon left. The odds that there is at least 1 shiny in those 29,999 pokemon is slightly less than 63.2%. If we continue to iterate through this process, we see that the % returns to 1/30,000. At no point did our odds get any better.

 

At each stage, we only ever had a 1/30,000 chance to find a shiny in any given encounter. And, it is quite literally a fallacy to think that next time the odds are more in your favor. The 36.8% (100%-63.2%) removes the notion of "process of elimination" and is why the odds do not increase as you go along. The only time that your argument works, is if it was a 100% chance that a shiny existed within the population of 30,000; because, then you know one exists, and the equation simply turns into: p(x) = 1/(30,000-x) which clearly gives you better odds as you go along, by simple process of elimination (aka, the next encounter does care about all previous encounters).

 

 

tl;dr if this was a meme, then yes I took the bait cause bad probability keeps me up at night.

uwu someone took statistics ;)

 

good ol coin toss 

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It's been a while since I stopped thinking that someone could ever appear. I've been playing for almost 2000 hours, more than 100 pokémon comp raised, uploaded to lvl 100, with its ev's, 3 regions, daily meetings and I haven't seen one yet, so ...

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Posted (edited)

Keep in mind this only works for truly independent odds, I personally think independent odds are damn near impossible to actually accomplish. Even a coin flip is dependent on various things like which side the coin was on before the flip, and how much force is applied to the coin on the flip. 

Edited by blankf4ce

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Posted (edited)

My average over 30 OTs is 30/993,900 aka 1/33,130 using donator status nearly the whole time. I record every encounter using the app. I am slightly unlucky but does that make it unfair?

Edited by awkways

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Posted (edited)
28 minutes ago, awkways said:

My average over 30 OTs is 30/993,900 aka 1/33,130 using donator status nearly the whole time. I record every encounter using the app. I am slightly unlucky but does that make it unfair?

What app do you use? I've been looking for a good one, as of right now I just use the calculator on my phone, but I don't exactly stay on top of keeping track as much as I should, but yea I'd still consider that fair, its close enough, unless you were using some secret trick, which there probably isn't

Edited by blankf4ce

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2 hours ago, blankf4ce said:

What app do you use? I've been looking for a good one, as of right now I just use the calculator on my phone, but I don't exactly stay on top of keeping track as much as I should, but yea I'd still consider that fair, its close enough, unless you were using some secret trick, which there probably isn't

wrong

Fair is getting 150k encounters for first shiny, 270k for second, and 100k+ for the third that has not yet been seen.

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6 hours ago, blankf4ce said:

but it does, for example, let's assume you have already done 30,000 encounters and haven't found a shiny yet. You do 10,000 more encounters. With your logic the odds of finding a shiny within those 10,000 encounters would be ~33%, however you did not do 10,000 encounters, you did 40,000, and the probability of finding a shiny within 40,000 encounters would be ~80%. This would apply to the entire set of data, and not just cut off at the present moment. This is a trick caused by the concept of time making you think your past attempts didn't matter. The data found in the past still belongs to the data set. The past present and future results all belong to the same set of data. You only know one of the three however.

Your odds of finding a shiny never increase due to simply encountering more.  You are no more likely to find a shiny at encounter 50,000 than you were at encounter 1. Probability just doesn’t work the way that you are explaining. I tried to articulate this in my reverse example, but it seems I failed to get my point accross.

 

Basically, what you are saying, is because the aggregated probability increases as you encounter more pokemon, your odds of finding a shiny increases. But, this is wrong. Each time you encounter a pokemon, you can’t add it to the aggregated probability, because that pokemon is no longer random; you know that it isn’t shiny and thus is not a part of the equation anymore (because they are independent like you keep saying). The appropriate way to tackle it is the reverse case in my previous post. And, that showed that as you encountered your 30,000th pokemon you had a 1/30,000 chance of it being shiny (aka no better than pokemon 1).

 

Specifically your comment that the past data belongs to the data set is false.

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Posted (edited)
6 hours ago, blankf4ce said:

What app do you use? I've been looking for a good one, as of right now I just use the calculator on my phone, but I don't exactly stay on top of keeping track as much as I should, but yea I'd still consider that fair, its close enough, unless you were using some secret trick, which there probably isn't

Dunno if this still works for everyone but it still works for me so it should still be up to date.

 

Edit: You are completely wrong about this topic - Gillan on point tho.

Edited by awkways

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Posted (edited)

I would like to expand a bit on what Gilan said.

Let's assume our goal is to achieve Heads on a coin toss.
on 2 consecutive coin toss results are:

HH
HT
TH
TT
 

 

So our chances of it happening is 2/4 which is 50%
After each failed toss let's add 1 more line to possible outcomes.

Toss 1, we failed

HH

HT
THH
THT
TTH

TTT

Our chances of having heads on next toss at this state is 2/4 which is 50%

Toss 2, we failed

HH

HT
THH
THT

TTHH

TTHT

TTTH

TTTT

Our chances of having heads on next toss at this state is 2/4 which is 50%

You see a pattern here? What confuses you is the calculation of independent odds, BEFORE it actually happens and in a FIXED AMOUNT of attempts.

You'll see a picture here,
Chance to get a heads in 3 consecutive tosses is:

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

1/8 which is 12,5% chance

Toss 1, heads;

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT



Now our chance is 1/4 which is 25% chance.

Toss 2, heads;

 

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT



Now our chance is 1/2 which is 50% chance.

So the math on your chances being increased after attempts relies on some sort of visible progress towards your goal. But in a binary result such as shiny or not, your chances are reset after each attempt, doesn't matter if it's successful or not.

Your theory is actually correct mathematically ONLY if people were making infinite amount of attempts, because in infinite amount of attempts the end result is expected to be extremely close to 1 shiny per 30,000 encounter. Since we are not having that chance realistically, those calculations of odds increasing are unfortunately incorrect.

Last month at work I was doing some sort of debug testing on a graphic displayer engine we were working on, the devices had a starting value of 25, every 5 seconds the value was randomly increased by 0.5, stayed same or reduced by 0,5. Given infinite amount of time every device should have the displayed value extremely close to 25 but what actually happening was, after 3 days or so, majority of the devices were showing values either above 100 or below -100. Also some of the devices showing values above 100 could go below -100 or sky rocket to above 250 in 2-3 more days. Hope you get the idea.


Moral of the story is, of course people with more time spent shiny hunting will encounter more shinies.
it's basicly like,
chance to get a heads in 1 toss is 50% chance.
chance to get a heads in 2 tosses is 75% chance.
chance to get a heads in 3 tosses is 82,5% chance.
and so on.

The person who has bought more lottery tickets have more probability to win the lottery.
But it doesn't mean a person with 1 ticket will lose it every time, or the person with a lot of tickets will win it every time.
 

Edited by PrincessDia

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23 minutes ago, PrincessDia said:



Moral of the story is, of course people with more time spent shiny hunting will encounter more shinies.
it's basicly like,
chance to get a heads in 1 toss is 50% chance.
chance to get a heads in 2 tosses is 75% chance.
chance to get a heads in 3 tosses is 82,5% chance.
and so on.

The person who has bought more lottery tickets have more probability to win the lottery.
But it doesn't mean a person with 1 ticket will lose it every time, or the person with a lot of tickets will win it every time.
 

This is essentially what I was getting at, maybe it was worded poorly. It is very unlikely to continue to flip a coin and to only get tails, or to only get heads. Since shiny hunting in similar in the aspect that you will only get one or the other, shiny or non-shiny, continuing to encounter raises your chances (I'll say chances rather than probability here) of getting lucky. Since you are experiencing one outcome over and over again (no-shiny), the numbers would say you are more likely to experience a shiny within that set of numbers. I did say that there is no way to get up to 100%, so at the end of the day it is about getting lucky or unlucky, it's just harder to get unlucky in a large number of attempts. If you were looking at this like a function, you could say that there is a limit at infinity of 1, meaning you are becoming increasingly more likely to get lucky as time goes on. Like you said the math is there and is calculated correctly, I was also taught this at an American University which I will not disclose the name of, since it appears I was taught wrong. You are not really telling me I'm wrong, but calling out the entire field of statistics as being a sham, which I said in an earlier post that I would agree with, since there isn't really a way to accomplish independent odds. Most if not all things are dependent on something else, whether or not we are able to quantify them, and therefore the probability would have to be altered to account for said variables. 

 

25 minutes ago, PrincessDia said:

I would like to expand a bit on what Gilan said.

Let's assume our goal is to achieve Heads on a coin toss.
on 2 consecutive coin toss results are:

HH
HT
TH
TT
 

 

So our chances of it happening is 2/4 which is 50%
After each failed toss let's add 1 more line to possible outcomes.

Toss 1, we failed

HH

HT
THH
THT
TTH

TTT

Our chances of having heads on next toss at this state is 2/4 which is 50%

Toss 2, we failed

HH

HT
THH
THT

TTHH

TTHT

TTTH

TTTT

Our chances of having heads on next toss at this state is 2/4 which is 50%

You see a pattern here? What confuses you is the calculation of independent odds, BEFORE it actually happens and in a FIXED AMOUNT of attempts.

You'll see a picture here,
Chance to get a heads in 3 consecutive tosses is:

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

1/8 which is 12,5% chance

Toss 1, heads;

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT



Now our chance is 1/4 which is 25% chance.

Toss 2, heads;

 

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT



 

This could tie into the concept of multiple realities/dimension and theories on time. While you are correct, based on current understanding of these subjects you can just cross out any possibilities that have been made impossible due to prior events, can they really be crossed out when considering your chances in the long term? Can you really just leave the past in the past, or does it affect the future? In theory each current attempt is being added to the data set as time goes on, thus increasing the size of the data set continuously, and I think you would agree that an increase in the volume of the data set increases the probability of experiencing all possible outcomes within the data set. Just food for thought.

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3 hours ago, Gilan said:

Your odds of finding a shiny never increase due to simply encountering more.  You are no more likely to find a shiny at encounter 50,000 than you were at encounter 1. Probability just doesn’t work the way that you are explaining. I tried to articulate this in my reverse example, but it seems I failed to get my point accross.

 

Basically, what you are saying, is because the aggregated probability increases as you encounter more pokemon, your odds of finding a shiny increases. But, this is wrong. Each time you encounter a pokemon, you can’t add it to the aggregated probability, because that pokemon is no longer random; you know that it isn’t shiny and thus is not a part of the equation anymore (because they are independent like you keep saying). The appropriate way to tackle it is the reverse case in my previous post. And, that showed that as you encountered your 30,000th pokemon you had a 1/30,000 chance of it being shiny (aka no better than pokemon 1).

 

Specifically your comment that the past data belongs to the data set is false.

I don't know if I would necessarily agree with that assumption, in theory one in every 30,000 players would encounter a shiny on their first encounter. Once you have not encountered a shiny on your first encounter, you could hypothetically separate them into two groups, those who have encountered a shiny already, and those who haven't. As time goes on you would find that the person who encountered a shiny one their first encounter would have already encountered their shiny and thus would have to do a lot more encounters to find another one, in theory at least. Those who didn't find a shiny on their first encounter would by default be more likely to find a shiny in the future, since they have not found one in the past. This would really come down to the way the game is coded. If the game is coded to give out one shiny per 30,000 encounters, and 1 shiny per 27,000 encounters to those with donators, the only thing that would matter would be the timing of your encounter, not how many encounters you've done. It would be predetermined which encounter would be a shiny, it would just come down to getting an encounter at that exact time. Why would you not add each encounter to the aggregated probability? Even if not for the next encounter, but for future encounters in general? As you are encountering pokemon the present is becoming the past and you are moving into the future, you seem to be agreeing that you can find the probability of past events occurring using the method I used, the disagreement is coming from the lack of a way to actually quantify time or to quantify what is possible or impossible in regards to it. I also have not been saying they are independent, this is just operating under the assumption that they are. 

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