Herodotus

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.
 

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.

Stopped reading at the 3/4 chances of getting tail in the coin flip. 
 
 
 

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.