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This might be a little out of the box for this sub, but I have been shiny hunting in Pokémon HeartGold. For anyone not in the know, a shiny Pokémon is a Pokémon with a different skin which is exceedingly rare. In later generations of the games, there were ways to "shiny chain" to raise the odds in the player's favor. AFAIK, this game is not one of those games.

To set the question up, when you get into battle with any Pokémon in this generation of the game, you have a 1/8192 chance of finding a shiny Pokémon rather than its normal colors. There are legendary Pokémon, meaning there is only one present in the entire game. For most, you are able to save your game in front of the Pokémon before battling it, just in case you end up defeating it, so you can "soft reset" back to your save point before starting the battle.

What I understand by this is that, no matter how many times you "soft reset" your game, you have a 1/8192 chance of the Pokémon being shiny. It could be the first time if you're radically lucky, it could be the 200,000th time if you've got terrible luck. Hell, if you have the worst luck possible, you will simply never find the shiny that you want.

tl;dr for anyone who already understands Pokémon:  
The question, ultimately, is, am I right? If the shiny spawn rate is always 1/8192, is it no more likely that you spawn a shiny Pokémon on the 5000th soft reset as the 1st or 200,000th?
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When you encounter the Pokémon, you have a **1/8192** chance that it will be shiny

Let's say you've **already restarted 200000 times** without encountering a shiny. Now you encounter the Pokémon again, and you have a **1/8192** chance that it will be shiny, again. The game doesn't remember the times you restarted, because 1. you didn't save afterwards, and 2. even if you saved, the game doesn't give you better odds if you were unlucky

Now imagine you are making plans for the future: you have a couple hours of time, and you are planning to restart up to 100 times as long as you're not getting a shiny. Then the **future 100** encounters have a (8191/8192)^(100) ≈ 98.79% chance of not being a shiny; that's approximately a 1.21% chance, a **1/82** chance of encountering a shiny, in the **future 100** attempts.

Now imagine you are midway through these attempts, you restarted 50 times, and have not yet encountered a shiny. Then the **future 50** encounters have a (8191/8192)^(50) ≈ 99.39% chance of not being a shiny; that's approximately a 0.61% chance, a **1/164** chance of encountering a shiny, in the **future 50** attempts.

TL;DR the past matters not, only the future. You can do this calculation yourself: in the **future N** attempts, you have a 1 in 1/(1-[8191/8192]^(N)) chance to encounter a shiny
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To satisfy the automod, I'm just copying and pasting, but I already explained myself above:

​

To set the question up, when you get into battle with any Pokémon in this generation of the game, you have a 1/8192 chance of finding a shiny Pokémon rather than its normal colors. There are legendary Pokémon, meaning there is only one present in the entire game. For most, you are able to save your game in front of the Pokémon before battling it, just in case you end up defeating it, so you can "soft reset" back to your save point before starting the battle.

What I understand by this is that, no matter how many times you "soft reset" your game, you have a 1/8192 chance of the Pokémon being shiny. It could be the first time if you're radically lucky, it could be the 200,000th time if you've got terrible luck. Hell, if you have the worst luck possible, you will simply never find the shiny that you want.

tl;dr for anyone who already understands Pokémon:  
The question, ultimately, is, am I right? If the shiny spawn rate is always 1/8192, is it no more likely that you spawn a shiny Pokémon on the 5000th soft reset as the 1st or 200,000th?
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No, that is a common misconception about probabilistic behaviour. But I like your question!

​

What is happening in the game is - according to your post - a independen probabilistic event. Means the outcome of the n-th event doesn't depend on the events that happen before. So with every try you've got the exact same probability of finding a shiny in that exact try.

But, and that is where the misconception lays: the consecutiv battling is a event on its own (if you want to name it so). And this has it's own probability.

Ps. I did a quick qualculation: it takes around 5700 for you having a averaged 50 % chance of having found a shiny. Means if you got a whole bunch of players if they all take attempts until 50 % found a shiny, it took them a average of 5700 attempts.

​

Example: you are hunting for shinys. Now the probability of finding two shinys in a row is (1/ 8192)². But the probability of finding a shiny on the first battle and finding one on the second battle where (1/8192) each.

Or in other words: you start you first battle. Your probability of getting a shiny is (1/ 8192). Then you start your second battle. Finding a shiny in that battle is (1/8192) again. The probability of the event "two shinys in a row" however is (1/8192)².

People tend to have trouble understanding this, as they think that the probability of finding a shiny at the second try aswell and the probability of finding a shiny two consecutive times is identical. But it isn't, because the events are independend. But we dumb humans prefer to think in dependend causal events.

There is a own class of logical flaws (yes multiple of them!) due to weird human thinking of such things.
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Perhaps to explain independent probabilities a little bit better, let's talk about averages, tails, and expected times. For simplicity, let's call the inverse of a chance an encounter is shiny with the letter x (i.e. x = 8192). For no other reason than it's shorter to write.

On average, you will encounter one shiny every x encounters. They come randomly so you won't get them exactly when you think you should. You might not get them at all. But on average, over everyone playing Pokemon, the total encounters divided by the total shines will be about x.

However, because all of these events are independent, if you don't get a shiny in, let's say, 6x encounters, that doesn't change anything at all. From this point on you're still only expecting a shiny about every x encounters. The universe doesn't increase the rate just because you previously got bad luck. Similarly, if you got 6 shinies in x encounters, you're still only expecting a shiny once about every x encounters.

The key takeaway is that shiny hunting doesn't remember the past. It has no history. That's both good and bad, because a lucky streak cannot harm you, but an unlucky streak cannot help you.

Finally, let's talk about the long tail.

Let's number your attempts. Attempt #1, attempt #2, and so on. And let's say you only want to catch one shiny, so you stop after finding it. Which attempt number do you have the greatest chance of catching that shiny?

Believe it or not, the answer is the Very. First. Attempt.

Let's see why.

The probability that any particular pokemon is shiny is 1/x, let's call this y (i.e, y = 1/8192). So the first pokemon being shiny has probability y. If you get to the second pokemon, it also has a probability of y, but that's only if you get there, which has a probability of (1 - y), so the total probability that you catch a shiny on your second encounter is y (1 - y). Attempt #3 is y(1 - y)^(2), and so on.

Let's say you tried to catch a shiny 200000 times. Which attempt has the highest chance of having the shiny? The very next attempt, of course, with probability y. Getting to this attempt without a shiny was indeed quite unlucky, but now that the past has already happened, the 200,000th attempt and the 1st attempt are the exact same.

That's what the other user was getting at with the 10 hour session. If you play for 10 hours you have a certain chance of getting a shiny. If you play for 1 hour you have a different chance. But the idea is that if you decide to play for 10 hours and you don't get a shiny in the first 9, then your chance of getting a shiny in the last hour is the same as your chance of getting a shiny in only 1 hour, rather than your original 10 hour chance.

Hope this helps.
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The answer to your question of whether the odds are different on subsequent attempts is actually yes, but for reasons that are extremely complicated and have more to do with computers than math. The basic math would say that the odds are the same on subsequent attempts as this is how probability works, but the odds actually get very slightly smaller over time because of human imperfection. Luckily, if there are 2 things I know it's math and Pokémon. The full shiny Pokémon calculation is a bit complicated so I'll run through it in steps. First, there are 3 values that count toward shininess and all of them are 16 bit unsigned integers (meaning they between 0 and 65635, but in binary they are written with 16 digits ex: 00010101 11011011. Your trainer ID is used. This is randomly chosen by the DS number generator every time you start a new save file and it can be easily viewed on the trainer card. The Secret ID is used and this is also generated each time you start a save file, but, unlike the Trainer ID, this cannot be viewed without hacking or glitching your device. Personality Value is then generated whenever you encounter a Pokémon this one is 32 bits, but it is immediately split into p1 and p2 which are just the first and second set of 16 bits respectively.

The final calculation is to take these 4 numbers and do an operation called xor on them all, so Trainer ID xor Secret ID xor p1 xor p2. xor is commutative so doing it on 2 will tell you how to do it on 4 as you can just go from left to right. The xor of two 16 bit integers will be another 16 bit integer where you look at each digit of each one and compare them. xor outputs a 1 if those digits are different and a 0 if they are the same

ex: 11101111 01010110 xor 01010101 11011011 = 10110110 10001101

Shininess occurs if the output of this operation is less than 8. That occurs only when the first 13 of the 16 digits are 0 (the last 3 can be anything). This all leads to the 1/8192 number you had. These odds will always be the same regardless of previous attempts assuming each attempt is randomly generated. This last condition is the one that isn't always the case. Computers can't be truly random without the help of quantum physics (but none of the Nintendo devices use this because very few devices have truly random number generators). The way around this problem is to generate pseudorandom numbers. These are generated in a way that is largely unpredictable as they are based on a grab bag of system data that is both inaccessible and put together in a way that makes it difficult to determine any pattern without an extremely large number of repeated trials (we're talking larger than astronomical numbers of trials here). The catch to all this is that, if you turn on your device and inputted all the same inputs on all the same frames, you will get the same result. Even though probability says your odds on any given attempt are 1/8192, we are human and so, over the course of thousands of trials, you may accidentally place in all the same inputs in the same places as a previous trial and in that case your odds of obtaining a shiny for that attempt were 0! This is because you already knew it wasn't shiny from a previous attempt.

This sounds irrelevant, but, with 1000s of attempts, it's pretty likely this will have happened at least once because of how we as humans are.

It's also worth noting that it is definitely possible to manipulate this form of pseudorandom random number generation. Without external devices though, it's almost impossible, but people will use this for routing speed runs all the time. The any% glitchless speedruns of pokemon, for example, will often use strategic saving and reseting combined with frame perfect inputs to manipulate the rng of aspects of their run such as getting a specific wild pokemon encounter typically with good ivs and nature or making it so they won't run into any wild pokemon for a while.

I know that was probably more information than you expected but I hope this helped. If you have any questions feel free to reply
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It may be easier to conceptualize with smaller numbers, so let’s go with rolling a 6 sided die (1d6). You want to roll a 1. If you roll once, it’s a 1/6 chance you get it, or a 5/6 chance you don’t.

If you roll it 2 times, you have to look at the probability of it NOT happening to properly calculate the odds. So (5/6)^2 which is (25/36), or just under 70%

If you roll 6 times it’s (5/6)^6 or (15625/46656, just over 33%). You aren’t guaranteed to get it but the more times you try, the higher odds it will happen at least once.

Any previous tries don’t factor into the calculation though. Only however many you do starting at this point.


Edit: for the shiny example, if you hatch exactly 5678 eggs, there’s almost exactly a 50% chance at least one will be a shiny.
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each encounter has the exact same chance.

1st or 200,000st attempt both have the same chance.

Now if you attempt something very often, it is possible that one of those 1/8192 chances is successful and you get the shiny.

You can compare this to flipping a coin with heads and tails. Every time you encounter the pokemon the game flips a coin which is either heads (=shiny) or tails (=normal). With the only difference that the coin the game is using is extreeeemly unlikely to get heads (assume there is a weight attached to the coin, which modifies its chances).

And same as with every new coin flip the chance of getting heads is always the same.

Although if you flip more often your chances of getting one successful flip is increased. Simply because trying something only once is worse than trying something lots of times. This is why people say that your chances increase when you try it more often. It does not mean the coin gets fairer, just that with more tries you will eventually suceed.
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