Anyone ever encounter a shiny Pokemon?

[Djing]

I hatched a shiny Staryu once. It had a bad nature though.

Also found a shiny Graveler on a double wild battle. Thank Arceus it didn't use Explosion.

[DodgeDusk]

Hmm, let me think...

Shiny Numel in Ruby (gone)

Shiny Mankey in Fire Red (gone)

Shiny Pelipper in Diamond (gone)

So yeah memory is telling me I got none at the moment

[Sylphid]

Shiny Drowzee in gold.

Shiny Carvanha in Sapphire, but it killed my team.

[Chrom-ulent]

You'd seriously ditch a shiny Rayquaza just because his nature is bad for a competitive set!? Do you know that some people do countless restarts just to get a shiny one?

Okay, bad phrasing on my part. What I meant was that I would have been okay with perfect Speed / wrong nature even if it was non-shiny.

If it's shiny, nature doesn't matter.

[DodgeDusk]

Just found a shiny Mr. Mime on X. Make that four now

[Freohr Datia]

Just found a shiny Mr. Mime on X. Make that four now

Whaaaaaaaaaaaaaaaaat

After several years I haven't found one

And you find one in less than a day ;n;

[DodgeDusk]

I've had it since Thursday, although I'm surprised I've already found one. Not surprised that the the pokemon I DO find shiny is one that I don't like <_<

[chuuco]

Is it just me or are shinies ALOT easier to find in xy. I've already found a shiny and alot of other people have as well ._.

[Espinosa]

Is it just me or are shinies ALOT easier to find in xy. I've already found a shiny and alot of other people have as well ._.

Well, encountering 5 Pokemon at once has to help.

[iHeartLibra]

It was PokeMon Emerald in the Safari Zone when a shiny Girafrig galloped before me in all it's blue-horned glory! I nearly dropped my GBA in my excitement, but did break it when I threw a Safari Ball and it ran...D:

[Skynstein]

Short list:

Gold: Gyarados, Slugma, Drowzee, pretty sure there's another one I'm forgetting.

Platinum: Hoothoot (stupid Cheryl killed it), Roselia

Black: Watchog

[BBM]

I don't know how often horde encounters pop up, but if there are 5 pokemon in a horde, the odds of at least one being shiny is 1 - ((8191/8192) ^ 5), which I think roughly translates in fractional terms to something like 1/1666. So that's quite a bit higher (as might be expected, approximately 5 times higher).

[Elli]

I caught my first ever Shiny Just yesterday in Pokemon X a shiny Loudred, though you can hardly tell the difference its just a few shades lighter than the normal Loudred lol

First Shiny I ever encountered though was back in Diamond on victory road, a Shiny Graveler that used Self destruct...

Edited October 15, 2013 by Shelie

[Chocolate Kitty]

i ran into a shiny geodude in a hordde battle yesterday

and because there was no gleam or shimmer i just ran without thinking

and then i noticed it

anger x10

[Fruity Insanity]

I don't know how often horde encounters pop up, but if there are 5 pokemon in a horde, the odds of at least one being shiny is 1 - ((8191/8192) ^ 5), which I think roughly translates in fractional terms to something like 1/1666. So that's quite a bit higher (as might be expected, approximately 5 times higher).

More like 5/8192. Why in the world would the chance grow exponentially?

Each one has a 1/8192 chance in appearing. Add them and that's 5.

EDIT: Never mind, I get it now.

Edited October 16, 2013 by Fruity Insanity

[Redwall]

you don't multiply 1/8192 by 5 because the pokemon in the horde of five are presumed to be shiny/un-shiny independently of each other. pokemon 1 (of five) being shiny does not preclude pokemon 2 (of five) from being shiny. bbm's calculation is consistent with what we understand of X/Y at this point, though it is entirely possible that, for example, the shininess rate has changed from 1/8192 for any given pokemon, or that the developers have coded something preventing multiple shinies from appearing in a horde, etc.

[Fruity Insanity]

Apparently, I wasn't thinking clearly, lol.

Edited October 16, 2013 by Fruity Insanity

[Nobody]

Exactly.

Which is why you add.

Because they are independent from one another.

Nessie is right. Think about it like this: there is a 8191/8192 probability that a pokemon isn't shiny. If you want 5 non shiny pokemon, you have to multiple 8191/8192 x 8191/8192 x 8191/8192 x 8191/8192 x 8191/8192. The probability of having at least one shiny is (1 - that). Believe me, I studied statistics.

Here:

http://en.wikipedia.org/wiki/Binomial_distribution

Edited October 16, 2013 by Nobody

[Redwall]

You're still wrong. You generally add things when they are mutually exclusive: for example, if I wanted to find the probability of getting a one, two, or four when rolling a fair die with six faces, in that case it would simply be 3/6 = 0.5. Rolling a one is mutually exclusive from rolling any other number. In this Pokemon example, having one Pokemon in a horde of five be shiny is not mutually exclusive from having another Pokemon in that horde of five be shiny.

Independent events are different; two events are independent if the occurrence of one doesn't influence the odds of the occurrence of the other. Events A and B are independent iff P(A and B) = P(A) * P(B), where P(blah) denotes the probability of blah occurring. Going back to the Pokemon example, then, if the probability of any given shiny Pokemon appearing is p_s, then the probability of a non-shiny occurring is 1-p_s. If we look at a set of five Pokemon, assuming that their shininess is independent, the probability that none are shiny is (1-p_s)^5 going by the P(A & B) = P(A) * P(B). The probability that at least one is shiny is 1 - (1-p_s)^5, which is what BBM has written.

For a simpler example, note that flipping a fair coin twice does not guarantee me a heads, even though heads will occur on 50% of attempts.

[Fruity Insanity]

You're still wrong. You generally add things when they are mutually exclusive: for example, if I wanted to find the probability of getting a one, two, or four when rolling a fair die with six faces, in that case it would simply be 3/6 = 0.5. Rolling a one is mutually exclusive from rolling any other number. In this Pokemon example, having one Pokemon in a horde of five be shiny is not mutually exclusive from having another Pokemon in that horde of five be shiny.

Independent events are different; two events are independent if the occurrence of one doesn't influence the odds of the occurrence of the other. Events A and B are independent iff P(A and B) = P(A) * P(B), where P(blah) denotes the probability of blah occurring. Going back to the Pokemon example, then, if the probability of any given shiny Pokemon appearing is p_s, then the probability of a non-shiny occurring is 1-p_s. If we look at a set of five Pokemon, assuming that their shininess is independent, the probability that none are shiny is (1-p_s)^5 going by the P(A & B) = P(A) * P(B). The probability that at least one is shiny is 1 - (1-p_s)^5, which is what BBM has written.

For a simpler example, note that flipping a fair coin twice does not guarantee me a heads, even though heads will occur on 50% of attempts.

Didn't really pay attention to the text, but I'll take your word for it. :P

My thinking was just: "Hm, if there are five Pokemon in a horde, and each one has a 1/8192 chance of appearing, then since there are five Pokemon, it must be 5/8192."

Edited October 16, 2013 by Fruity Insanity

[Redwall]

Didn't really pay attention to the text, but I'll take your word for it. :P

You shouldn't just take my word for it. You should attempt to confirm or falsify my claims by performing some analogous experiments: for example, try rolling a die six times. There is only about a 1.5% chance that in that set of six rolls, all six numbers (1 through 6) will be observed.

My thinking was just: "Hm, if there are five Pokemon in a horde, and each one has a 1/8192 chance of appearing, then since there are five Pokemon, it must be 5/8192."

The actual odds are very close to 5/8192; you can see why if you write a Taylor expansion of BBM's expression. The first-order term is 5/8192. The higher-order terms make the probability, well, not quite 5/8192 (but still close).

Edited October 16, 2013 by Nessie

[Fruity Insanity]

You shouldn't just take my word for it. You should attempt to confirm or falsify my claims by performing some analogous experiments: for example, try rolling a die six times. There is only about a 1.5% chance that in that set of six rolls, all six numbers (1 through 6) will be observed.

The actual odds are very close to 5/8192; you can see why if you write a Taylor expansion of BBM's expression. The first-order term is 5/8192. The higher-order terms make the probability, well, not quite 5/8192 (but still close).

Too lazy.

So... I'm right and wrong at the same time?

[Alfred Kamon]

Whoo, just found a shiny Jigglypuff.

It's a little grey with green eyes. Cute!

Now that it's a Fairy, it might also be useful.

[Freohr Datia]

Hey, Rein told me that they boosted the chance of encountering a shiny now =o

He said they made it about 10x more likely now? Around 1/800 now or so

[Orpheon]

Hey, Rein told me that they boosted the chance of encountering a shiny now =o

He said they made it about 10x more likely now? Around 1/800 now or so

I would like to believe this is true, and I don't want to be that guy, but.... source?

I haven't found anything that neither confirms nor denies this statement. However, I'd be ecstatic if this were the case :P