Will double encounters in Eterna Forest help me find shines faster in any way?

Question

I'm playing Pokemon diamond for the DS, and I'm at Eterna forest with that one person. I'm using this to my advantage to find shines, but do the double encounters make finding shines 2/8192 odds?

Comments

The odds aren’t increasing by any means, but you’re doubling the amount of Pokémon you see within the time period.

Assuming you do one encounter per 30 seconds, you’re find two Pokémon per encounter at a rate of 30 seconds, rather than one.

Mathematically speaking, it isn’t the odds you would change in the formula, but rather the variables within the time period.

So yes it is faster, but it also isn’t. Faster in terms of how many Pokémon you encountering, but it’s all up to luck.

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I think it's still faster to use the Poke radar.

Answer 1

Technically, yes.

You are encountering two Pokemon at once, and both Pokemon have the same 1/8192 shiny chance. This means that when you have a dual encounter, there is a 2/8192 chance of a shiny being either one of the two Pokemon. This is 1/4096 odds. This theoretically doubles your odds of finding a shiny. Finding these odds are similar to Horde Encounters in XYORAS. If you have the shiny charm in these games, base shiny odds are 1/1365. Because you are encountering 5 Pokemon at once, you divide 1365 by 5 to get 1/273, which are your odds of encountering a shiny Pokemon in a Horde.

However, this does not necessarily mean you will find one any faster, because the odds are still absurdly low. Your odds are just as great looking for one Pokemon at a time in Gens 6-8.

Tldr: theoretically, yes, but not guaranteed.

Hope I could help

Comments

This is not exactly how probability works. If each coin has a 1/2 chance of landing on heads, flipping 2 coins doesn't double your chance and guarantee that you get heads.

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Your understanding of shiny probablities is a bit flawed. Firstly, the chance of finding a shiny on an encounter is completely independent of other encounters. The shiny chance of all encounters is 1/8192 pre-Gen 6, and not finding a shiny on an encounter doesn't increase the shiny chance of the next encounter to 2/8192, because both encouters are independent. This also means you aren't guaranteed a shiny every 8192 encounters, and could go for millions of encounters without finding a shiny or keep finding one every encounter, though the odds of both are extremely low.

Secondly, Ty's comment explains the situation better in that you're effectively cutting the time of an encounter by half, but not increasing shiny chances.

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You both misunderstand. I am talking about theoretical probability, not experimental probability.

sumwun, obviously flipping two coins does not guarantee you get heads, but because the odds are 1/2, *theoretically* the odds are much greater.

Psyduck, I completely agree and understand what you said.

 "The shiny chance of all encounters is 1/8192 pre-Gen 6, and not finding a shiny on an encounter doesn't increase the shiny chance of the next encounter to 2/8192, because both encouters are independent. This also means you aren't guaranteed a shiny every 8192 encounters, and could go for millions of encounters without finding a shiny or keep finding one every encounter, though the odds of both are extremely low."

This is obviously true when you are talking about chaining encounters, but not when we are talking about a single double encounter. Every single Pokemon you encounter has a 1/8192 chance of being shiny. So, theoretically, if you are encountering two Pokemon at once, the odds are 1/8192+1/8192. This is similar to Horde Encounters in XYORAS. If you have the shiny charm in XY, base shiny odds are 1/1365. Because you are encountering 5 Pokemon at once, you divide 1365 by 5 to get 1/273, which are the odds of encountering a shiny Pokemon in a Horde encounter.

If you follow my logic, the same principle applies to encountering two Pokemon at once in Eterna Forest. You are encountering 2 Pokemon at 1/8192 odds, so your odds become 1/4096 for any of them to be shiny.

Again, keep in mind we are talking about theoretical probability.

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Oh wait, so this is a double encounter where you encounter two Pokémon at once? Because the question said "with that one person," I thought OP was in the forest with a friend, and both of them were encountering Pokémon on their respective devices (I haven't played the game so I thought it could be possible). If its a double encounter, then what you said might be true (though I'll first read if double encounters somehow affect shiny chance in order to not jump at conclusions again).

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answer is a bit misleading in terms of wording, but it's still correct. it's like saying the starters i hgss are 3/8192, since youre finding three at once at the rate of 8192. does it increase? no. but youre still doing 3 pokemon per the encounter of 8192. same applies here, and that's what Baron means.

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Correct

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I never said it doesn't increase. I said it doesn't double. When you flip 2 coins, the chance of getting heads is 1 - (1 - 1/2)^2 = 3/4, which is not twice as much as 1/2. Similarly, when you encounter 2 wild Pokemon, your chance of getting a shiny is 1 - (1 - 1/8192)^2, which is not 2/8192.

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@sumwun (1 - 1/2)^2 is 1/4. Did you mean 1 - (1/2)^2?

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edited

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Your correct, but the difference is unbelievably marginal. 1-(1-1/8192)^2= ~0.000244125723. 2/8192= ~0.000244140625. As you can see, the first seven digits of the decimal are the same, so while the two percentages are not identical, they are extremely close