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

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