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# What is the chance of finding a Shiny Boldore with Everstone in just 4 or 5 encounters?

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I was at Chargestone Cave and I randomly found a Shiny Boldore, caught it and it was holding a Everstone. I'm pretty sure it was 4 or 5 encounters.

Did you have a shiny charm? Also what generation? BW or BW2
Doesn't Gen 2-5 games have a 1/8192 chance?
I'm talking about the chance of finding a Boldore there in general, you have to factor that in as well.
Boldore has a 5% chance to appear in B2W2.

First, the other answer is correct about how the chance of finding a shiny everstone Boldore in one encounter is 1/327,680.
The chance of anything else is 1 - 1 / 327,680.
The chance of anything else twice in a row is (1 - 1 / 327,680) * (1 - 1 / 327,680), or (1 - 1 / 327,680)^2.
The chance of anything else n times in a row is (1 - 1 / 327,680)^n.
The chance of not finding anything else n times in a row is 1 - (1 - 1 / 327,680)^n. This is the same as the chance of a shiny everstone Boldore at least once in n encounters.
1 - (1 - 1 / 327,680)^4 is close to 1/81920.
1 - (1 - 1 / 327,680)^5 is close to 1/65536.

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Chance of Boldore 1/20 x chance of Everstone 1/2 x chance of shiny 1/8192 = 1/327,680
Please correct me if math is wrong.

Chance of Boldore 1/20 x chance of Everstone 1/2 x chance of shiny (1/8192-4 from 4th encounter) = 1/327,520
Is this what you mean?

edited
That's the chance of finding a shiny Boldore with everstone in one encounter, not 4 or 5 encounters.
If you flip a coin for the third time the chance of it being heads after the last two being tails is 50%. shiny chance isn't affected by time.
You're missing the point.
Are you more likely to see at least one "heads" if you flip a coin once, or if you flip it five times?
yes you are more likely but it doesn't directly influence the numbers. If it did it would be more like a countdown.
Because there are 4 encounters, the odds are 1/2048 in that example which would make the final outcome would be 1/81920
Do either of you know binomial probability?
@DarkVoid If it's more likely, then why don't you update the answer? It absolutely does affect the final result. (It would not be like a countdown, either -- the chance of finding at least one shiny trends toward 100% with infinite trials, but it is never guaranteed.)
@GmaxWaluigi Repeated trials don't have this effect. Following the logic that you just multiply the chance of success by the number of trials, you could guarantee seeing a heads with only two coin flips, like so: 2*(1/2)=1. You need to use the binomial distribution.
According to the formula you have now, the chance of finding a shiny Pokemon in exactly 8192 encounters is undefined because it divides by 0. If your formula is wrong for 8192, then it's probably also wrong for 4.
@Fizz sorry I do not know how to do binomial distribution I have heard of it

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