2
$\begingroup$

I don't really know the process of doing the math here (this is actually the chance of Shiny Pokemon, if anyone's curious). Once you've battled 500, a 3% chance of getting 1/512 odds is added on.

So 1/1365 is your chance of getting a Shiny, but each encounter with a Pokemon has a 3% chance of changing the odds to 1/512, which are much higher. I'm curious what my total odds are then, basically.

Normally (I know it doesn't really work this way), I would say that if my odds are 1/1365 and I go over 1365 then I'm "over odds." I'm curious what would be the expected "odds" when you factor in my "3% chance of 1/512."

Thanks.

Improve this question
$\endgroup$
3
  • $\begingroup$ Hi: So your new "1 in whatever number is the transformation of 0.97 ( 1/365) + 0.03 (1/512). So, you calculate that number and then convert it to odds. $\endgroup$
    – mlofton
    Feb 1, 2021 at 17:31
  • $\begingroup$ What was your source for these stats? $\endgroup$
    – R Carnell
    Feb 1, 2021 at 17:38
  • 1
    $\begingroup$ It is a bit weird to speak about odds of 1:1365 plus 3% chance for odds of 1:512 how exactly are they blended? You probably want some sort of odds for a number of battles (or at least the number of battles should matter since it seems that the 1:512 bonus only kicks in after 500 battles) for how many battles do you wish to compute? The odds are per single battle or per 500? The bonus odds of 1:512 are for every battle after 500 or only for some? Do you have the bonus for one battle or do you get the bonus for ever? Depending on such considerations the computation might need to be different... $\endgroup$
    – Sextus Empiricus
    Mar 5 at 15:53

1 Answer 1

Reset to default
6
$\begingroup$

Just starting with the headline question:

$$P_{total} = \frac{1}{1365}(1-0.03) + \frac{1}{512}(0.03) = 0.000769$$ or 1 in 1300.

This is the probability per catch. Your long run probability would need to also account for whether the change to 1/512 odds is permanent.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks very much. $\endgroup$
    – Bob
    Feb 2, 2021 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.