Which is mathematically better - 6 Pokeballs or 1 Ultra Ball?

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Pleaselog inorregisterto add a comment.

Since I’m lazy, I’m going to use Mew as the subject of the question (first Pokemon that popped into my head).Catch rate of 45, 5.9% chance of catching at full health with a Pokeball.Ultra Ball increases the catch rate by 2x so the catch rate moves up to 90, and there is a 11.8% chance of catching the Mew at full health (Yes, if catch rate doubles, chance of catching will double, go check something like Bronzong which has 90 catch rate, 11.8% chance of catching).

So given this, we know there is an 11.8% chance of catching the given Mew with an Ultra Ball, or 59/500 as a fraction.

How about not 1, but 6 Pokeballs?The chance of catching the Mewat least oncewith the given Pokeballs is equal to;1 - (chance that you will not catch Mew all 6 times)or100% - (% you will not catch Mew all 6 times)So we need to solve the chance that you will not catch Mew all 6 times.That is equal to(Chance of not catching Mew once)^6If you don’t understand why, don’t worry, you’ll learn in Probability or Combinations and Permutations later.Chance of CATCHING Mew = 5.9% or 59/1000So the chance of NOT CATCHING Mew is 94.1% or 941/1000

So now you want to solve (941/1000)^6This number is too long for me to be bothered putting in fractional form, so in a rough % form it would be 69.42849331%

100% - 69.42849331% = 30.57150669

There is a 30.57% chance that you will catch the given Mew in those 6 Pokeballs