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

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A masterball
Not a master ball, don't have it yet

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 Mew at least once with the given Pokeballs is equal to;
1 - (chance that you will not catch Mew all 6 times)or 100% - (% 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)^6
If 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 59/1000
So the chance of NOT CATCHING Mew is 94.1% or 941/1000

So now you want to solve (941/1000)^6
This 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

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Maaaaaath :D

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