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**

## TL:DR 6 Pokeballs > 1 Ultra Ball.