### Probability of hitting 1 or more times when performed p% loot box n times

Probability of hitting 1 or more ＝ 100% － Probability of not hitting even once

$P=1-(1-p)^n$#### e.g. Probability of hitting 1 or more times when performed 1% loot box 1 times

$1-0.99=0.01=1\%$

-> It hits about 1 in 100 people.

#### e.g. Probability of hitting 1 or more times when performed 1% loot box 100 times

$1-0.99^{100}=1-0.36603...=0.63397...\approx63.4\%$

-> It hits about 2 in 3 people.

### Probability of hitting m or more times when performed p% loot box n times

If you want to hit more than 2 times, the calculation will be difficult.

- Probability of hitting more than 2 times ＝ 100% - Probability of not hitting even once - Probability of hitting only once
- Probability of hitting more than 3 times ＝ 100% - Probability of not hitting even once - Probability of hitting only once - Probability of hitting only twice

“Probability of hitting only x times” can be calculated by the following formula.

$P=nCx \times p^x \times (1-p)^{(n-x)}$nCx is number of combinations to choose x from n different.

$nCx={\dfrac {n!}{x!(n−x)!}}$#### e.g. Probability of hitting 2 or more times when performed 1% loot box 100 times

Probability of not hitting even once:

$P_0={}_{100}C_0 \times 0.01^0 \times (1-0.01)^{(100-0)}\\
=1 \times 1 \times 0.99^{100}\\
\approx 0.36603$

Probability of hitting only once:

$P_1={}_{100}C_1 \times 0.01^1 \times (1-0.01)^{(100-1)}\\
=100 \times 0.01 \times 0.99^{99}\\
\approx 0.36973$

Probability of hitting 2 or more times:

$P=1-P_0-P_1\\
=1-0.36603-0.36973\\
=0.26424\\
\approx 26.42\%$