[notafraudster]

Let's say you roll a dice with 100 sides. If you roll a 1, you die. If you roll anything else, you live. We want to know the probability you will die if you roll the dice 100 times.

One way we could do this is look at the probability you'll roll it on the first roll... then the probability you won't roll it on the first roll but you will on the second roll... and so on. But that's a lot of math.

The probability of an event (death) and its complement (not death) totals 1.0. So one way we can get the probability of death is 1.0 - the probability of life.

Okay, so the only way you'll live if is if survive all 100 rolls. Each dice roll is independent (surviving the first dice roll doesn't affect the second dice roll which doesn't affect the third). So each individual dice roll has probability 0.99 of survival. For joint probability, we can multiply these together. The probability of getting heads on a coin twice is 0.5 * 0.5 = 0.25. So in our situation here, p(survival) = 0.99, and 100 times means 0.99^100, to get the probability of survival. 0.99^100 = 0.36. 36% chance of survival.

The probability of death is thus 1 - 0.36 = 0.64. 64% chance of death.

[dmos62]

Counter question. How to calculate how many tries it would take to reach a specific probability for an outcome? For example, how many times do I have to roll the dice to have 90% chance of death? Due to my field of work, I'd solve everything by bruteforce, but I wonder what a more elegant solution could be.

[wutbrodo]

Just solve the equation above.

0.99^n = (1-0.9)

n*log(0.99) = log(0.1)

n = log(0.1)/log(0.99) = ~229

[mark-r]

The comment that prompted the question stated the opposite probability - 99% chance of death, not life. That changes the odds quite drastically.

[dmos62]

Clear and concise. Thanks!