1% win rate played over 1000 times, how to calculate probabilty of results?

Question

With a coin toss, you have the probability of 99% that you will get 1 head in 10 coin tosses.

If you play a game where there is a 99% failure rate and a 1% win rate, how many times would you need to play to be sure that you will get one win?

Answer 1

Before to start playing the probability of winning in $n$ rounds is: $$p_n=1-(1-p)^n$$ where $p\le 1^-$ is probability of losing at any coin toss. The to make sure of winning you must have (!): $$1-(1-p)^n=1 \\ (1-p)^n=0 \\ n=\infty \quad(!!!)$$ Then to make sure of winning you must toss $\infty$ times . So start form right now!! Are you patient enough?!?!

Answer 2

You'll never be sure you win. It's always possible to fail every time. Call the amount of wins $X$, then $X \sim bin(1000, 0.01)$.