I'm trying to calculate the probability to win in a slot machine. Return-to-Player is 96%. I have used the sum of the Binomial Coefficient to calculate it.
Since Return to Player is 96%, I make the assumption that the chances to win is 48% given that you're paid 1:1. (Please correct me if I'm wrong)
To win money over 1000 spins you need to win at least 501 of them.
My equation:
$$ \sum \left(\,^nC_r (0.48^X)(1-0.48)^{y-x},x,1000,501\right) = 0.0927$$
which is about 9.3%
In 9.3% of the cases, the player would have more money than he started with after 1000 spins.
Is this the correct way to do it?
How would I instead calculate the probability that he has 75% left of what he started with?
Let $N$ be an amount that the player started with.
He has 75% left thus after the 1000 spins he have to lose $0.25 N$. If $k$ is the number of wins then $k-(1000-k)=-0.25 N \rightarrow k = 500 - 0.125 N $ (let's assume that $N$ is such that $k \in \mathbb{N}$).
Assuming that $N>=1000$ then \begin{equation} P(X=k) = {1000 \choose k } 0.48^k(1-0.48)^{1000-k} \end{equation}