# Calculate probabilities to win an a slot machine

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:

$$\sum \left(\,^nC_r (0.48^X)(1-0.48)^{y-x},x,1000,501\right) = 0.0927$$

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?

• I have attempted to type out your equation in MathJax, however I'm not sure I understand exactly what you meant. take a look and edit it if I have done it wrong :) Commented Mar 15, 2019 at 12:09
• Hey Lioness, what part is it you don't understand? What I want to achieve or the equation as such? Commented Mar 21, 2019 at 21:02
• The equation, why does it have commas in it? Commented Mar 21, 2019 at 23:08

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 $$$$P(X=k) = {1000 \choose k } 0.48^k(1-0.48)^{1000-k}$$$$
• Such player has to win $375$ times out of $1000$ (Wins 3750, loses 6250, net loss = $6250-3750 = 2500$). $P = {1000 \choose 375} 0.48^{375}0.52^{625} \simeq 4.97 * 10^{-12}$ Commented Mar 15, 2019 at 13:23
• Yes, there is. Notice that in your reply it is first time that you wrote "at least 75%". In your post and in comments you wrote 75% left (exactly 7500) which for given parameters is quite unlikely (as you spotted). You want $P(X=7500)$ or $P(X\geq 7500)$? Commented Mar 15, 2019 at 15:25