# Slot machine probabilities?

Probability of net gain given 100 pulls of slot machine with 3 slots, each numbered 1 through 9, and 1 dollar per push with average winning for any given winning push equal to 2 dollars (happening when there are either 2 of the same number, or 3 of the same number)?

First I considered the probability of getting either 2 of the same number, or 3 of the same number. I came up with P(two 8s) = (3C2)(1/9)^2(8/9) = 24/729. And, P(three 8s) = (3C3)(1/9)^3 = 1/729. So, P(two 8s, or three 8s) = 25/729 = 3.4%

Then considering 100 pulls of the slot machine, I got (100 C 50)(3.4%)^50(96.6%)^50 which came up to be a very small number, which I do not think is right. I considered winning 50 times with 2 dollars means you gained 100 dollars, and put in 100 dollars. So this probability, I think, represents the probability of breaking even. Is this correct? If so, how to compute probability of a net gain?

• Please use MathJax to format your questions. – saulspatz Sep 2 '18 at 6:35

Your probability of getting $2$ or $3$ $8$s is correct but, you can win for any other number than $8$ too, so you need to take the union of those. Luckily the event are disjoints so in the end it is just $9\times \frac{25}{729}=\frac{25}{81}$. The probability of breaking even is then as you mentioned ${100 \choose 50} \left(\frac{25}{81}\right)^{50} \left( \frac{56}{81} \right)^{50}$ which is of course pretty small because we consider a very large space of events. Something more interesting would be to compute (or estimate) the average gain (loss), the probability of wining. Here is the probability of not loosing on wolfram, pretty small. And the average gain is about $-38.3$. Conclusion : don't play.
• Well, I'm sorry I mess it up, just edited my post, so the gain would be $2 i-100$ and the average loss is around 38 – P. Quinton Sep 2 '18 at 18:02