At a roulette table, there is a wheel containing 37 possible pockets a ball can land on (each are marked 0, 1, 2, 3, etc. through to 36) and it is equally likely the ball could land on any of these spots.
If I were to say that there is a (roughly) 81% chance that the ball will fall on one of the numbers between 4 to 33 would I be correct? And, what is the chance it will not land on any of these numbers (what is the chance it will land on 0, 1, 2, 3, 34, 35, and 36?).
In a possibly less confusing way of saying this, would it be correct to equate these probabilities to 30/37 and 7/37?
*Also, this is not a homework assignment or anything like that. I was thinking of how I could come up with a good strategy for roulette that would minimize losses and maximize profits. Gambler, yes. Cheater, no.
Yes, that's the correct interpretation. Since in roulette, all of the outcomes are mutually exclusive (the ball has to land on a single value, it can't land in multiple spaces at the same time), you can just add up the probabilities associated with each number.
So, $P(0 \text{ or } 1) = P(0) + P(1)$, and $P(0 \text{ or } 1 \text{ or } 2) = P(0) + P(1) + P(2)$, etc.
Since the probabilities for each space are the same, that is, $P(0) = P(1) = P(2) = ... = P(36) = \frac{1}{37}$, you can just multiply by the number of spaces you're looking at:
$$P(0 \text{ or } 1) = P(0) + P(1) = \frac{1}{37} + \frac{1}{37} = \frac{2}{37}$$
$$P(0 \text{ or } 1 \text{ or } 2) = P(0) + P(1) + P(2) = \frac{1}{37} + \frac{1}{37} + \frac{1}{37} = \frac{3}{37}$$
Generally, if you want to know the probability that the roulette ball lands on one of $n$ spaces, it's $n\cdot \frac{1}{37} = \frac{n}{37}$. Plug in $n = 1$, and that's the probability for any one space, and plug in $n = 37$, and that's the probability it lands anywhere on the roulette wheel.
In your question, you're asking for the probability of $n = 30$ (there are 30 spaces between 4 and 33, inclusive), so that's $\frac{30}{37} \approx 81\%$.
Note, many roulette tables have 38 spaces, with $00$ as an option also. In that case, the probabilities for each space are $\frac{1}{38}$, and the probability of landing in 1 of $n$ spaces is $\frac{n}{38}$.
To calculate probabilities of specific strings, you can just multiply the probabilities for each individual roll, because rolls of the roulette wheel are independent of each other (the result of one spin doesn't affect what happens on the next spin).
So, if you have bets placed on $n$ spaces, you have the probability of winning is $\frac{n}{37}$, and the probability of losing is $\frac{37-n}{37}$ for each spin.
Then, if you want to know the probability of winning two spins in a row, that would be $\frac{n}{37}\cdot \frac{n}{37} = \left(\frac{n}{37}\right)^2$. For winning three spins in a row, it's $\frac{n}{37}\cdot \frac{n}{37}\cdot \frac{n}{37} = \left(\frac{n}{37}\right)^3$. In general, the probability of winning $p$ in a row is $\left(\frac{n}{37}\right)^p$. Similarly, the probability of losing $q$ spins in a row is $\left(\frac{37-n}{37}\right)^q$.
This type of probability distribution is called a binomial distribution in case you'd like to read more about it.