How to calculate the probability of at least one success from a number of scenarios with varying probabilities

To illustrate what I'm asking, let's imagine a scenario where you have varying numbers of tables with cups on them. Each table has 1 ball hidden under a single cup and a varying number of cups with nothing. You, as the player are allowed to lift 1 cup at each table. How do you work out the probability that at least one of those lifted cups will have a ball there?

Let's take this scenario as an example:

• Table 1: 2 cups
• Table 2: 5 cups
• Table 3: 20 cups
• Table 4: 4 cups

I've managed to come up with this formula:

$$ \frac 1 2 + \frac 1 2 \cdot \frac 1 4 + \frac 1 2 \cdot \frac 1 4 \cdot \frac 1 5 + \frac 1 2 \cdot \frac 1 4 \cdot \frac 1 5 \cdot \frac 1 {20} = 65.125\% $$

I believe that this correctly calculates the probability but it has the problem that it's only correct if you start with the most probable event and work linearly towards the least probable.

Is there an alternative calculation which doesn't care about the order?

That is not the correct answer. The correct answer using your approach will be,

$ \displaystyle \frac 12 + \frac 12 \cdot \frac 14 + \frac 12 \cdot \color {blue} {\frac 34} \cdot \frac 15 + \frac 12 \cdot \color {blue} {\frac 34 \cdot \frac 45} \cdot \frac 1 {20} = \frac {143}{200}$

Also the order does not matter. Say we start with the table with $20$ cups and go in reverse order. Then it would be,

$ \displaystyle \frac 1{20} + \frac {19}{20} \cdot \frac 15 + \frac {19}{20} \cdot \frac 45 \cdot \frac 14 + \frac {19}{20} \cdot \frac 45 \cdot \frac 34 \cdot \frac 12 = \frac {143}{200}$

However, a simpler approach is to find probability $p$ that none of the lifted cups will have a ball and then subtract from $1$.

$p = \displaystyle \frac 12 \cdot \frac 34\cdot \frac 45 \cdot \frac {19}{20} = \frac{57}{200}$

Desired probability is $ \displaystyle 1 - p = \frac{143}{200}$