Problem thought of at an actual football game:
Say we are in a football stadium with each successive row elevated (slightly) above the previous row. The rows are numbered starting with $1$ at the bottom and continue upward to some number $n$ at the top. If one person decides to stand up, then every person directly behind that first person (in the same "column", if you will) will stand up in a domino-like fashion.
Let's say that the probability that a given person will stand up is a fixed probability $p$ in the interval $[0, 1]$. Given $p$, what is the probability that a person sitting at row $k$ (in $[1, n]$) will stand up?
Concretely, if the probability for each person of standing up is $0.02$, what is the probability that the person in row $37$ will stand up?
For the person in row $1$, the answer would obviously be $0.02$. And it clearly becomes more and more likely that you will stand up the higher you sit. But beyond that, I don't know how to think about this problem. My probability has gotten a bit rusty; is there a type of random variable that models this problem?
This problem setup seems like you could model it with a geometric probability distribution.
And the trick is to consider the probability that someone in the $k$th row doesn't stand up. Let $p=0.02$, the probability that someone spontaneously decides to stand up.
Hence, if each person will spontaneously stand up with probability $p$, and everyone behind you will automatically stand up after you in a domino effect, the probability that the person in the $k$th row does stand up is:
$$p_k = 1-(1-p)^k$$
Concretely, when $p=0.02$, and $k=37$, we have that the person will stand up with probability $1-(1-0.02)^{37} \approx 0.526$.