Problem:
On average, how many cards in a normal deck of 52 playing cards do you need to flip over to observe your first ace?
To solve this problem, we can change the question to:
For any card which is not an ace, what is the probability it comes before all the aces?
Now consider the case where we have 1 non-ace and 4 aces, that means the probability of the non-ace coming before all the aces is going to be one fifth, this is the simplest case.
For this smaller problem, the expected number of cards to flip over to observe the first ace is:
$$E[X] = \frac{1}{5} + 1 = 1.2$$
We have 48 non-aces to consider and the probability of any one of them coming before all the aces is always going to be one fifth. This means the expected number of cards to flip over to observe the first ace is:
$$E[X] = 48 \cdot \frac{1}{5} = \frac{48}{5} = 9.6 + 1 = 10.6$$
The 9.6 is the expected number of non-aces we need to flip over and the additional 1 is the ace that we flip over.
This took awhile to make sense to me but once I understood more about linearity of expectation, it makes solving problems like this very easy.