This question appears in Zhou's A Practical Guide To Quantitative Finance Interviews.
What is the expected number of cards that need to be turned over in a regular 52-card deck in order to see the first ace ?
In his solution Zhou labels the non-ace cards $1$ to $48$ and for each $i\in \{1,\ldots,48\}$ he defines the event $A_i$ "card $i$ is turned over before all the 4 aces". With these notations, we need to compute $1+\sum_{i=1}^{48} E[1_{A_i}] = 1+48P(A_1)$ by symmetry. Zhou then argues
As shown in the following sequence, each card $i$ is equally likely to be in one of the five regions separated by 4 aces: $$1 \quad A \quad 2 \quad A \quad 3 \quad A \quad 4 \quad A \quad 5$$
hence $P(A_1) = \frac 15$.
I don't understand why it is obvious that $P(A_1) = \frac 15$. The locations of the aces are themselves random. Is Zhou using some kind of independence ?
Can someone provide additional insight, or a less handwavy argument ?
