First Ace
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?
Answer
Label the 48 non‐ace cards as \(1, 2, \ldots, 48\). For each non‐ace card \(i\), define
\[
X_i =
\begin{cases}
1, & \text{if card \(i\) appears before \emph{all} 4 aces},\\
0, & \text{otherwise}.
\end{cases}
\]
Then the total number of cards seen before the first ace (including that ace itself) is
\[
X \;=\; 1 \;+\;\sum_{i=1}^{48} X_i,
\]
because we add 1 for the ace that actually appears.
In this case, only \(5\) cards are relevant: the fixed non-ace card and the \(4\) aces. All orders of these \(5\) cards are equally likely, so the probability that the fixed non-ace card comes before all \(4\) aces is
\[
\frac{1}{5}.
\]
Hence,
\[
\mathbb{E}[X_i] = \frac{1}{5},
\]
and
\[
\mathbb{E}[X] = 1 + \sum_{i=1}^{48} \mathbb{E}[X_i]
= 1 + 48 \times \frac{1}{5}
= 1 + \frac{48}{5}
= 10.6.
\]
Notes and comments
Comment 1: To get two aces in a row we just multiply by 2.
Comment 2: The answer would be the same for any card.