The following question has appeared elsewhere on the site:
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?
The correct answer is $10.6$. However, I got something different from the following approach of conditional expectation:
Let $N$ denote the random variable for the number of cards to be turned over to see the first ace.
Also, let $R$ denote the random variable for the « value » of the card in the first round, i.e. the four aces have values $1$ to $4$ respectively and the other $48$ cards admit values $5$ to $52$ respectively.
Therefore, by the Tower’s property of conditional expectation,
\begin{eqnarray}
\mathbb{E}[N] & = & \sum_{i=1}^4 \mathbb{E} [N| R=i] \mathbb{P}(R=i) \\
& & + \sum_{i=5}^{52} \mathbb{E} [N| R=i] \mathbb{P}(R=i) \\
& = & \sum_{i=1}^4 1 \big( \frac{1}{52} \big) + \sum_{i=5}^{52} \Big( 1 + \mathbb{E}[N] \Big) \Big( \frac{1}{52} \Big) \\
& = & \frac{4}{52} + \frac{48}{52} \Big( 1+ \mathbb{E}[N] \Big).
\end{eqnarray}
This gives
$$ \mathbb{E} [N] = 13. $$
I fail to see any problems with this approach of conditional expectations, yet this does not give the correct answer. Any ideas?
Best Answer
You can work out for smaller number of cards to see a pattern.
For $\color{red}5$ cards with $4$ aces: $$E(N)=\sum_{k=1}^5k\cdot P(k)=1\cdot \frac45+2\cdot \frac15\cdot \frac44=\frac{\color{red}6}5.$$ For $\color{red}6$ cards with $4$ aces: $$E(N)=\sum_{k=1}^6k\cdot P(k)=1\cdot \frac46+2\cdot \frac26\cdot \frac45+3\cdot \frac26\cdot \frac15\cdot \frac44=\frac{\color{red}7}5.$$ For $\color{red}7$ cards with $4$ aces: $$E(N)=\sum_{k=1}^7k\cdot P(k)=1\cdot \frac47+2\cdot \frac37\cdot \frac46+3\cdot \frac37\cdot \frac26\cdot \frac45+4\cdot \frac37\cdot \frac26\cdot \frac15\cdot \frac44=\frac{\color{red}8}5.$$ Hence, for $\color{red}{52}$ cards with $4$ aces: $$E(N)=\sum_{k=1}^{52}k\cdot P(k)=\cdots=\frac{\color{red}{53}}{5}.$$