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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 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?

evilpegasus
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Richard
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2 Answers2

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The 10.6 answer is sampling without replacement. 13 would be the correct answer if you replaced each card after checking it.

Expected value of sums

  • Just to show the math behind this assertion, we have that the expectation can be given by $$1\cdot\frac{4}{52}+2\cdot\frac{48}{52}\frac{4}{51}+3 \cdot \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{4}{50} + ... + 49 \cdot \frac{48}{52} \cdot \frac{47}{51} \cdot ... \cdot \frac{1}{5} \cdot \frac{4}{4} = \frac{1}{13} + \sum_{n=1}^{49}\left(n\cdot\frac{4}{53 - n}\cdot\prod_{k=0}^{n-2} \frac{48 - k}{52 - k}\right)=10.6$$ where the first equality is found by computing the probability of not picking an ace for the first $n-1$ turms and picking an ace on turn$n$, and the last can be done numerically. – Cade Reinberger Aug 23 '19 at 03:07
  • There is also a very good intuitive argument given in the link I gave. If you added a fifth ace and arranged the 53 shuffled cards in a circle, the expected gap between aces including one of the aces would clearly be ${53 \over 6}$. –  Aug 23 '19 at 03:09
  • That is quite nice – Cade Reinberger Aug 23 '19 at 03:51
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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: $$E(N)=\sum_{k=1}^{52}k\cdot P(k)=\cdots=\frac{\color{red}{53}}{5}.$$

farruhota
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  • Say that $E(N)=\frac{n+1}{5}$ when you play with $n$ cards, you want to prove $E(N+1)=\frac{n+2}{5}$ for the case you are using $n+1$ cards to play. This can be simply seen from $E(N+1)=\frac{4}{N+1}+\frac{N+1-4}{N+1}(E(N)+1)$, where $E(N)=\frac{N+1}{5}$. With basic algebra you end up proving the final result by induction. – EmG Mar 20 '21 at 09:56