4.3.9 problem 47

(a)

Let $X={\sum }_{i=1}^{52}{I}_i$ be the number of cards that are called correctly. $\text{E}\left(X\right)={\sum }_{i=1}^{52}P(I_i=1)=52\frac{1}{52}=1$.

(b)

Source: https://math.stackexchange.com/a/1078747/649082

Let $X={\sum }_{i=1}^{52}{I}_i$ be the number of cards that are called correctly. $\text{E}\left(X\right)={\sum }_{i=1}^{52}P(I_i=1)$. To find $P(I_i=1)$, consider the first $i$ cards, with the $i$-th card correctly guessed. Let $k$ be the number of correctly guessed cards within the $i$ cards. For instance, for $i=5$, $k=2$, $Y$ representing a correctly guessed card and $N$ representing an incorrectly guessed card, one possible sequence of $i$ draws is $\mathrm{NY}\mathrm{NNY}$.

$$P\left(\mathrm{NY}\mathrm{NNY}\right)=\frac{51}{52}\frac{1}{51}\times \frac{50}{51}\frac{49}{50}\frac{1}{49}=\frac{1}{52}\times \frac{1}{51}$$

Notice that the second $N$ in the sequence has probability $\frac{50}{51}$, because the second card is guessed correctly. The only piece of information we have is that the third card is not the card that was correctly guessed, leaving a total of $51$ possibilities. Generalizing, the probability of a string of length $i$ with $k$ $Y$s is $\frac{(52-k)!}{52!}$. There are $\left(\frac{i-1}{k-1}\right)$ strings of length $i$ with $k$ $Y$s that end in a $Y$, and since $1\le k\le i$,

$$P(I_i=1)=\sum _{k=1}^i\left(\frac{i-1}{k-1}\right)\frac{(52-k)!}{52!}$$

Thus,

$$\begin{array}{cccc}\text{E}\left(X\right)& ={\sum }_{i=1}^{52}{\sum }_{k=1}^i\left(\frac{i-1}{k-1}\right)\frac{(52-k)!}{52!}& & \\ & ={\sum }_{k=1}^{52}{\sum }_{i=k}^{52}\left(\frac{i-1}{k-1}\right)\frac{(52-k)!}{52!}& & \\ & ={\sum }_{k=1}^{52}\left(\frac{(52-k)!}{52!}{\sum }_{i=k}^{52}\left(\frac{i-1}{k-1}\right)\right)& & \\ & ={\sum }_{k=1}^{52}\left(\frac{(52-k)!}{52!}\left(\frac{52}{k}\right)\right)& & \\ & ={\sum }_{k=1}^{52}\frac{1}{k!}& & \end{array}$$

Note that $e^x={\sum }_{i=0}^{\infty }\frac{x^i}{i!}\Rightarrow e^1\approx 1+\text{E}\left(X\right)+10^{-15}$. Thus,

$$\text{E}\left(X\right)\approx e-1$$

.

(c)

Since at any given time, we know all the cards remaining in the deck, the probability of the $i$-th card being the card guessed correctly is $\frac{1}{52-i+1}$. Thus, $\text{E}\left(X\right)={\sum }_{i=1}^{52}\text{E}\left(I_i\right)={\sum }_{i=1}^{52}P(I_i=1)={\sum }_{i=1}^{52}\frac{1}{52-i+1}={\sum }_{i=0}^{51}\frac{1}{52-i}\approx 4.54$.