Expected value of number of different cards

Question

I have this question:

From a deck of $10$ cards, Tom draws two cards and places them back in the deck. Now Jerry draws two cards from the deck.

Let $X$ be the number of cards that was chosen by only one of Tom or Jerry. What is $\Bbb E[X]$ ?

I've figured that the range of $ X$ is $\{0,2,4\}$, and calculated the probability for zero to be: $\frac 2{10} \frac 19$, as he needs to pick the same two cards.
The probability of two cards to be : $\frac 1 {10}$$\frac 89$ , as he needs to pick the same card once and a different card after.
The probability of $4$ cards : $\frac 8{10}$$\frac 79$, as he needs to pick two different cards.
$$\Bbb E[X]=4\frac 8{10}\frac 79+2\frac 1{10}\frac 89=\frac {8}{3}$$

But I have a mistake and I can't figure out what it is. Can anyone help?

Answer 1

The probability of picking two cards should be multiplied by $4$, because there are two cards to match with, and we can either match-miss or miss-match.

This gives the numbers of hands as $\{X=0:2, X=2: 32, X=4:56\}$

and gives the revised expectation as:

$$\Bbb E[X]=4\frac 8{10}\frac 79+2\frac 4{10}\frac 89=\frac {16}{5}$$

Answer 2

Here is another approach.

Number the cards from $1$ to $10$, and let $$X_i = \begin{cases} 1 \qquad \text{if card i is chosen by exactly one of Tom and Jerry} \\ 0 \qquad \text{otherwise} \end{cases}$$ for $i = 1,2,3,\dots,10$. Then $$P(X_i=1) = 2 \cdot \frac{2}{10} \cdot \frac{8}{10} = \frac{8}{25}$$ so the expected number of cards chosen by exactly one person is $$E \left( \sum_{i=1}^{10} X_i \right) = \sum_{i=1}^{10} E(X_i) = 10 \cdot \frac{8}{25} = \frac{16}{5}$$ where we have used linearity of expectation.