This is a tough worded question involving discrete probability.
$52$ people are in a room, and each tell the dealer their favourite card. They pay the dealer $1$ dollar to participate. The dealer shuffles the deck and deals $1$ card to each player. If their card is their chosen favourite, they receive $10$ dollars. What is the dealer's expected net income?
I have so far figured out that the first player has a $\frac{1}{52}$ chance of receiving their favourite card. The next player has a $\frac{1}{51}$ chance, however, does it not follow that unless the first player received their favourite card, there is a $\frac{1}{52}$ chance that they received player 2's favourite card? This would mean that player $2$ has zero chance of receiving their favourite. Does this need to be taken into account?
I just need to know roughly what direction to set off in, because at this point I am just wringing my hands over this problem.
Just use linearity of expectation. Let $E_i$ be the amount of money the dealer expects to make on the $i^{th}$ player. Then, of course $$E_i=\frac {51}{52}\times 1-\frac {1}{52}\times 10$$
The answer is just $$E=\sum E_i=52\times E_1=51-10=41$$
Note: it makes no difference in the expectation if you require your players to choose different favorite cards, though of course the probability distribution is different. To take an extreme example, if every player likes the Ace of Spades best then it is guaranteed that the dealer will lose $\$10$ once and win $\$1$ fifty-one times, so the distribution is constant at $\$41$.
Edit: I read the rules as saying that the dealer wins $1$ on a single player if the cards don't match and loses $10$ if they do. On revision, the rules might actually say that he wins $1$ if they don't match but loses only $9$ if they do (if the dealer keeps the $1$ that is). Doesn't change the concept. All that changes is that we must replace $10$ by $9$ in the calculation and the final answer would then be $51-9=42$.