Your method looks perfectly good to me.
If I subtract 1 from all the prize values except \$99,999 and follow your procedure, I get the book's answer. It may be the original table had the gross prize amounts (not counting the \$1 it cost to buy the ticket), and the author was going to convert them to net prizes by subtracting 1, but for some reason only did the first and last values. As written, it's inconsistent in either interpretation.
You might see if you can find an errata list for the book that corrects this error; if not, it would be nice to contact the author (or the publisher) and let them know of the error.
We give several approaches. The first one is closest in spirit to yours.
You have bought your $3$ tickets. Now the lottery corporation is choosing the $4$ winning tickets. These can be chosen in $\binom{n}{4}$ ways. If the lottery is well run, the process makes more or less sure that the choices are all equally likely.
How many ways can they choose the $4$ winning tickets so that none of them are yours? Clearly $\binom{n-3}{4}$. So the probability you win no prize is
$$\frac{\binom{n-3}{4}}{\binom{n}{4}}\tag{A}.$$
(The expresssion in (A) can be considerably simplified.) The probability you win at least one prize is therefore $1$ minus the answer of (A).
Another way: Imagine the lottery corporation picks the winning tickets one at a time. The probability the first ticket it picks is bad (not one of yours) is $\frac{n-3}{n}$.
Given that the first ticket picked was bad, there are $n-4$ bad left out of $n-1$, So given that the first ticket was bad, the probability the second is bad is $\frac{n-4}{n-1}$. Thus the probability the first two are bad is
$\frac{n-3}{n}\cdot\frac{n-4}{n-1}$.
Continue two more rounds. The same reasoning shows that the probability all our tickets are bad is
$$\frac{n-3}{n}\cdot\frac{n-4}{n-1}\cdot\frac{n-5}{n-2}\cdot\frac{n-6}{n-3}.$$
Still another way: Let's switch points of view.
Imagine that the winning tickets have already been determined. So there are $4$ "good" tickets and $n-4$ bad. We will find the probability that you pick all bad.
There are $\dbinom{n}{3}$ ways that we could choose our three tickets. There are $\binom{n-4}{3}$ ways to choose them so they are all bad. So the probability that we choose all bad is
$$\frac{\binom{n-4}{3}}{\binom{n}{3}}.\tag{B}$$
The probability of at least one good is $1$ minus the number in (B).
Remark: We could use a strategy like yours: Find the probability of winning exactly $1$ prize, exactly $2$ prizes, exactly $3$ prizes, and add up,
We calculate the probability of winning exactly one prize. The other calculations are roughly similar. So you have bought $3$ tickets. We find the probability exactly one is good,
There are $\binom{n}{4}$ ways for the corporation to choose $4$ tickets. How many ways are there to choose $1$ that you have and $3$ that you don't have?
The one you have can be chosen in $\binom{3}{1}$ ways. For each such choice, the $3$ you don't have can be chosen in $\binom{n-3}{3}$ ways. Thus the probability of exactly one winning ticket is
$$\frac{\binom{3}{1}\binom{n-3}{3}}{\binom{n}{4}}.$$
Best Answer
I think your value of $.3825$ is correct for not taking into account the amount paid for the ticket.
So I think your expected value is $.3825-1=-\$0.6175$