Yes it is. The probability of winning $k$ times out of $n$ lotteries is determined from a binomial distribution:
$$P(K=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
where $n=300$, $k=1$, and $p=1/300$. The answer is about $0.368494$, or $36.8\%$. The probability of winning at least once is
$$P(K \ge 1) = 1- P(K=0) = 1-\left( \frac{299}{300} \right )^{300} \approx 0.632735$$
or about $63.3\%$ chance of winning something. Not bad, but not $100\%$.
Your optimal strategy is to offer 0, i.e. don't buy the car. Intuitively, you can see this because if you pay $p$ to buy the car from X, there's a 50% chance that the car's value $v$ is less than half of $p$, but Y will only pay you $1.5\times$ $v$. You would need Y to pay you at least $2\times$ $v$ to break even on average.
You can make this more formal by defining $V$ to be the price of the car, uniformly distributed between 0 and 1 (working in units of 1000s is easier). Let $p$ be the price you offer.
Then (1) the probability of the price being more than the value is
$$P(p \ge V) = p$$
by the definition of the uniform distribution.
The profit $r$ in this case is the money Y pays you, $1.5v$, less the price you paid $p$. If we let $Q$ be a random variable for the value of the car given we only know the value is $\le p$ then we can see that $Q$ is uniformly distributed on $(0,p)$ and its expectation is therefore $p/2$. Hence the expectation of the profit $R$ is $1.5p/2 - p$.
Case (2) the probability of the price being less than the value is
$$P(p < V) = 1-p$$
and the profit $R$ in this case has an expectation of 0 because we're neither buying nor selling the car.
Hence by the law of total expectation, the expected profit is given by the profit in each case times the probability of the case
$$ \begin{align} E(R) &= E(R|\mathrm{bought})P(\mathrm{bought}) + E(R|\neg\mathrm{bought})P(\neg\mathrm{bought}) \\
&= (1.5p/2 - p) \times p \;-\; 0 \times (1-p) \\
&= -0.25p^2
\end{align}$$
This is negative so you wouldn't want to make an offer! With a price $d$ in your original units of currency your expected loss would be $0.25(d/1000)^2\times1000 = d^2/4000$.
Re your update, the probabilities look correct. However it is the amount of money you expect to make that is important, not the probability of making it. For example given these two games which one would you play?
You have a 90% chance of winning \$1, and a 10% chance of losing \$1million.
You have a 10% chance of winning \$1million, and a 90% chance of losing \$1.
I hope you can see from this that the probability of ending up with more money isn't the most important thing: it's how much more money you end up with on average (the expectation) that's important.
In your case, it's easy to see that the expectation associated with cases 1 and 3 is relatively low, because you've included case 1 where no money changes hands. If you assumed that the payoffs are similar with winning and losing, and wanted to look at probabilities alone, you would compare case 2 (buy the car and get less) with case 3 (buy the car and get more), and notice that $$P(\mathrm{case 2/lose}) = m/1500 > P(\mathrm{case 2/win}) = m/3000$$
Best Answer
Let random variable $X$ represent our (gross) win if we buy one ticket of each kind. Then $X$ has mean $8$. We can also compute the variance of $X$. For example, let $W_1$ be our winnings on the first type of ticket. Then $E(W_1^2)=\frac{(1200000)^2}{600000}=2400000$. Thus $W_1$ has variance $2400000-4$. Do the same for the other types of ticket, and add up. (We are asssuming the four lotteries are independent.) Thus $X$ has variance $4800000-16$.
Now let $Y$ be our total gross winnings on the $1$ million trials. This has mean $8000000$ and variance $4.8\times 10^{12}$. Our outlay has been $4000000$, so we want the probability that $Y\lt 4000000$.
Now comes the crossing of the fingers part. Assume that $Y$ has a distribution close enough to normal for our computations to be not too far off.
Then we want the probability that a normal with mean $8000000$ and variance $4.8\times 10^{12}$ is less than $4000000$. This is the probability that $Z\lt -4/\sqrt{4.8}$, where $Z$ is standard normal. That turns out (tables) to be about $0.039$.
Remark: The (very) weak point of the calculation is the normal aapproximation. The greatest contributor to the weakness is the $1200000$ prize. So I would suggest redoing the computation as follows. It is easy to calculate to calculate the probability $p_0$ that we win no big prize, the probability $p_1$ that we win exactly one, and so on. For practical calculation we only need these two, and the probability $p_r=1-p_0-p_1$ that we win two or more big prizes.
Calculate the probability $l_0$ that we lose money given that we got no big prizes, the probability $l_1$ that we lose money given that we got exactly one big prize. These can be estimated by the same sort of normal approximation as before, but the approximation is substantially more reliable.
For all practical purposes, we will lose money only if we win no big prize or only one, and the smaller prizes are not enough to cover our big prize losses. The probability we lose money is well approximated by $p_0l_0+p_1l_1$.