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 $$P(\mathrm{case 2/lose}) = m/1500 > P(\mathrm{case 2/win}) = m/3000$$
Step 0: Work backwards from the end
Step 1: If the score is $3-3$ you want the net position so far to be $0$, and you bet $1000$
Step 2: If the score is $3-2$ you want the net position so far to be $+500$, and you bet $500$; if the score is $2-3$ you want the net position so far to be $-500$, and you bet $500$
Step 3: If the score is $3-1$ you want the net position so far to be $+750$, and you bet $250$; if the score is $1-3$ you want the net position so far to be $-750$, and you bet $250$; if the score is $2-2$ you want the net position so far to be $0$, and you bet $500$ etc.
You want the position at a particular score to be the average of the two possible positions after the next game. You want to bet half the difference of the two possible positions after the next game. You should end up with this:
Bets:
=====
Score 0 1 2 3
0 312.5 312.5 250 125
1 312.5 375 375 250
2 250 375 500 500
3 125 250 500 1000
Positions:
==========
Score 0 1 2 3 4
0 0 312.5 625 875 1000
1 -312.5 0 375 750 1000
2 -625 -375 0 500 1000
3 -875 -750 -500 0 1000
4 -1000 -1000 -1000 -1000
Best Answer
If you bid $a$, the probability you win the item is $a/1000$. The expected amount you win is $(a/1000)\times (a/2)$ where $a/2$ is the mean of the value given it is less than $a$.
Then you receive $3/2$ multiplied by this as a prize. This is bounded by $3a/4$ so on average you make a loss of $\ge a/4$. Of course, there is a possibility you win something, but it's probably not large enough to make it worth it. (This is a matter of opinion, not maths.)