Hard Auction Game Theory Question: Optimal Strategy with Asymmetric Info There is a box containing $X$ dollars. $X$ is a random variable which is uniformly distributed between $0$ and $1000$. 
The box is auctioned via a first price sealed bid auction. There are two participants in the auction. Player 1 only knows that $X$ is uniformly distributed between $0$ and $1000$, whereas player 2 also knows whether $X$ is greater than or less than $500$ dollars.
To clarify, $X$ and the values of the bids can be arbitrary real numbers; they do not have to be integers.
What is the optimal strategy for player 1?
I've been hitting my head against this problem for a few days, but I haven't gotten much further than that one would probably want to play a random strategy where you bet a dollars y% of the time and b dollars z% of the time.
But what would the optimal values for a,b,y,z be?
 A: So the grounding assumption here is that both players are only trying to maximise their own profit, and both players will avoid strategies that do not pursue this goal. Essentially both players want the largest positive EV (Expected value, or expected gain from the bid). We will also be talking about the average outcome of the bet, or the outcome of an arbitrarily large number of bets.
We quickly note that in the symmetric case with no additional information, both players will bid $499$, because the average value of the box is $500$. We reach $499$ as this is the first and only stable equilibrium for both players. i.e. $499$ beats all bets from $0-498$ with a payoff of 1, draws with itself at 0.5, and loses only to illogical bets with a zero or negative EV. Visualising how this equilibrium stabilises is easier if you consider the same question but with the option to place subsequent bids. If the first player bids 1, the other will bid 2, the first will bid 3, ...
The EV for both players is ~$0.5$ or $0$, depending on what happens in the tiebreak. It is debatable perhaps that the best bid is $500$, but since the game is not competitive and players are only looking to maximise their money, no player will choose a bid with an EV of $0$.
The important conclusion is that crucially, without additional information, any bet above $500$ has a negative EV.
The asymmetric case is clearer to understand from the perspective of player 2 (who has more information).
Player 2 knows that the box is either in the range $[0,500]$ or $[500,1000]$. We shall call the $[0,500]$ case $L$ and the second $[500,1000]$, $H$. Clearly for $L$ the average value of the box is $250$ and for $H$ it is $750$. Thus, player 2 will never bid more than $249$ and $749$ respectively. And indeed, player 2 will bid $249$ for $L$.
However player 2 is also logical and understands that any bet from player 1 above $499$ is illogical, since player 1 has no information on the state ($L$ or $H$). Thus, player 2 is guaranteed to win with a bid of $500$. Player 2's overall EV is $0.5$ for L and $250$ for H.
Player 1's best strategy is to bid $249$ always. Any bid larger than $249$ but smaller than $500$ can only beat player 2 in the situation $L$, and in that case player 1 has a negative EV. 
As the game is not competitive, player 1 should never bid more than $499$. This is because, from player 1's perspective, we have a $50\%$ chance to be in $L$ and a $50\%$ chance to be in $H$. If we are in $L$, then player 1 loses $250-B$, where $B$ is the bid. If we are in $H$, player 1 gains $750-B$. The EV is $\frac{500-B}{2}$ for a bid of $B$. Since both $L$ and $H$ are equally likely, the only values of $B$ which return a positive EV here are 250-499. But, player 1 is aware that player 2 is able to deduce all of this, and thus player 2 will never lose $H$ to a bet under $500$. Thus, to win in $H$, player 1 is forced to bid more than $500$, but this returns a negative EV, which contradicts player 1's goal to only maximise their own money.
Even if the game was competitive, nothing actually changes. Player 1 cannot hope to beat player 2 with any strategy. Betting less than $250$ is illogical since player 2 will always in the boxes. Always betting 250 still loses to a 250,500 strategy in the long term. Always/ sometimes betting exactly 500 gives a negative EV for player 1 in $L$ and still allows player 2 to get a positive EV in $H$ with larger bets. Any strategy involving bids of more than $500$ from player 1 automatically have a negative EV for reasons previously discussed.
