What to bid for this treasure chest? (puzzle) Suppose you are given the opportunity to bid for a treasure chest, which you know to be priced anywhere between 0-1000 dollars inclusive). Treasure price is uniformly distributed. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest and get your money back. Now, also suppose you have a friend who is willing to buy the treasure chest from you for 1.5 times the price of the treasure chest (should you obtain the chest). What should your bid be?
I don't know how to approach the problem. How can I calculate expected value?
 A: 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.)
A: There is no good answer without making some assumption about the probability distribution of the price.  If the price were exactly $0$ it would fit the $0-1000$ range and you should not bid.  If you want to assume a uniform distribution, the problem can be solved.  Let my bid be $y$ and the price of the chest be $x$.  My gain is $$g(x,y)=\begin {cases} 0 & y \lt x \\1.5x-y & y \ge x \end {cases}$$  So for a bid $y$ the expected gain is $$\frac 1{1000}\int_0^{1000} g(x,y) dx=\frac 1{1000}\int_0^y(1.5x-y)\;dx=\frac 1{1000}\left.\left(\frac {3x^2}4-y^2\right)\right|_0^y=-\frac {y^2}{4000}$$
So I should still not bid.  The basic reason is that there is too much room from $x=0$ to $x=\frac 23y$ where I take a loss.
