# Expected value of profits in auction question

You have a box filled with cash. Cash value is uniformly randomly distributed from 1 to 1000. You are trying to win the box in an auction: you win the box if you bid at least the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box, you can resell it for 150% of its value. How much should you bid to maximize the expected value of your profit (resale of box minus bid)?

I'll assume the amount of cash in the box is an integer (and the bids as well).

If you bid $n$, you win the box with probability ${\large{\frac{n}{1000}}}$, with expected revenue $${\small{\frac{3}{2}}}\left(\frac{1 + 2 + \cdots + n}{n}\right)$$ and guaranteed cost $n$ (the bid).

So your expected profit is $$f(n) = \frac{n}{1000} \left( {\small{\frac{3}{2}}}\left(\frac{1 + 2 + \cdots + n}{n}\right) -n\right)$$ You need to maximize $f(n)$ for $n \in \{0,1,2,3,...,1000\}$.

Can you finish it?

Followup question for the OP . . .

How much should you bid if you are competing against other bidders (assume a single simultaneous bid).

Let X be the amount of cash in the box. It is a random variable with pdf $f(x) = \frac{1}{999}$ for x between 1 and 1000 and 0 otherwise.

Fix your bid amount $B$.

The profit is a random variable, $P$, defined piecewise. $P = 1.5X - B$ when $B \ge X$ and $P(X) = - B$ when $B < X$.

Since $P$ is a function of $X$, you can calculate its expectation as

$$\int_1^{1000} p(x) f(x) dx$$

Where $p(x)$ is the profit function defined above. Once you solve for this with a fixed $B$, use calculus to determine which $B$ maximizes it.