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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)?

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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).

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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.

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