Sealed bids, three bidders

Idea. First find the equilibrium bid by maximising one bidder's expected profit against the others. Then the revenue is that fraction of the expected highest value.

Step 1: best response. The other two bid $\beta V_2$ and $\beta V_3$. A bid $b\le \beta$ wins when both are below it, with probability ${\left(\frac{b}{\beta }\right)}^2$. With value $v$, the expected profit is $$(v-b){\left(\frac{b}{\beta }\right)}^2.$$ Setting the derivative in $b$ to zero, $2b(v-b)-b^2=0$, so $b=\frac{2}{3}v$.

Step 2: equilibrium. So bidding $\beta =\frac{2}{3}$ of one's value is the best reply when the others do the same.

Step 3: revenue. The winning bid is $\frac{2}{3}\max (V_1,V_2,V_3)$, and the maximum of three uniforms has mean $\frac{3}{4}$. So the expected revenue is $$\frac{2}{3}\cdot \frac{3}{4}=\frac{1}{2}.$$

Revenue equivalence. In a second-price auction, everyone bids their value and the winner pays the second highest value, whose mean is also $\frac{1}{2}$. That is no accident: under these conditions every standard auction gives the seller the same expected revenue. With $n$ bidders the equilibrium bid is $\frac{n-1}{n}v$ and the revenue $\frac{n-1}{n+1}$.