Bayes-Nash equilibrium in Auction Consider a seller offering a single object for sale to two buyers with
  independent valuations – each bidder’s value is uniformly distributed
  on [0,1]. Assume that buyer 1 submits a bid that buyer 2 observes.
  Buyer 2 then has the ‘first refusal’ right to buy the object at that
  price. If she refuses, buyer 1 gets the object at the price he bid. What is the optimal strategy for buyer 1?


I have attempted the following:

$$\pi_1(v_1,v_2)=p(v_2&lt;\beta(v_1))(v_1-\beta(v_1))$$
$$=(1-\beta(v_1))(v_1-\beta(v_1))$$

If this is an equilibrium, we must have that

$$=(1-\beta(v_1))(v_1-\beta(v_1)) \geq (1-\beta(w))(v_1-\beta(w))$$

for all $w\in[0,1]$. Hence,

$$\left.\frac{\partial \pi_1(v_1,v_2,w)}{\partial w}\right|_{w=v_1}=-v_1\beta'(v_1)-\beta'(v_1)+2\beta(v_1)\beta'(v_1)=0$$

$$\iff\beta(v_1)=\frac{1+v_1}{2}$$

But this strikes me as clearly wrong, if $v_1=0$ why should the player overbid? What am I missing?
I think your one mistake was writing $\Pr(v_{2} &lt; \beta(v_{1})$ incorrectly. It should be:

$$\begin{split}
\pi_1(v_1,v_2)&amp;=\Pr(v_2&lt;\beta(v_1))(v_1-\beta(v_1))\\
&amp;=F_{2}(v_2&lt;\beta(v_1))(v_1-\beta(v_1))\\
&amp;= \beta(v_1)(v_1-\beta(v_1))\\
\end{split}$$

$$\begin{split}
\left.\frac{\partial \pi_1(w,v_{2})}{\partial w}\right|_{w=v_1} =v_1\beta'(v_1) - 2\beta(v_1)\beta'(v_1)&amp;=0\\
\beta(v_1) &amp;= \frac{v_{1}}{2}\end{split}$$

Loosely we can look at, say, type $1$ of player $1$. If he bids according to this strategy his expected payoff is $1/4$. Suppose he picks any other bid amount $a$. Then his expected payoff is $(1-a)F(a) = (1-a)a$. This is clearly maximized at $a = 1/2$.