Auction with "buy-it-now" option Consider the buy-it-now price often included in online auctions. Suppose that 2 bidders in an ascending oral (English) auction bid for an object. Each has values i.i.d uniformly on $[0,1]$. Suppose the auction has a buy-it-now price of $B \ge 1/2$; either bidder can end the auction by paying $B$. Assume that there's an equilibrium when the bidding reaches $p(x)$ that a bidder with value $x\geq B$ will pay $B$ (i.e. $p(x)$ is the cutoff when a bidder with value $x$ pays the buy it now price). Also assume also that $p'(x)&lt;0$. i) What price, conditional on winning, does a buyer with valuation $x$ expect to pay (as a function of $x$ and $p(x)$)? 


It seems reasonable to assume that bidders will not bid more than $p(x)$. Hence, $b_1,b_2\in[0,p(x)]$. Then, it makes sense to consider two cases: where $x&lt;p(x)$. Here the expected payment is simply the expected payment of the lower valuation bidder, namely, $\frac{p(x)}{2}$. 

Secondly, we should consider the case where $x&gt;p(x)$: the expected payment is here, the probability of facing a below-$p(x)$-valuation opponent, times their bid, plus the probability of facing an above-$p(x)$-valuation opponent times the probability of winning (ties are broken randomly) times $p(x)$

$$p(x)*\frac{p(x)}{2}+(1-p(x)*\frac{1}{2}*p(x))=\frac{1}{2}p(x)$$


  ii) Assuming risk neutrality, using the fact that the expected payment
  with valuation $x$ is the same without the buy-it-now option (namely,
  $1/2x$), what is $p(x)?$ (Note: you should get a quadratic equation; one root will be $p(x)=x$, but this is not the solution, given $p'(x)&lt;0$)


I'm not sure how to solve this part, and I'm doubting my answer to the first part given that I don't know how it enables me to derive $p(x)$.
Here is how I would approach the first question:


We are conditioning on the fact that you have won. Given the symmetry, this implies that your valuation $x$ is greater than your opponent's valuation $y$.
Given that you have won, you either pay the 'auction price' or $B$. If you don't pay $B$, you pay somewhere between $0$ and $p(x)$. The expected auction price is the expected value of the second highest draw, so actually $p(x)/3$ - not $p(x)/2$ as you seem to be saying (intuitively, the expected valuation of the lowest bidder must be below the unconditional expected valuation).
The probability that you pay the auction price is the probability that your opponent's valuation $y$ is below your cutoff $p(x)$. This is because your opponent will drop out of the bidding as soon as the price exceeds her valuation. $Pr(y &lt; p(x)) = p(x)$ since $y$ is uniform on $[0, 1]$.
Therefore, assuming you win, the probability that you pay $B$ is $1 - p(x)$.
Putting these claims together, I conclude that your expected payment conditional on winning equals: $(p(x)/3)p(x) + B(1 - p(x))$.
To check the plausibility of this, set $p(x) = 1$ so effectively the buy it now option disappears, and yields an expected payment of $1/3$. This is the standard result for $2$ buyers and a uniform distribution.