\documentclass[12pt,epsf]{article}

\parskip 10pt




\topmargin  -.8in

 \oddsidemargin -.5in


 \textheight  10in

\textwidth 7.6in  


\begin{document}
\baselineskip 16pt



\parindent 24pt

 \parskip 10pt

\titlepage


\begin{LARGE}
               
\begin{center}
    {  \bf  Picking a Sealed Bid,  Two Bidders, April 28, 2002 \\ }
\end{center}
 

 Suppose you think each person’s value is equally likely to be
anywhere between 0 and 10.  You know your own value is X.

Start with just 2 bidders, you and your rival.  It is a Nash
equilibrium for each of you to follow a strategy of bidding half your
value--.5X for you and .5R for your rival.   If you bid b (which we
will show should be .5X), your payoff is

$$
\begin{array}{l}
Probability (you\; win) (X-b) + Probability (you\; lose) (0)\\
 \\
= Probability (he \;bids \;less\; than\; b) (X-b)\\
\\
= Probability (.5R < b)  (X-b) \\
\\
= Probability (R<2b) (X-b)\\
\\
= \frac{2b}{10} (X-b)\\
\\
= \frac{bX}{5} - \frac{b^2}{5}
\end{array} 
$$


The payoff’s  derivative  is  
$$
\frac{ d \;payoff}{db}=  \frac{ X}{5} - \frac{2b  }{5}  
 $$
Setting this derivative equal to 0 yields $b=X/2$ as your payoff-
maximizing bid.

\titlepage

\begin{center}
    {  \bf  Picking a Sealed Bid,  $N$ Bidders \\ }
\end{center}
   Suppose you think each person’s value is equally likely to be
anywhere between 0 and 10.  You know your own value is $X$.  There are
$N$
bidders, including you.   It is a Nash equilibrium for each of you to
follow a strategy of submitting a bid of your value shaded by fraction
$1/N$, so if $N=6$ you would bid $(5/6) X$.
 
      Let us denote by  $V_i$  the value of bidder $i$, for  the other
$N-1$ bidders. Your payoff is
$$
\begin{array}{l}
Probability (you \;win) (X-b) + Probability (you \;lose) (0)\\
 \\
= Probability (everyone \;else\; bids \;less\; than\; b) (X-b)\\
\\
= Probability (  V_i <b, for \;all\; i) (X-b)\\
\\
= Probability ( V_i <b  )^(N-1) (X-b)\\
\\
= (b/10 )^{N-1} (X-b)\\
\\
= (1/10 )^{N-1} b^{N-1} (X-b)\\
\\
= (1/10 )^{N-1}  [ b^{N-1} X - b^N]\\
\end{array}
 $$
The derivative of the payoff with respect to $b$  is 
 $$
 \frac{D\; payoff}{Db}= (1/10)^{N-1}  [(N-1) b^{N-2} X - Nb^{N-1}].
$$
Setting the derivative equal to 0 and simplifying yields
$0 =  [(N-1) X - Nb]$,  so $b = (\frac{N-1}{N} ) X$. 


 

 



\end{LARGE}
 
 \end{document}