Best bid given the other's bid
by fancyfairy, Mar 30, 2025, 10:04 PM
Consider sets
(player) and
(items), with valuations given by matrix
![$
V = \left[\begin{matrix}
v_{1a} & v_{1b} & v_{1c} & v_{1d} \\
v_{2a} & v_{2b} & v_{2c} & v_{2d}
\end{matrix}\right]
$](//latex.artofproblemsolving.com/d/4/6/d465348852f12739e1fc2a04a9c2e5fdad0dbcb4.png)
where
.
A player’s total valuation is the sum of item values won, with half-value for half-won items.
Given player
's bid, player
's bid
is strictly better than another possible bid
of his if the total valuation derived from
is strictly more than that derived from
.
Each player simultaneously bids on items with a bid vector
such that
and
. The highest bidder wins an item; ties split ownership equally.
Given valuation matrix
, can we always find bid vectors
and
such that given
, the second player can not submit a bid strictly better than
and vice-versa?


![$
V = \left[\begin{matrix}
v_{1a} & v_{1b} & v_{1c} & v_{1d} \\
v_{2a} & v_{2b} & v_{2c} & v_{2d}
\end{matrix}\right]
$](http://latex.artofproblemsolving.com/d/4/6/d465348852f12739e1fc2a04a9c2e5fdad0dbcb4.png)
where
![$v_{ij} \in [0, \infty]$](http://latex.artofproblemsolving.com/7/8/8/788bffd362b46a318a2cb0703a9e2ae77089e4e7.png)
A player’s total valuation is the sum of item values won, with half-value for half-won items.
Given player






Each player simultaneously bids on items with a bid vector
![$B_i = [b^i_1, b^i_2, b^i_3, b^i_4]$](http://latex.artofproblemsolving.com/7/b/2/7b233ca3e9056aa374e937b1133cd9c81726f407.png)
![$b^i_j \in [0,1]$](http://latex.artofproblemsolving.com/b/9/6/b96c2a7618767934c8a44d9238c485fed39b8510.png)

Given valuation matrix




