Compute mixed Nash equilibria when objects are to be won by bidding

Question

Consider a complete information game being played between two players given five objects. Players have certain preferences over the objects and each simultaneously places a sealed bid-vector to win them. For each object, the highest bidder wins it (ties are broken by giving each player exactly half of the tied object except when both bid $0$, in which case the object is not distributed).

• Preferences: Suppose player $i$ derives a payoff of $u_{ij} \in \mathbb{R}_{\geq 0}$ when he wins object $j$. If he wins half of $j$, then his payoff is $\frac{1}{2} u_{ij}$. The values $u_{ij}$ are given beforehand.

• Payoffs: Define $1_{ij}$ as the indicator function that is assigned the value $1$ if player $i$ entirely wins object $j$, value $0.5$ if player $i$ wins half of $j$ and $0$ otherwise.

  Player $i$’s total payoff after the game concludes is given by $\displaystyle \sum_{j \in M} \left(1_{ij} \cdot u_{ij}\right)$ where $M$ is the set of five objects. That is, the sum of payoffs over all objects won.

• Strategies: Each player begins with $\$ 1$. Player $i$’s strategy bid-vector is given by $B_i = [b_{i1}, b_{i2}, b_{i3}, b_{i4}, b_{i5}]$ such that $\sum_{j=1}^{5} b_{ij} \leq 1$ where $b_{ij}$ denotes the (non-negative) amount he bid for object $j$.

If the preferences $u_{ij}$’s are given beforehand, can we compute the set of all mixed strategy Nash equilibria?

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Just to clarify, a Nash Equilibrium is when no player can place a (mixed-strategy) bid-vector that strictly increases his overall payoff.

In the earlier version of this post (pre-edit), I had included some of my thoughts and approach. However, that turned out to be completely incorrect and hence is removed from the post.

The goal obviously is to work out for the case where the number of players and objects are arbitrary. If anyone is interested in that direction, you are welcome.

Answer 1

First note that your payoff is not continuous, because of the zero-payoff when bid is zero. There might happen that no Nash equilibrium exists.

Let's start with a simpler situation, with only 2 objects. We can assume also WLOG that the sum of bids is $1$ for all players.

• If a player has equal payoffs, they can play any bid-vector provided that all bids are nonzero (for nonzero payoffs), e.g. $0.5$ on each. The other player choose its preferred object (or plays $0.5$ with equal payoffs).
• If a player has a preferred object with a payoff greater than the double of the other object, they bid $1$ on it. If both players are on this case with the same preferred object, nobody will gain the other object, else the other player bids on the remaining object.
• If players have different preferred objects, then each one bids only on it.
• In the last case both players have the same preferred object with a payoff less than the double of the other object. This case needs more details below.

If we allow only bids of the form $\frac kn$ with $n$ fixed, then the strategies with $1$ or $\frac{n-1}{n}$ on the preferred object dominate the others.

The expected value of the payoff for player $1$ becomes $$(1-a_{1,n})a_{2,n}u_{1,2}+(1-a_{1,n})(1-a_{2,n})\frac{u_{1,1}+u_{1,2}}{2}+a_{1,n}a_{2,n}\frac{u_{1,1}}{2}+a_{1,n}(1-a_{2,n})u_{1,1}$$

The first order conditions give: $$a_{2,n}u_{1,2}+(1-a_{2,n})\frac{u_{1,1}+u_{1,2}}{2}=a_{2,n}\frac{u_{1,1}}{2}+(1-a_{2,n})u_{1,1}\\ \iff a_{2,n}=\frac{u_{1,1}-u_{1,2}}{u_{1,2}}$$

Symmetrically, we get $$a_{1,n}=\frac{u_{2,1}-u_{2,2}}{u_{2,2}}$$

But you cannot let $n\to+\infty$, because that would lead each player to bid arbitrarily near $1$ on their preferred choice.

Nevertheless you keep two equilibria when players bid $1$ on different objects (while they prefer the same), but it won't be chosen, since both will try to play the greedy one.