1
$\begingroup$

I am having some problems with an exercise that showed up in my game theory course. Consider the two player game where each player bids a non-negative integer multiple of five cents. The highest bidder wins two dollars, and if the two bids are equal, neither player receives the two dollars, each player pays their own bid (even the loser).

We consider the players' payoffs to be their net winnings. I am interested in constructing a mixed strategy equilibrium where each bid less than 2.00 dollars has a positive probability. I think that the best way to go about this problem would be to construct a strategy for player 1 such that for any two arbitrary player 2 strategies $s_{21},s_{22} \in S_2,$ $$E[s_{21}] = E[s_{22}].$$ However, I am having a hard time deriving the conditions that would make this expected value work. Is there a better way to go about this problem?

$\endgroup$
1
$\begingroup$

You are on the right track.

Notice, in any mixed equilibrium where bidding zero receives positive weight, the expected value of playing any bid must be zero since the expected value of bidding zero is zero.

Let $b_1$ be the bid of player 1. Bidding $\$1.95$ for 2 is a win if $P(b_1<1.95)$ it is a tie if $P(b_1=1.95)$. The expected value is:

$$(0.05)P(b_1<1.95)-(1.95)P(b_1=1.95)$$

In equilibrium it must be:

$$(0.05)P(b_1<1.95)-(1.95)P(b_1=1.95)=0$$

Further:

$$P(b_1<1.95)=1-P(b_1=1.95)$$

Thus:

$$(0.05)(1-P(b_1=1.95))-(1.95)P(b_1=1.95)=0$$

$$P(b_1=1.95)=0.025$$

You can now solve $P(b_1=1.90)$ and then $P(b_1=1.85)$ and so on.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.