# Mixed Strategy Equilibrium of a Game

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?

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.