I took the following game from the Peter Winkler collection (chapter "Games"):
Two numbers are chosen independently at random from the uniform distribution on [0,1]. Player A then looks at the numbers. She must decide which one of them to show to player B, who upon seeing it, guesses whether it's the larger or smaller of the two. If he guesses right, B wins, otherwise A wins. Payoff to a player is his/her winning probability.
To be clear, let me define a strategy for B as a (measurable) function ${f}_{B}:[0,1]\longmapsto \{larger, smaller\}$, i.e., a strategy for B specifies "larger" or "smaller" for every real in [0,1].
Similarly, A's strategy is a function ${f}_{A}(\{x,y\})$ $=x$ or $y$, i.e., a strategy for A specifies x or y for every $\{x,y\}$ where $x,y\in [0,1]$.
A strategy of A (or B) is called a candidate (for equilibrium), if it prevents the opponent from achieving a winning probability strictly greater than 1/2 no matter what strategy he (or she) adopts. By this definition, the following strategy of A is a candidate (check it yourself):
" ${f}_{A}(\{x,y\})=x$ iff $|x-1/2|<|y-1/2|$"
My question: Are there any other candidates for player A, except those that differ only by a measure zero set from the above?
(P.S., One can show that the candidate for player B is unique, except by a difference of measure zero set. Hence if candidate for A is unique, too, there's a unique (pure) equilibrium.)
Edit: Here's how I prove that candidate for B is unique. I don't know if a simpler argument can be made, or similar reasoning can be used to prove or disprove the case of A.
By definition, B's strategy is to choose measurable $B_L\subseteq[0,1]$ such that he reports larger for $x\in B_L$ and smaller for $x\in [0,1]/B_L$. Now if $m(B_L)=a>1/2$, A can adopt the following strategy:
"Show the smaller number if both $x,y\in B_L$, otherwise show the larger number".
which guarantees her winning probability $\geq a^2+(1-a)^2>1/2$. Hence $m(B_L)>1/2$ can't be a candidate for B. Reversing the strategy shows that $m(B_L)<1/2$ can't be candidate, either. Hence $m(B_L)=1/2$.
Now let $m(B_L)=1/2$. Define $B_S=[0,1]/B_L$. Consider the following incomplete specification of a strategy for A:
"Show the smaller number if both $x,y\in B_L$; show the larger if both $x,y\in B_S$"
which guarantees her winning probability $=1$ in those situations. What about the remaining situations, i.e., $x\in B_L$ and $y\in B_S$?
For any measurable $B\subseteq B_L$ with $m(B)>0$, we can define $C=\{x\in B_S|x>y, \forall y\in B\} $. Suppose there exists such a $B$ such that $m(C)>0$, then A can adopt the following strategy:
"Show the smaller number if both $x,y\in B_L$, otherwise show the larger number"
which will guarantee her winning probability:
P(win)$=$ P(both $x,y\in B_L$ or both $x,y\in B_S$ and win)$+$ P($x\in B_L$ and $y\in B_S$ and win) $\ge$ P(both $x,y\in B_L$ or both $x,y\in B_S$ and win)$+$ P($x\in B$ and $y\in C$ and win) $=1/2+2m(B)m(C)>1/2$
Hence if a strategy of B is to be a candidate, we must have $m(C)=0$. Because $B\subseteq B_L$ was arbitrary, it is true that the set $\{x\in B_S|x>y, \forall y\in B_L\} $ has measure zero. Hence, to conclude, necessary conditions for a strategy of B to be candidate:
$m(B_L)=m(B_S)=1/2$.
$\{x\in B_S|x>y, \forall y\in B_L\} $ has measure zero.
There is only one strategy of B satisfying these conditions (up to measure zero difference), i.e., $B_L=[1/2,1]$. For sufficiency it is easy to check this is indeed a candidate. Hence it is the only candidate for B.