Alice and Bob play the following game. They write the expressions $x + y$, $x - y$, $x^2+xy+y^2$ and $x^2-xy+y^2$ each on a separate card. The four cards are shuffled and placed face down on a table. One of the cards is turned over, revealing the expression written on it, after which Alice chooses any two of the four cards, and gives the other two to Bob. All cards are then revealed. Now Alice picks one of the variables $x$ and $y$, assigns a real value to it, and tells Bob what value she assigned and to which variable. Then Bob assigns a real value to the other variable. Finally, they both evaluate the product of the expressions on their two cards. Whoever gets the larger result, wins. Which player, if any, has a winning strategy?
Problem
Source: 2019 Baltic Way P6
Tags: Game Theory, combinatorics
grupyorum
19.11.2019 18:28
Alice. I'll use A to denote Alice and B to denote Bob, and will also use $A,B$ to denote the product/pairs they have. Here is the strategy: If the revealed card is $x-y$ or $x+y$, then A hands it to B, otherwise gets to keep it. This ensures that $A$ does not end up with $(x-y,x+y)$. In particular, at the end are there the following possibilities: $i)$ $A=(x-y,x^2+xy+y^2)$. In this case $A$ has product $x^3-y^3$, whereas $B$ has product $x^3+y^3$. $A$ then selects $y$ to be a negative number. No matter what $B$ selects afterwards, $A$ always wins. $ii)$ $A=(x-y,x^2-xy+y^2)$. The product $A$ computes is $x^3-2x^2y+2xy^2-y^3$, and the product that $B$ computes is $x^3+2x^2y+2xy^2+y^3$. Then, $A-B=-4x^2y-2y^3=-2y(y^2+2x^2)$. $A$ then chooses a negative $y$, and no matter what $B$ chooses, $A-B>0$. $iii)$ $A=(x+y,x^2-xy+y^2)$. Then, $A-B=2y^3$. Thus, $A$ chooses a positive $y$ and ends the game. $iv)$ $A=(x+y,x^2+xy+y^2)$. Then, $A-B=2y(y^2+2x^2)$, which $A$ wins by choosing $y>0$. $v)$ Finally, let $A=(x^2-xy+y^2,x^2+xy+y^2)$. Then, $A=x^4+x^2y^2+y^4$ and $B=x^2-y^2$. But now, $A$ simply chooses $x>1$ and thus $x^4>x^2$, concluding $A>B$. We have exhausted all cases, and thus the proof is complete.