TITLE: What is the (mixed strategies) equilibrium of this game?
QUESTION [5 upvotes]: Given a weight vector $w\in [0,1]^d$ such that $\sum w_i=1$, the game goes as follows:
Two players, $X,Y$ choose strategies $x,y\in [0,1]^d$ such that $\sum x_i = \sum y_i = 1$.
The utility (profit) for player $X$ is given by $$u_X=\sum_{i:x_i>y_i} x_i\cdot w_i$$
That is summing over all coordinates in which $x$ is larger than $y$, and the profit for the coordinate is given by $x_i\cdot w_i$.
Y's utility is symmetric ($u_Y=\sum_{i:y_i>x_i} y_i\cdot w_i$).
This also interests me for low dimension (say $d=3$) if it makes it easier.

We assume w.l.o.g that $i>j\implies w_i\geq w_j$.
If $w_1\geq \frac{1}{2}$ then this is trivial, so assume this is not the case.
If needed we may assume that if $x_i=y_i$ then they both get $\frac{x_i\cdot w_i}{2}$ utility for that coordinate.


Is there a known name for this game?
Is the equilibrium of this game computable or is it $PPAD$-hard?

REPLY [6 votes]: If you change the payoff to 
$$\sum_{i:x_i>y_i} w_i $$
then this type of game is called a Colonel Blotto game or a Blotto game. It was studied by Borel. Your assumption that $\sum x_i = \sum y_i$ is not always a requirement.
In a 1950 paper, Gross and Wagner analyzed a few variants and contributed the name. They covered the cases of 

$d=2$ in your notation, $n=2$ in theirs.
$d=3$, $\sum x_i = \sum y_i.$
$\sum x_i = \sum y_i$ and $w_j=1/d.$

Blotto games have been identified and analyzed in a few other papers.