Mixed strategy Nash equlibria (defending territory) 
Hi I am trying to figure out the MSNE of this game. I am confused wether there are 2 MSNE or 3. Player 2 would put (1/2) on xx and yy and then player 1 would either put (1/4)on all or (1/2) on xxx and yyy or (1/2) on xxy and yxx. I am not sure if (1/2) on xxx and yyy is possible as this only hold becaise you can eliminate xy as player 2 would never play this in MSNE as it is strictly dominated.
If anyone could help it would be greatly appreciated. Thanks in advance 


 A: One fact about Nash equilibria is that no game can have an even number of mixed Nash equilibria, so it is impossible for the game you described to have 2 NE.
You are correct that Player B will choose to play strategy XX with probability 1/2 and strategy YY with probability 1/2 because of the argument you gave in your writeup.
Now for player A playing (1/2) on XXX and (1/2) on YYY wouldn't be an equilibrium strategy since Player B would want to switch to XY.
In order for player B to not want to switch to XY, it has to be that player A assigns at least as much probability on strategies XXY and YXX as on XXX and YYY. (condition 1)
Moreover, player A must place the same probability on strategies XXX, XXY as on strategies YYY, YXX since otherwise player B would want to switch strategy. (condition 2)
So the two conditions are :


*

*p(XXX) + p(YYY) $\le$ p(XXY) + p(YXX) and 

*p(XXX) + p(XXY) = p(YYY) + p(YXX)


Any strategy for player A that satisfies these properties gives a NE, so there is a continuum of different equilibrium strategies. 
For example, p(XXX) = 0.1, p(YYY) = 0.3, p(XXY) = 0.4 and p(YXX) = 0.2 is a valid mixed Nash equilibrium, as no player wants to switch strategies.
