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
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 :
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.