Heres the payoff for player one.
I'm searching for mixed strategies of player two. However I do the algebra, i get:p=4/5, q=3/10 and z=1-p-q=-0,1.
Could anybody please explain, how negative probabilities are possible in mixed strategies, or, and I hope, where did I the error?
-p+5q+2=-2p+q+4=3p+q
results in:
-2p+q+4=3p+q p=4/5
-(4/5)+5q+2=3(4/5)+q q=(16/5-2)/4 q=0,3
Thanks in advance
No, probabilities can't be negative. If an attempt to calculate a mixed strategy produces negative numbers, it means there is no mixed strategy that does what you're attempting to get it to do. In this case it looks like you're trying to make the first player's average payoffs for each of his three pure strategies be equal.
That can't be done.
Instead, with the mixed strategy $(4/5, 0, 1/5)$ the second player can ensure the first player's average payoff is at most $12/5$ (namely the average payoff would be $6/5$ with strategy A and $12/5$ with B or C). In a zero-sum game, this would in fact be an optimal strategy for the second player. I obtained this by solving the linear programming problem
$$ \eqalign{\text{minimize} & \ z\cr \text{subject to} & \ z \ge p + 7 q + 2 r\cr & z \ge 2 p + 5 q + 4 r \cr & z \ge 3p + q\cr & p + q + r = 1\cr &p,q,r \ge 0 }$$