The table above represents payoff for three players game accordingly. The probability for z1 is z and z2 is (1-z), r1 is r and r2 is (1-r) meanwhile for p1 is p and p2 is (1-p).
Firstly, I would like to know whether the solution of mixed strategy for three players game is possible for my payoff matrix?
Secondly, I have extracted the mixed Nash equilibrium strategy as follows by using payoff value from player Z:
if player $P$ plays,
$p_{1} = $50(z)(r)+0(z)(1-r)+100(1-z)(r)+50(1-r)(1-z)$
$p_{2} = $0(z)(r)+0(z)(1-r)+50(1-z)(r)+50(1-r)(1-z)$
Then equate $p1=p2$
$50(z)(r)+50(1-z)(r)=0$
if player $R$ plays,(note: the payoff value of Z is read vertically)
$r_{1} = $50(z)(p)+100(1-z)(p)+0(z)(1-p)+50(1-z)(1-p)$
$r_{2} = $0(z)(p)+50(1-z)(p)+0(z)(1-p)+50(1-z)(1-p)$
Then equate $r1=r2$
$50(z)(p)+50(1-z)(p)=0$
which after I have solved the equation $p = 0$
if player $Z$ plays,(note: the payoff value of Z is read horizontally)
$z_{1} = $50(r)(p)+0(1-r)(p)+0(r)(1-p)+0(1-r)(1-p)$
$z_{2} = $100(r)(p)+50(1-r)(p)+50(r)(1-p)+50(1-r)(1-p)$
Then equate $z1=z2$
$-50(r)(p)-50(1-r)(p)-50(r)(1-p)-50(1-r)(1-p)= 0$
which after I have solved the equation $-50=0$
Hence, I realized that it is impossible to solve for probability of p,z and r. I want to know whether the equations are correctly performed or not for all players?
Thirdly, is there any other method that will be able to solve for the probability of p,z, and r.