Principle of Indifference Contradicting with Nash Equilibrium?

Question

I have come across a game related to a bimatrix, after IESDS it is like this

$$\begin{bmatrix} (10,4)&(5,3)\\ (11,1)&(4,6) \end{bmatrix}$$ First off, by using the principle of indifference, we can establish \begin{align*} 10q+5(1-q)&=11q+4(1-q)\\ 4p+1-p&=3p+6(1-p) \end{align*} which gives \begin{align*} p&=5/6\\ q&=1/2 \end{align*} But, if we approach by the definition of mixed-strategy Nash Equilibrium, i.e., a $\sigma$ such that $\forall i $ $$u_i(\sigma_1,\sigma_2) \geq u_i(s_i,\sigma_2), \forall s_1 \in S_1$$

To assess this, we calculate Player 1 first, by definition of $u_i(\sigma)$ \begin{align*} u_1(\sigma_1,\sigma_2) & = \sum_{s\in S} (\sigma_1(s_1)\sigma_2(s_2)) u_1(s)\\ &=(pq)10 + p(1-q)5+(1-p)q11+(1-p)(1-q)4\\ &= -2pq+p+7q+4 \end{align*} and it has to be the case that \begin{align*} -2pq+p+7q+4 &\geq u_1(U,\sigma_2) \tag*{U as the upper row}\\ &\geq u_1(L,\sigma_2) \end{align*} We can calculate, again by definition \begin{align*} u_1(U,\sigma_2) &= \sum_{s_2\in S_{2}} \sigma_2(s_2)u_1(U,s_2)\\ &=10q+5(1-q)\\ u_1(L,\sigma_2)&=11q+4(1-q) \end{align*} Hence, we have that \begin{align*} -2pq+p+7q+4&\geq5q+5\\ -2pq+p+7q+4&\geq 7q+4 \end{align*} Here we can have that $$q\leq 1/2$$ We can calculate for player 2 using the same method, I got that $$p\geq 5/6$$

My questions are:

1. How does the definition not give an exact value unlike the principle of indifference?
2. Or, those steps above are not wrong at all, as $p=5/6$ and $q=1/2$ are just a Nash Equilibrium. i.e., mixed-strategy equilibria are not unique.
3. Generally, does the principle of indifference guarantee a mixed-strategy Nash Equilibrium? If yes, is the proof simply just by equating the first two inequalities after the definition?

Answer 1

Let me answer your questions by solving this particular NE for example.

First of all, that's correct how you establish the two equations under the principle of indifference. Nevertheless, it will be more beneficial to establish inequalities rather than equalities.

For the column player, it is better off to play purely left, if the row player is not mixing her own strategies perfectly, in the sense that $$4p + (1-p) \geq 3p+ 6(1-p) \Rightarrow p\geq \frac{5}{6}$$

Similarly, the row player will deviate to play a pure strategy up if the column player is not mixing her own strategies perfectly, in the sense that $$10q+5(1-q) \geq 11q +4(1-q) \Rightarrow q\leq \frac{1}{2}$$

Then, let's see what happens after we obtain these two quantities. By definition, the two above case discussions reveal how each player will best respond to the other one. We can draw the following best response curve one against the other. In fact, the intersection point will be the mixed NE and you can clearly spot there is no pure NE.

                               

And another remark here would be, if the vertical part of the green moves to the leftmost/rightmost or the horizontal part of the orange moves to the top/bottom, there will be infinitely many intersections between these two best response curves and that means there are infinitely many mixed NE. Mathematically, a NE (no matter pure or mixed) is nothing but a fixed point.