I am trying to solve this and used the following:
\begin{bmatrix}(0,1)&(0,0)&(10,4/5)\\(3,1)&(1,2)&(0,1)\\(1,5/2)&(2,3/2)&(0,2)\end{bmatrix}, where the columns represent Miles and the rows represent Stevie. Note that, the first column is L, second column is C and the third column is R. The first row isT, the second row is H and the third row is B.
I hope that this is understandable. I do not know how to insert matrix like this in math stack exchange.
Moving on to my analysis:
Consider R and consider s_2 = (p, 1-p, 0)
Then we have:
1p + 0(1-p) > 4/5
1*p + 2(1-p) > 1
5/2 *p + 3/2 *(1-p) > 2
Solving this, we obtain that 1/2 < p < 1.
Therefore, we can eliminate R and thereafter we can eliminate T.
By doing this, I am left with a 2 by 2 matrix and in my opinion, these are the strategies that survive the iterative deletion of strictly dominated strategies.
My question is, am I correct?
For the second part of the question, I defined a mixed strategies for Stevies (a, 1-a) and for Miles (b, 1-b).
Solving the following equations:
3a + 1(1-a) = 1a + 2*(1-a) --> a = 1/3
and
1b + 2(1-b) = 5/2 *b + 3/2 *(1-b) --> b = 1/2
Am I doing the right thing now? I can observe that there are no pure strategies here. Therefore, I went on determining the mixed nash equilibrieum. Now that I have the values for a and b. What is the next thing left to do to show that there exist a mixed NE or not?