Finding a stable equilibrium I'm trying to determine which strategy is stochastically stable.

In my lecture notes, we assign probabilities $a$, $b$ $1-a-b$ to player 2 strategies $T,M,B$ resp.
Hence we have the payoffs
$$u(t)=2a+1-a-b$$ $$u(m)=b+2-2a$$ $$u(b)=4-2b-3a$$
And it is inferred from this that $b,b$ is stochastically stable. I tried using the utilities to derive a system of three equations, and showing that $B$ yields the highest payoff for the largest range of $a$ and $b$, but this did not seem to succeed.
As a caveat, disclaimer, I am not too familiar with Evolutionary Game Theory and it is quite outside my research area/main areas of expertise. However, I can show here that range of $(a,b)$ for which $B$ is a best response is strictly greater than for either of the other two strategies. To that end,

You can characterize the range of $(a,b)$ for which $B$ is a (strict) best response by solving the system

$$\begin{split}
4−2b−3a &amp;\&gt; b+2-2a\\
4−2b−3a &amp;&gt; 1+a-b\\
\end{split}$$

Or,

$$\begin{split}
2 &amp;&gt; 3b + a\\
3 &amp;&gt; 4a + b\\
\end{split}$$

Or,

$$\begin{split}
a &amp;&gt; 7/11, b&lt;3-4a\\
a &amp;\leq 7/11, b&lt;\frac{2-a}{3}\\
\end{split}$$



We can solve for the area of this quadrilateral:

$$\begin{split}
B &amp;= \int_{0}^{7/11}\frac{2-a}{3}da + \int_{7/11}^{3/4}(3-4a)da\\
&amp;= \frac{259}{726} + \frac{25}{968} = \frac{101}{264} \approx .382\bar{57}
\end{split}$$

Through similar analysis for $M$, we obtain

$$\begin{split}
a &amp;&gt; 7/11, b&gt;\frac{3a-1}{2}\\
a &amp;\leq 7/11, b&gt;\frac{2-a}{3}\\
\end{split}$$



Hence,

$$\begin{split}
M &amp;= 1 - \int_{0}^{7/11}\frac{2-a}{3}da - \int_{7/11}^{1}\frac{3a-1}{2}da\\
&amp;= 1 - \frac{259}{726} - \frac{32}{121} = \frac{25}{66} \approx .3\bar{78}
\end{split}$$

Finally,

$$T=1-\frac{25}{66}-\frac{101}{264} = \frac{21}{88} \approx .238\bar{63}$$

Since $$\begin{split}
\frac{101}{264} &amp;&gt; \frac{25}{66}\\ \frac{101}{264} &amp;&gt; \frac{21}{88}
\end{split}$$
the result is shown.