Is there a measure of how well a Markov chain allows movement between states? Define
$$ A = \left( \begin{matrix} .5 & .5 \\ .5 & .5 \end{matrix} \right),\; \;  B = \left( \begin{matrix} .99 & .01 \\ .01 & .99 \end{matrix} \right), \; \; C = \left( \begin{matrix} .01 & .99 \\ .99 & .01 \end{matrix} \right)$$
Taken as Markov chains, clearly $A$ permits movement between its states better than $B$, and $C$ permits movement between its states better than $A$. 
Is there some sort of measure on the space of Markov chains that measures how well they permit movement between their states? Some kind of congestion or accessibility statistic? 
Edit: I know about mixing rates, but I'm wondering about a statistic that works for absorbing Markov chains too.
 A: Maybe the conductance of the Markov chain is the right notion to look at. Let $P\in[0,1]^{n\times n}$ be a transition matrix with stationary distribution $\pi$ (in your cases, $\pi$ is always the uniform distribution). The conductance of $P$ is 
$$\Phi(P):=\min_{S\subset [n], \pi(S)\le\frac{1}{2}}\frac{\sum_{i\in S,j\in S^c} \pi(i)\cdot P_{i,j}}{\pi(S)}.$$
See for example "Conductance and Rapidly Mixing Markov Chains" by James King.
Your examples yield $\Phi(A)=0.5$, $\Phi(B)=0.1$, and $\Phi(C)=0.99$.
A: If the transition graph is strongly connected (i.e. given an initial state, any other state is reachable with p>0, possibly via intermediate states) , then as time goes to infinity the probability to find the system in a given state does not depend on the initial state. That is to say, there is a chance $$p_i(X)$$ to find the system in state X after i steps, which converges to some constant $$p(X)$$ which is a function only of the transition matrix (The stationary distribution $\pi$ in Tobias's answer). 
For all 3 your examples, this $\pi$ is simply (.5, .5) as both states are equally likely. This makes sense: $$\left( \begin{matrix} .5 \\ .5 \end{matrix} \right) \left( \begin{matrix} .5 & .5 \\ .5 & .5 \end{matrix} \right) = \left( \begin{matrix} .5 \\ .5 \end{matrix} \right)$$ but also $$\left( \begin{matrix} .5 \\ .5 \end{matrix} \right) \left( \begin{matrix} .9 & .1 \\ .1 & .9 \end{matrix} \right) = \left( \begin{matrix} .5 \\ .5 \end{matrix} \right)$$ but in general this need not hold. Not all states have to be equally likely. Simple example: $$\left( \begin{matrix} .5 & .5 & 0 \\ .25 & .5 & .25 \\ 0 & .5 & .5 \end{matrix} \right) $$ with probabilities (.25, .5, .25). You can think of this as a Left<->Middle<->Right triplet of states, with a 50% chance of moving but not directly from left to right. Since you always have to go through the middle, it's most likely.
Now, as the comments on the question already indicated, you can use this probability to weigh the chances of staying in each different state. 
In your simple examples, the respective results would be 0.5, 0.99 and 0.1, simply because the chances to stay in the same state (values of the diagonal) are the same on the diagonal. For non-trivial matrices, it would be a weighted average of the diagonal. 
This means that the exact off-diagonal values do not matter. I believe this reflects the intent of the question, which does not differentiate between different kind of state transitions either.
