Improve Markov Chain results
Let $\mathcal{S}$ be the state space of your Markov chain, and assume that we start at some state $i \in \mathcal{S}$. Let $\mathcal{Y}_i$ be the set of states (pages) that are ranked lower than $i$ by two of the three engines. According to your description, the probability of transition to state $j$ can be defined as follows:

\begin{equation}
P_{ij} =
 \begin{cases}
  1 - \frac{|\mathcal{Y}_i|}{|\mathcal{S}|} &amp; \text{if } j = i, \\
  \frac{1}{|\mathcal{S}|} &amp; \text{if } j \in \mathcal{Y}_i, \\
  0 &amp; \text{Otherwise}
 \end{cases}
\end{equation}

If I have understood your question correctly, this Markov chain does not always produce credible results due to the following issues. First, your chain allows repetitive comparisons. In other words, it is likely that a state $i$ is compared against some state $j$ multiple times. This, of course, does not make trouble as long as your chain is irreducible. However, in some cases, you will not get the correct result. For example, consider the following scenario.

Imagine for some state $i \in \mathcal{S}$, no state (page) can be found that is ranked lower than $i$ by two other engines, that is $\mathcal{Y}_i = \varnothing$. Therefore, $P_{ii} = 1$ making $i$ an absorbing state. In this case, your chain will be trapped in state $i$ in finite time, and from then on, it will only compare state $i$ against other states repeatedly. If chain is trapped in such a state and you let it run for a long enough time, then your results would clearly be wrong. Similar problem occurs if your chain contains recurrent sub-classes.

Assuming that no such states exist, then you would have an irreducible chain. In that case, if you let the chain run for sufficiently long time, the counting process would result in a good approximation of correct rankings (according to your algorithm).

P.S. This answer is just based on my understanding of your problem and may not be 100% correct.