MSNE and the probability of encountering each payoff I'm currently setting up a model using data from the game "Hearthstone". Those familiar with the game know that it can (for the most part) be described as a zero-sum game, where players choose a deck before being randomly matched against another player. These decks have over-all winrates against all players, as well as winrates for each specific matchup (Deck A vs Deck B).

Let's assume there are only 5 decks that can be played (there are a lot more). We could then form a 5x5 expected payoff matrix for each matchup using the respective winrates of all decks.

My question is this: how would the probability of encountering a certain deck factor in? We could technically calculate an optimal strategy from the payoff matrix, but it doesn't seem like that takes into account the likelihood of encountering a certain deck more than another.

At first I thought it could be a type of signaling game, where the matchmaking process was a "type" drawn randomly, but that isn't the case here, since the players are making their decisions before this occurs. 

Edit

I should have specified. I mention MSNE since I'm trying to see if you can better your chances of winning statistically by playing a number decks in a number of sequential games, rather than just choosing one deck and sticking with it.
Denoting your five decks by $d_1, d_2,...,d_5$, you will get a $5 \times 5$ matrix
$$
\begin{array}{|c|c|c|c|c|}
\hline
a_{11} &amp; a_{12} &amp; a_{13} &amp; a_{14} &amp; a_{15} \\
\hline
a_{21} &amp; a_{22} &amp; a_{23} &amp; a_{24} &amp; a_{25} \\
\hline
a_{31} &amp; a_{32} &amp; a_{33} &amp; a_{34} &amp; a_{35} \\
\hline
a_{41} &amp; a_{42} &amp; a_{43} &amp; a_{44} &amp; a_{45} \\
\hline
a_{51} &amp; a_{52} &amp; a_{53} &amp; a_{54} &amp; a_{55} \\
\hline
\end{array}
$$
where $a_{ij}$ denotes the the expected payoff of deck $i$ against deck $j$. I am guessing this is actually the probability of deck $i$ beating deck $j$ but I am not familiar with Heartstone or your exact intentions. If it is in fact a probability then the payoff of deck $j$ against deck $i$ is $a_{ji} = 1-a_{ij}$.

Which desk yields the highest expected payoff?
 There is no answer this question without some assumptions about your opponent. E.g. in rock-paper-scissors there is no best move unless you specify what your opponent will do. If your opponent plays $d_j$ with probability $p_j$ then your expected payoff for playing $d_i$ is
$$
\sum_{j=1}^5 p_j \cdot a_{ij}.
$$
Whichever $i$ (or $i$'s) maximize this is (are) the best deck(s) to play.

How does one derive the probabilites $p_j$?
 There is no clear answer to this. You can use the concept of Nash-equilibrium to calculate mutual best responses. Or you could rely on experience by calculating the actual frequency of opponents playing $d_j$.