# 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 $$\begin{array}{|c|c|c|c|c|} \hline a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\ \hline a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\ \hline a_{31} & a_{32} & a_{33} & a_{34} & a_{35} \\ \hline a_{41} & a_{42} & a_{43} & a_{44} & a_{45} \\ \hline a_{51} & a_{52} & a_{53} & a_{54} & a_{55} \\ \hline \end{array}$$ 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 $a_{ji} = 1-a_{ij}$.
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 $d_j$.