<p>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 $a_{ji} = 1-a_{ij}$.</p>

<p>Which desk yields the highest expected payoff?
<br> 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.</p>

<p>How does one derive the probabilites $p_j$?
<br> 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 $d_j$.</p>

CC BY-SA 3.0
2017-01-21T13:10:36.180Z
2017-01-21T13:10:36.180Z
Giskard
<p>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).</p>

<p>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.</p>

<p>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.</p>

<p>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. </p>

<h3>Edit</h3>

<p>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.</p>

<game-theory>
MSNE and the probability of encountering each payoff
CPerkins
# MSNE and the probability of encountering each payoff

(Asked by: username_0 on 2017-01-20)

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.

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2017-01-21. Score: 3

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 $a\_{ji} = 1-a\_{ij}$.

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 $d\_j$.


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