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There are 18 different types in pokemon with different strengths and weaknesses like rock-paper-scissors. For example, Fire beats Grass, Grass beats Water, and Water beats Grass. My goal is to determine the best type of pokemon (ignoring stats, special abilities, move sets, turn order, etc.) In this simplified game of pokemon, you pick one pokemon type and the pokemon will use that pokemon type attack as well (if you pick fire, then you get a generic fire pokemon that only uses a generic fire move and likewise for your opponent for their chosen type).

I have 2 approaches to the problem.

  1. Make a matrix such that each column is a type and each entry in that column is probability that it loses to a different type (as indicated by row, and each probability is either 0 or uniform). Find the eigenvector for this matix. Whichever type has the higher probability is the best type.

  2. Solve the nash equilibrium for this game. Find where the opponent is indifferent to the probability distribution to how you pick your type. This would be a simultaneous game.

I tried both approaches but got different results. Why?

Michael
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    It's hard to say why something is not the case when you haven't explained why you think it should be the case. – joriki Feb 05 '20 at 19:22
  • I tried making pokemon into a marcov chain. As I understand it, a marcov chain locates a steady state vector.

    A Markov chain is a mathematical system that experiences transitions from one state to another according to certain probabilistic rules. The defining characteristic of a Markov chain is that no matter how the process arrived at its present state, the possible future states are fixed.

    This differs from a Nash equilibrium which should also provide a steady state vector as well but has different values of how likely I should pick a certain type.

     – Michael Feb 05 '20 at 19:23
  • I don't understand. Your comment explains a difference between the two approaches. But you're asking us why they produce different results. – joriki Feb 05 '20 at 19:42
  • In rock paper scissors these approaches lead to the same result. They lead to the result that you should mix your strategy with 1/3 chance for rock, paper and scissors. The matrix created is a compilation of many implications. If your opponent throws paper, you should choose scissors 100% of the time. Extending this idea to pokemon, create a matrix that follows the same sort of logic. If opponent chooses fire, then choose water (or something else that beats fire). By finding a vector that doesn't change based off this matrix, shouldn't this the same as the Nash Equilibrium? – Michael Feb 05 '20 at 20:06
  • It seems a bit adventurous to infer this from a simple example with three choices with high symmetry. I don't see why it should be the same in general. Have you tried it for a small non-symmetric example e.g. with $3$ or $4$ options? – joriki Feb 05 '20 at 20:10
  • I've done a 6x6 matrix which worked. No symmetry in this either. But it has a pure Nash Strategy. – Michael Feb 05 '20 at 20:13
  • Aha! Now there's something to build on. Could you share that with us? – joriki Feb 05 '20 at 20:14
  • 1 2 2 1 2 2 0 1 0 2 2 2 0 2 1 2 1 2 1 0 0 1 2 0 0 0 1 0 1 0 0 0 0 2 2 1

    Sorry about the formatting. It's a 6x6 matrix.

     – Michael Feb 05 '20 at 20:17
  • The 2's represent a win, the 1's represent a tie, and the 0's a loss. – Michael Feb 05 '20 at 20:19
  • Let us continue this discussion in chat. – Michael Feb 05 '20 at 20:26

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