Hello everyone, this post is a little bit off-topic, but I would like to attempt to apply a bit of game theory to the current state of the metagame, particularly when it comes to match-ups in the tournament scene.
For the sake of simplicity, and given that players generally have preferences when it comes to the playstyles they are comfortable with, I will assume a scenario where 2 different players, paired against each other, must choose between the 2 playstyles they are most comfortable with.
Perhaps an approach that bears some similarities with reality could be:
Player 1:
-Grassy Terrain BO
-Rain
Player 2:
-Sun
-Stall
Let´s assume that there is a certain average Win Rate for the 4 possible match-ups, and that both players must play a multitude of rounds, say a Bo5.
Our objective is to determine the Probability Distribution between which playstyle to bring, that maximizes one´ Win Rate, assuming the opponent does the same. Both of the players are rational and have access to information about the average Win Rate for all 4 match-ups.
Consider the following Win Rates from Player 1´s perspective
GTerrain vs Sun - 35%
Gterrain vs Stall - 60%
Rain vs Sun - 70%
Rain vs Stall - 30%
Assuming that there is no possibility of a tie in either of the 4 match-ups, the addition of Player 1´s winrate with Player 2´s winrate for a particular match-up must equal 1.
Now, let´s put ourselves in Player 1´s shoes, if we bring Grassy Terrain, we are favored to win in case Player 2 brings Stall, but probably lose in case they bring Sun. If we bring Rain, we probably lose when Player 2 brings Stall, but probably win when Player 2 brings Sun.
Does this mean the ideal probability distribution is a 50/50?
It would for sure, in case the Win Probabilities were all either 100% or 0%, but in this case there is a slight discrepancy.
Let´s use some math to figure the ideal probability distribution.
In order to do this, we must find a function that describes Player 1´s Win Rate, where the variables are his Probability of Bringing GTerrain, his Probability of bringing Rain, the probability of Player 2 bringing Sun, and the probability of Player 2 bringing Stall.
W = Win Rate
P(G) = Probability of bringing Grassy Terrain
P(R) = Probability of bringing Rain
P(Su) = Probability of Player 2 bringing Sun
P(St) = Probability of Player 2 bringing Stall
Our Win Rate Function is :
W = (P(G) x P(Su) x 0.35 ) + (P(G) x P(St) x 0.6 ) + (P(R) x P(Su) x 0.7 ) + (P(R) x P(St) x 0.3 )
Confusing? It´s pretty simple actually. This Function consists of a sum of 4 parts. Each part corresponds to a different match-up. The probability of each match-up is the product of the probability of each playstyle, for example, the probability of a Grassy Terrain vs Sun Match-up is the product of the probability of Player 1 bringing Grassy Terrain with the probability of player 2 bringing Sun. After we do this, we can multiply each individual match-up probability with it´s expected Win Rate, sum the obtained values and obtain the final Win Rate.
There are still a few problems, the most notable one, is that this Function still has 4 variables, but worry not, we can get rid of 2 immediately.
How? Well, since Player 1 can either bring Grassy Terrain or Rain, this means that:
P(G) + P(R) =1
or
P(R) = 1- (PG)
The same thing applies to Player 2
P(Su) + P(St) = 1
or
P(St) = 1 - P(Su)
Let´s go back to our Win Rate Function and simplify it!
W = (P(G) x P(Su) x 0.35 ) + (P(G) x (1-P(Su)) x 0.6 ) + ((1-P(G)) x P(Su) x 0.7 ) + ((1-P(G)) x (1-P(Su))x 0.3 )
After doing some math, this can be simplified to:
W = (-0.65 x P(G) x P(Su)) -0.4 x P(G) + 0.4 P(Su) +1
Alright, so we have simplified our Win Rate Function the most we could, but what does this all mean?
Currently we have a function that describes our Win Rate based on 2 variables, the Probability of us bringing Grassy Terrain, which is under our control, and the probability of Player 2 bringing Sun, which is not. This means that we can not simply calculate the maximum value of the function and claim that this is our maximum possible Win Rate. It would be, in case Player 2 was stupid, but Player 2 is also trying to maximize it´s own Win Rate, meaning we need to do a little bit more work to find our Nash Equilibrium.
Let´s start with calculating Player 2´s own Win Rate Function.
Since Player 2 Wins Whenever Player 1 Loses W2 = 1-W
1-W = (P(G) x P(Su) x 0.65 ) + (P(G) x (1-P(Su)) x 0.4 ) + ((1-P(G)) x P(Su) x 0.3 ) + ((1-P(G)) x (1-P(Su))x 0.7 )
(=) (After Some Math)
W = (-P(G) x (13/20 P(Su) - 3/10)) + 2/5 P(Su) + 0.3
Now that we have 2 functions that describe Player 1´s Winrate, one directly, and the other via the ammount of times player 2 loses, we can match one to the other:
(-0.65 P(G) P(Su)) -0.4 P(G) + 0.4 P(Su) +1 = (-P(G) (13/20 P(Su) - 3/10)) + 2/5 P(Su) + 0.3 (=)
P(Su) = (3/10 P(G) -0.1) : (-1/4 + 1/4P(G))
This gives us the relation between the Probability of Player 2 bringing Sun in the Nash Equilibrium and the probability of Player 1 bringing Grassy Terrain in the Nash Equilibrium. Now, we can simply substitute P(Su) in the Win Rate Expression to get a one variable functon.
W = (-0.65 x P(G) x P(Su)) -0.4 x P(G) + 0.4 P(Su) +1
(=)
W = (-0.65 x P(G) x ((3/10 P(G) -0.1) : (-1/4 + 1/4P(G)))) -0.4 x P(G) + 0.4 ((3/10 P(G) -0.1) : (-1/4 + 1/4P(G))) +1
(=) (After a bit of math)
View attachment 603249
This means we can now find the ideal probability distribution for player 1 by graphing the function!
View attachment 603251
Looking at this graph, and considering the maximum Win Probability is 1, while the Maximum probability of choosing a Grassy Terrain team is 1, we conclude that Player 1 must choose the Grassy Terrain Option, approximately 46% of the time, chosing Rain, the other 54%.
In theory, this probability distribution between the playstyles he should bring maximize his winning chances against Player 2, however, we also conclude that there is some mistake in our calculations, since the Win Probability can not be 1, or 100%, as shown in the graph.
Analysing this result, it seems logical that the probability of bringing Rain is superior to the probability of bringing Grassy Terrain, since the sum of Grassy Terrain´s Win Rates, 35% vs Sun and 60% vs Stall equals 95%, a little bit less than the sum of Rain´s Win Rates, 30% vs Stall and 70% vs Sun equals 100%.
Regardless, I am pretty sure I made some mistake along the way, and it would be really interesting if someone could expand on the subject and perhaps find my mistakes.
I think applying Game Theory to Pokemon is not a waste of time, and it could expand our knowledge about the metagame.
Thanks for Reading!