Today, we are lucky enough to have a guest post from none other than a finalist of a nation-wide Pokemon Championship, who has agreed to write on the maths behind this game, and a perfect strategy to capture the game’s hardest Pokemon. We’re sure that if you love either Pokemon or Maths (or both!), then you’ll love Bryn Brunnstrom’s article.

Like many other people born in the late 90s, the Pokemon franchise was a large part of my childhood. I watched the show, bought the toys, and played the video games. And like most people who played the Pokemon games, I was constantly infuriated by how many pokeballs it took to catch a legendary Pokemon. It was when I was trying to catch a mewtwo, somewhere around my 30th ultra ball that I stopped and thought “there has to be a better way.” Now, most people at this point would just go out, buy a gameshark, and give themselves infinite master balls, but I was guided by some misplaced sense of honour to do something infinitely geekier, and instead tried to figure out a perfect strategy.

In Pokemon, whenever capture is attempted the game runs through a simple formula to calculate the odds of capture. The game then generates a random number between 1 and 255. If this number is lower than the calculated value, then capture is successful. The formula is as follows.

p={[(3×HpMax-2×HpCurrent)×rate×BallBonus]÷3×HpMax}×StatusBonus

The variables are defined as follows.

• p=odds of capture
• HpMax=The Pokemon’s maximum number of hit points
• HpCurrent=The Pokemon’s current hit points
• rate=the Pokemon’s catch rate (this has a minimum of 1 and a maximum of 255)
• BallBonus=The bonus provided by the type of pokeball you are using
• StatusBonus=The bonus provided by the Pokemon’s status affliction (if there is one)

This formula is created so that the value of p can never fall below 1 (making the Pokemon impossible to capture.) The minimum value of p, when the Pokemon’s Hp is full, and it has no status affliction, can be said to be (rate×BallBonus)÷3, as

3HpMax-2HpCurrent=HpMax
HpMax÷3HpMax= 1÷3

Because the random number only ranges between 1 and 255, if we can make the calculated value exceed 255 then capture will always be certain. For example, if we take a Pokemon with a catch rate of 255, caterpie for instance, then calculate it’s minimum catch value, which in this case is (255×BallBonus)÷3, then using a pokeball with a bonus of ×3, like the lure ball, will guarantee the Pokemon’s capture.

Thus, in order for there to be a truly “perfect” capture strategy, there must be a way to make the calculated value for the hardest Pokemon to catch to exceed 255.

However, in order for us to figure this out, we must first find which Pokemon is the hardest to capture. Looking at the formula above, we can see that this is determined by two factors: the Pokemon’s catch rate, and the Pokemon’s Hp. For these purposes, I have decided to treat the Pokemon’s Hp stat as if it were at it’s maximum level, as some wild Pokemon have been known to reach level 100.

This turned out to be Giratina, a legendary Pokemon from generation 4, which has a maximum Hp stat of 504, and an incredibly low catch rate of 3.

Now, for the perfect capture, we can assume that the Pokemon’s current Hp is 1, which can be achieved using the move “False Swipe.” This means that the maximum chance you have of catching Giratina will be:

{[(3×504)-(2×1)×3×BallBonus]÷3×504}×StatusBonus.

In order to continue, we must find the maximum Ball and Status bonuses.

Finding the maximum status bonus is fairly easy. Each of the status afflictions in Pokemon have a set value for their bonus. Sleep and freeze have a value of 2; poison, paralyze, and burn have a value of 1.5; and confused, infatuated or healthy have values of 1.

The maximum ball bonus is also fairly easy to find. The three standard balls, poke, great and ultra, have values of 1, 1.5 and 2 respectively. However, with special balls, this value can become much greater. Currently, the record is 8, which is given to the level and love balls, albeit under very specific conditions. The love ball will yield its bonus of 8 only if the Pokemon you are trying to capture is of the same species and gender as your Pokemon, and the level ball will only do this if the Pokemon you are facing has a level four times greater than that of your Pokemon. These balls can be found in the generation 2 games and their remakes.

So, if we plug in these values, this gives us the ultimate capture, the highest possible chance you can get of capturing the hardest possible Pokemon.

p={[(3×504)-(2×1)×3×8]÷3×504}×2
p=[(1512-2)×24÷1512]×2
p=(36240÷1512)×2
p=47.936(3dp)

Unfortunately, Nintendo have done their job well. This gives about an18.8% chance of capture. However, I was still quite happy with these results. This is the hardest possible Pokemon to catch, and there’s a 1/5 chance of capturing it. Although I wasn’t successful, I feel confident that this should be a pretty safe strategy for any Pokemon that would actually be encountered.

Bryn Brunnstrom

We hope you’ve enjoyed this post! If you did then please check out our last two posts:

Pursuit and Evasion: the Maths of Cops and Robbers - We were lucky enough to attend a lecture given by Imre Leader, Professor of Combinatorics at Trinity College, Cambridge, on Pursuit and Evasion games, and discuss some of the problems he posed.

What is Dark Matter? And what happened to 96% percent of the universe? - We talk about one of the greatest unsolved mysteries in physics and possible solutions.

What we are posting on next – How Felix Baumgartner broke the sound barrier – we discuss the physics behind the remarkable feat of the Austrian skydiver last week, and the measures in place to make sure he made it down in one piece.