Research Pokemon Question Challenge

[Joshua Elkington]

Hey,

I'm doing a bit of research on the dopamine and behavioural reinforcement within games, namely Pokemon, and I wanted to really try to work out the depth of the customization possibilities within Pokemon relating to how much it captivates us as fans, by coming up with a number for...

Every possible different variation of one chosen Pokemon, within the possibilities in the current generation. Let's say, Charizard. I've seen some math for every possible Pokemon team, but i'd like to try and work out how many different possibilities there are for one single Pokemon.

This includes every different possibility of a Charizard of different natures, IVs, EVs, abilities, moves, etc.

So what i'm trying to assess is, just how many 'different' Charizards can exist.

I just wondered if a similar question had been answered, and if anyone more mathematically minded could help me try and crack this if it hadn't.

Thanks!

 

[Codraroll]

Let's see...

The number of possible IV combinations is 32^6 = 1,073,741,824

According to the Smogon dex, Charizard's can learn 112 different moves. 127 if you count all the Hidden Power variants (16 types). However, since Hidden Power type is tied to IVs, you cannot just multiply the IV combinations with 127. For simplicity's sake, let's just say there are 112 moves, and Charizard can have either one, two, three or four moves in its moveset.

So the number of possible Charizard movesets is 112C1 + 112C2 + 112C3 + 112C4 = 112 + 6,216 + 227,920 + 6,210,820 = 6,445,068. I can't remember enough statistics to tell if this calculation method allows for movesets of several identical moves, but if it does, we'll just say that it weighs up for the different Hidden Power types.

Charizard has two different abilities. So far, so simple.

However, the EVs is what inflates the number ridiculously. Five different stats, up to 510 EVs to be allocated. And I'm not entirely sure how to calculate it. In a Reddit thread I found, somebody said you can just take 6^510 (Where the sixth "stat" is EVs that are not allocated), and subtract any combination where more than 252 EVs are allocated in any stat, that is 5^(510-252) = 5^258.

So the number of possible EVs would be 6^510-5^258 = 7196771197622729713213463542152052821798011755454212336994963801444687424229388254559379032982162625954062004695679540577875601847708847682258651023417768407649370912612113780861130127981766520525324038910360177378698374953264513643487474836704530043030972218231269108913400592415194948749194864745841191693118003225738328775299150300460289327273766126498748526787066122631481375832141580262040551.

Or 7.20*10^396, to make it simple.

EDIT since this post seems to keep getting likes. The above calculation is wrong, see post below for details.

There are also Trainer IDs and Secret IDs. Both have 65535 possible combinations.

Natures, gender, shininess and such are incorporated in the Pokémon's Personality Value (or at least it used to be, I'm not sure if it still is), so we don't need to account for the different combinations of those, we can just take a look at the possible PVs. 4,294,967,295 different combinations there too.

So... 1,073,741,824 * 6,445,068 * 2 * 7.20*10^396 * 65535^2 * 4,294,967,295. WolframAlpha says 1,828*10^432. This, by extension, is also wrong.

I'm not going to bother with catch/hatch locations, Pokéballs, OT names or nicknames. Other people might, as an extra exercise. Some quick estimates suggest there are some 10^30 possible combinations of nicknames alone, but I could be off by a factor of quintillions. Trainer names should be in the same ballpark. Catch/hatch locations, roughly 1000. So you can add something in the vincinity of 10^30 - 10^60 more combinations to the final number, depending on the number of legal characters used for nicknames (my estimate was 200, could be off).

So in total, the number of different Charizard would be vastly larger than the number of atoms in the universe (estimated to be 10^84, if I recall correctly). I'm fairly confident in all the numbers except those for the EVs. Perhaps people better at statistics than me could answer?

 

[sb879]

The reddit user is massively overcounting the number of EV combinations, as they suspect. To illustrate the issue, say a Pokemon could only have 4 EVs total. Two examples:

EV 1: HP
EV 2: Nothing
EV 3: HP
EV 4: Speed

EV 1: Speed
EV 2: HP
EV 3: Nothing
EV 4: HP

According to their logic, both of these are distinct combinations. But all we care about in the end is the EV value for each stat, not the stat for each EV value. And both of these are just 2 HP/1 Speed, so they're actually the same.

As for the correct answer, I'm not 100% sure. To start, there's 253^6 possible combinations, 0-252 for each of the 6 stats. But that still needs to factor in the 510 EV cap. We have 6 uniformly distributed and discrete values (the EVs for each stat), and need to find the probability distribution of their sum so we can find out how many of the 253^6 combinations are within the 510 cap. The Irwin-Hall distribution describes how the sum of n uniformly distributed continuous values between 0 and 1 is distributed, which is a bit different. Still, I tried to apply it here. The value of x as described in the above link is anywhere between 0 and n (6 in this case). The values of our sum could range from 0 to 1512, and the x we'd want to check for would be 510. So the analogous value for x, I think, would be 510/252: the value between 0 and 6 that's roughly the same percent of the way through the range.

Using this value with the cumulative distribution function in the link, I get that 0.085833124727123 of the combination's sums are within 510. This lines up fairly closely with some simulations I ran, which found an average of 0.08719918 over 10 samples with 10^7 random EV combinations. Which in terms of the actual number of EV combinations is around ~2.251 * 10^13 or ~2.286 * 10^13, respectively. Someone better versed in statistics could try to parse this, which describes our exact case (P(Y<=510) = P(Y=0)+...P(Y=510)) on page 2, but I have no idea what subscripts on the binomial coefficients are supposed to mean.

Anyways, in addition to what Codraroll mentioned, ribbons and happiness are also individual aspects of a Pokemon.

 

[Codraroll]

To top it off, it seems like I made a pretty bad mistake in my calculations. For some reason, I must have mixed natures and EVs, and assumed that the HP stat was unaffected by EVs. This is, of course, very wrong, the HP stat is unaffected by Natures. So there are six stats both affected by IVs and EVs.

To make up for the terrible mistake, let's try to estimate the number of characters that are possible to use in Trainer names and nicknames:
- Latin letters: 26*2 = 52
- Digits: 10
- Other symbols (counted on Bulbapedia): 47
- Accented letters (various languages): 57 (there are lots of duplicates, but I hope I managed to skip them all and still count every separate letter)
- Other symbols in various language versions: 11
These characters may be used in nicknames up to 12 letters long. So the total number of available characters (discounting the fact that some diacritics are exclusive to certain languages - I presume they could be added together using cheats) is 177. This adds up to 177^12 = 9.46*10^26 combinations.

The Eastern languages only allow for 6-letter nicknames. The available characters are, if I understand Bulbapedia and Wikipedia correctly:
- Latin letters: 52
- Digits: 10
- Other symbols: 47
- Hiragana: 46
- Katakana: 48
- Small vowel kana: 10
- Yoon: 6
- Soukon: 2
- Other kana with "dakuten" and "handakuten": Heck if I know or understand this system, let's guesstimate 20
- Other symbols: 9
- Hangul: 40
- Hangul Jamo: 94, if I understand correctly.

So... across the various Japanese, Chinese and Korean languages, you have 384 characters eligible for 6-letter nicknames only. That's 384^6 = 3.21*10^15. This is just below one third of one hundred-billionth of the number of possible Western nicknames, so they won't even make a noticeable difference on the total amount of nicknames: 9.46*10^26. Even if I vastly miscounted by a factor of ten and there are 3840 available Eastern characters in total, the six-character limit still means the number of combinations is 100,000 times smaller than it is for Western nicknames. Then again, if I understood the Wikipedia article on Hangul correctly, a properly implemented system would have thousands of eligible letter blocks, which would have a noticeable impact on the number of nicknames.

...anyway, nine point forty-six to the twenty-sixth power. We actually have to square that number, since there are as many possible Original Trainer names as there are Pokémon nicknames. So, 8.95*10^53.

Then there are 256 values for Happiness, and 109 Ribbons. Let's assume a hacked Charizard still is a distinct Charizard, and that it can have both Gen III ribbons and a 12-letter OT name.

With that info in mind, the new estimate on the number of possible Charizard is 1,073,741,824 * 6,445,068 * 2 * 2.251*10^13 * 65535^2 * 4,294,967,295 * 8.95*10^53 * 256 * 109 = 1.435*10^107. Not as mind-boggling as the previous number, but still quite impressive. Still more than twenty orders of magnitude more than the number of atoms in the universe.