DeathsSpook

Damn. Looks like I'm going to need to actually run the data. Grumble grumble grumble.

Thank you for this. This was super helpful for the different types of point counts (and clarifying that the wiki saying "Battle Count / (Wins)" is NOT a division, but a clarification). I'm going to go through and do the math to check these later, but I'm pretty sure these are correct. However, I just did some testing vis a vis the pokemon encounter rates: (Emphasis mine) I actually know that time is NOT the deciding factor for encounters. It's clearly a step based counter: sit on a map for an hour in the same spot, and no encounters will appear. You have to move around to get an encounter. In fact, in tall grass, you can tell if a movement counts or not: if grass flies up, the game counts that one step. Experiments on Route 16 at night in tall grass that I just conducted seem to confirm that this wait is NOT probabilistic though, like you said: I very consistently got 5 official steps between encounters, at least for common Pokemon. What seems to happen is that after 4 counted steps, the pokemon disappears, while after step 5, the new pokemon appears. That said, I only tested this for common pokemon. I'm a bit skeptical at differences in variant numbers: the wiki insists that all the non Normal variants are equal, as does the encounter table I generated for my thread here, which is based on this post here. The 1/25 number for variants against the different numbers in the spreadsheet for the different variants, but the source for the encounter probabilities is starting to look a little sus in light of the very consistent number of official steps per encounter after a common pokemon. I think I'm going to have to break down and do a full trial of encounters and generate my own results. Dammit. I was hoping to avoid that. EDIT: I know we're not allowed to use bots to PLAY the game, but are we allowed to use bots that will simply record us playing the game? EG, I set up a bot that simply reads in pokemon in a route, their variant, their rarity, and the time between encounters, but I am still the one actually doing the moving around? EDIT TO EDIT: I did NOT do this, and will NOT do this unless I get explicit confirmation that it's fine.

My questions come from this post, but I'll reiterate them here for convenience. I know some of this probably runs up against what the devs want public, but I'm hoping at least some of them will be answered: With regards to the Ranking Points formula, the two questions I have are a) The current formula seems to reward number of battles but penalize the number of wins. Is that correct? Further, b) What is the base of the logarithm in the Ranking Points formula? The Ranking Points formula as I have it, after some algebra, is Ranking Points = (t * √(u/n) * Log(b)) / 1000, where: t = Total Experience of all of your Pokemon, summed together. n = Your total number of Pokemon u = Your total number of unique Pokemon. Note that variants of the same species count as different pokemon, e.g. Dark Mawile, Shiny Mawile, and Mawile are three unique pokemon. b = (Battle Count (presumably now only in the battle tower))/ (Wins). How is the probability of catching a Pokemon calculated? I could not find Probabilities of catching a Pokemon at all. I'm assuming it's based on ball modifiers, HP, level, status conditions(?), and so forth, but I don't know what relationship any of these variables have to the final probability. How are encounter probabilities decided? I see a couple of separate possibilities. For clarity, I will use the term "variant-rarity" as a noun meaning "the co-occurrence of a variant and rarity": e.g. "Shiny-rare is a variant-rarity." For all examples, I will use a theoretical route with Common Starly and Swablu, Rare Fennekin and Staravia, Legendary Rayquaza, and UB Kartana, assuming the player has all badges. Does the RNG first decide if an encounter happens at all and THEN what type of encounter that is, or does it roll to decide between types of encounters and no encounter in one go? The former would be "First check if an encounter happens this step. IF an encounter happens, decide the variant-rarity". The latter would be "For each variant-rarity, check if it occurs at this step. If none of them occur, then no encounter occurs." (I suspect the former is the case: the probability of an encounter happening at all divides out a bit too cleanly from the probabilities of encountering a specific variant-rarity and of encountering a specific variant-rarity given an encounter occurs at all.) Are variants and rarities rolled separately or together? For example, does RNG decide first that I'm getting a shiny Pokemon, then a rare pokemon, or does it roll in one go and decide that I'm getting a Shiny-rare? Are Pokemon appearances in a particular variant rarity uniformly distributed? EG, is my probability of getting a shiny Swablu given that rolled into the Shiny-common pool 1/2? Are variant-legends and variant-UBs pooled together into a single variant-veryRare pool? EG, to get a Shiny Kartana, do I have to roll Shiny-veryRare and then there's a 1/2 probability of getting it, or did I roll to get into a separate Shiny-UB pool and then Kartana appears because it's my only option?

Does this mean that regional variants will only ever be accessible in a single playthrough? EG, once I went past Alola in sidequests, I'd never be able to get Alolan forms ever again?

Will sprites ever be added for pokemon with obvious gender differences that don't cross into form differences but that are still very obvious? EG, Jellicent. I imagine it's way too much to ask for gender different sprites for every pokemon with gender differences, but it's slightly bizarre to look at my female Jellicent who looks like the Pringles Man.

Made some important edits to the friendship formula, in that it is now accurate whereas it was not before. Also, I will be very grateful to anybody who can give me formulae for catching probabilities, with citations. EDIT 2022-07-20: I'm compiling a list of questions here for some of the developers (I'm hoping @Patrick will answer even though they're probably super busy) With regards to the Ranking Points formula, the two questions I have are a) The current formula seems to reward number of battles but penalize the number of wins. Is that correct? Further, b) What is the base of the logarithm in the Ranking Points formula? The Ranking Points formula as I have it, after some algebra, is Ranking Points = (t * √(u/n) * Log(b)) / 1000, where: t = Total Experience of all of your Pokemon, summed together. n = Your total number of Pokemon u = Your total number of unique Pokemon. Note that variants of the same species count as different pokemon, e.g. Dark Mawile, Shiny Mawile, and Mawile are three unique pokemon. b = (Battle Count (presumably now only in the battle tower))/ (Wins). How is the probability of catching a Pokemon calculated? I could not find Probabilities of catching a Pokemon at all. I'm assuming it's based on ball modifiers, HP, level, status conditions(?), and so forth, but I don't know what relationship any of these variables have to the final probability. How are encounter probabilities decided? I see a couple of separate possibilities. For clarity, I will use the term "variant-rarity" as a noun meaning "the co-occurrence of a variant and rarity": e.g. "Shiny-rare is a variant-rarity." For all examples, I will use a theoretical route with Common Starly and Swablu, Rare Fennekin and Staravia, Legendary Rayquaza, and UB Kartana, assuming the player has all badges. Does the RNG first decide if an encounter happens at all and THEN what type of encounter that is, or does it roll to decide between types of encounters and no encounter in one go? The former would be "First check if an encounter happens this step. IF an encounter happens, decide the variant-rarity". The latter would be "For each variant-rarity, check if it occurs at this step. If none of them occur, then no encounter occurs." (I suspect the former is the case: the probability of an encounter happening at all divides out a bit too cleanly from the probabilities of encountering a specific variant-rarity and of encountering a specific variant-rarity given an encounter occurs at all.) Are variants and rarities rolled separately or together? For example, does RNG decide first that I'm getting a shiny Pokemon, then a rare pokemon, or does it roll in one go and decide that I'm getting a Shiny-rare? Are Pokemon appearances in a particular variant rarity uniformly distributed? EG, is my probability of getting a shiny Swablu given that rolled into the Shiny-common pool 1/2? Are variant-legends and variant-UBs pooled together into a single variant-veryRare pool? EG, to get a Shiny Kartana, do I have to roll Shiny-veryRare and then there's a 1/2 probability of getting it, or did I roll to get into a separate Shiny-UB pool and then Kartana appears because it's my only option?

I added a wish list to my top post: there are certain things related to wild encounters that I would quite like to know, because then I can generate a formula of "expected steps to get a particular pokemon of a particular variant" and "optimal strategies to catch Pokemon" Also Probability that you need at least k battles to take a pokemon from 0 happiness to at least X happiness Transition matrix of target friendliness Let a probability of happiness vector be an X+1 dimensional vector such that its ith value denotes the probability that a Pokemon's happiness is i-1. To transition happiness, we define an X+1 x X+1 dimensional transition matrix M s.t. for a probability of happiness vector v, Mv is the probability of happiness vector after one battle. If we’re talking about a state where we start with some probability vector of friendliness scores from 0, 1, 2, …, X, where the ith entry of the vector denotes the probability of that friendliness score exactly i-1 unless i = X+1, in which case it’s the probability that the score is at LEAST X, then the transition matrix M has the form: M_{i+1,i} = 1/2, i < X M_{i+2,i} = 1/2 i < X M_{X+1,X} = 1 M_{X+1, X+1} = 1 M_{i,j} = 0 otherwise Observe that M^k_{j,1} is the probability that a Pokemon starting at 0 friendliness ends up at friendliness j-1 after k iterations if j ≤ X. If j= X+1, it’s the probability that a Pokemon that is completely unfriendly gets to X+1 friendliness after at least k iterations. M^k_{j,1} = 0 if j ≤ k or j ≥ 2k+2. (Excepting M^k_{X+1,1}). Observe that if 2k+1 ≥ X-1, then M^k_{X+1,1} > 0. Observe also that if k ≥ X, then M^k_{X+1,1} = 1. In fact, M^k_{X+1,1} encodes the CDF of a Pokemon reaching friendliness at least X after k iterations. IE, M^k_{X+1,1} = P(Pokemon reaching friendliness at least X after at least k battles) If 2k+1 ≥ X-1 AND k < X, then M^k_{X+1,1} = 1-sum_{j = k+1}^{X}(k choose j-(k+1))(1/2)^j P(Pokemon reaching friendliness at least X after at least k battles) = 0 if k < X/2-1 1 if k ≥ X 1-sum_{j = k+1}^{X}(k choose j-(k+1))(1/2)^j if X/2-1 ≤ k < X Applied to X = 219 (three hearts): P(Pokemon reaching friendliness at least 219 after at least k battles) = 0 if k < 108.5 1 if k ≥ 219 1-sum_{j = k+1}^{219}(k choose j-(k+1))(1/2)^j if 109 ≤ k < 219 EDIT: I am like 90% sure I did the math here wrong somewhere. I will correct over the next couple of days EDIT 2: Definitely did the math wrong. Revised math below. Observe that the probability that we get to at least X happiness in k battles exactly is P(happiness ≥ X after k battles AND happiness < X after the k-1 battles before). NOTE THAT THE LAST BATTLE IS THE ONLY PLACE WHERE ORDER MATTERS ONCE Observe that these two event can co-occur in only mutually exclusive two scenarios: Either k-1 battles before gave exactly X-1 happiness, OR the k-1 battles before gave X-2 happiness AND the kth battle gave 2 more happiness. Therefore, since the kth battle is independent from the k-1 battles before it: P(happiness ≥ X after k battles AND happiness < X after the k-1 battles before) = P(happiness after k-1 battles = X-1) + P(happiness after k-1 battles = X-2)*P(kth battle gives two happiness) N battles can only give exactly M happiness iff N ≥ M/2 AND N ≤ M. In this case, 2N-M battles would have to give 1 experience and M-N battles would have to given 2 experience. Therefore, there are N choose M-N ways for this set of outcomes to occur (saying there are N choose 2N-M is exactly equivalent). We can exclude the the k-1 = X possibility: we would not be able to get exactly X happiness in X+1 battles. Therefore, if k-1 ≥ M/2 AND k-1 < M: P(it takes exactly k battles to get at least X happiness) = P(happiness after k-1 battles = X-1) + P(happiness after k-1 battles = X-2)*P(kth battle gives two happiness) = ((k-1) choose (X-k)) *(1/2)^(k-1)+ ((k-1) choose (X-k-1)) *(1/2)^(k-1)*(1/2) = (1/2)^k*(2*((k-1) choose (X-k))+((k-1) choose (X-k-1)) If k = X/2 exactly, then every battle must be 2 happiness battle, which has a probability of occurring of (1/2)^(X/2). Analogously, if k = X exactly, every battle must be a 1 happiness battle, which has probability (1/2)^X Therefore: P(it takes exactly k battles to get at least X happiness) = 0, if k < X/2 OR k > X; (1/2)^(X/2) if k = X/2 ; (1/2)^k*(2*((k-1) choose (X-k))+((k-1) choose (X-k-1)) if k-1 ≥ X/2 AND k-1 < X (1/2)^X if k = X. For 219 happiness (three hearts), this comes out to E[number of battles for 219 happiness] = ~146. Variance is ~16.32, and the PDF looks fairly normally distributed, so you can approximate this as a normal distribution with mean 145.888.. and sd ~ 4.04 For 254 happiness (four hearts), this comes out to E[number of battles for 254 happiness] = ~169. Variance is ~18.92, so again use the central limit theorem to approximate this as normal with mean 169.222... and sd ~ 4.35

NOTE: an alternative way to fulfill the battles with 35 experience is to fight a trainer with 6 level 7 pokemon using exactly 6 level 100 pokemon: O = 6*7 = 42 L = 6*100 = 600 E = round(500*42/600) = round(5*42/6) = round(5*7) = 35

This is actually EVER so slightly wrong. Basically, the rounding function makes this complicated. I'm going to solve the inequalities necessary to REALLY nail the experience you get from a single battle here: E = round(500*O/Y) <=> 500*O/Y - 0.5 ≤ E < 500*O/Y + 0.5 <=> (Y(E - 0.5))/500 < O ≤ (Y(E + 0.5))/500 First, because you're rounding after every single battle, you can only try to nail an exact number for the experience you want if you get the remainder you need down below a certain value. Let's say you want to earn a single level 100 pokemon to have exactly 69420 EXP (because duh). Currently, it has 50000 EXP exactly, so you need to earn 19420 EXP in a single battle. (100(19420 - 0.5))/500 < O ≤ (100(19420 + 0.5))/500 (19420 - 0.5)/5 < O ≤ (19420 + 0.5)/5 (Note: this means that, for a level 100 pokemon, you can ALWAYS earn any amount of experience points divisible by 5 EVENTUALLY) (3884 - 0.1) < O ≤ (3884 + 0.1) I could simplify further, but all you need is to find an integer that's between (3884 - 0.1) and (3884 + 0.1), and there's exactly one that fits the bill: 3884. However, this is blatantly impossible in a single battle: no battle with six pokemon in it, ever, has a combined level greater than 150*6 = 900. To try to figure out when to start fine tuning, note that the following is true: E < 500*O/100 + 0.5 => E < 5O + 0.5 The maximum value for your opponent's levels in practical, on demand settings is 6*135 = 810 (any of the big boss team leader special battles other than Lysandre). Therefore, you should not even THINK of fine tuning, at all, unless: E < 5(810)+0.5 => E ≤ 4050 (since E is an integer) FURTHER, it's not really practical to assume that any arbitrary pokemon can solo one of the big boss team leader special battles. However, there ARE the Type[Blank] trainers that are specifically training accounts for any Pokemon type. They are designed so that O = 600. So therefore, you can reasonably be expected to be able to train one level 100 pokemon in chunks of: E < 5(600)+0.5 = 3000.5 => E ≤ 3000 You should not even THINK of fine tuning for single EXPs unless your exp remaining is less than or equal to 3000. To train 1 level 100 pokemon: E = round(500*O/100) = round(5O) = 5O Therefore, if you're training ONE level 100 pokemon, you can only add experience points that are divisible by 5. The minimum number of experience points you can add to this is 5*6 (minimum wild pokemon battle) = 30. For certain reasons I'll get into later, the number of experience points you can get that really matters is when you fight a level 7 wild pokemon with one Level 100 Pokemon, which is 5*7 = 35 Because of this, I'm going to introduce the following notation: E %% n = The remainder when you divide an integer E by another integer n. If E is divisible by n, E %% n = 0 Observe that if you use 5 level 100 pokemon: E = round(500*O/500) = O Therefore, you can theoretically easily earn any arbitrary number of experience points between 6 (lowest wild type pokemon level) and 600 if you use 5 level 100 hundred pokemon in each battle. You can also easily earn 725 experience (the N battle) or 750 experience (any of the big bosses, again other than Lysandre), But this isn't quite practical for REALLY small numbers (can YOU send out 4 level 100 pokemon to die against a level 6 Corphish before killing it with another one?). If there exist training accounts that only have 6 level 6s (help: which HAVE to exist even if I don't know where, someone give me a link to one, please), then you can practically get 36 EXP per battle. One more ingredient we need to create a practical guide, here: we need a way to get a number of experience points relatively prime to 36. This is done most easily if O = 7 (attainable with a wild pokemon battle), so now we have a way to get 35 EXP per battle. For an arbitrary whole number x of integers n_i, GCD(n_1, n_2, ..., n_x) is the largest integer that divides all the n_i (i.e. you can multiply the GCF by some integer f_i so that f_i * GCD = n_i for each n_i. Similarly, LCM(n_1, n_2, ..., n_x) is the smallest positive integer that all of the n_i divide evenly (i.e. you can multiply each of the n_i by some integer c_i so that n_i*c_i = LCM for each n_i). Observe the following: GCD(35, 36) = 1 LCM(35, 36) = 35*36 = 1260 METHOD TO NAIL EXPERIENCE POINTS. (1225 ≤ Your remaining Experience Points ≤ 2484) For any experience you want to get under this method, E = 1260I+R, where I is a positive integer and R is an integer between 0 and 1259, inclusive. Equivalently, E = 1260Z+r, where Z is a nonnegative integer and r is between 1225 and 2484, inclusive. To get a positive number of battles while minimizing how many battles you want to do under this paradigm, you want Z = 0 so that E = r. Therefore, you want your experience left to nail whatever number to be between 1225 and 2484. Battle until you reach this point. DO NOT OVERSHOOT. Calculate r %% 35 = a. If r %% 36 = 0, let b = 36. Otherwise, b = r %% 36. With our restrictions on r, the Chinese Remainder Theorem ends up telling us that a*36+(36-b)*35 = r. Therefore, you want to fight (36-b) level 7 wild pokemon with your target level 100 and a trainers with 6 level 6 Pokemon using 5 level 100s including your target level 100.

Thank you both very much. Updated.

I'll post this later, but would you please edit to include screenshots of the discord as well? I want people looking at the sources to be able to trace it pretty easily, and not all of them are on the discord. (I'm aware that I'm being a little persnickety, but I want this thread to have the formulas in a way that's as beyond reproach as possible)

Yup. These are good enough for me! Adding now. Thank you!!

It would be, mostly, except none of your screenshots are showing up, at all. I'm pretty sure you're right, but I would like the actual screenshots to confirm.

Do you have a source for this?

Formulas I'm looking for: POKEMON CATCHING PROBABILITIES