I've seen lots of questions about probability since I've been here. I thought I'd make an FAQ about it. I'm open to suggestions: Things that need to be explained more/less. Also other questions I haven't included, although this isn't meant to be comprehensive (like every single thing you can think of with inheriting IVs) but is meant to give enough information that people can do similar problems themselves. Anyway, enjoy!
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Q: What is the Probability of getting a shiny?
A: 8/65536 ways to get a shiny
Q: What is the probability of getting Pokerus?
A: 3/65536 ways to get Pokerus.
Q: But I got Pokerus before getting a shiny! How is that possible?
A: just because something is more probable, doesn't mean it will happen first. For example, rolling a number bigger than 1 is more probable than rolling a 1. But that doesn't mean you can't roll a 1 your first turn: you still have a 1/6 chance.
3 ways to get pokerus, 8+3=11 ways to get shiny or pokerus.
3/11 = 27.27% probability of getting Pokerus before a shiny.
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Q: Two parents have a 31 in one gene. What is the probability the child will have it also?
A: In order to solve any probability problem, take the # of success/# total success and faliure. In this situation, a success is the # of ways to inherit the specific gene, divide by the # of ways to inherit any genes. We also will need to add in the probability of not inheriting it, but still getting it at random.
6C3 + (6C2 * 4C1) + (6C1 * 5C2) + 6C3 = 160 ways to inherit genes.
5C2 + (5C1 * 4C1) + 5C2 = 40 ways to inherit one specific gene from one parent. same 40 ways to inherit same gene from different parent. Leaving 80 ways for random gene.
80/160 + 80/160(1/32) = 51.56%
(note: 6C3 (C for Combination, reads as "6 choose 3") means there are 6 things, how many ways are there to chose 3 of them?
If it was 6P3 (Permutation, also reads "6 choose 3") its the number of ways to choose 3 of 6 things, but now order matters.)
formulas are:
kPn = (k!/(n-r)!).
kCn = (kPn)/n!
x! (read 5! as "5 factorial") means multiply by itself, and all the whole numbers down to 1. ie, 5! = 5*4*3*2*1 = 120
However, in the actual game, sometimes less than 3 genes are inherited. The game apparently was not programmed to tell whither it has selected a specific gene already, wich allows the 2nd gene possibly overwriting previous gene. What is the real probability? How much does this matter?
In this situation, the total number of ways to inherit genes is 12P3 = 1320
(10*9)+(10*9 + 1*10)+(11*10) = 300 ways to inherit from one parent. 600 if both parents have the gene.
720 ways to leave it to chance --
600/1320 + 720/1320(1/32) = 47.16%
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Q: What is the probability of winning the different prizes at the lottery?
A: This is also something that you wouldn't expect. ID numbers do not go 0-99999, but from 0-65535. In the first case, all numbers would be eaqually likely for less than 1st plase. But since all numbers don't go 0-9, some ID numbers are more probable than others.
1/65536 probability of winning the 1st place lottery. per ID num.
at least 2nd place: wins with #xxxx (x=match, #=no match) add to 5535, then add 10000 6 times, so 6/65536 (.009%) per ID where ending four>5535, 7/65536 (.01%) otherwise.
at least 3rd place: wins with ##xxx add to 535, then add 1000 65 times, so 65/65536 (.099%) per ID where ending three>535, 66/65536 (.10%) otherwise.
at least 4th place: wins with ###xx add to 35, then add 100 655 times... so 655/65536 (.999%) per ID where ending two>35, 656/65536 (1.0%) otherwise.
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Q: What is the probability of mirage island appearing?
A: 1/65536 probability of seeing mirage island, per pokemon in party.
Q: How many times do I have to check the guy to see mirage Island?
A: We know the probability that we can see the island next time we check. (6/65536 for a full party) So we know the probability we won't see it next time also. We can multiply the probability that we won't see it until it drops to a probability of failiure that we like. For example, if we want 50% chance of success, we'd have to check 7570 times. (1-(65530/65536)^7570 = 50%) The same math can be applied to find out how many times to look for a shiny, etc.
Q: I've checked a lot more times than that, and I still haven't seen mirage island!
A: Deciding to check 7570 times means you have a 50% chance of seeing it during this period. You are never guarenteed success (or failiure). It's just like flipping a coin. If you get heads the first time dosen't guarentee you'll get a tails next time. You still have the same 50% chance at the next toss as you always will. There's no number you can be given to guarentee you success, but you can be given a number of times for a certain probability of success.
Q: but what if when checking mirage island, 2 pokemon in your party is the same? what is the probability of 2 or more pokemon being the same?
A: right; unlike the lottery, we don't know our Pokemon's hidden number. It is easiest to solve for the probability of 6 being different, and then subtract from 1.
1 - (65536 P 6 / (65536^6)) = .02%
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Q: What is the probability of having a pokemon with all 6 IVs as 31?
A: different combinations of Pokemon IVs: 6 genes each having values of 0-31 = 36^6 = 1,073,741,824
so, about 1 in a billion chance.
Q: So that means my friend cheated!
A: Not nessisarily. Just because it is improbable, dosent mean it is impossible.
My Sableye has IVs of HP:28 - Atk:9 - SpAtk:19 - Speed:29 - SpDef:23 - Def:17
What are the odds of getting that? The same 1 in a billion chance. I didn't cheat to get this! Actually, every combination of IVs has exactly the same probability. However, that being said, for someone who does cheat, they will be more likely to set it to max IVs than something weaker. Also, someone who is bad at math (or forgets to factor EVs in) could be likely to get max IVs. Knowing something about the person will help you decide if their IVs are correct.
Q: But I heard that IVs closer to 15 are more probable ...?
A: What you are thinking is averages. The SUM of all the IVs forms a probability curve. Sums of 0 and 186 aren't likely, as there is only one way each to get those sums. But there are many ways to get a sum of 90 without using all 15s.
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Q: What's the probability that two pokemon are the same if you DON'T do the emerald cloning glitch? Is it small enough that a Nintendo program could detect someone who used it?
A: As said before, the number of uniqe IVs is 32^6, which is the same as 2^30.
Natures, gender, and other things are determined by its personality number, a 32-bit number = 2^32
Total: (2^32) * (2^30) = 2^62 = 4,611,686,018,427,387,904
However, we want to know not just what the probability that 2 pokemon are the same, but ANY two pokemon in all your boxes. This is the same sort of problem as above, with mirage island, and this uses the same formula. Let's say we have all of our boxes filled with the same Pokemon. If I remember right, that's 420. 426 counting your party.
1 - ((2^62) P 426 / ((2^62)^426) = .0000000000019629% Which is almost 1 in 51 trillion.
in fact, you'd need to have a 100 million Pokemon before you had a even a tenth of a percent chance of two being the same.
---------------------------------
Q: What is the Probability of getting a shiny?
A: 8/65536 ways to get a shiny
Q: What is the probability of getting Pokerus?
A: 3/65536 ways to get Pokerus.
Q: But I got Pokerus before getting a shiny! How is that possible?
A: just because something is more probable, doesn't mean it will happen first. For example, rolling a number bigger than 1 is more probable than rolling a 1. But that doesn't mean you can't roll a 1 your first turn: you still have a 1/6 chance.
3 ways to get pokerus, 8+3=11 ways to get shiny or pokerus.
3/11 = 27.27% probability of getting Pokerus before a shiny.
---------------------------------------
Q: Two parents have a 31 in one gene. What is the probability the child will have it also?
A: In order to solve any probability problem, take the # of success/# total success and faliure. In this situation, a success is the # of ways to inherit the specific gene, divide by the # of ways to inherit any genes. We also will need to add in the probability of not inheriting it, but still getting it at random.
6C3 + (6C2 * 4C1) + (6C1 * 5C2) + 6C3 = 160 ways to inherit genes.
5C2 + (5C1 * 4C1) + 5C2 = 40 ways to inherit one specific gene from one parent. same 40 ways to inherit same gene from different parent. Leaving 80 ways for random gene.
80/160 + 80/160(1/32) = 51.56%
(note: 6C3 (C for Combination, reads as "6 choose 3") means there are 6 things, how many ways are there to chose 3 of them?
If it was 6P3 (Permutation, also reads "6 choose 3") its the number of ways to choose 3 of 6 things, but now order matters.)
formulas are:
kPn = (k!/(n-r)!).
kCn = (kPn)/n!
x! (read 5! as "5 factorial") means multiply by itself, and all the whole numbers down to 1. ie, 5! = 5*4*3*2*1 = 120
However, in the actual game, sometimes less than 3 genes are inherited. The game apparently was not programmed to tell whither it has selected a specific gene already, wich allows the 2nd gene possibly overwriting previous gene. What is the real probability? How much does this matter?
In this situation, the total number of ways to inherit genes is 12P3 = 1320
(10*9)+(10*9 + 1*10)+(11*10) = 300 ways to inherit from one parent. 600 if both parents have the gene.
720 ways to leave it to chance --
600/1320 + 720/1320(1/32) = 47.16%
--------------------------------
Q: What is the probability of winning the different prizes at the lottery?
A: This is also something that you wouldn't expect. ID numbers do not go 0-99999, but from 0-65535. In the first case, all numbers would be eaqually likely for less than 1st plase. But since all numbers don't go 0-9, some ID numbers are more probable than others.
1/65536 probability of winning the 1st place lottery. per ID num.
at least 2nd place: wins with #xxxx (x=match, #=no match) add to 5535, then add 10000 6 times, so 6/65536 (.009%) per ID where ending four>5535, 7/65536 (.01%) otherwise.
at least 3rd place: wins with ##xxx add to 535, then add 1000 65 times, so 65/65536 (.099%) per ID where ending three>535, 66/65536 (.10%) otherwise.
at least 4th place: wins with ###xx add to 35, then add 100 655 times... so 655/65536 (.999%) per ID where ending two>35, 656/65536 (1.0%) otherwise.
---------------------------------------------
Q: What is the probability of mirage island appearing?
A: 1/65536 probability of seeing mirage island, per pokemon in party.
Q: How many times do I have to check the guy to see mirage Island?
A: We know the probability that we can see the island next time we check. (6/65536 for a full party) So we know the probability we won't see it next time also. We can multiply the probability that we won't see it until it drops to a probability of failiure that we like. For example, if we want 50% chance of success, we'd have to check 7570 times. (1-(65530/65536)^7570 = 50%) The same math can be applied to find out how many times to look for a shiny, etc.
Q: I've checked a lot more times than that, and I still haven't seen mirage island!
A: Deciding to check 7570 times means you have a 50% chance of seeing it during this period. You are never guarenteed success (or failiure). It's just like flipping a coin. If you get heads the first time dosen't guarentee you'll get a tails next time. You still have the same 50% chance at the next toss as you always will. There's no number you can be given to guarentee you success, but you can be given a number of times for a certain probability of success.
Q: but what if when checking mirage island, 2 pokemon in your party is the same? what is the probability of 2 or more pokemon being the same?
A: right; unlike the lottery, we don't know our Pokemon's hidden number. It is easiest to solve for the probability of 6 being different, and then subtract from 1.
1 - (65536 P 6 / (65536^6)) = .02%
-----------------------------
Q: What is the probability of having a pokemon with all 6 IVs as 31?
A: different combinations of Pokemon IVs: 6 genes each having values of 0-31 = 36^6 = 1,073,741,824
so, about 1 in a billion chance.
Q: So that means my friend cheated!
A: Not nessisarily. Just because it is improbable, dosent mean it is impossible.
My Sableye has IVs of HP:28 - Atk:9 - SpAtk:19 - Speed:29 - SpDef:23 - Def:17
What are the odds of getting that? The same 1 in a billion chance. I didn't cheat to get this! Actually, every combination of IVs has exactly the same probability. However, that being said, for someone who does cheat, they will be more likely to set it to max IVs than something weaker. Also, someone who is bad at math (or forgets to factor EVs in) could be likely to get max IVs. Knowing something about the person will help you decide if their IVs are correct.
Q: But I heard that IVs closer to 15 are more probable ...?
A: What you are thinking is averages. The SUM of all the IVs forms a probability curve. Sums of 0 and 186 aren't likely, as there is only one way each to get those sums. But there are many ways to get a sum of 90 without using all 15s.
--------------------------------------
Q: What's the probability that two pokemon are the same if you DON'T do the emerald cloning glitch? Is it small enough that a Nintendo program could detect someone who used it?
A: As said before, the number of uniqe IVs is 32^6, which is the same as 2^30.
Natures, gender, and other things are determined by its personality number, a 32-bit number = 2^32
Total: (2^32) * (2^30) = 2^62 = 4,611,686,018,427,387,904
However, we want to know not just what the probability that 2 pokemon are the same, but ANY two pokemon in all your boxes. This is the same sort of problem as above, with mirage island, and this uses the same formula. Let's say we have all of our boxes filled with the same Pokemon. If I remember right, that's 420. 426 counting your party.
1 - ((2^62) P 426 / ((2^62)^426) = .0000000000019629% Which is almost 1 in 51 trillion.
in fact, you'd need to have a 100 million Pokemon before you had a even a tenth of a percent chance of two being the same.
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