Flawless or Bust: How to Catch Perfect Legendary Pokémon in Omega Ruby & Alpha Sapphire

76 Responses to Flawless or Bust: How to Catch Perfect Legendary Pokémon in Omega Ruby & Alpha Sapphire

1. P3DS says:

   December 18, 2014 at 8:17 am

   Something to note for IV checking when you don’t have access to the IV judge (for the cosplay pikachu and the lati twins on southern island) is to confirm that you have the stats shown, and then use the super training to check the IVs. For pikachu, it takes a few EVs, but lati twins only need 12 EVs to confirm 30 or 31. When you get one with all the flawless stats, EV train it with 12 EVs. If the stat increases, it is 31, else it is 30. Don’t do on all stats, and the one you know is flawless (from characteristics)
2. crobert says:

   December 18, 2014 at 8:47 am

   Hey all – I am new to NB, but rather old to competitive battling, and I have to say that I find this article really misleading. The raw probabilities of catching a flawless legendary (meaning 5 perfect IVs in every relevant stat), legitimately, are really low. First of all it isn’t about 1/4000, unless I am really mistaken, yet I doubt it. The probability of getting 5 perfect IVs (3 are guaranteed) are:

   Pr(one irrelevant stat gets the non perfect IV) * Pr(#1 stat gets 31 for values [0..31]) * Pr(#2 stat gets 31 for values [0..31]) = 1/6 * 1/32 * 1/32 = 1/6144

   Furthermore, the article is correct where it says that even 4000, or 6144 rather, tries doesn’t guarantee you getting a flawless one. See the following:

   Pr(Getting a flawless legendary after 4250 SRs) = 1 – Pr(getting non-flawless ones for 4250 SRs) = 1 – (6143/6144)^4250 ~= 50%

   That means that after 4250 SR’s it is still a coin toss as to whether you will end up with a flawless legendary. Notice that all these calculations assume that you get the correct nature from your synchroniser, but even this isn’t given and makes these probabilities even lower.

   TLDR; If you have all the time in the world by all means go for 5 31 IVs, but chances are you will have a very rough time getting them. Personally I would settle for 27/28+ IVs in every relevant stat, if the nature is correct and the 31 IVs are assigned to the most important ones (e.g. Speed / S Att for Latios).

   The rough probability of landing a 5IV is
   (5 choose 3)/(6 choose 3) * (1/32)^2 which is around 2000. Couple that with synchronize it’ll be around 4000.
3. Sol64 says:

   December 18, 2014 at 8:47 am

   The most informative thing I read was the Final Gambit part which I’m gonna start calling “The Gambit Checker”

   Most pokemon I’ve been getting before Landorus, Heatran and Cobalion are rather decent, only Raikou is a huge letdown (Iv 30 speed…) I have both games so I’m gonna be brutal on Omega Ruby for my “2nd chance legends” Other bits of information are very helpful right on the page.
4. VenBombadil says:

   December 18, 2014 at 8:53 am

   Ray did the calculations for me but I’ll ask him about later. Regardless, the actual probability number isn’t the point. The point was to provide tips on how you can cut down the time to get “perfect” Pokemon and give some ideas on what set ups to go for.
    
   Additionally, there is a whole section explaining why you should go “flawless or bust”. Even if the odds were say 1/20000 I’d still be going for flawless. I played VGC for about 6 years , and I’ve won or lost many many games to low probability damage rolls on both sides. Using imperfect Pokemon opens the chance for this to happen even more. Using imperfect Pokemon is the equivalent of using less than 508 EV’s. Sure it might take a long time, but it’s worth it, and I know that all of the top players will settle on nothing less than Perfect.

    
   I understand by re-reading my post that it might sound aggressive; it really wasn’t my intention, I wanted it to be strictly informative. The article is very helpful in terms of tips and ideas of how to capture the legendaries and I believe anyone reading the site will be thankful for that. The “misleading” part was only referring to the part of how realistic it is to actually encounter a flawless legendary. Again I did these calculations on foot, but I expect them to be correct. 
    
   As I said on my last sentence, it is really up to anyone to decide how much time he will invest in order to find the flawless legendary – I am just providing a little perspective of what that means exactly. If you are committed enough to try these 4250 SR’s (which are more in fact, maybe double if you consider the chance of synchronising the nature) in order to have a 50% chance to encounter that 5 IV legendary, then by all means 
5. TM Gold says:

   December 18, 2014 at 10:54 am

   Ok, So let me drop some math on you.

   There are two scenarios that are being examined,  One is where 3 of the given 31s are in the stats that we want and the other is where one of the three  31s are in the stat we do not care about.

   So, let us find out the probability of the first scenario.

   We have 1/2 from Synchronize, 10/20 (The possible outcomes where all 3 ivs are in a stat we want) 1/32 (the probability of getting a flawless IV in one of the remaining stats) and 1/32 (The probability that the other of the last remaining stats is a 31)
   We have 1/2 * 10/20 * 1/32 * 1/32 =   1/4096  

   Now we have to take in the account where one of the IVs are in the unfavorable spot

   We still have the 1/2 from synchronize,  10/20 (Just a coincidence because the amount of combinations given that one of the stats is in an IV we do not care about is equal to the other outcome) And (1/32)^3 for the remaining 3 ivs

   We have 1/2 * 10/20 * 1/32 * 1/32 * 1/32 =  1/131072

   Since we do not care which scenario gives our flawless pokemon, it is situation 1 or situation 2.

   1/4096 + 1/131072 – (1/4096 * 1/131072)

   Which exactly, is     135167/536870912

   Or an estimated fraction is  1/3972  

   Which is basically 1/4000
    
6. crobert says:

   December 18, 2014 at 11:28 am

   And if you want to be super precise, you can still land the desired nature after failing synchronize check so the probability of getting the correct nature is 26/50.
7. TM Gold says:

   December 18, 2014 at 11:31 am

   And if you want to be super precise, you can still land the desired nature after failing synchronize check so the probability of getting the correct nature is 26/50.

   Crap, :/  forgot about that situation.  

   Give me like 5 minutes
8. VenBombadil says:

   December 18, 2014 at 11:56 am

   Ok, So let me drop some math on you.

   There are two scenarios that are being examined,  One is where 3 of the given 31s are in the stats that we want and the other is where one of the three  31s are in the stat we do not care about.

   So, let us find out the probability of the first scenario.

   We have 1/2 from Synchronize, 10/20 (The possible outcomes where all 3 ivs are in a stat we want) 1/32 (the probability of getting a flawless IV in one of the remaining stats) and 1/32 (The probability that the other of the last remaining stats is a 31)
   We have 1/2 * 10/20 * 1/32 * 1/32 =   1/4096  

   Same question with before still stands. Let me make it even more clear by focusing in the easiest of the two cases:
    
   I agree that this is a probability to get a 5IV legendary with 3 of the desired stats being 31 on any stat other than the undesired one. 
    
   I don’t necessarily agree that this is the probability that the random “not 31” IV will go to the single undesired stat (e.g. Att for Latios or S. Att for Terrakion) because I believe you are not taking this into account.
9. VenBombadil says:

   December 18, 2014 at 12:25 pm

   Let me break it down even more to put this to bed. We search for the probability of having a 31/*/31/31/31/31 legendary (assuming without loss of generality that Att is the undesired stat). 
    
   We have two buckets:
    
   Bucket 1 : 31 31 31
   Bucket 2 : x y z 
    
   Case 1: Undesired stat is in Bucket 2 (Prob  = 1/2 or 10/20 = (5 choose 3)/(6 choose 3) as you correctly say)
    
   Without loss of generality assume that x = y = 31, with probability 1/32*1/32
    
   We further need the probability that z is indeed the one undesired stat which gives us an extra 1/3 factor to multiply.
    
   All together makes it 1/2 * (1/32*1/32) * 1/3 = 1/6144
    
   Or 1/12288 if you want to include also the probability of the synchroniser giving the desired nature. (taking the sync prob 1/2 rather than 26/50, which is indeed the correct one, due to laziness)
    
   Case 2: Undesired stat in Bucket 1 (Prob = 1/2) – this prob will be negligible at the end but still lets compute it
    
   Since the undesired stat is in Bucket 1, the only way we get a 31/*/31/31/31/31 legendary is getting a 31/31/31/31/31/31 one. That means x = y = z = 31 with probability 1/32*1/32*1/32
    
   All together makes it 1/2 * (1/32*1/32*1/32) = 1/65536
    
   Or 1/131072 for synced nature. 
    
   Both cases: 1/12288 + 1/131072 ~= 1/11235
    
   See that this is really far from the ~1/4000 you are suggesting. I feel also that I hijacked this thread enough with basic mathematics, if you still disagree please reread carefully this post and I will come back to say that I am wrong if I realise that I am. But again, I doubt it. 
10. TM Gold says:

    December 18, 2014 at 12:46 pm

    Let me break it down even more to put this to bed. We search for the probability of having a 31/*/31/31/31/31 legendary (assuming without loss of generality that Att is the undesired stat). 
     
    We have two buckets:
     
    Bucket 1 : 31 31 31
    Bucket 2 : x y z 
     
    Case 1: Undesired stat is in Bucket 2 (Prob  = 1/2 or 10/20 = (5 choose 3)/(6 choose 3) as you correctly say)
     
    Without loss of generality assume that x = y = 31, with probability 1/32*1/32
     
    We further need the probability that z is indeed the one undesired stat which gives us an extra 1/3 factor to multiply.
     
    All together makes it 1/2 * (1/32*1/32) * 1/3 = 1/6144
     

    I think this is where you are confusing yourself.  You cannot separate the events into ‘Buckets’ because they are independent of each other.  And of each independent stat roll there is a 1/32 chance for each stat from 0-31.  If we do not care about the stat roll, say in ‘z’ the outcome of that roll does NOT effect the outcomes of the others and can be ignored (reason why we say stuff like 31/31/31/x/31/31 as x can be variable).   We do not need to locate the stat which is undesired because we already have that given.  

     
11. crobert says:

    December 18, 2014 at 12:54 pm

    I will do a brute force calculation just to make it super clear. Let us label the five stats as HABCDS (because that’s the Japanese convention and I’m used to it) and say we want 31 in all stats other than A.

     

    Lets say the three guaranteed 31 falls on H, B, and C. By brute force, the number of configurations that give us 31-x-31-31-31-31 is (# of choices for A ranging from 0 to 31) * (# of choices for D = 31) * (# of choices for S = 31). Obviously the first term equals 32 while the other two equals 1 so there are 32 possible configurations. The total number of configurations is 32^3 so the probability of getting an acceptable configuration is 32/32^3 = 1/32^2 so the probability of getting 31-x-31-31-31-31 given that the three guaranteed 31s lie on HBC is 1/32^2. For this even to occur you need to multiply by the probability that the three guaranteed 31s lie on HBC which is 1/(6 choose 3), because these are independent events.

     

    Now the problem is exactly the same for any other choice of three guaranteed 31s landing on desired stats. The total number of configurations is 32^3 and the number of acceptable configurations is 32 hence the probability of landing an acceptable configuration is 1/32^2. Multiplied by the probability of getting the three guaranteed 31s to land on where you had them is 1/(6 choose 3). You know sum over all acceptable choices of guaranteed 31s of which there are 5 choose 3. Therefore the total probability is (5 choose 3)* ((1 / (6 choose 3)) * (1 / 32^2)) = ((5 choose 3)/(6 choose 3))*(1/32^2) = 1/2048.
12. Ray says:

    December 18, 2014 at 12:57 pm

    did you guys know when you’re flying to the clouds you can press y and go faster ? i think that should be in the guide
13. TM Gold says:

    December 18, 2014 at 1:00 pm

     

    I will do a brute force calculation just to make it super clear. Let us label the five stats as HABCDS (because that’s the Japanese convention and I’m used to it) and say we want 31 in all stats other than A.

     

    Lets say the three guaranteed 31 falls on H, B, and C. By brute force, the number of configurations that give us 31-x-31-31-31-31 is (# of choices for A ranging from 0 to 31) * (# of choices for D = 31) * (# of choices for S = 31). Obviously the first term equals 32 while the other two equals 1 so there are 32 possible configurations. The total number of configurations is 32^3 so the probability of getting an acceptable configuration is 32/32^3 = 1/32^2 so the probability of getting 31-x-31-31-31-31 given that the three guaranteed 31s lie on HBC is 1/32^2. For this even to occur you need to multiply by the probability that the three guaranteed 31s lie on HBC which is 1/(6 choose 3), because these are independent events.

     

    Now the problem is exactly the same for any other choice of three guaranteed 31s landing on desired stats. The total number of configurations is 32^3 and the number of acceptable configurations is 32 hence the probability of landing an acceptable configuration is 1/32^2. Multiplied by the probability of getting the three guaranteed 31s to land on where you had them is 1/(6 choose 3). You know sum over all acceptable choices of guaranteed 31s of which there are 5 choose 3. Therefore the total probability is (5 choose 3)* ((1 / (6 choose 3)) * (1 / 32^2)) = ((5 choose 3)/(6 choose 3))*(1/32^2) = 1/2048.

    ^ This is exactly correct  without factoring in nature
14. VenBombadil says:

    December 18, 2014 at 2:18 pm

    Ok guys – I can see now both of your points and I stand corrected indeed. Apologies for being a bit reckless on this and not giving the calculations some more thought. 
15. Missed says:

    December 18, 2014 at 3:12 pm

    Thundurus Timid should be 144 speed
16. dodgson says:

    December 24, 2014 at 9:50 am

    I believe Timid Heatran only reaches 106 speed (not 126).
17. Bort says:

    December 26, 2014 at 9:48 am

    I’m SRing for Latios in the story…Ugh. Where is the closest point I can save? I’m currently SRing from one step away from triggering the cutscene on Route 118. Can I do a SR point on Southern Island instead?
18. Werford says:

    December 26, 2014 at 10:04 am

    You have a couple of steps after you walk into the clearing before the cutscene triggers. You can save there.
19. Bort says:

    December 26, 2014 at 10:33 am

    Thank you! I’ve just caught this Timid Latios with the ‘Capable of taking hits’ characteristic:

    HP 95
    Atk 56
    Def 62
    Sp.Atk 92
    SpDef 80
    Spd 88
20. Bort says:

    December 26, 2014 at 12:45 pm

    Something to note for IV checking when you don’t have access to the IV judge (for the cosplay pikachu and the lati twins on southern island) is to confirm that you have the stats shown, and then use the super training to check the IVs. For pikachu, it takes a few EVs, but lati twins only need 12 EVs to confirm 30 or 31. When you get one with all the flawless stats, EV train it with 12 EVs. If the stat increases, it is 31, else it is 30. Don’t do on all stats, and the one you know is flawless (from characteristics)

    Could you explain this further please? I want to know how I can check the Latios in my previous post.
21. Bort says:

    December 26, 2014 at 4:04 pm

    Hm, my topic on the forum has gone missing so I’ll have to carry on here.

    I’m getting a little confused. Why does the Speed stat only raise by 1 after putting 12 Super Training points or EVs into it? I thought 4 EVs = 1 point stat gain.
22. voodoo pimp says:

    December 26, 2014 at 5:11 pm

    Hm, my topic on the forum has gone missing so I’ll have to carry on here.

    I’m getting a little confused. Why does the Speed stat only raise by 1 after putting 12 Super Training points or EVs into it? I thought 4 EVs = 1 point stat gain.

    4 EVs only equal 1 point at level 100.  At lower levels, it takes more EVs between points.
23. Bort says:

    December 27, 2014 at 5:48 am

    I didn’t know that! Is there any way to work the needed EVs out other than levelling to lv50 or 100?
24. voodoo pimp says:

    December 27, 2014 at 1:00 pm

    I didn’t know that! Is there any way to work the needed EVs out other than levelling to lv50 or 100?

    This calculator will tell you how many EVs it needs to narrow down the range.  Or you could put it into Showdown or another teambuilder and play around with the numbers for a while.
25. Bort says:

    December 27, 2014 at 2:30 pm

    This calculator will tell you how many EVs it needs to narrow down the range.  Or you could put it into Showdown or another teambuilder and play around with the numbers for a while.

    Thank you! I’ll have a look with my next Pokemon.
26. keyno says:

    December 28, 2014 at 7:36 am

    I think the Speed Stat of the Timid Heatran is incorrect at level 50 with 31/30 Iv´s it reaches 106 not 126.
27. redtown says:

    December 28, 2014 at 10:34 am

    I think it’s worth bearing in mind that you could settle’ for 30 IVs in 2 stats, as long as you don’t plan on putting more than 200 EVs into that stat.
28. Reinier says:

    December 28, 2014 at 12:06 pm

    I think it’s worth bearing in mind that you could settle’ for 30 IVs in 2 stats, as long as you don’t plan on putting more than 200 EVs into that stat.

    As long as you invest EVs in a stat a 30 iv is inferior to a 31. This is true for  any number of EVs, not some arbitrary cutoff.
29. voodoo pimp says:

    December 28, 2014 at 12:13 pm

    A 30 prevents you from getting the absolute max value with 252 EVs.  At any value lower than that, it doesn’t matter how many EVs you invest, it’s always going to cost you 4 more to hit a stat value than it would otherwise.
30. Tanner Hurley says:

    December 29, 2014 at 9:40 am

    does anyone have a cresselia mirage spot that i can trade with