Breeding Perfect Pokemon in Pokemon X&Y

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You only need Dittos for genderless Pokemon.
I've only been using the dittos to transfer natures to the initial mother, much faster than trying to synchronize for parents each time.

But Smeargle is indeed the ultimate example of why safari dittos aren't as great as many believe, because of his ability to have any egg move combination via sketch.

And being able to spread IVs from one egg group to another is indeed very easy now.

Chingling can spread IVs to the Water 1, grass, mineral and fairy groups. Trevenent (grass) can breed with bulbasaur who in turn can spread to many egg groups via his connections in the monster group.
 
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interesting yet frustrating update
new batch
x/x/x/31/31/31
31/x/x/x/31/31
31/x/x/x/31/31
31/x/x/31/31/31
31/x/x/x/31/31 (how am I getting this set so frequently...)
31/x/x/31/31/31
31/x/x/31/x/31
31/x/x/31/x/31 attack is awful
x/x/x/31/31/31
31/x/x/31/31/31
31/x/x/31/31/31

and finally 1
31/x/31/31/31/31
after 9 hours=/ srsly this was absurd, i spent less time hunting a .05% chance card on rag than this x.x
 
I luckily found a 31/x/31/31/x/31 ditto so Its been very easy breeding special attackers. Just breed the nature and SPD in with other ditto and then I'm pretty much set from there.
 
quick question! if a left a pkm at the daycare and as he levels up one egg move he has previously learned gets overwritten, can he learn it back at the move reminder?
 
This is not directly related to breeding, but i have a question.

if i have a pokemon with cute charm as my 1st pokemon
is there any way to stack that with synchronize?
 
If i put a destiny knot on 1 parent, and a power item on another, what should be the effect? Or is it more productive to do destiny knot+everstone?
 

Agonist

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Power item forces one IV to be inherited, and the other four would be random. Personally, I prefer to use an everstone for the 100% nature.
 

Jimera0

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Wow, this is a much more intelligent way of getting your final flawless Pokemon. Using power items towards the end makes loads more sense. I'll add this final bit to the guide shortly, thank you!
Hold up Roarke, I'm not sure that the probabilities actually calculate out favorably that way. Though I can't figure it out for myself... we never stayed long on probability calculation in my math class and the last time I did any work with probabilities was back in like grade 10 or 9 (so around 6-7 years ago). Anyway, here's my reasoning.

Cuz you have to both get the 1/10 chance of getting your replacement parent and the 1/12 chance of getting the perfect right? You have to hit two outcomes instead of one, and while both of those outcomes might be more likely than the one, getting both is less likely. Except I'm not sure if I'm putting this together right. Actually I'm pretty sure I'm not. It's the difference between rolling a pair of dice and trying to get snake-eyes and trying to roll a single die until you get a one twice. I'm not sure if there's a difference in the probability calculations between the two. The goal here is to figure out on average how many rolls it'd take to get two ones. The first example (using a pair of dice, rolling both simultaneously) definitely uses the simple multiplication method we used to arrive at the other probabilities (in this instance, 1/6 * 1/6 = 1/36). The second one though (which is similar to the breeding scenario in question), I'm pretty sure it doesn't? But I don't remember how you calculate the average number of roles it'd take in that scenario. Does it just add together? That would make it an average of 22 eggs for McGrrs method. But I'm not entirely certain that's the way it works.

Anyone know the math to figure this out? I want to make sure that we got the best method up there. Also, it'd just be good for me to finally get a grip on how to calculate probabilities again XD.
 
Just out of curiosity, what do you guys do to those pokes that you got that had 3-4 IVs, when you were aiming for 5? They are flooding my boxes right now. lol
 
Just out of curiosity, what do you guys do to those pokes that you got that had 3-4 IVs, when you were aiming for 5? They are flooding my boxes right now. lol
Most people probably release them. It's what I've been doing.

Assuming you've gotten that 5 and keep a 4 mate for it. Just incase you want to breed another down the line.
 
Hold up Roarke, I'm not sure that the probabilities actually calculate out favorably that way. Though I can't figure it out for myself... we never stayed long on probability calculation in my math class and the last time I did any work with probabilities was back in like grade 10 or 9 (so around 6-7 years ago). Anyway, here's my reasoning.

Cuz you have to both get the 1/10 chance of getting your replacement parent and the 1/12 chance of getting the perfect right? You have to hit two outcomes instead of one, and while both of those outcomes might be more likely than the one, getting both is less likely. Except I'm not sure if I'm putting this together right. Actually I'm pretty sure I'm not. It's the difference between rolling a pair of dice and trying to get snake-eyes and trying to roll a single die until you get a one twice. I'm not sure if there's a difference in the probability calculations between the two. The goal here is to figure out on average how many rolls it'd take to get two ones. The first example (using a pair of dice, rolling both simultaneously) definitely uses the simple multiplication method we used to arrive at the other probabilities (in this instance, 1/6 * 1/6 = 1/36). The second one though (which is similar to the breeding scenario in question), I'm pretty sure it doesn't? But I don't remember how you calculate the average number of roles it'd take in that scenario. Does it just add together? That would make it an average of 22 eggs for McGrrs method. But I'm not entirely certain that's the way it works.

Anyone know the math to figure this out? I want to make sure that we got the best method up there. Also, it'd just be good for me to finally get a grip on how to calculate probabilities again XD.
I didnt want to think too hard on this and I didn't take it too seriously to get out a piece of paper to make sure it was correct, but i think he was unclear in saying that he would use an everstone on the 31/x/x/31/31/31(call it parent A) with the newly bred 31/x/31/31/31/31 parent (call it parent B).

parent A i would assume had the original correct nature, while parent B does not, but since everstone negates that by putting everstone on parent A and destiny knot on parent B. So at this point he's corrrect in saying he has a 1/12 chance (1/6 * 1/2)

He then says once he has two 31/x/31/31/31/31 parents he has a 1/6 chance to make a new one.

I believe his calculations are correct UP UNTIL the point where you start calculating Gender rates, which actually do matter. It gets a bit more complicated when you start including gender and it would also depend on which gender you managed to find first, and i would need a pen and pencil and calculator to find out the exact rates.

The reason You are incorrect in using the simple multiplication method is because we do not have to multiply the 1/10 and 1/12 for EACH Scenario of the 10 (hard to explain in words at 2am) im saying there are 10 different scenarios that can happen, but if you get the first 9 unfavorable scenarios, you are not using those scenarios to start breeding for the next stage of breeding (1/12 probability scenarios)

you would wait for that 1/10 scenario to occur in which you can then start searching for the 1/12 scenario. A better estimate would be to add the 10 and 12 together and say your probability is 1/22 which is ABSOLUTELY INCORRECT, but it would be more accurate then multiplying 1/10 and 1/12 which is 1/120. I would need a pen and paper, and still probably get the correct probability incorrect, just cause it starts getting confusing.

I dunno if any of this made sense, im sure it wasnt formatted or worded well, because im tired and kind of rambling, but i hope it helps
 
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I'm breed Honedge's right now, I have a mother and father both with 31 HP, Def and Spe. I'm trying to get a 31 Atk and Sp.Def as well, what would be the best way to do so? Keep breeding with the two Honedge's without a destiny knot or is there an alternative?
 
Just out of curiosity, what do you guys do to those pokes that you got that had 3-4 IVs, when you were aiming for 5? They are flooding my boxes right now. lol
I had an entire box full of 3-4 IV squirtles that I wonder traded away, got a few gems back that helped fill up my pokedex and will help for new breeding projects. It also helps fill up them pokemiles.
 

TheMantyke

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I have hatched 43 eggs using 2 froakies that have 31/X/31/31/X/31 and 31/X/31/X/31/31 Iv's and I can't seem to get a 5 Iv'ed froakie.
If you're having a lot of difficulty with this, consider giving the female froakie a Power Item (Power Lens if it's the first one, Power Band if it's the second) and have your Male Froakie hold the Destiny Knot. Now, you won't be aiming for a final product, but a 31/x/31/31/31/31 froakie with the wrong nature, which has a significantly lower 1/10 chance of being bred. Once you have that flawless parent, give it the Destiny Knot and the Froakie of the opposite gender with the right nature and everstone. Your chances have been cut in half from 1 in 24 to 1 in 12.
 
I am breeding synchronizers to use to catch parents in the safari with. So far, I have timid, jolly, adamant and modest for the dedicated ATK/SPA sweepers and calm, careful, impish and bold for my bulky pokemon. I am not sure what 4 natures are best for mixed sweepers though, when I check the B/W sets on mixed sweepers I see a ton of different natures. Any recommendations on what the top four natures for mixed sweepers are?
 
I'm seriously having an issue here, I've breed 40+ Honedge's and I'm having no luck getting a 31 IV in either Atk or Sp.Def.

I'm breeding with just an everstone, nothing else. Am I doing something wrong, or am I just being impatient and unlucky?

The chance of me getting a 31 in either stat is less than 3% with how inheritance works, though I would have though I'd have got at least one by now?
 

TheMantyke

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I am breeding synchronizers to use to catch parents in the safari with. So far, I have timid, jolly, adamant and modest for the dedicated ATK/SPA sweepers and calm, careful, impish and bold for my bulky pokemon. I am not sure what 4 natures are best for mixed sweepers though, when I check the B/W sets on mixed sweepers I see a ton of different natures. Any recommendations on what the top four natures for mixed sweepers are?
Mixed sweepers traditionally enjoy +Atk, +SpAtk, or +Speed / -SpDef natures. Naughty, Rash, and Naive are probably what you should look out for.
I'm seriously having an issue here, I've breed 40+ Honedge's and I'm having no luck getting a 31 IV in either Atk or Sp.Def.

I'm breeding with just an everstone, nothing else. Am I doing something wrong, or am I just being impatient and unlucky?

The chance of me getting a 31 in either stat is less than 3% with how inheritance works, though I would have though I'd have got at least one by now?
There's your problem, lol. Use the Destiny Knot too and be sure to follow the IV chaining strategy I've laid out in the guide.
 
If you're having a lot of difficulty with this, consider giving the female froakie a Power Item (Power Lens if it's the first one, Power Band if it's the second) and have your Male Froakie hold the Destiny Knot. Now, you won't be aiming for a final product, but a 31/x/31/31/31/31 froakie with the wrong nature, which has a significantly lower 1/10 chance of being bred. Once you have that flawless parent, give it the Destiny Knot and the Froakie of the opposite gender with the right nature and everstone. Your chances have been cut in half from 1 in 24 to 1 in 12.
Ok thanks.
 
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