None of those Pokemon are in the No Egg group. It has been confirmed that all No Egg group pokemon (that are catchable in the game right now at least) have three 31 IVs if caught in the wild. http://www.serebii.net/pokedex-xy/egg/noeggs.shtml
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*Big* News for breeding in XY.
- Thread starter Reikai
- Start date
Yes. Breeding with Ditto can now pass on Hidden Abilities as long as the parent that isn't Ditto has it.
Does Ditto having its hidden ability give access to hidden abilities of the Pokemon it breeds with? I was breeding a Galvantula with Compound eyes with a Ditto with Imposter, and hatching Swarm Joltic eggs. The Galvantula was from a Friend Safari w/access to Swarm, so I'm not sure if that's a factor.
In BW, having a Ditto as one of the parents made it impossible to have DW offspring. It's possible they didn't change that.
EDIT-- Wait shit Swarm is a DW ability. Just to be sure, what was your Galvantula's gender?
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No, what I'm saying is I'm getting DW Abilities on Pokemon that didn't have their DW Abilities from a DW Ability Ditto. Swarm is Joltik's DW Ability, and somehow my Coumpoundeyes Joltik was hatching Swarm Joltik. My only theory is Ditto having Imposter opens up Pokemon to hatching offspring with their respective DW abilities.
EDIT: Nevermind. It must have to do with that Galvantula coming from a Friend Safari with access to Swarm. Hatched a bunch of Charmander with that same Ditto, they all had Blaze.
EDIT: Nevermind. It must have to do with that Galvantula coming from a Friend Safari with access to Swarm. Hatched a bunch of Charmander with that same Ditto, they all had Blaze.
How many Eggs did you hatch for both pokémon? And what were the parents' genders?
I don't know why they would make Safari pokémon work that way if that's right but hey, the easier the better.
Oh, wow, I'm dumb. Thought the one I bred was the one with Compoundeyes, but apparently I had gotten another Galvantula with Swarm along the way.
Ignore me - this was all a mistake. I could have sworn I'd only caught one Galvantula with Compoundeyes and used it, but I must've forgotten something.
Ignore me - this was all a mistake. I could have sworn I'd only caught one Galvantula with Compoundeyes and used it, but I must've forgotten something.
Ok. Using a flawless ditto (which should not be that hard to obtain, just take some time), I believe the highest chance to get two IVs that you want down the line can be as high as 50%. Here's how I reckon this will work...
First, synchronise something that you want to breed egg moves to lock in a nature, then catch something from the friend safari that you are looking to breed with one of the IVs you are eventually looking for maxed. If you're not breeding egg moves, you can skip this step. Otherwise, breed the thing with the egg move holding an everstone and the parent with a power item (example: find an adamant aipom with baton pass and a torchic with a power anklet in 31 speed. This will create a torchic with adamant, baton pass, and 31 speed 100% of the time).
Then breed that pokemon with an everstone and a flawless ditto and the chance of specific IVs can be calculated using a fixed formula.
Ditto: 31/31/31/31/31/31
Other: x / x / x / x / x /31
This will pass the egg moves and nature 100% of the time, and the chances of getting different desired 31s, assuming the "other's" IV of 31 is a "desired IV", is...
1: 83.3% or 5/6
2: 33.33% or 1/3
3: 12.5% or 1/6
4: 4.16% or 1/24
5: 1.041% or 1/96
The chances of getting a flawless poke from one flawless and another with one 31 IV? 1/3072, or .0325% chance. Low, but a far cry from what it used to be--9*10^-8%, or 1 in a billion chance.
What I can conclude from this is that the chance to get "sweeper" IVs, as in Speed and an offensive stat of your choice, following this method, is 33%. Amazingly high. The chance for "mixed sweeper" or "wall" stats, with maxes in both offenses and Speed or HP and both defenses is still very high at 12.5%.
Why is the first stat different from the last? If you have one flawless poke and something with 31 in speed, the destiny knot will chose five out of six IVs, or 5/6. The chance of it choosing the stat where both IVs are 31 is 5/6, very high! For the next case, the chance of the destiny knot choosing two stats you want is 4/6, which looks like this:
31/31
x /31
The chances of getting double 31s here is 50%. Multiply that by 4/6 and you get 33.3%.
First, synchronise something that you want to breed egg moves to lock in a nature, then catch something from the friend safari that you are looking to breed with one of the IVs you are eventually looking for maxed. If you're not breeding egg moves, you can skip this step. Otherwise, breed the thing with the egg move holding an everstone and the parent with a power item (example: find an adamant aipom with baton pass and a torchic with a power anklet in 31 speed. This will create a torchic with adamant, baton pass, and 31 speed 100% of the time).
Then breed that pokemon with an everstone and a flawless ditto and the chance of specific IVs can be calculated using a fixed formula.
Ditto: 31/31/31/31/31/31
Other: x / x / x / x / x /31
This will pass the egg moves and nature 100% of the time, and the chances of getting different desired 31s, assuming the "other's" IV of 31 is a "desired IV", is...
1: 83.3% or 5/6
2: 33.33% or 1/3
3: 12.5% or 1/6
4: 4.16% or 1/24
5: 1.041% or 1/96
The chances of getting a flawless poke from one flawless and another with one 31 IV? 1/3072, or .0325% chance. Low, but a far cry from what it used to be--9*10^-8%, or 1 in a billion chance.
What I can conclude from this is that the chance to get "sweeper" IVs, as in Speed and an offensive stat of your choice, following this method, is 33%. Amazingly high. The chance for "mixed sweeper" or "wall" stats, with maxes in both offenses and Speed or HP and both defenses is still very high at 12.5%.
Why is the first stat different from the last? If you have one flawless poke and something with 31 in speed, the destiny knot will chose five out of six IVs, or 5/6. The chance of it choosing the stat where both IVs are 31 is 5/6, very high! For the next case, the chance of the destiny knot choosing two stats you want is 4/6, which looks like this:
31/31
x /31
The chances of getting double 31s here is 50%. Multiply that by 4/6 and you get 33.3%.
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Ok, I've found a way to simplify my formula. If you have a Pokemon with IVs in the stats you want and another pokemon with similar IVs, here's a calculation you need to do for the %chance of the baby having ideal IVs...
First, determine how many IVs you are looking to be passed down. One stat is 5/6, two is 4/6, three is 3/6, and so on. Then, create a simple chart with the IVs of the pokemon...
atk1/satk1/spe1
atk2/satk2/spe2
or
31/31/31
x /x /31
The chances of getting all 31's here...50% chance for the first set of IVs to be 31, 50% chance for the second, and 100% chance for the third. Take all those percentages and multiply it by the starting fraction--3/6*.5*.5=12.5% chance. The chance of passing three 31's with one flawless and one garbage, or 31/31/31 and x/x/x is 3/6*.5*.5*.5, or 6.25%. Two 31's and a garbage and two garbages and a 31, or 31/x/31 and x/31/x is the same, 3/6*.5*.5*.5. The most important part is the initial fraction, while everything else is just 50% chances. If you only want to pass one IV, the chance of the Destiny Knot choosing in one of its five passes to be that IV is 5/6. This decreases linearly, which makes it easy to find the chances and thus the ~average number of eggs at time needed.
First, determine how many IVs you are looking to be passed down. One stat is 5/6, two is 4/6, three is 3/6, and so on. Then, create a simple chart with the IVs of the pokemon...
atk1/satk1/spe1
atk2/satk2/spe2
or
31/31/31
x /x /31
The chances of getting all 31's here...50% chance for the first set of IVs to be 31, 50% chance for the second, and 100% chance for the third. Take all those percentages and multiply it by the starting fraction--3/6*.5*.5=12.5% chance. The chance of passing three 31's with one flawless and one garbage, or 31/31/31 and x/x/x is 3/6*.5*.5*.5, or 6.25%. Two 31's and a garbage and two garbages and a 31, or 31/x/31 and x/31/x is the same, 3/6*.5*.5*.5. The most important part is the initial fraction, while everything else is just 50% chances. If you only want to pass one IV, the chance of the Destiny Knot choosing in one of its five passes to be that IV is 5/6. This decreases linearly, which makes it easy to find the chances and thus the ~average number of eggs at time needed.
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And, just a few hours later, I just bred a female Scyther: Technician, Adamant, 31/31/31/x/31/31. It goes so fast now!
I think I am going to use it to try for a shiny one with the Masuda method now. Hopefully one that inherits most of momma's good genes (unfortunately the foreign male Scyther I got is IV crap).
Also something that I've noticed is that if you breed an ordinary Pokémon obtained in-game from one of the various routes with a member of the same species then the baby Pokémon will have a higher chance of obtaining it's first ability.
I noticed this when I was breeding ralts. Whenever I start playing a new gen game I always spend some time breeding synchronizers to use when catching legends and wild Pokémon. In gen 4 &5 it was a pain, ralts would have trace 50% of the time and synchronize 50% of the time. However in gen 6 I've hatched around 25 something eggs so far and only like 2 have had trace as the ability. This might just be luck but I thought I would mention it anyway.
I noticed this when I was breeding ralts. Whenever I start playing a new gen game I always spend some time breeding synchronizers to use when catching legends and wild Pokémon. In gen 4 &5 it was a pain, ralts would have trace 50% of the time and synchronize 50% of the time. However in gen 6 I've hatched around 25 something eggs so far and only like 2 have had trace as the ability. This might just be luck but I thought I would mention it anyway.
So if I'm breeding 2 mons each with 4 perfect IVs to try and get a 31/31/31/x/31/31 spread and the parents have the 5 desired IV's covered between them, what's the % chance of getting the quint-perfect baby?
also Ausgirl - babies have an 80% chance of inheriting the mothers ability.
Is everything set in stone, or only the IVs?
Oh well that explains it then. The mother was adamant with synchronize.
This depends on which IV's they have...it may be a 100% chance or 50% chance for each IV. First off, you're desiring 5 IVs, so you'll take your initial modifier as 1/6. Then make a chart with each IV...I'm assuming it looks something like this
31/x /31/31/31/x
31/31/31/x /x /31
Which would wind up being 1*1/2*1*1/2*1/2*1/2. Multiply this by 1/6 to get 1/96, 1% chance! On average, you'll need to get about 100 eggs for this to work. However, after you get that quint-child, you can replace one of the parents to increase your chances.
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Should I get another Ditto for this breed?
Male - 31/31/31/x/31/x (Destiny Knot)
Ditto - 31/x/x/x/x/31 (Ever stone, might change it to the power item)
I think both are Adamant.
Male - 31/31/31/x/31/x (Destiny Knot)
Ditto - 31/x/x/x/x/31 (Ever stone, might change it to the power item)
I think both are Adamant.
OmegaForte It depends how necessary your special attack is. As neither parent has a good IV for that, the chances of getting a 31 in special attack are abysmal. However, if you're breeding a physical sweeper and are only looking for attack and speed...you've got a 16% chance of that happening.
Yeah just a physical sweeper. Still low, but I'll take my chances. Thanks.
16% is roughly 7 eggs until you get a desired poke, that's hardly low. NOTHING like it used to be.
K then. I'm a person with horrible luck in video games, so I thought 16% was low imo. Good to know it won't take too long.
Just bred a quint-Vivillon using this method. I'm so happy that breeders' Pokemon can now match RNGers'. :D
I think the chances are actually better than 1/96. I'm sorry if I'm wrong, if so please correct me.
Parent 1: 31 / 31 / 31 / x / x / 31
Parent 2: 31 / x / 31 / x / 31/ 31
Desired: 31 / 31 / 31/ x / 31 / 31
You're shooting for 5 perfect IVs and one that doesn't matter. The chance of you NOT inheriting the meaningless IV (which is what you want, so you inherit the others) is 1/6.
1/6
Of the remaining IVs, 3 are shared among the parents, and 2 are not. Those three are guaranteed, but the for the other two the RNG has a 50% chance of selecting the correct parent.
1/6 * (1/2) * (1/2)
RESULT: 1/24 chance (Plus the small off-chance of your desired IV being randomly set to 31 in a failed case)
..which is veeeery feasible if you work up to it. :)
Hex-stats are harder because you need the last stat to be randomly generated, but that's only a concern if you're going mixed (or to reduce confusion/foul play damage :P )
If my math is wrong please feel free to correct, I'm no expert at this.
EDIT: We might actually both be right. Saying that "both parents have the desired IVs covered between them" leaves room for interpretation.
I think the chances are actually better than 1/96. I'm sorry if I'm wrong, if so please correct me.
Parent 1: 31 / 31 / 31 / x / x / 31
Parent 2: 31 / x / 31 / x / 31/ 31
Desired: 31 / 31 / 31/ x / 31 / 31
You're shooting for 5 perfect IVs and one that doesn't matter. The chance of you NOT inheriting the meaningless IV (which is what you want, so you inherit the others) is 1/6.
1/6
Of the remaining IVs, 3 are shared among the parents, and 2 are not. Those three are guaranteed, but the for the other two the RNG has a 50% chance of selecting the correct parent.
1/6 * (1/2) * (1/2)
RESULT: 1/24 chance (Plus the small off-chance of your desired IV being randomly set to 31 in a failed case)
..which is veeeery feasible if you work up to it. :)
Hex-stats are harder because you need the last stat to be randomly generated, but that's only a concern if you're going mixed (or to reduce confusion/foul play damage :P )
If my math is wrong please feel free to correct, I'm no expert at this.
EDIT: We might actually both be right. Saying that "both parents have the desired IVs covered between them" leaves room for interpretation.
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