Part II of the Breeding Guide

Peterko

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#51
actually, even if it worked, people should not care about average total IVs but max individual IVs...you´ll get better results with the numbers in part III, anyway

@suicune137: is there any difference between the parent you switched and the one you used before? I don´t know, ID, which generation, which game etc., I´ve started with another pkmn, too, and I don´t think I´m at 230 yet after like 15 eggs, though I´m using the running method and get an egg every 765 steps (it´s an internal step counter 20 pkmn)
 
#52
@suicune137: is there any difference between the parent you switched and the one you used before? I don´t know, ID, which generation, which game etc., I´ve started with another pkmn, too, and I don´t think I´m at 230 yet after like 15 eggs, though I´m using the running method and get an egg every 765 steps (it´s an internal step counter 20 pkmn)
I actually got 36 eggs before I changed the parents since I could do IV battle at least three times. Thought I posted that but can't find it in my post so sorry. They all hatched at 2805 steps.

I picked the best parent out of those and it was the same ID tangela (obviously) and it was the second generation. As soon as i got the egg, it hatched in 2530 steps. My father, Roselia, remained the same.
 

Tangerine

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#53
I did some breeding of my own to test this claim.

Raw Data: 25 Magikarps bred from 2 Perfect Timid Parents. Male held the Lucky Egg, Female held the Everstone.

Timid 28/31/12/31/4/31 - 137
Rash 9/31/31/31/2/27 - 131
Lonely - 20/31/31/23/28/0 - 142
Timid - 29/31/22/31/17/31 - 161
Timid - 31/31/31/19/11/24 - 147
Timid - 31/27/31/31/1/31 - 152
Timid - 0/31/2/31/16/31 - 111
Serious - 29/31/11/7/31/11 - 120
Rash - 21/21/13/26/31/31 - 143
Timid - 31/31/29/31/2/10 - 134
Timid - 31/31/0/31/17/1 - 111
Timid- 31/31/31/15/5/1 - 114
Sassy - 18/31/30/31/18/12 - 140
Quirky - 12/2/7/31/31/31 - 114
Modest - 31/31/25/31/14/4 - 136
Timid - 26/31/3/31/2/2 - 95
Timid - 3/15/14/31/24/31 - 121
Timid - 18/31/31/31/23/10 - 144
Serious - 30/31/1/31/8/31 - 132
Impish - 11/11/31/9/26/31 - 119
Timid - 22/12/2/31/31/31 - 129
Timid - 5/2/31/13/31/31 - 113
Hasty - 12/15/22/31/26/31 - 137
Timid - 31/19/24/9/31/31 - 145
Timid - 3/31/14/30/27/31 - 136

Expected Total based on the Research in this topic - 135.4917
Mean of Sample: 130.56
Standard Deviation: 15.5485

I would say myth debunked.
 
#54
Please can I refer you to this? My friend and I, LingFungJPN have been writing THE, FAQ regarding OUR "Discovery" (please note the word "speculative" in the title), in this case I would say that our FAQ (incomplete as yet, due to the inordinate amount of Batch tables we need to format from 6 months of breeding and the best part of 2000 eggs, EACH). "Myth debunked"? No - not finished publishing our results yet - we are still writing the Speculative FAQ, however our results have been finished for months - we are continuing to update this particular FAQ with every section of it we finish explaining.

http://boards.gamefaqs.com/gfaqs/genmessage.php?board=924664&topic=36645671

Please, read at your own discretion, and not post the link anywhere else. We've had some trouble with "new users" in the past.

Lastly, 25 Magikarp eggs hatched, with parents holding a Lucky egg, is exactly what we were trying to avoid when we first started publishing sections of the FAQ in the first place - we wanted people to read through our results, make up their minds, then try all of the methods we used in refinement of the IVs for the hatched pokemon overall. Not that I mind, or that LingFungJPN minds, but we wanted everyone to read it through carefully, and at their own discretion.

Thank you for your time.
 

X-Act

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#55
I added a few Ditto, Slugma and Everstone locations, thanks to Paperfairy.

@simierskiUK: I've looked at that link, but I have difficulty in following exactly what you are researching. Also, in which game are you breeding those Larvitars? Because many of those Larvitars don't even share one IV with any of the parents, which is either wrong, or a discovery in itself.
 
#56
This is my first time every posting here, but I found this guide very useful. I don't know if this is a dumb question or not I may have skipped something in the guide about it. Is there any way to accurately determine the IVs of level 1 Pokemon that hatch from the eggs in Diamond/Pearl.
 

david stone

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#57
You could level them up with a bunch of rare candies, or get into a level 100 battle, but aside from that, no. Too many IVs share the same stat.
 

Tangerine

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#58
This is my first time every posting here, but I found this guide very useful. I don't know if this is a dumb question or not I may have skipped something in the guide about it. Is there any way to accurately determine the IVs of level 1 Pokemon that hatch from the eggs in Diamond/Pearl.
http://www.smogon.com/forums/showthread.php?t=24032

Some base stats you can tell at level 1, some you are not able to, but by level 5 you should be able to tell if it's any good or not.
 

X-Act

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#60
We also listed various methods of finding the IVs of your Pokemon, as well as the maximum stats of all the baby Pokemon at Level 5 in the breeding guide.
 
#61
I started the using a synchronizer to catch alot of Ditto over the weekend. I just wanted to say that since I only have Diamond/Pearl to work with that I found it easier to catch one Ralts and then catch alot of Ditto (Ditto just seemed less rare and I was able to get huge Ditto streaks), and then breed the Ralts with the Ditto with Everstone to get the Ralts of the nature I wanted.
 

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#62
Yes, Ditto is not that rare in Diamond and Pearl, if you know what you're doing (i.e. use chaining with the Pokeradar).
 
#63
Erm, in regards to the Egg glitch/trick (Where you can preview the type/nature/ability of an unhatched egg) - there is also a no-release version.

Instead of releasing a pokemon, put the Egg in a PC slot that has a pokemon in the same position before or behind it (Say, Egg in Box 1, Slot 1, and any pokemon in Box 2, Slot 1)

Put the cursor hand over the pokemon. As the preview menu in the lower right hand corner rolls over, switch with the L or R button to the box with the Egg. You'll see the Egg's information instead of the Pokemon, if done correctly.

I'd also like to add my greatest appreciation for the Breeding Guide in all it's parts, particularly the probabilities. *bows*
 
#64
I finally get it! Thanks for this guide...I realized what I was doing wrong when I was breeding for my Porygons, I was breeding with a Japanese Porygon-Z with my Ditto, which is why my Everstone didn't work..
 
#65
Excellent guide, very easy to understand, but I have just one question.

Using the guide's example of obtaining a Timid Gastly with 31 Speed IV, wouldn't it be much faster if you chain-breeded through two Ditto (one with 31 Speed IV and the second w/ Timid Nature + an Everstone), instead of one Ditto with both the desired nature and IV? Wouldn't that save you the time of having to Synchronize and catch six boxes worth of Ditto? Pls correct me if I'm wrong.
 

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#66
EDIT: Okay I understand your question better now.

So you're saying, you first breed a Gastly with a 31 Speed IV Ditto, getting a Gastly with 31 Speed IV within 4 eggs. Then you breed this 31 Speed IV Gastly with a Timid Ditto with Everstone, getting a Timid Gastly with 31 Speed IV within 8 eggs. So you'll need to breed 12 eggs in all using this approach.

It works, but having a Timid Ditto with 31 Speed IV reduces the number of eggs you need to hatch for your Gastly to 8 eggs.

Your method though lets you search for Dittos quicker (since it's easier to find a Timid Ditto with any IVs and another Ditto with 31 Speed IV, rather than to find one Ditto having both) so you get the tradeoff there. You'll just need to breed more eggs afterwards.
 

david stone

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#67
Yeah, I think the guide is looking at it from a perspective of "If you're doing a lot of breeding, having a Jolly 31 Speed IV Ditto is helpful, so you should get one of those then get your Gastly", whereas you are thinking "I want a max Speed Gastly as fast as possible, and I do not have my Dittos".
 
#68
first post!

Hi there.
Now correct me if i'm wrong but your math is incorrect.
I'm referring to the 170 dittos required for 6 dittos, each one with a differing maxed IV.

the chance of finding a ditto with any given iv of 31 is, as you say, INITIALLY 17.3%. to have a 99% chance of finding that ditto though, with that percentage, you need to catch 25 dittos.
now then, because one stat is already cared for, there are only 5 remaining stats left to be caught, reducing the 17,3% to 14,6% and increasing number caught required to 30.

follow this up and you get.

4 stats, 12%, 36 dittos.
3 stats, 9%, 49 dittos.
2 stats, 6%, 75 dittos.
last stat, 3%, 152 dittos.

add it all up and you need around 367 dittos.
which is a bit of a drag if you ask me ;)
 

david stone

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#69
You can have 0-32 in each IV. Assuming random distribution of IVs, that's a 1/32 chance of any given IV being perfect. The probability of getting a perfect IV in at least one stat, therefore, is

1 - (31/32)^6 = 17.34477868%

(1 minus the probability of getting imperfect IVs in every stat)

This works out to 17.3% chance of getting at least one perfect IV. However, this probability includes the probability of getting 1, 2, 3, 4, 5, or 6 perfect IVs, which is where a simple multiplication doesn't fit. The distribution for these is

1 perfect IV: (1/32) * (31/32)^5 * 6 = 15.99778477%

2 perfect IVs: (1/32)^2 * (31/32)^4 * (6 * 5) / 2 = 1.290143933%

3 perfect IVs: (1/32)^3 * (31/32)^3 * (6 * 5 * 4) / (3 * 2) = .05549006164%

4 perfect IVs: (1/32)^4 * (31/32)^2 * (6 * 5) / 2 = .001342501491%

5 perfect IVs: (1/32)^5 * (31/32) * 6 = .00001732259989%

6 perfect IVs: (1/32)^6 = .0000000009313225746%, or 9.313 * 10^-8 %

The last two probabilities are almost entirely negligible. And to double check, adding all of those up does indeed get 17.3%. So the number of Dittos required to get a 31 IV for each stat is lower than you calculate, because some Ditto can have more than one 31 IV stat.

To find the actual probability, you'd need to find the probability of six Dittos at 1 perfect IV each, OR five Dittos of 1 and one Ditto of 2, OR... Then there's also the fact that finding a 4 flawless Ditto when you already have 5 1 flawless Dittos is the same as finding a 1 flawless Ditto.
 
#70
I can't quite tell who your response is directed at.

the 17.3% is a best case scenario because all it requires it that at least ONE of the IVs is maxed. anything above 1 is a bonus.

the way you want to calculate it simply breaks down the 17.3% into its constituents. the logic applied above still applies, you still need to calculate how many dittos per case per IVs remaining need to be caught.
With your method you'd end up having to catch more, because now you're going for 2 IV maxed dittos, 3 IV maxed dittos, etc. you can't really suggest that with a lower chance (15.9% <-> 17,3%) you're going to need fewer ditto to be successful?

I'd prefer to catch two dittos, each with a differing matched stat and try hatch a 2 IV maxed ditto.

Finally:

"To find the actual probability, you'd need to find the probability of six Dittos at 1 perfect IV each, OR five Dittos of 1 and one Ditto of 2"

no, because the 17.3% expects at least 1. anything with more than 1 can be considered a bonus.

"Then there's also the fact that finding a 4 flawless Ditto when you already have 5 1 flawless Dittos is the same as finding a 1 flawless Ditto."

The chances of winning the lottery is 50:50, either you do or you don't. no, wait, that isn't right...
statistics has no memory.

chance of finding a ditto with perfect speed, for example, is roughly 3%.
chance of finding a ditto with 4 perfect random IVs=

6!/4!(6-4)! * 1/32^4 * 31/32^2 = 0,000013425

which comes works out at 0,001343%.

which is greatly smaller than 3%!
 

X-Act

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#71
first post!

Hi there.
Now correct me if i'm wrong but your math is incorrect.
I'm referring to the 170 dittos required for 6 dittos, each one with a differing maxed IV.

the chance of finding a ditto with any given iv of 31 is, as you say, INITIALLY 17.3%. to have a 99% chance of finding that ditto though, with that percentage, you need to catch 25 dittos.
now then, because one stat is already cared for, there are only 5 remaining stats left to be caught, reducing the 17,3% to 14,6% and increasing number caught required to 30.

follow this up and you get.

4 stats, 12%, 36 dittos.
3 stats, 9%, 49 dittos.
2 stats, 6%, 75 dittos.
last stat, 3%, 152 dittos.

add it all up and you need around 367 dittos.
which is a bit of a drag if you ask me ;)
Hello there. Welcome to Smogon.

Okay, let's ask the question differently. How many dice rolls should be thrown so that, on average, all the numbers from 1 to 6 appear?

The first dice roll has a guarantee of producing a number from 1 to 6.

The second dice roll has a 5/6 chance of producing a number differing from the first one. That means that, on average, we'll need 6/5 dice rolls. (5 times out of 6, one dice roll is enough; 1/6 x 5/6, two dice rolls are enough; 1/6 x 1/6 x 5/6, three dice rolls are enough, etc. So we get 5/6 x 1 + 5/6 x 1/6 x 2 + 5/6 x (1/6)^2 x 3 + ... Adding these up to infinity gives us 6/5).

The third number requires 6/4 dice rolls on average to appear, working similarly. The fourth, fifth and sixth numbers require 6/3, 6/2 and 6/1 numbers to appear, on average. The answer, then, is 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7 dice rolls.

Now this problem is very similar to ours. We need the average number of Ditto we need to catch to get one with 31 HP IV, one with 31 Atk IV, one with 31 Def IV, one with 31 SpAtk IV, one with 31 SpDef IV and one with 31 Spd IV. So there are six Dittos we want to catch, just like there are 6 numbers in the dice we want to roll. The only difference is that the probability of getting a Ditto with 31 IV in any stat is 1 - (31/32)^6 = 0.173, not 1. Adding the fact that we want the Ditto to have the same nature as our synchronizer (50%) means that the probability of getting a Ditto with 31 IV in any stat with the appropriate nature is 0.173 x 0.5 = 0.0865.

This means that getting the first Ditto requires 1/0.0865 caught Dittos on average (11.56). This contrasts with the one dice roll needed to produce the first number.

After catching another 11.56 Dittos on average, you'll get another good Ditto, but this time it could have the 31 IV in the same place as the first one. As we said before, in the dice analogy, this number must be multiplied by 6/5 to get the Ditto having a 31 IV in a different place as the first one.

Similarly for the third, fourth, fifth and sixth Dittos, we multiply the 11.56 by 6/4, 6/3, 6/2 and 6/1 respectively. Thus, the number of Dittos required is

(1 / (2 x (1 - (31/32)^6))) x (6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1) = 169.5 Ditto, which I rounded to 170 Dittos.

I didn't want to include this calculation in the guide, but, since you asked from where I got that number, here it is.

Also, so that I'm 100% sure of my calculations, I made a program that simulated Ditto catches and calculated how many catches are required to get all 6 Dittos 100,000 times, and found the average value. The answer was very near 170, thus confirming my working.

EDIT: Here is the code in Java:
Code:
public class Test {
  public Test() {
    int n=0,c=0,d=0;
    long a=0;
    boolean[] already = new boolean[6];
    for (int j=0; j<100000; j++) {
      for (int i=0; i<6; i++) already[i]=false;
      while (d<6) {
        int IV1 = (int) (Math.random()*32);
        int IV2 = (int) (Math.random()*32);
        int IV3 = (int) (Math.random()*32);
        int IV4 = (int) (Math.random()*32);
        int IV5 = (int) (Math.random()*32);
        int IV6 = (int) (Math.random()*32);
        int ab = (int) (Math.random()*2);
        n=-1;
        if (IV1==31 && ab==1) n=0;
        if (IV2==31 && ab==1) n=1;
        if (IV3==31 && ab==1) n=2;
        if (IV4==31 && ab==1) n=3;
        if (IV5==31 && ab==1) n=4;
        if (IV6==31 && ab==1) n=5;
        if (n>=0 && !already[n]) {
          already[n]=true;
          d++;
        }
        c++;
      }
      a += c;
      c=d=0;
    }
    System.out.println((double) a/100000);
  }
  public static void main(String args[]) {
    new Test();
  }
}
 
#72
The second dice roll has a 5/6 chance of producing a number differing from the first one. That means that, on average, we'll need 6/5 dice rolls.
you've lost me here already. if i want to hit a number other than the one I already hit then 5/6 = 0.83% of doing so. not hitting = 0.17%, chance of this happening at 1% (therefore 99% that the 0.83% will occur) is log0.01/log0.17 = 2.51 -> 3 rolls. 5/6 gives 1.2.

Also, so that I'm 100% sure of my calculations, I made a program that simulated Ditto catches and calculated how many catches are required to get all 6 Dittos 100,000 times, and found the average value. The answer was very near 170, thus confirming my working.
I don't mean to be rude but if one of your base assumptions is wrong and you build that into a program that then in turn validates your false assumption you can't state that you're right. circular argument and all.
 

X-Act

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#73
you've lost me here already. if i want to hit a number other than the one I already hit then 5/6 = 0.83% of doing so. not hitting = 0.17%, chance of this happening at 1% (therefore 99% that the 0.83% will occur) is log0.01/log0.17 = 2.51 -> 3 rolls. 5/6 gives 1.2.
Well, to be honest, I'm not understanding your working, either.
I don't mean to be rude but if one of your base assumptions is wrong and you build that into a program that then in turn validates your false assumption you can't state that you're right. circular argument and all.
If you had looked at the code (which, presumably, you didn't), you would have noticed that the program does not use my working at all. It just generates 6 random IVs, a random nature (either 0, meaning different from the one of the synchronize Pokemon, or 1, meaning the same as that of the synchronize Pokemon), and checks whether there is a 31 in one of the IVs and if the ability is 1. It continues to do this until six Dittos, each with a 31 IV in every one of the stats, all of which of the same nature of the synchonizer, are caught, and takes note of how many Dittos were caught in order to get to these six Dittos. Then I repeated the procedure 100,000 times, and found the average number of Dittos required to be caught. The program finds the average amount of Dittos empirically, not mathematically. It always produced a number between 169 and 171, confirming my mathematical working.
 
#74
Well, to be honest, I'm not understanding your working, either.

If you had looked at the code (which, presumably, you didn't), you would have noticed that the program does not use my working at all. It just generates 6 random IVs, a random nature (either 0, meaning different from the one of the synchronize Pokemon, or 1, meaning the same as that of the synchronize Pokemon), and checks whether there is a 31 in one of the IVs and if the ability is 1. It continues to do this until six Dittos, each with a 31 IV in every one of the stats, all of which of the same nature of the synchonizer, are caught, and takes note of how many Dittos were caught in order to get to these six Dittos. Then I repeated the procedure 100,000 times, and found the average number of Dittos required to be caught. The program finds the average amount of Dittos empirically, not mathematically. It always produced a number between 169 and 171, confirming my mathematical working.
I did review your code but the leap from my basic C knowledge to JAVA was a bit too much in 2 lines of code. Also i'm not using the natures in the equation to keep it simple.

ok, my train of thought is this:

you have, lets say, a 50% chance of hitting heads or tails, right?
how many times do you have to flip a coin to ensure that you will definitely hit, say, tails? It isn't two, because stats don't add up like that. And definitely doesn't work either in stats because 100% doesn't exist. In school we always agreed on 99% being acceptable.

SO, chance of hitting tails is 50%. chance of not hitting tails, i.e. heads, is 50%.

0.5 to the power of x equals 0.01 (which is the opposite of 99%, 1-0.99).

log0.01/log0.5 = 6.64. because you can't throw a coin .64 times, 7 is the answer.

I use the same principle when calculating my odds. chance of hitting 31 in any given stat. how many catches required to hit 99% success.
then the same again. chance of hitting 31 in the remaining stats, because one stat is already cared for.

p.s.: I don't bother with natures because the prospect of having to catch 21 different natured abras, leveling them to 20 only to use them to try capture different dittos is a bit silly. heck, if i have to catch 367 dittos i'm sure all the natures will drop once ;)
 

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#75
Okay, I finally start to understand a bit what you're saying.

Let's continue with your analogy. How many times do you have to flip a coin to get tails? You could flip a coin and get heads forever, and you could flip it once and get tails immediately. However, the average amount of flips needed is definitely not infinite.

You are finding the number of flips required so that you are 99% sure of getting a tails, and your result is 7. Indeed, (1/2)^7 < 0.01 and (1/2)^6 > 0.01.

The average amount of coinflips required, however, is not 7, but much less.

You have:
1/2 chance of requiring one coinflip;
1/2 x 1/2 = (1/2)^2 chance of requiring two;
1/2 x 1/2 x 1/2 = (1/2)^3 chance of requiring three;
1/2 x 1/2 x 1/2 x 1/2 = (1/2)^4 chance of requiring four;

and so on. What is the average amount of coinflips required? The average is:

S = (1/2 x 1) + ((1/2)^2 x 2) + ((1/2)^3 x 3) + ((1/2)^4 x 4) + ...

This summation continues forever, but we can still add it. Multiplying everything by (1/2), we get

(1/2)S = ((1/2)^2 x 1) + ((1/2)^3 x 2) + ((1/2)^4 x 3) + ((1/2)^5 x 4) + ...

Subtracting the second equation from the first one, we get

S - (1/2)S = (1/2 x 1) + ((1/2)^2 x 1) + ((1/2)^3 x 1) + ((1/2)^4 x 1) + ...

(1/2)S = (1/2) + (1/2)^2 + (1/2)^3 + (1/2)^4 + ...

The right hand side is an infinite geometric progression, whose sum to infinity is (1/2) / (1 - (1/2)) = 1. Hence,

(1/2)S = 1

and hence S = 2. This means that the average number of coinflips required to get a tails is 2, and not 7.

You can actually try this out yourself. Flip a coin until you get tails, and count how many flips you did before you get tails. Repeat this for a number of times and find your average value. You will find that the average amount of flips required should be approximately 2.

Your answer of 7 means that 99% of the time, you'll need 7 or less flips to get a tails. It doesn't mean that you need 7 flips on average to get a tails.

Similarly with the Ditto thing, your 99% result means that you'll need to catch that amount of Dittos or less 99% of the time. But it doesn't mean that is the average amount of Ditto you need to catch. Barring you are one of the unlucky 1%, you'll need much less than that.

Hope I was clearer now.
 

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