When breeding, either the parent or the father passes down two of its IVs to the child while the other parent provides 1 more.
Well, you'd want to know the chance of that kind of stuff happening, don't you?
The chance to pass down 1 wanted IV onto the child Pokemon is a 25% chance.
[50% * (1/6)] + [50% * (2/6)]
Meaning, there is a 50% chance that the selected parent with the wanted IV will pass 1 IV, and out of that 50% chance, the IV has a 1/6 chance to be the IV you want. The same applies for the other, when the parent provides 2 IVs, in which case it has a 2/6 chance to pass down the IV you'd like.
Therefore, the average chance to produce a certain favorable IV to pass on from a parent is about 25%.
The chance for a parent with a single favorable IV to be passed on, with another parent with a favorable IV to be passed on is calculated a bit differently. While one parent has a 25% chance to pass down the IV, the other does not. This is because the both parents can't pass down the same IV. This is like a probability problem where you draw two cards and don't put back the first card after picking it. 1/52 x 1/51 = Answer.
25% x 25% would be too easy of an answer... The theoretical answer should be a bit higher than 6.25%. This is because after 1 parent chooses that certain IV, it makes it an easier chance for the other parent to give the other stat. For example...
If the 1 IV giving parent passes down the Speed IV, that means the 2 IV passing parent can't give a Speed IV anymore. Therefore, the chance for them to pass the HP and Attack IV becomes 2/5 instead of the previous 2/6.
A harder answer to figure out would be having a parent pass down two specific IVs and the other pass 1 specific IV. This means you essentially half it down the chance in a way. In the previous equation, the criteria could be met if 1 of the parents passed down two IVs, one being the favored IV and the other parent passing on their favored IV. In this, it has to be that a certain parent passes a certain number of IVs.
50% chance that the right parent picks the right amount of IVs. 1 parent passes the favored 1 IV at the chance of 1/6. Therefore 1/12 percent so far.
50% chance that the right parent picks the right amount of IVs. 1 parent passes the two favored IVs at the chance of at best 1/20. There are 5 IVs left, in the best situation, because this parent could choose the first 2 IVs out of a fresh batch of IVs without the eliminated IV from the other parent. 1/20 x [(1/6) * (1/5)] = 2.51666%. Therefore, the chance of you having a single parent pass down 2 select IVs is 2.5%.
Continuing on from having one select parent with 1 IV and the other select parent with 2 IVs would be 2.5% * (1/12) = .208%... This is unlikely.
I'm usually good on probability, but the probability involved in this is absolutely massive. I do believe it would be easier for my brain to list out every milliionth possibility when breeding, but that'd take too long.
Is there pre-established percentages? Or does no breeder care, because all that matters is luck and determination?
Well, you'd want to know the chance of that kind of stuff happening, don't you?
The chance to pass down 1 wanted IV onto the child Pokemon is a 25% chance.
[50% * (1/6)] + [50% * (2/6)]
Meaning, there is a 50% chance that the selected parent with the wanted IV will pass 1 IV, and out of that 50% chance, the IV has a 1/6 chance to be the IV you want. The same applies for the other, when the parent provides 2 IVs, in which case it has a 2/6 chance to pass down the IV you'd like.
Therefore, the average chance to produce a certain favorable IV to pass on from a parent is about 25%.
The chance for a parent with a single favorable IV to be passed on, with another parent with a favorable IV to be passed on is calculated a bit differently. While one parent has a 25% chance to pass down the IV, the other does not. This is because the both parents can't pass down the same IV. This is like a probability problem where you draw two cards and don't put back the first card after picking it. 1/52 x 1/51 = Answer.
25% x 25% would be too easy of an answer... The theoretical answer should be a bit higher than 6.25%. This is because after 1 parent chooses that certain IV, it makes it an easier chance for the other parent to give the other stat. For example...
If the 1 IV giving parent passes down the Speed IV, that means the 2 IV passing parent can't give a Speed IV anymore. Therefore, the chance for them to pass the HP and Attack IV becomes 2/5 instead of the previous 2/6.
A harder answer to figure out would be having a parent pass down two specific IVs and the other pass 1 specific IV. This means you essentially half it down the chance in a way. In the previous equation, the criteria could be met if 1 of the parents passed down two IVs, one being the favored IV and the other parent passing on their favored IV. In this, it has to be that a certain parent passes a certain number of IVs.
50% chance that the right parent picks the right amount of IVs. 1 parent passes the favored 1 IV at the chance of 1/6. Therefore 1/12 percent so far.
50% chance that the right parent picks the right amount of IVs. 1 parent passes the two favored IVs at the chance of at best 1/20. There are 5 IVs left, in the best situation, because this parent could choose the first 2 IVs out of a fresh batch of IVs without the eliminated IV from the other parent. 1/20 x [(1/6) * (1/5)] = 2.51666%. Therefore, the chance of you having a single parent pass down 2 select IVs is 2.5%.
Continuing on from having one select parent with 1 IV and the other select parent with 2 IVs would be 2.5% * (1/12) = .208%... This is unlikely.
I'm usually good on probability, but the probability involved in this is absolutely massive. I do believe it would be easier for my brain to list out every milliionth possibility when breeding, but that'd take too long.
Is there pre-established percentages? Or does no breeder care, because all that matters is luck and determination?
