Indeed. That is maybe another possible use for a calculator: helping people understand that when you are dealing with poss hets, you very quickly get into really small odds of proving out.
If the gene is important to you it’s far better to get a visual, or at the very least a 100% het from someone you trust.
And 100% of the combos you want to be female are male, yup.
I know this thread is probably done, but if a recessive trait is valuable enough they will track the 25%s and such. Albino gtps were the last ones I remember seeing. So it does happen in an acceptable way, but rarely and I’ve never seen it with bps. But if you’re dealing with animals like that, you shouldn’t need a calculator to tell you what’s up.
With that said unless it’s that rare case, it would be seen as a marketing ploy and perceived as a bad thing to even list.
As far as the calc is concerned, I think it would be more clutter than useful and potentially confusing depending on the typos someone types in. I can’t see it being a good thing to put in.
Hey, I already anted in a plethora LOL
I am sorry if this has been answered but I can not find it. Is there a way to input a percentage in the boa calculator. For example a Motley that is 66% (any morph) bred to a Normal 50% (any morph). How to do these percentages. Thank you
Welcome @zebtax,
I’ve moved your post here as it is the same question but for a different species 
enjoy!
Im sure my post will be merged here as i didnt see this topic when i made mine, but id like to chime in, usually i find myself ageeeing with wyman on many topics, but on this one i believe it isnt as cut and dry as he states.
When you allow for “possible hets” you open the market to probability. In reality, the animal is either het or it isnt. So to claim 66% and 50% het exist while saying 33% and 25% dont doesnt make sense to me when you are speaking of probabilities.
Lets use the cube example. I have 3 boxes. The first 2 boxes contain 1 red and 1 green cube. These are the " grand parents ". Now i take 1 cube from each box, and put them in the 3rd box. Now there are 3 possible outcomes 2 red, 2 green, or 1 of each.
Now, lets assume that if the boxes will change to the color of the cube if it is 2 red cubes, but if its 2 green or 1 of each, the box doesnt change. The box has not changed color, this means you cannot have 2 reds in the box. Now, mathematically this means there is a 66% chance of there being 1 of each cube, and a 33% chance of it being 2 green cubes. You have no way of knowing anything more without opening the box, or repeating the experiment with more boxes and seeing if the new box changes color.
Now lets say you now use this box in this way, but the second box now contains 2 green cubes. You take 1 cube from each “parent” box, and put them into the “offspring box” since you have know the contents of only 1 box, you know there will be at least 1 green cube, but you dont know the second boxes contents, but you do know there is a 66% chance of there being a red cube in it. This means the “offspring box” mathematically has a 33% chance of having a red cube in it, or half the total probability of the “parent box”.
Mathematically and genetically all of this is true about the probability of the outcome of subsequent breeding. Just because the parent has unknown genetics doesnt negate the possibilities.
You either allow for probabilities, in which case, it follows mathmatical logic (modified once possibilities are proven which would update the problem) or you say an animal is het or not, and dont allow for the possibilies. You cannot do both. Ill try to write out the equation if i can figure out how to get a good formula that makes sense.
This is the easiest way to write it out. The 33% GG posibility is already factored in by saying its a 66% chance for the RG edit: i botched it a little bit for explanations sake. Ill fix it tomorrow. The idea is there, but its not a perfect proof because i skipped some steps.
Edited to add the more accurate/complete version for maximum clarity 
