Possible Het Input to Morph Calculator [#1189]

To go along with how I’m understanding your “trick” coin analogy: 2 consecutive coin flips. One with a heads/tails coin. The second flip is with either another heads/tails coin or with a tails/tails coin. If the first flip is heads, we get the fair coin, if the first flip is tails we get the “trick” coin. Some honest referee is flipping the coins as prescribed, but we can’t see the results of the flips.

The first flip is our 100% het bred to a normal. The second is the 50% pos het to another normal.

At the end of the second flip in the game above, what is the probability that the result of that flip is heads? Is that impossible to quantify?

For that second flip the conditional probability of Heads given the second coin is the “trick” coin is 0%
Also, for the second flip the conditional probability of Heads given the fair coin is 50%
But we are interested simply in “what is the probability the second flip is heads.”

We don’t know the exact condition we are in after the first flip though, so we can’t rely completely on the conditional probabilities. But we do know the probability for being in each of the conditions for those conditional probabilities: in this game it’s 50/50.

The way the math works out for this:
probability of H2 = (probability of H2 given H1) * (probability of H1) + (probability of H2 given T1) * (probability of T1) = 0.5*0.5 + 0*0.5 = 0.25

So yes, breeding a 50% het to a normal means that every hatchling in that clutch has a 25% chance of having inherited the het gene. Further observations of genetic interactions can update this belief for all the animals involved. But as it stands after that pairing: it is certainly honest, theoretically, and mathematically correct to state that there is a 25% probability, chance, odds, that those snakes have the gene. *assuming we can ignore the probabilities for other genetic mechanisms; which I am certainly no expert on

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I have a genuine question because I have been reading through this and following but am still SO confused.

Breaking this down to basics: If I have a Normal female and a het pied male. I breed them. Their offspring will be 50% het pied, right? So 2/4 babies will be het pied, statistically.

What if I never prove out the het pied babies and breed one to a Hypo a few years later? Offspring from that clutch would then be 100% het Hypo, “Pos” het pied, right? Then later on next season, if I prove out the 50% het pied parent from the first clutch as 100% het Pied, the baby would then become 100% het hypo, 50% het pied, right?

My question is what % would the “pos” het baby be before being proven? Just keep it “pos” het or call it 33% het Pied?

I’m just trying to put this in a way I could understand better. haha

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In the case you mentioned, before you proved that the first 50% het pied was in fact 100% het pied: the first clutch from that one would have had a 100% chance of het hypo and a 25% chance of being het pied. Once you proved the 50% het pied was actually 100% het pied, then the updated probability for the hypo clutch would be 100% chance het hypo and 50% chance of being het pied.

As to what you would call it, my interpretation is that the meaning of “x% het” means the probability this animal carries that heterozygous gene is x%. So it would follow that initially calling the animal 100% het hypo, 25% het pied would be correct.

Even with 50% and 66% hets though, the topic can often trip people up as they learn it. I’ve seen 50% hets listed as 66% het, etc. Talking about multi-generation odds, and odds once parents have proven or failed to prove makes it even trickier. So it’s key to ask about the specific pairing and history for any “poss het” so that you can understand and “check the work” so to speak.


Although I may not be the sharpest tool in the shed when it comes to the complexities of genes I will however say that I don’t think this topic is very complex nor should it have as many responses as it did. It’s really not hard to understand, poss hets are exactly that. They are just POSSIBLE HET for a certain gene. I also don’t think having a poss het option should even be considered because just why??? There really isn’t a way to pick out a poss het over another in terms of ones you think may carry that gene so why even have that option in the calculator? Really thought this topic would just be answered with a response or two but I guess not. Some of you may or may not agree with me which is completely fine but that is just my two cents on the matter.


Oh I’m with you 100% (proven, not pos.). There’s absolutely no need to have possible hets in the calculator, it doesn’t add anything useful. People just need to run the calculation to see what they’d get if the snake was a het, then run it again to see what they’d get if it wasn’t a het, then they’ll know what to expect either way. Basically, they’ll either get visuals or they won’t


Oh haha I see what you did there. Got a good laugh.


I actually agree that for predictive planning purposes that pos het in the genetics calculator seems not super practical and possibly confusing.

Understanding the odds that the offspring you’ve already made have het or other hidden genes I think is actually useful in the long run though. If you can say one snake is 25% het it could influence your best choice for pairing.


Obviously we all know that an animal is either het for a gene or it is not, that was never in question and is very black and white… and although I agree that not everyone cares to learn about probability of hets, I do. And since this is a forum for discussion, I will discuss when I see the opportunity to do so just as I have seen you do as well. :slight_smile:

I do not believe there should be any changes to the genetic calculator… However, when pairing up for breeding, not matter how much you think probability doesn’t play a part in pairings, it absolutely does. So, some people may want to know for themselves the probability of one paring over the next. If I have a Super pastel Enchi Ghost female and a Mojave Clown pos het ghost male that I am wanting to pair, I’m pairing them thinking about the probable outcomes of the paring. I get that not everyone works like that, but some of us do. :slight_smile:

Thanks, crypticoils-pythons for answering my question. :slight_smile:

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You continue to ignore the fact that this is not an iterative process. The probability of an animal is only applicable when choosing animals out of a single specific clutch. It cannot be carried forward. So let us try this looking at the actual Punnett square and how the probabilities are determined.

A basic Punnett so we all know what we are taking about here.


To start with, we make a 100% het with this breeding. All the offspring are phenotype normal but they carry a single copy of the recessive gene (h)


If we take one of these 100% het animals (Hh) and breed it to another of these 100% het animals (Hh), three of the offspring from this pairing are phenotype normal.


When we look at the probable genetic outcome of this clutch we can very plainly see that, statistically, 2 of the 3 normal looking animals will carry a copy of the recessive gene (h). However, because we cannot tell by appearance alone which two are the carriers, we then label all of these normal looking animals as 66% possible het because there is a 66% probability of any given animal being a het (Hh).

With me so far?

Next, we take one of these 66% probably het animals (genotype H?) and breed it to a WT (HH). All of the offspring from this pairing are phenotype normal.

When you look at the Punnett square what is important to note is that, even without determining the het status of the unproven animal, we can very plainly see that half of these animals are statistically going to be WT (HH) while the other half are an unknown. Do you see the relevant word there? HALF. So when considering the probabilities of this clutch, you have a statistical 50/50 chance of getting a WT (HH) or a statistical 50/50 chance of getting an undetermined (H?). This is the same statistical probability as the 100% het x WT breeding was because there are two copies of the gene in play making this a binary matter

What you do not have, correction, what you cannot have, is a Punnett square where only one of the four possible genetic outcomes is the unknown (H?) while the other three are WT (HH) As such, you can NEVER have a 33% probable het from any type of breeding.

If the original unknown animal proves out you can end up with offspring that are 50% probable hets


Or, if the original unknown animal does not prove out you end up with 0% probable hets


You cannot, ever, create a Punnett square where a 33% probable het or a 25% probable het is an outcome


I am in no way trying to tell you not to discuss a topic i’m just saying that this topic is being treated like a more complicated topic than it really is.


Again, very genuine question. As that last rely was very informative!!

I’m not trying to drag this out, simply just trying to understand for my own personal gain. How do we get 66% hets? Or should we only do het vs 50% het?

you would get 66% het by breeding two proven hets to each other and whichever baby is not a visual of the gene then it would be 66% het for that recessive.


If you look at my first Punnett square (the one without any animals pictured in it) you will see there are three genotypes: HH, Hh, and hh

If I threw pictures on that square the hh would have an Albino phenotype, and both the HH and Hh would have a normal phenotype. So, out of four potential outcomes, three look normal. Statistically, two of those three will carry the recessive gene (genotype Hh) so 2/3 = 66%

Make sense?

And if you were to take that 66% probable animal and plug it into a breeding with a WT then that would be exactly the same as the fourth slide (HH x H?)


Oh, I’m sorry! I understand that part. :slight_smile: i have never had a problem with understanding how genes work when it comes to basics like het to het, visual to het and visual to visual. That’s all fairly simple.

I meant where do 66% hets fall on a Punnett square. And that was answered!!! Thanks,t_h_wyman . :slight_smile: makes total sense now. Haha

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:+1:t4: :+1:t4:


Sorry for editing the hell out of that last one. I’m walking to my car on lunch break and was replying haphazardly haha.

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Correct. But what are the odds that the “?” is either H or h for a 50% possible het?

If Punnett squares are the preferred tool for reasoning out the probability, then you could conceptually make 2 Punnett squares to represent that half of the time “?” is h and half of the time it is H. Then count up all 8 results, and see the total odds for the “H?” being Hh.

Yet another way to frame this is “what are the odds it both inherits the “?” AND the question mark is h.” Both of those must be satisfied for the snake to carry the het. Since each condition does not change the probability for the other (independent probabilities), the math for this is:

P(? and “?” is h) = P(?) * P(“?” is h)

@t_h_wyman, responding to your earlier analogy (thanks for sharing that), what I hear you saying is that once the “pos hets” in generation 1 are proved out (ie the light was turned on), you can’t use those probabilities in generation 2. I wouldn’t disagree with that at all – to do so would be the jerkwad you referred to, and dishonest. This is what I referring to in the last part I wrote.

I think this the point that somewhat satisfies me is this: If we are talking about an actual animal, the 2nd generation so called “pos 33%”, it implies that the pos het parent has laid a clutch. And if they have done so there is additional information about what its true genetics are (assuming it was bred to an animal with the same gene), i.e., the light has been turned on to some extent. So practically speaking a “pure pos het” 2nd gen may be unlikely. That being said, if you just bred it to a normal, you’d have no additional info and “pos 33%” still seems accurate.

However, if we are talking about a feature for the calculator where we are just letting users make predictions, then even the 1st generation pos het parent may not exist in reality, so there is no additional information since they haven’t laid a clutch. I still think “pos het” input to the calculator makes sense, though we can question the ultimate utility.

FYI, I scanned but didn’t read all the intervening posts since this message above that I’m responding to. Also I’m ready to agree to disagree, and I’m tapping out. :wink:


This was my whole point for the topic. Not to debate hets or to advertise potential percentages of hets. But for potential pairings

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I think a lot of the current interpretation of output from the calculator is “what % of the clutch will have x gene.” Most of the time that is analogous to “what is the chance an egg from this clutch will have x gene.” Once uncertainty of the genetics of the parents is introduced however, those two interpretations diverge. My opinion is that the latter interpretation is actually the most useful in terms of optimizing breeding plans.

I think what you’re saying @jalderman, for example: you’d like to understand “What is the probability that a visual hatches from an egg when you pair, say, a 66% het to a 100% het?” It’s 1/6 chance btw. And there is a 1/2 chance that a het hatches. And a 1/3 chance that a normal hatches. So a total of 5/6 chance that they are “normal” looking.

But a “normal” looking one from that clutch won’t be a 50% het. They won’t be 60% het either. Each egg that hatches will change what % chance that a “normal” looking offspring carries the het gene. (And will also even change the % chance that the poss het parent actually carries the het gene.)

So a big question is how to avoid confusion for how people interpret the output for different uses.