DuxorW
Derailer of Threads
This year, I crossed an ultramel lavender motley to a sunkissed (redcoat?) het lavender to start an ultramel orchid motley project. I generated normals het sunkissed lavender motley and ultra OR amel, and lavenders het sunkissed motley and ultra OR amel.
Since my goal are ultramel orchids (I will treat ultra orchids as synonymous for simplicity), it's important that at least one of my holdbacks contain the ultra allele. If I hold back 2 males and 6 females, and only one animal contains ultra, that will be good enough in terms of giving me the possibility to create ultramel/ultra orchids. So I ask myself, "If I hold back x number of snakes, what are the odds that AT LEAST one of them will contain ultra?"
This is also important for when I sell the other babies. I want my customers to be confident of what their odds are, if they purchase a pair or a trio, of being able to produce the coveted yet currently nonexistent ultramel orchid.
Say I hold back a trio (a male and two females). What are the odds that ONLY one of them contains ultra? The calculation is not as simple as you might think. You might think it is 1/2 (odds of one being ultra) x 1/2 (odds of one being amel) x 1/2 (odds of another being amel)=1/8. This is incorrect as it does not account for the fact that there is more than one way of getting only one het ultra - the birth order does not matter. What you have really just calculated are the odds of the first snake being het ultra and the second and third being amel, kind of like calculating the odds that if you flip a coin you will get heads the first time and tails the second and third time. What you really want to calculate are the odds that the first snake is het ultra OR the odds that the second snake is het ultra OR the odds that the third snake is het ultra, and add these probabilities together.
1/2(het ultra) x 1/2(het amel) x 1/2(het amel)=1/8
1/2(het amel) x 1/2(het ultra) x 1/2(het amel)=1/8
1/2(het amel) x 1/2(het amel) x 1/2(het ultra)=1/8
Since any of these three possibilities represent the odds of one of three possible ways of getting only a SINGLE het ultra, adding them up gives us the total probability that we will only get ONE het ultra, that is 3/8.
But we are more interested in the odds of getting AT LEAST one het ultra. If 2/3 or 3/3 are het ultra, that will be just fine too, assuming that I am just as ok getting ultra orchids as I am getting ultramel orchids.
What are the odds of getting ONLY two het ultras or getting ALL 3 het ultras?
For two het ultras
1/2(het amel) x 1/2(het ultra) x 1/2(het ultra)=1/8
1/2(het ultra) x 1/2(het ultra) x 1/2(het amel)=1/8
1/2(het ultra) x 1/2(het amel) x 1/2(het ultra)=1/8
So the odds of getting two het ultras is 3/8, the same as getting only one het ultra!
There is only one way to get 3 het ultras out of a group of 3
1/2(het ultra) x 1/2(het ultra) x 1/2(het ultra)=1/8
3/8 + 3/8 +1/8=The odds of getting 1 or more het ultras in our trio=7/8, pretty good!
The only remaining possibility is getting all three het amels, which as you can probably see is 1/8. Those are the only 4 possible outcomes in a group of 3 snakes, and you can see that they all add up to 1 or 100%, which makes sense.
But what if I hold back 2 males and 6 females, for a total of 8 snakes? Lets calculate the odds that ONE OR MORE will be het ultra:
1/2(het ultra) x 1/2(het amel) x 1/2(het amel) x 1/2(het amel)...oh screw it!
There is an easier way. You will notice if you write out the whole thing that for a group of 8 snakes there are 255 arrangements, all with the same probability of occurring, where at least one of the 8 snakes is het ultra, and there is one and only one arrangement where all 8 are het amel. In fact, no matter how many snakes we are considering, there is only one combination we don't want, and that is the combination where all snakes are het amel. So we just need to figure out a quick way to calculate the number of possible combinations, then subtract the undesirable one.
The formula for the number of combinations is pretty simple, it is 2^n where n is the number of snakes. So for 3 snakes it is 2^3=8. But one of those combinations is the combination where all snakes are het amel. So there are 7 out of 8 where at least one snake is het ultra, so our odds of getting at least one het ultra out of 3 snakes is 7/8.
For the 8 snake example above, it is 2^n=256. Subtract the one undesirable combination to get 255/256.
Another way to think of it is "What are the odds that out of 8 snakes, zero would be het ultra, and 8 would be het amel?" The odds are (1/2)^8=1/256. Thus the other 255/256 possibilities must involve at least one het ultra.
You can see that just picking a trio gives you very good odds, 7/8, but if you were paranoid you could do pick 2 males and 6 females and be virtually assured you will get the desired result.
Another practical application of this formula is to possible hets. Say I cross a Lava het cinder to a tessera het lava, all of the resulting offspring will be 50% possible het cinder. I want to know whether I will have good odds of getting tessera lava cinders in the subsequent generation if I hold back a trio.
Unlike the previous scenario with ultra and amel, this time gender matters. It is not just important that at least two holdbacks are het cinder, but the male needs to be het cinder and at least one female needs to be het cinder. So the probability calculation is
(odds that male holdback is het cinder) x (odds that AT LEAST one female holdback is het cinder)
The odds that a single male is het cinder is simply 1/2, since it either inherited the cinder allele from the father or it didn't, and both have equal probability.
For the female part of the scenario, we just want to calculate the total number of combinations where at least one female is het cinder, then subtract the combination (there is only one) where both females are not het cinder.
Since we have two females we have 2^2=4 combinations. Subtract the single undesirable one and you get 3/4 combinations where at least one female is het cinder.
Multiply the odds that at least one female is het cinder by the odds that the male is het cinder and you get (1/2)(3/4)=3/8...less than 40%, not so good!
If you hold back two males your odds of one of them being het cinder are (3/4), just like the odds were with the two females. Say we hold back 2 males and 6 females, so that each male has 3 females to breed with.
For the females we have 2^6 combinations=64, minus the undesirable one =63/64 odds of at least one female being het cinder.
For the males we have 3/4 odds, just like we did when we considered the case of the two females.
(3/4)(63/64)=73.8% odds that at least one male and at least one female will be het cinder, giving us the potential to generate lava cinder tesseras in the next generation.
However, what you will probably do is mate one male to 3 females and the other male to 3 females. So in this case you should calculate the odds to account for that. I will not do that here.
I am thinking about constructing a table that just tells you what the odds will be if you hold back x amount of males and x amount of females. But I hope I have at least driven home the point that with het ultra and het amel, a trio gives you great odds that you will get ultra or ultramels from your animals that are het ultra OR amel, but for 50% possible hets a trio gives you quite poor odds.
More to come...
Since my goal are ultramel orchids (I will treat ultra orchids as synonymous for simplicity), it's important that at least one of my holdbacks contain the ultra allele. If I hold back 2 males and 6 females, and only one animal contains ultra, that will be good enough in terms of giving me the possibility to create ultramel/ultra orchids. So I ask myself, "If I hold back x number of snakes, what are the odds that AT LEAST one of them will contain ultra?"
This is also important for when I sell the other babies. I want my customers to be confident of what their odds are, if they purchase a pair or a trio, of being able to produce the coveted yet currently nonexistent ultramel orchid.
Say I hold back a trio (a male and two females). What are the odds that ONLY one of them contains ultra? The calculation is not as simple as you might think. You might think it is 1/2 (odds of one being ultra) x 1/2 (odds of one being amel) x 1/2 (odds of another being amel)=1/8. This is incorrect as it does not account for the fact that there is more than one way of getting only one het ultra - the birth order does not matter. What you have really just calculated are the odds of the first snake being het ultra and the second and third being amel, kind of like calculating the odds that if you flip a coin you will get heads the first time and tails the second and third time. What you really want to calculate are the odds that the first snake is het ultra OR the odds that the second snake is het ultra OR the odds that the third snake is het ultra, and add these probabilities together.
1/2(het ultra) x 1/2(het amel) x 1/2(het amel)=1/8
1/2(het amel) x 1/2(het ultra) x 1/2(het amel)=1/8
1/2(het amel) x 1/2(het amel) x 1/2(het ultra)=1/8
Since any of these three possibilities represent the odds of one of three possible ways of getting only a SINGLE het ultra, adding them up gives us the total probability that we will only get ONE het ultra, that is 3/8.
But we are more interested in the odds of getting AT LEAST one het ultra. If 2/3 or 3/3 are het ultra, that will be just fine too, assuming that I am just as ok getting ultra orchids as I am getting ultramel orchids.
What are the odds of getting ONLY two het ultras or getting ALL 3 het ultras?
For two het ultras
1/2(het amel) x 1/2(het ultra) x 1/2(het ultra)=1/8
1/2(het ultra) x 1/2(het ultra) x 1/2(het amel)=1/8
1/2(het ultra) x 1/2(het amel) x 1/2(het ultra)=1/8
So the odds of getting two het ultras is 3/8, the same as getting only one het ultra!
There is only one way to get 3 het ultras out of a group of 3
1/2(het ultra) x 1/2(het ultra) x 1/2(het ultra)=1/8
3/8 + 3/8 +1/8=The odds of getting 1 or more het ultras in our trio=7/8, pretty good!
The only remaining possibility is getting all three het amels, which as you can probably see is 1/8. Those are the only 4 possible outcomes in a group of 3 snakes, and you can see that they all add up to 1 or 100%, which makes sense.
But what if I hold back 2 males and 6 females, for a total of 8 snakes? Lets calculate the odds that ONE OR MORE will be het ultra:
1/2(het ultra) x 1/2(het amel) x 1/2(het amel) x 1/2(het amel)...oh screw it!
There is an easier way. You will notice if you write out the whole thing that for a group of 8 snakes there are 255 arrangements, all with the same probability of occurring, where at least one of the 8 snakes is het ultra, and there is one and only one arrangement where all 8 are het amel. In fact, no matter how many snakes we are considering, there is only one combination we don't want, and that is the combination where all snakes are het amel. So we just need to figure out a quick way to calculate the number of possible combinations, then subtract the undesirable one.
The formula for the number of combinations is pretty simple, it is 2^n where n is the number of snakes. So for 3 snakes it is 2^3=8. But one of those combinations is the combination where all snakes are het amel. So there are 7 out of 8 where at least one snake is het ultra, so our odds of getting at least one het ultra out of 3 snakes is 7/8.
For the 8 snake example above, it is 2^n=256. Subtract the one undesirable combination to get 255/256.
Another way to think of it is "What are the odds that out of 8 snakes, zero would be het ultra, and 8 would be het amel?" The odds are (1/2)^8=1/256. Thus the other 255/256 possibilities must involve at least one het ultra.
You can see that just picking a trio gives you very good odds, 7/8, but if you were paranoid you could do pick 2 males and 6 females and be virtually assured you will get the desired result.
Another practical application of this formula is to possible hets. Say I cross a Lava het cinder to a tessera het lava, all of the resulting offspring will be 50% possible het cinder. I want to know whether I will have good odds of getting tessera lava cinders in the subsequent generation if I hold back a trio.
Unlike the previous scenario with ultra and amel, this time gender matters. It is not just important that at least two holdbacks are het cinder, but the male needs to be het cinder and at least one female needs to be het cinder. So the probability calculation is
(odds that male holdback is het cinder) x (odds that AT LEAST one female holdback is het cinder)
The odds that a single male is het cinder is simply 1/2, since it either inherited the cinder allele from the father or it didn't, and both have equal probability.
For the female part of the scenario, we just want to calculate the total number of combinations where at least one female is het cinder, then subtract the combination (there is only one) where both females are not het cinder.
Since we have two females we have 2^2=4 combinations. Subtract the single undesirable one and you get 3/4 combinations where at least one female is het cinder.
Multiply the odds that at least one female is het cinder by the odds that the male is het cinder and you get (1/2)(3/4)=3/8...less than 40%, not so good!
If you hold back two males your odds of one of them being het cinder are (3/4), just like the odds were with the two females. Say we hold back 2 males and 6 females, so that each male has 3 females to breed with.
For the females we have 2^6 combinations=64, minus the undesirable one =63/64 odds of at least one female being het cinder.
For the males we have 3/4 odds, just like we did when we considered the case of the two females.
(3/4)(63/64)=73.8% odds that at least one male and at least one female will be het cinder, giving us the potential to generate lava cinder tesseras in the next generation.
However, what you will probably do is mate one male to 3 females and the other male to 3 females. So in this case you should calculate the odds to account for that. I will not do that here.
I am thinking about constructing a table that just tells you what the odds will be if you hold back x amount of males and x amount of females. But I hope I have at least driven home the point that with het ultra and het amel, a trio gives you great odds that you will get ultra or ultramels from your animals that are het ultra OR amel, but for 50% possible hets a trio gives you quite poor odds.
More to come...
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