A lesson on probabilites with double recessives

[towelie4365]

So, I saw the thread on the odds of getting 4 out of 5 albinos in a het x het clutch, and figured I'd give a lesson on how to calculate odds.

Let me start first with an example of what I want to do, and look into double recessives:

I have an albino het pied male, and a pied female. When I finally get into breeding them, 50% will be pied het albino, 50% will be double hets. For my purpose, I want a female pied het albino.

The chance I get a female in each egg is 50%, and the chance I get a pied het albino in each egg is 50%, so in order to find out the chance per egg of getting a female pied het albino, I multiple the odds together:
.5 x .5 = .25
Therefore, each egg has a 25% chance of being what I was, and thus, a 75% chance of not being what I want.

In order to figure out the odds of getting at least one female het pied, I need to look at the odds of not getting a female het pied in the clutch. So, if there are "n" eggs in a clutch, the odds of me getting one is:
1 - (.75)ⁿ
To explain that further, if I have 3 eggs, then there is a .75 x .75 x .75 = 42% (or .75³) chance that I don't get one, which means there is a 58% chance that I do get one, which is why you subtract it from one.
With each egg, you have a higher chance of getting what you want, just plug in the number for "n":
3 eggs = 58% chance
4 eggs = 68%
6 eggs = 82%
10 eggs = 94%... etc

So, now that I have the odds of getting the first snake I want, we'll say its an 82% chance with 6 eggs, now I can calculate the odds of getting the albino pied once I raise her up:
Albino het pied x pied het albino: There is a 25% chance I get an albino pied, 25% chance I get an albino het pied, 25% chance I get a pied het albino, 25% chance for a double het
Again, I have a 25% chance at getting what I want, so the odds are the same as before. This means that, if both clutches are 6 eggs, I have an 82% chance at getting what I want each time.

Now I have to calculate the odds of getting what I want both times, so I multiply those together again:
.82 x .82 = .676 = 67.6%

If everything goes right, and each clutch is 6 eggs, I have a 67.6% chance at getting an albino pied in two generations.

Things get messy if you start to look at if you'll get it within 3 or more generations, but if you add the odds of not getting it 3/3 times and not getting it the first and second time, or the first and third time, or the second and third time (we'll just multiply the odds of not getting it 2/3 times by 3), and subtract it from one, you get the odds of getting it at least 2/3 times: 1 - (.18³ + 3*.18²*.82) = 1 - (.00583 + 3*.0265) , or about a 91.5% chance of getting one within 3 generations. You could also do it by adding up the chances of getting what you want 2/3 or 3/3 times: (.82³ + 3*.82²*.18) = .9145, or 91.5% again.

Since some people deal with double hets alone, I'll cover that quickly:
The odds of getting an albino pied from two double hets are 1/16, or 6.25%. So, if you have 6 eggs in a clutch, your odds of getting it are 1 - (1-.0625)⁶, which is a 32% chance. So, the odds of not getting an albino pied two years in a row is .68 x .68, which is 46.2%, making your chances of getting it within two years 53.8% (you can also calculate it based off of number of eggs you get from the two years, so using 12 instead of 6 yields the same results).

Of course this is all assuming you know how many eggs you will have. Now, if you knew the odds of how many eggs you'll get, you could calculate the exact chance you have... but that depends on the snake and other factors I believe we don't understand.

If you want me to explain anything in further detail, I'd be more than happy to.