Reptile & Amphibian Forums

Maybe everyone else has already figured this out, but it was certainly a surprise to me. Here goes...

Looking at multiple descriptions of the Punnett Square for simple recesive (HET X HET) would lead you to believe that there is a 25 percent chance that you would get, lets say, an albino. Well if you look at the way the square works, it says that there is a 25 percent chance that any given egg will contain an albino. Since it is widely accepted that the average clutch size is four and there are four parts to the square, one in four will be albino, hence 25 percent. That is not what the square says, every egg has a 25 percent chance for the recesive trait, not every clutch. If you do the math, the clutch has a 68 percent chance for a single recessive offspring. If you bump the "average" clutch size up to 6 the percentage for one albino per clutch goes up in the 80's.

WOW!!!!!! Those are pretty good odds!

Now I know some of you are saying "lets see the math", the math is not that straight forward, at least no to me, but here it is

1-(3/4)^4=.68359375 or roughly 68 percent. 3/4 is the chance of a normal or a het and 1/4 is the chance for an albino. The exponent of 4 is the clutch size. Why do you have to compute the chance of it not being an albino and then subtract that from one? Like I said, the math is not that straight forward.

Well I didn't believe I was doing the math right, so I wiped up a quick perl script using a random number generator and ran 10,000 trials. You guessed it, 68 percent. I plan to run this out over a hetXnormal project and see where it ends up.

Cheers,
Al

I believe each egg has a 25% chance of albino.
It can vary each time. Like a coin flip. You will never come out exactly the right amount.
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RicK @ BbI

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And on the dark side...each offspring has a whopping 75% chance at looking completely normal. Although the Punnett Square can be loosely interpreted as a whole clutch ratio, by strict interpretation it is a miracle that any visual animals are produced at all.

Chris
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mean people suck

This is true!
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RicK @ BbI

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No, totally the opposite. It is very likely visual animals will show up. Not 25 percent likely, but 68 percent likely. Those are really good odds. On a clutch of 6 it is over 80 percent you will get at least one visual. Those are awesome odds.

Al

Nope 1.5
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RicK @ BbI

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You lost me... No those are not good odds? Reasonable odds of getting one per clutch sounds great when I thought I was up against 75 percent chance of nothing per clutch.

To turn it around, there is on a 17 percent chance that I would walk away empty handed on a clutch of 6 hetXhet. I thought the odds were much worse than that before I did the math.

I guess it is all perspective.

Al

the chances be of getting 2 homozygous from a het to het breeding in a clutch of 3 eggs? We had such a clutch and I have wondered what the overall chances would be for the clutch.
Thanks,
Justin

Australian Addiction Reptiles (and ball pythons)

Clutch of 3 and ended up with two pied's and a normal.

15 percent.

I would say you got lucky.

Al

Same clutch of 4 hetXhet yeilds

1 visible 68% chance
2 visible 25% chance
3 visible 5% chance
4 visible 0% chance

Obviously there is actualy a small chance that two hets could produce all visibles, but it is less than one percent and got rounded.

clutch of 6 hetXhet

1 visible 82% chance
2 visible 46% chance
3 visible 16% chance
4 visible 3% chance
5 and up visible 0% chance

Almost even odds of two visibles in a clutch of six and 4 to 1 that you will get a visual.

cheers,
Al

I'm not really sure the formula is correct.

To evaluate the probability, we first need to find the Fractional Probability of a visual Albino. This is found in the formula: (total # of ways to get the desired outcome) / (total # of possible outcomes).

Given that the Punnett Square suggests only one way to get an Albino (aa) and a total of four possible outcomes (AA, Aa, aA, aa), the Fractional Probability of a visual Albino is 1/4.

This Fractional Probability is for a single trial only. (Think: odds per egg.)

The expected outcome of a visual Albino within a whole clutch is found in the formula: (total # of eggs in the clutch) x (Fractional Probability) = Expected Outcome.

Let us assume the clutch is four eggs. This formula would read: 4(1/4) = Expected Outcome. And the Expected Outcome equals one.

We use different math.

Chris
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mean people suck

Chris you and me think the same. It is .25 X N.
N = number in clutch!
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RicK @ BbI

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"Chris you and me think the same."

That's right. And I think you should loan me a few adult PH females for my PDH Albino Piebald male. I'm glad we think the same.

Chris

PS -- And I prefer delivery via FedEx Priority Overnight. Thanks.
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mean people suck

hummm
In 2099
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RicK @ BbI

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(It was worth a shot.)

Chris
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mean people suck

Sure
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RicK @ BbI

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Here is a pretty good intro to probability text from Dartmouth College. The first 15 pages get our issue covered.

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/pdf.html

It is essential this, if there is a 25% chance of something and it happens four times, there is more than a 25% chance that it will happen at least one of the four times. If I role a dice there is a one in six chance that I will get a six. If I role it 10 times, there is more than a one in six chance that I will get a six, not 10 sixes, a six. An 83.8% chance that I will get one six. In one role there is a 16.6% chance that I will get one six. The more I role the better the odds get.

In one clutch of four there is more than a 25% chance that you will get one visible. The one in four thing is a good way to keep it simple, it is just miss leading about the chances when you start figuring the odds of something on 30 pairings.

I don't want to come off like a prick, just trying to make sure I explained it right.

Cheers,
Al

one in six for any number. This is die!
1
2
3
4
5
6

You used the word dice which is plural of die. Two die = dice!
2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1
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RicK @ BbI

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I stand corrected again!! Man, I better get lost

Thanks,
Al

Nah...no problem
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RicK @ BbI

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I think we are all attacking the problem at different angles. Correct. The more offspring produced, the higher probability of producing a single visual offspring. Correct. The total number of offspring does not influence (read: change) the 1:4 expected ratio of visual to non-visual offspring.

For instance: "The more I role the better the odds get."

You interpret this statement as: "The more I role, the better the odds get...at rolling my first six."

We interpret this statement as: "The more I role, the better the odds get...at exceeding the expected 1:4 ratio."

So you are correct. Out of 36 rolls you are virtually guaranteed a six. And we are correct. Out of 36 rolls we should expect approximately 10 sixes.

Chris
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mean people suck

So you are correct. Out of 36 rolls you are virtually guaranteed a six. And we are correct. Out of 36 rolls we should expect approximately 10 sixes.

NOPE!!!!!
1
2
3
4
5
6
You will get 6 with one die in 36 rolls
2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1

Five sixes in 36 rolls with two die [a pair dice]
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RicK @ BbI

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LMAO Oops! I'm a silly goose. I guess my eyes are crossing. LOL Thanks. =)

Chris
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mean people suck

I can not afford a visual and was killing myself thinking I
would never get one with hets. But given several normal females
and a few het males, I am guaranteed to get a visual within five
years. Several visuals as a matter of fact

Thanks for the discussion - Al

I totally agree. With enough females and PH offspring, you should without a doubt eventually produce a beautiful little Albino hatchling.

Good luck! And congratulations in advance! =)

Chris
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mean people suck

"With enough females and PH offspring, you should without a doubt eventually produce a beautiful little Albino hatchling."

Actually it never really reaches the "without a doubt" point. The doubt just gets smaller and smaller. Remember that a probability of 1 is 100% - a certainty. That's why your formulas must be wrong. What if it was a 5 egg clutch? Would you have a 125% chance of producing an albino? Something can't be more certain than 100%. The more possible homozygous albino eggs you hatch the lower the chance is that you will not hatch an albino but here is always still some remote chance, be it eventually very very very small.

I know that Randy! LOL

I didn't mean LITERALLY without a doubt. But with enough girls, the odds are better for producing an Albino than winning the 290MILLION JACKPOT. So I feel safe saying "without a doubt."

LMAO

Chris
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mean people suck

Even with one pair of hets and only one egg they odds are way better than the lottery.

I guess we are true geeks to be arguing statistics online on a Friday night ...

Chris,

"(total # of ways to get the desired outcome) / (total # of possible outcomes)"

The problem is that there are 4 ways of getting 1 albino in the clutch of 4 eggs. 1 for each egg that could be the one albino.

...

"Let us assume the clutch is four eggs. This formula would read: 4(1/4) = Expected Outcome. And the Expected Outcome equals one."

In probability 1 is 100% - a certainty. I think we would agree you aren't guaranteed an albino in a clutch of 4 from het X het.

The way you add up chances of multiple independent things happening is by multiplying them together. You could find the odds of getting all albinos with .25 * .25 * .25 *.25 = 1 in 256. The original formula from this thread starts by adding up the possibilities of not getting an albino in each of the 4 eggs (.75 * .75 * .75 * .75) to come up with the probability of striking out. Since the only other possibility is to produce 1 or more albinos we know that the two outcomes combined equals 1 (100%) so 1 - .75^4 = the chance of getting at least one albino. The chance of getting exactly one is a much more complicated formula as shown in the post below but the odds of not striking out are of interest to most.

"The problem is that there are 4 ways of getting 1 albino in the clutch of 4 eggs. 1 for each egg that could be the one albino.

In that case the formula should also include the 16 total possible combinations -- four combinations (AA, Aa, aA, aa) per egg. Then the formula would read: 4(4/16) = Expected Outcome. The expected outcome would still equal 1.

"In probability 1 is 100% - a certainty. I think we would agree you aren't guaranteed an albino in a clutch of 4 from het X het. "

True. We are not guaranteed an Albino. However, we are guaranteed the expectation of an Albino. The actual ratio is not the same as the expected ratio.

For instance, we can logically EXPECT a penny flipped 100 times to land 50 times on heads. (All things being equal using a fair penny.) Yet the ACTUAL results may only be 30 heads. However, the more pennies we flip, the closer the actual results move to the expected results.

Chris
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mean people suck

Chris,

Expected value of random variables in this instance is a no brainer.

I know the terminology is different, but it’s all the same:
Expected value, mean, average or even “expected ratio”

It is formalized as E(N)= the summation of n*P(n) for all n in N
(Here P(n) is the probability of n occurring)

All expected value is, is the sum of all the probabilities times the number of times they give the desired outcome of a random trial.

In our case as you stated our outcomes are N={AA,Aa,aA,aa}
And n1=AA,n2=Aa……….

We are looking for albino here, so we investigate each outcome.

AA is the only outcome to give albino so all others can be discarded, since they will give a zero as the number of times they give the expected outcome.

So we sum our results from one trial(egg):
AA happens once with probability of ¼……1*1/4
Aa,aA,aa do not result it albino outcome hence we discard.

Or E(N)=1*1/4 plus 0*1/4 plus 0*1/4 plus 0*1/4 = ¼

Ok a long drawn out way to what we all know, each egg has a ¼ expectation of being albino.

Now you said we had 4 eggs that’s 4 trials with the same parameters hence as you stated we expect one albino out of a four egg clutch.

What does this mean??? Nothing really, like I said a no brainer.

What the main post was referring to was Bernoulli trials, based on independent probability.

Whole nother story, but I’ve gone on long enough.

Sorry for my long windedness, I worked 12 hours today, then had a few brews(too many).

TomChambers

yea.....25% PER egg...........right......

and odds ,nature, and luck are good and bad......

last year 6 eggs het to amel ....1 het and 5 amels.......

this year 3 good eggs and 1 amel and 1.1 hets.......yep.....good luck........

of all the balls she made me.....total 9....8 male and 1 female..........(het).......other odds.......LOL

it all evens out sooner or later.......and stuff like Katrina can even you out even quicker........
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"If you do the math, the clutch has a 68 percent chance for a single recessive offspring"

math is a very precise science, one word changes everything sometimes.

The above statement is untrue.

your simplified equation is true for calculating at least one simple recessive offspring.

here is the formula I wrote for a clutch of three a few months back.

What if someone asked what are the odds of exactly 1(as your statement indicates), it's not the same percentage, so you need the general formula.

(n choose k) * (p^k)* (q^(n-k))

I stand corrected. And you are right; I did mean at least one.
I was beginning to feel like Columbus arguing that the world was
round.

I would be greatfull if there is anything else you would like
to share. Feel free to shoot me an email.

Thanks,
Al

Columbus?

He got lost!
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RicK @ BbI

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I dont know if I agree.

I have writen a program that will put to test just the chances of this breeding. I say add 25% chance for each egg. every 100% means you should get a homo animal.

Download the program here:
www.reptilestudbook.com/downloads/random.exe
http://www.reptilestudbook.com/downloads/random.exe
It was writen in C#, so you will have to have Framework 1.1 installed. If you dont know if you do, try it anyhow, it it wont run, you dont have it. Get it from microsoft.com Dont run too many at a time, or your computer will take a while, limit X to 1000, and it will do better.

I guess I could have told you how the program works. Each time you press the GO ONCE button, it will generate a random number between 1 and 4, for each of the 3 eggs. Lets say 1 is a normal (aa) and 2 is a het (Aa) and 3 is a het (aA) and 4 is a homo (AA). It will add each trait to the totals, and then calculate the percent of that type of offspring produced so far. If you hit the GO X TIMES, it will do the same thing, only X times. Dont put too many or you might wait a while. I didnt build it for speed.

I think this program clearly illistrates the probability of each clutch. Each time you press that button you hatch X clutches of 3 eggs and it figures the results. It really doesnt have to be as hard as some of you are making it.

Ben

Man, looks like to my surprise 58% was about right on the money. I put up a new version that will calculate weather a clutch had on of the animals, not just how many as a whole. Looks like 58% for a clutch of 3 is right on the money. I updated the program if any of you want to see for your self.

Ben

Ben,

I downed your first program, but it didn't run on my computer, as you said I need something from M/S.

I can only tell you that I have a vast background in Math(near PHD).

My focus wasn't probability, but this isn't high end stuff.

I am quite confident in the formula's I posted, but it is just probability, not actual outcomes.

I understand your skepticism; I always need to see things for myself as well.

TomChambers

here is a link to the Framework 1.1 you have to install. Its about 24MB. Click Here

For the genetics program, Click Here Its only 52KB.

To get a new set of data, close the program and reopen it.

Ben

The results...

3 eggs each clutch, 1 million clutches.

571790 (57.179%) clutches had atleast one trait 1 animal.
571202 (57.1202%) clutches had atleast one trait 2 animal.
577139 (57.7139%) clutches had atleast one trait 3 animal.
576245 (57.6245%) clutches had atleast one trait 4 animal.

Can yah tell I have nothing to do today?

Ben