Possible het's

I've asked around and I can't get a definitive answer from anybody. Since I have two poss het's what are my odd's of one being het? I remember some one saying if you had three the odd's were 87.5% but i don't know if that's right either? My worst subject in school was math but I gave it a try and came up with 59.1% am I right?:rolleye2: One more thing they are 50's.

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Originally Posted by Emilio

I've asked around and I can't get a definitive answer from anybody. Since I have two poss het's what are my odd's of one being het? I remember some one saying if you had three the odd's were 87.5% but i don't know if that's right either? My worst subject in school was math but I gave it a try and came up with 59.1% am I right?:rolleye2: One more thing they are 50's.

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If they are both 50% poss hets, what are you asking?

I know that what I'm asking is what are the odd's together that one will be a het?

think of it as 4 eggs/clutch..that helps me out

http://www.ballpythons.ca/what_get/recessive.html also helps me out a ton

The numbers you are saying, i've never heard of...all i know of is 100%, 66%, 50%, 25%

If you've got 2 possible hets, that means each one came from a clutch where some were normals and some are hets.

If i give you 2 snakes that look normal, but one carries a morph gene, then you have a 50/50 chance (or 50%) chance of choosing the right het.

You don't acctually add/multiply/divide the numbers to figure out what your odds are.

I think you're making it more than it is, 50% hets have a 50% chance of being Het, don't think there is anymore math you can do.

From what I remember from taking statistics, you could think of possible hets as flipping a coin. So, a simple formula applies like this:

Nh / N

with N = number of trials (in your case 2) and Nh being the number of times you have a het (or "land heads). The law of large numbers says that the more trials you run, the closer that ratio equals 50%.

Brad is right for the chance of say ..... seeing 1 head out of 2 coins, however there is the possibility of seeing 2 heads. You have two independent coins, so the "odds" of coin flips work like this.

Lets say you are intersested in the "odds" of seeing heads while flipping 2 coins. The possibilities and odds of different combinations are:

2 Heads 25% (1:3)
1 Head 1 Tail 50% (1:1)
2 Tails 25% (1:3)

If you are interested in the "odds" of seeing AT LEAST 1 Head, they are 75% or 3:1 (50% of seeing 1 plus 25% of seeing 2).

This is a pure statistical look at it .... remember you are dealing with nature:) .

Neil

Emilio,

I think I understand what you are asking. If you have 2 50% hets, each one has a 50% probability of being a het. But the probability that you have at least one het is calculated differently, though I'm not sure off the top of my head how to do it.

Think of it this way. If you have a single 50% het, the probability that you have a het is 50%. What if you have TEN 50% hets? What is the probability that you have at least one het? Off the top of my head, I think if you have 10 50% hets, the probability that you have at least FIVE hets would be 50%. The probability that ALL TEN are hets is much, much lower, and the probability that AT LEAST ONE is a het is much, much higher. This is why sometimes a person will try to buy an entire clutch of possible hets; to increase their odds of getting at least one het.

Considering the previous paragraph, if you have TWO 50% hets, what is the probability that AT LEAST ONE is a het?

Steve

Neil this is exactly what I was looking for.;) Brad it's one of those question's many of my co-worker's did not want to burn brain cell's on.Thank you

Not a problem. The rule that Brad used is correct when looking at how many would be het out of a group. Meaning if you had 10 50% hets, you woul d probably get 4, 5, or 6. But if you had 100 50 % hets your range would probably be between 45 and 55. So .45 and .55 are closer to .5 than .4 and .6 are ....... then you could get into the probabilities of ranges, like a 95% chance that the range would be 47 - 53, etc ... I hate statistics, but I had it two of my last three semsters ... one class left and I will have my degree (whew ... before I turn 40):laughing:

Neil

Yeah, I took two semesters of statistics and promply left all the knowledge I gained at the door when I left the final exam! T-tests and Chi squares OH MY....

I hope this isn't what I have in store for me next semester in Business Statistics. Statistics is supposed to be the easier side of math but I'll take calculus anyday.

That's exactly what you have in store Rich. I actually enjoyed statistics to a certain degree, at least the upper level zoology courses that dealt with it, not so much the stat class I took. Excel becomes your friend fast, as well as the other statistical programs designed to do all the work for you. You just have to be able to understand what the results mean.

-Evan

VCU's statistics program is horrible...they teach all of the statistical calculations on this program called Minitab...mind you this is Statistics for business majors....while almost every business out there is using Excel or an excel-like format.

I had to re-learn everything in order to be able do it on excel for other classes...not cool...

ewwwww Minitab, I loathe it !!!!!!

Neil

Minitab works great for fake, pre-designed textbook examples. For real world data...excel is much more useful.

Not to mention, while VCU statistics teaches stat in minitab, professors in the upper level classes where you use stat calculations want it done in excel. Go figure?

Emilio, dont overload your brain thinking about percentages. when it comes to bp hets anything goes. heres an example. I produced a bunch of 66% het.albinos, there were 3 females. I kept 2 females and sold a female to my buddy Ted Thompson. we raised them up and both of us bred the poss. het. females to albino males. all 3 females produced albinos, we then knew that they were indeed 100% hets. heres the other end of it. cinnamon pastel X cinnamon pastel 8 eggs that hatched out 8 normal ball pythons. you would expect to see a super or 2 in this breeding. thats just bad luck.I dont even like to think about percentages when breeding hets, they will make you look like a fool sometimes. :P

8 normals out of a cinny x cinny pairing? With that kind of luck, I would never step foot outside the house.

Okay maybe I'm just a simple Canadian chick here but with possible hets isn't it just either it is or it ain't sort of situation folks? In other words whether I buy 1 possible het or 100 possible hets....each het in and of itself has a 50/50 chance of either being a het or being a normal. So I could get lucky on that one purchase or extremely unlucky on that 100 purchase but in the end it's still about each snake and it's own individual chances of being het right?

Yes you are correct ... and the same thing COULD happen with coin flips. It is possible to flip 100 coins and have them all end up heads. That is not likely, but neither is getting 100 "true" 50% possible hets and having none of them actually be het.

In statistics they use the term "probabilities" to explain the likelyhood of something happening. So really correct terminology would be, if you have two 50% possible hets the probability that at least one is a het is 75%.

Neil

I guess i was trying to reinforce my decision on going the possible route LoL!!!!Great info guy's!!! 75% probability is awesome.

You know everytime I think about probabilities....I just can't help myself but to go to the Hitchhiker's Guide to the Galaxy and their improbability driven spaceship (bet that says something about me :rolleye2: )

You know it!!!

:P

Neil,
(or any of you other statistics gurus)

To turn this around and look at it from a slightly different angle...

Lets say you breed a homozygous (visible) albino to a 50% probable het albino and get 6 eggs. If the probable het IS het, you would expect to get 3 albino babies (statistically 50%).

If you get any albino babies at all, your 50% probable het has just been proven, and is now 100% probable.

Lets say you don't get any albinos; all babies appear normal. Obviously, all babies are hets due to the sire being homozygous, but thats not the point of my post. Your confidence level in that 50% probable het just went down. It doesn't mean she's not a het, just less likely (see Ed's post re: cin x cin). How would you QUANTIFY your decreased confidence?

Steve

In sobs per second :P




dr del

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Originally Posted by hoo-t

Lets say you don't get any albinos; all babies appear normal. Obviously, all babies are hets due to the sire being homozygous, but thats not the point of my post. Your confidence level in that 50% probable het just went down. It doesn't mean she's not a het, just less likely (see Ed's post re: cin x cin). How would you QUANTIFY your decreased confidence?

Steve

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Well first of all I am faaaaarrrrrr from a stats guru :D but, there are formulas for figuring confidence intervals (CI) based on related events and sequences, so I believe there is way to figure the CI that possible het is not a het based on the example you have given. I would have to dig the books out, but the answer would be formatted something like this:

"Based on 6 non-homozygus babies being produced, there X% confidence (lets say 90%) that the possible het is indeed not a het."

Again, the chance still exists, so you would have to breed again, and each successive breeding with no homozygus animals would increase that confidence interval.

Any breeders have a rule of thumb when proving possible hets as to when you finally say "she is not a het"?

This is starting to make my hair hurt ......

:rockon:

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Originally Posted by dr del

In sobs per second :P

dr del

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I agree!!!!

Neil

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Originally Posted by aaajohnson

Any breeders have a rule of thumb when proving possible hets as to when you finally say "she is not a het"?

Neil

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Thanks, Neil. Hmm. I'm starting to wonder if I have a stats book laying around somewhere! Its been 15 - 20 years since I took stats, but I kept most of those types of books. Anyone else know the formula?

Personally, if no homozygous animals were produced after 2 clutches, I think I'd give up and consider the "possible het" to be normal. My interest in this stems from the fact that we have 1.12 unproven "100%" het albinos (in addition to our homozygous pair), 1.0 unproven 66% het albino (daughter's pet), 1.1 unproven "100%" het pieds, and 0.6 unproven 66% het pieds. The males are all ready to go, but none of the females are big enough to breed yet. Just thinking about the future!

Steve

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Originally Posted by Emilio

I guess i was trying to reinforce my decision on going the possible route LoL!!!!Great info guy's!!! 75% probability is awesome.

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75% probability is good, however, it's wrong in your case. ;)

What you are asking for is a probability of two independant events, meaning whether one snake is het or not has no regard on whether the other is het or not. So here's your probabilities:

Both Normal = 25%
Both Het = 25%
One het, one normal = 50%

Here's the formula for determining the statistical probability of the same outcome from two independant events (two normals or two hets). We use .5 because there is a 50% chance of het and 50% chance of a normal:

P(A and B) = P(A) * P(B)
P(.5 and .5) = P(.5) * P(.5)
.5 * .5 = .25

.25 or 25% probility of both being the normal, .25 or 25% probability of both being het and of course, 50% probability that one is het one isn't :).

True, but that means 75% probability that AT LEAST 1 is het (50% + 25%)


Neil

This is a great thread!!!.... great info from everyone. Neil sorry about the hair ache. LOL

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Originally Posted by aaajohnson

True, but that means 75% probability that AT LEAST 1 is het (50% + 25%)

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Oy vay, you are indeed correct.
My earlier numbers read:

Both Normal = 25%
Both Het = 25%
One het, one normal = 50%

When they should have been:

Both Het or Both Normal = 25%
One het, one normal = 75%

Feeling a bit sheepish now :).

Now, let's try it with 3 hets:

P(A and B and C) = P(.5) * P(.5) * P(.5)

In this case you have a 12.5% chance of all three being the same (3 hets or three normals). Or, have a 87.5% chance of at least 1 being het.

4 hets:

P(A and B and C) = P(.5) * P(.5) * P(.5) * P(.5)

6.25% chance of them all being the same or a 93.75% chance of at least one het.

I'm by no means an expert, but I believe your original numbers are correct.

Both normal 25%
Both het 25%
one het, one normal 50%

The same distribution applies when breeding a het to het -

normal (both genes normal) 25%
homozygous (both genes morph) 25%
het (one normal gene, one morph) 50%

The problem is interpreting "AT LEAST ONE" in the original question.

"AT LEAST ONE" includes BOTH possibilities: "both hets = 25%" AND "One het, one normal = 50%". So you must add the two together to get the probability of "AT LEAST ONE". 25% + 50% = 75% probability of AT LEAST ONE het.

Right????
Steve

To add to my previous post, I think your calculations are working properly, but you have an incorrect assumption. You are assuming that you are calculating the probability of both being the same. In reality, you are calculating the probability of there being NO hets. In the sample of 4, there is a 6.25% probability that you get NO hets, leaving a 93.75% probability of AT LEAST ONE, which includes 1, 2, 3, 4, 5, or 6 hets.

Steve

imo, the more possible hets you get from the same clutch, the greater chances you have of one of them being het

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Originally Posted by hoo-t

I'm by no means an expert, but I believe your original numbers are correct.

Both normal 25%
Both het 25%
one het, one normal 50%

The same distribution applies when breeding a het to het -

normal (both genes normal) 25%
homozygous (both genes morph) 25%
het (one normal gene, one morph) 50%

The problem is interpreting "AT LEAST ONE" in the original question.

"AT LEAST ONE" includes BOTH possibilities: "both hets = 25%" AND "One het, one normal = 50%". So you must add the two together to get the probability of "AT LEAST ONE". 25% + 50% = 75% probability of AT LEAST ONE het.

Right????
Steve

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Yes, that is correct. I realized my mistake after reading this thing 100 times and finally figuring out I simply forgot to add the two that had hets :D.

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To add to my previous post, I think your calculations are working properly, but you have an incorrect assumption. You are assuming that you are calculating the probability of both being the same. In reality, you are calculating the probability of there being NO hets. In the sample of 4, there is a 6.25% probability that you get NO hets, leaving a 93.75% probability of AT LEAST ONE, which includes 1, 2, 3, 4, 5, or 6 hets.

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I'm actually calculating the probability that all animals will be the same (all hets or all normals) and the probabilities I listed show (in your example) a 6.25% probability of their either being ALL hets or ALL normals. Which, it obviously follows that subtracting .0625 from 1.00 will give you the total probability of the original equation (all hets or all normals) NOT being true (which, as you stated, would be having a distribution of 1+ normals and 1+ hets).

The irony in all this, I'm currently working on a project where given a set of visitor demographics, our equation is to determine the likely "peak times" of customer trends (purchasing, calling the number, contacting support, etc). And yet, I screwed up (twice no less) on a simple independant probability equation :D. Too many statistics for one day me thinks.

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Originally Posted by Kizerk

imo, the more possible hets you get from the same clutch, the greater chances you have of one of them being het

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That's not true ... a possible het is a possible het ... period. You have the exact same chances of proving or not proving 3 female possible hets whether they all come from the same clutch or from 3 different clutches ... Psycholigically, it "feels better" to have all the girls from a clutch for example, but in reality it gains you no advantage what so ever.

I've been producing, buying, and proving possible hets for many many years and the key is quantity ... nothing more. ;)

-adam

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Originally Posted by Adam_Wysocki

I've been producing, buying, and proving possible hets for many many years and the key is quantity ... nothing more. ;)

-adam

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Bingo ... and statistically speaking that can be proven ..... :D

Neil

Statistics suck...but useful in any discipline...we're breeding snakes for crying out loud, and all those core statistical concepts still manage to creep in...

good info guys!

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Originally Posted by jhall1468

Yes, that is correct. I realized my mistake after reading this thing 100 times and finally figuring out I simply forgot to add the two that had hets :D.



I'm actually calculating the probability that all animals will be the same (all hets or all normals) and the probabilities I listed show (in your example) a 6.25% probability of their either being ALL hets or ALL normals. Which, it obviously follows that subtracting .0625 from 1.00 will give you the total probability of the original equation (all hets or all normals) NOT being true (which, as you stated, would be having a distribution of 1+ normals and 1+ hets).

The irony in all this, I'm currently working on a project where given a set of visitor demographics, our equation is to determine the likely "peak times" of customer trends (purchasing, calling the number, contacting support, etc). And yet, I screwed up (twice no less) on a simple independant probability equation :D. Too many statistics for one day me thinks.

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When I posted, I was rushing to get out the door to go to work. But, as I was thinking about your post, I realized that you had answered my question about diminishing confidence. I just now got home from work, and I still gotta think about it some, but I know the answer is there!

Steve

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Originally Posted by elevatethis

Statistics suck...but useful in any discipline...we're breeding snakes for crying out loud, and all those core statistical concepts still manage to creep in...

good info guys!

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Its amazing, when I was taking statistics type classes, I figured I'd never, ever use it. Now here I am! And while I found biology as a whole very interesting, I never paid a whole lot of attention to genetics. Punnett squares? I didn't even THINK about them again until my kids needed help with their school work. Now I'm fascinated enough with genetics, that I'm trying to learn which traits are dominant/recessive in rats and mice! Who cares what color the snake food is? Well, for one, I do!!!!

Steve

Lets use X for a dom trait, and O for normal.
Animals have two copies of each gene. During gamete (sperm/egg) production, only one copy of each gene is passed on. Actually one cell splits into 4 gametes technically, but that doesnt matter in this example, so for the sake of making it simpler lets just say one cell forms one gamete.

lets say dad ha OX and mom has XX
Dads sperm will be randomly O or X.
Moms eggs will be X
If dad only formed and released 4 sperm at a time and fertilized 4 eggs in the mother, then the outcome would be guarenteed to be XX, XX, XX, and OX. The problem with that is that dad makes many sperm and out of that group only some fertilize the eggs. Lets say dad makes X, X, X, X, O, O, O, O. That group of sperm follows the "rules" and is half and half. But, only part of that group gets to mom's eggs. That group could be any random group of 4 (to stick with the example, but it would be how every many eggs are fertilized) That means the possible combinations for the genes that make it to the eggs are XXXX, XXXO, XXOO, XOOO, and OOOO. There are more possibilities than that because in XXOO each X could be any of the 4 X's and each O could be any of the 4 O's. That is why its possible for several in a clutch to prove out, or for none of them to prove out. I hope this made sense.

PS, this was meant to go after what is now post #44
in this thread.

hrm...

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Originally Posted by Adam_Wysocki

That's not true ... a possible het is a possible het ... period. You have the exact same chances of proving or not proving 3 female possible hets whether they all come from the same clutch or from 3 different clutches ... Psycholigically, it "feels better" to have all the girls from a clutch for example, but in reality it gains you no advantage what so ever.

I've been producing, buying, and proving possible hets for many many years and the key is quantity ... nothing more. ;)

-adam

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oh ok, i know what you mean

psychologically , i would feel better buying all the possiblet hets, in a clutch, because i feel like it increases my chances of getting a het, which isn't true...b/c what's there is there, thanks for helping me sort that out

Oy!! This thread makes my head hurt! LOL


Great information - we really do have such a great group of contributors here!

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Originally Posted by piranhaking

He's right. We usually use 4 eggs as an example, so say there is a 25% chance of an egg having the gene, then 1 out of 4 of the babies will have the gene. On average in a large number of groups of eggs, 1 in 4 would have the gene but it is still completely random, and there is no guarentee that one of any clutch of four would have the gene. However, the way it is often discussed it would be easy to think it means that out of 4 eggs one is guarenteed to have the gene. If we KNEW that one out of four eggs in a clutch has the gene, then you more you buy from one clutch the better your chances would be, but since there is no guarentee that the gene is passed on to one of the eggs buying more from the same clutch doesnt affect your chances. I hope that made sense....

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That's incorrect, both technically and in real-life use. When we discuss "50%" it literally means that any one individual snake has a 50% chance of being het.

So regardless of whether you pick up an entire clutch or pick and choose your snakes that statistics won't vary. Again, we have to realize that nothing we can do when picking hets will increase the probability of getting "more hets" because whether each snake is het or not is completely independant of whether the next snake is.

Whether we pick and choose in a closed system (only from the same clutch) or an open system (regardless of whether it's from the same clutch) the variables don't change, so the odds remain the same.

edit: sorry after re-reading it I realized you agreed with Adam :).

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Originally Posted by hoo-t

When I posted, I was rushing to get out the door to go to work. But, as I was thinking about your post, I realized that you had answered my question about diminishing confidence. I just now got home from work, and I still gotta think about it some, but I know the answer is there!

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Well at least something good came out of it. We can really get creative with this as well, think double hets, double homozygous :).

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Originally Posted by Adam_Wysocki

That's not true ... a possible het is a possible het ... period. You have the exact same chances of proving or not proving 3 female possible hets whether they all come from the same clutch or from 3 different clutches ... Psycholigically, it "feels better" to have all the girls from a clutch for example, but in reality it gains you no advantage what so ever.

I've been producing, buying, and proving possible hets for many many years and the key is quantity ... nothing more. ;)

-adam

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He's right. We usually use 4 eggs as an example, so say there is a 25% chance of an egg having the gene, then 1 out of 4 of the babies will have the gene. On average in a large number of groups of eggs, 1 in 4 would have the gene but it is still completely random, and there is no guarentee that one of any clutch of four would have the gene. However, the way it is often discussed it would be easy to think it means that out of 4 eggs one is guarenteed to have the gene. If we KNEW that one out of four eggs in a clutch has the gene, then you more you buy from one clutch the better your chances would be, but since there is no guarentee that the gene is passed on to one of the eggs buying more from the same clutch doesnt affect your chances. I hope that made sense....

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Originally Posted by piranhaking

If we KNEW that one out of four eggs in a clutch has the gene, then you more you buy from one clutch the better your chances would be, but since there is no guarentee that the gene is passed on to one of the eggs buying more from the same clutch doesnt affect your chances. I hope that made sense....

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Very small word, very important meaning in this case sorry for the mix up, i should have done that the first time through. :P Did you see my other post that is out of order that goes into a little more detail on that?