Odd's

Alright who's good with odd's calculations? What are the odd's of me getting a pied from a hetpied x hetpied breeding with 3 egg's?:please: Math was never my favorite subject.. I know each egg has a 25% chance thats not what I'm looking for.

My dad is a GENIUS. Listen to this: Each egg has a 75% normal, 25% pied, so thats 3:1 odds AGAINST you. Multiply by three (3 eggs), that's 9:3, 3:1 33% chance you'll get pied.

And of course this is if she proves out.:gj::please::tears: Either way more poss het girls are always good to.

uhhhh goodluck lol

It's exciting to say the least.

Trust me, I'm horrible at math. But the ultimately wise Dad of mine is surprisingly good at it. :D

So more than likely you may not get the pied but then I think you get 1 normal and 66 percent het pieds on average but since it is less than four eggs it will kind of be harder to calculate it.

I'm gonna call it right now. One pied.

I figure it a little different. Starts out the same, each egg from het pied X het pied has a 1/4 chance of being a homozygous pied so a 3/4 chance of not being visible pied. Then to find the odds of all three eggs not being pied you multiple the chance of each not being one together, so (3/4)^3 = 27/64 (so instead of multiplying by 3 you multiply by it's self three times, or how ever many eggs you have times). So if 27/64 is your odds of getting none then 37/64 is your chance of getting at least one. So I'm showing 57.8% chance of getting at least one pied from three eggs from het X het. So a good bit more likely to get at least one than to get none if you hatch all three eggs. However, as long as you are sure the parents are both hets all normal offspring are 66% chance hets.

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Originally Posted by RandyRemington

I figure it a little different. Starts out the same, each egg from het pied X het pied has a 1/4 chance of being a homozygous pied so a 3/4 chance of not being visible pied. Then to find the odds of all three eggs not being pied you multiple the chance of each not being one together, so (3/4)^3 = 27/64 (so instead of multiplying by 3 you multiply by it's self three times, or how ever many eggs you have times). So if 27/64 is your odds of getting none then 37/64 is your chance of getting at least one. So I'm showing 57.8% chance of getting at least one pied from three eggs from het X het. So a good bit more likely to get at least one than to get none if you hatch all three eggs. However, as long as you are sure the parents are both hets all normal offspring are 66% chance hets.

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You just boggled my mind.
:O:O

Here's hoping the odd's Gods bless you with a pied :) or two

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Originally Posted by RandyRemington

I figure it a little different. Starts out the same, each 57egg from het pied X het pied has a 1/4 chance of being a homozygous pied so a 3/4 chance of not being visible pied. Then to find the odds of all three eggs not being pied you multiple the chance of each not being one together, so (3/4)^3 = 27/64 (so instead of multiplying by 3 you multiply by it's self three times, or how ever many eggs you have times). So if 27/64 is your odds of getting none then 37/64 is your chance of getting at least one. So I'm showing 57.8% chance of getting at least one pied from three eggs from het X het. So a good bit more likely to get at least one than to get none if you hatch all three eggs. However, as long as you are sure the parents are both hets all normal offspring are 66% chance hets.

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57% don't you think thats a little high.

I only had one semester of college statistics and I did find the hot lady professor from Iran a little distracting but I believe my formula is correct.

The earlier posted formula of taking the odds for one egg and multiplying the ration by the number of eggs results in the odds always staying the same regardless of the number of eggs. 3:1 X 3 = 9:3 which reduces to 3:1 just like 3:1 X 100 = 300:100 which also reduces to 3:1. Doesn't it make sense that the more dice you roll (i.e. eggs you hatch) the better your chance of hitting what you want? With only 1 egg you only have a 25% chance, so surely 2 eggs should give you a better chance and 3 better than that.

You also can't just add odds because then you would quickly get over 100%. For example, if each egg has a 25% chance of hatching pied if you could just add them you would be 100% guaranteed a pied every 4 eggs (25% + 25% + 25% + 25% = 100%). We also know it doesn't work that way because people some times miss on producing the visual one year and hit later.

But with the formula I used of multiplying the odds of not getting any together we see odds of striking out getting smaller with the number of eggs hatched but never reaching 0%. There is always some chance of not getting any but with a large number of eggs that chance gets very small. Then when you subtract the chance of none from 100% you get the chance of getting one or more. You are never 100% guaranteed to get any (i.e. this formula never quite reaches 100%). But the more eggs you hatch the closer it gets.

The general formula for hatching 1 or more of what you want is 1 - p^n where p is the percentage chance of missing per egg (i.e. 0.75 or 75% for het X het) and n is the number of eggs. If this clutch had 5 good eggs and they all hatched the odds of getting at least one piebald would be 1 - 0.75^5 = 1 - 0.2373 = 0.7627 or 76.27%. With 10 eggs the odds of getting at least 1 piebald go up to 94.37%. With 100 eggs 99.99999999997%; there is still a 1 in 3 trillion chance of not getting any pieds if you are very very very unlucky.

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Originally Posted by RandyRemington

With 100 eggs 99.99999999997%; there is still a 1 in 3 trillion chance of not getting any pieds if you are very very very unlucky.

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i think id be ready to say, dude its not a het.

Luck > statistics :)

i'd say u have 0 if your unlucky, 3 if ur a cheapass lucky basterd :d

:banana::banana: post results ! :d :gj:

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Originally Posted by RandyRemington

I figure it a little different. Starts out the same, each egg from het pied X het pied has a 1/4 chance of being a homozygous pied so a 3/4 chance of not being visible pied. Then to find the odds of all three eggs not being pied you multiple the chance of each not being one together, so (3/4)^3 = 27/64 (so instead of multiplying by 3 you multiply by it's self three times, or how ever many eggs you have times). So if 27/64 is your odds of getting none then 37/64 is your chance of getting at least one. So I'm showing 57.8% chance of getting at least one pied from three eggs from het X het. So a good bit more likely to get at least one than to get none if you hatch all three eggs. However, as long as you are sure the parents are both hets all normal offspring are 66% chance hets.

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Exactly right, Randy. Odds are 1-(3/4)^3= 1-(27/64)=37/64 just as you said, a tad over a 1 out of 2 chance.

JonV