I have four fertilized eggs in an incubator. Assuming that each egg, if it produces a chick, has a 50% chance of being a rooster, what are the odds that all four will be roosters, assuming all four hatch? What are the odds that three will be? What are the odds that two will be? What are the odds that only one will be? Zero roosters? How do you arrive at that? Thanks.
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
"Social order at the expense of Liberty is hardly a bargain” de Sade "He who'll not reason is a Bigot, he who cannot is a Fool, and he who dares not is a Slave."SirWilliamDrummond
Each egg has a 1/2 chance of being male. The odds are the same as flipping a coin the same way four times in a row.
You have eight possible outcomes. Coin 1 either H or T Coin 2 either H or T Coin 3 either H or T Coin 4 either H or T
So zero H and four T Or one H and three T Or two H and two T Or three H and one T Or four H and zero T
If I remember properly you will have odds 1/8 that all will be male and 1/8 that all will be female.
Then 1/4 that one will be male or 1/4 that one will be female.
And 1/2 that you will have an even split.
I fail to remember the calculations. I am sure there is a derivative to calculate the curve. But that was too many years ago.
Maybe someone will correct me if I have erred.
ETA, I think Ballistic has the correct formula. 1 X 2 to the fourth for all male and 1 x 2 to the fourth for all female. 1 x 2 to the third for three males and 1 x 2 to the third for three females 1 x 2 to the second for two males and 1 x 2 to the second for two females
Last edited by Idaho_Shooter; 04/15/21.
People who choose to brew up their own storms bitch loudest about the rain.
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
You have it twice as likely to get four of the same sex as to get an even split.
And you have the odds of getting three females at 15/16 while odds of getting three males is 3/16.
Ballistik had it right. You have to get the common denominator first. 2x2x2x2, then you factor it out. There are two possibilities for each egg. You're only accounting for one with the eight denominator.
if you take your 4H x 4T = 16, you get the same correct common denominator you need to run the formula correctly.
"He is far from Stupid"
”person, who happens to have an above-average level of intelligence”
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
I’m going with this until mathman sees this! 🤣
1/2 x 1/2 x 1/2 x 1/2 for all four to be roosters.
Last edited by ironbender; 04/15/21. Reason: Friggen autocorrect
If you take the time it takes, it takes less time. --Pat Parelli
American by birth; Alaskan by choice. --ironbender
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
I’m going with this until mathman sees this! 🤣
1/2 x 1/2 x 1/2 x 1/2 for all four to be roosters.
1/2 x 1/2 x 1/2 x 1/2 = 6.25% (1 rooster)
.5 x .5 x .5 x .5 = 6.25% (1 being the numerator for 1 rooster gives you the correct % there)
2/2 x 1/2 x 1/2 x 1/2 = 2/16 = 12.5% ( 2 roosters)
3/2 x 1/2 x 1/2 x 1/2 = 3/16 = 18.75% (3 roosters)
4/2 x 1/2 x 1/2 x 1/2 = 4/16 = 25% (4 roosters)
Last edited by ElkSlayer91; 04/15/21.
"He is far from Stupid"
”person, who happens to have an above-average level of intelligence”
highest probability: two of each second highest: one of either sex and three of the other...equal probability either way lowest probability: all of one sex, either male or female
People who choose to brew up their own storms bitch loudest about the rain.
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
Bingo, or Yahtzee
Originally Posted by mauserand9mm
Originally Posted by mauserand9mm
Originally Posted by Raspy
Whatever you said...everyone knows you are a lying jerk.
That's a bold assertion. Point out where you think I lied.
I believe I was correct. I wrote down all the possibilities. This is a well known formula for getting r successes in N independent trials with probability of success p.
See https://www3.nd.edu/~rwilliam/stats1/x13.pdf
The difference between genius and stupidity is that genius has its limits.- Albert Einstein
I believe I was correct. I wrote down all the possibilities. This is a well known formula for getting r successes in N independent trials with probability of success p.
See https://www3.nd.edu/~rwilliam/stats1/x13.pdf
Almost. Your probabilities don't sum to 1 since you left out the event of zero roosters (heads).
I believe I was correct. I wrote down all the possibilities. This is a well known formula for getting r successes in N independent trials with probability of success p.
See https://www3.nd.edu/~rwilliam/stats1/x13.pdf
Almost. Your probabilities don't sum to 1 since you left out the event of zero roosters (heads).
Originally Posted by mathman
Originally Posted by Milwroad
I believe I was correct. I wrote down all the possibilities. This is a well known formula for getting r successes in N independent trials with probability of success p.
See https://www3.nd.edu/~rwilliam/stats1/x13.pdf
Almost. Your probabilities don't sum to 1 since you left out the event of zero roosters (heads).
Ah yes, you are correct! I left that one out.
The difference between genius and stupidity is that genius has its limits.- Albert Einstein
A few years back when we still had some laying hens we had one in particular that was pretty broody. I would pull several eggs out from under her every day, her's and a few of the other hen's. One day I decide to let her go ahead and set them out so I marked the 9 eggs she was sitting on with a Sharpie. Each day after I would pull any eggs that weren't marked. After 3 weeks they started hatching, 7 of the 9 hatched, I tossed the other 2 after several days and it was apparent they weren't gonna hatch. I was hoping for 4 or 5 new layers, but all 7 of them darn chicks turned out to be roosters.
Why would two roosters only have a 3/8th chance? If one egg has a 50% chance of producing a rooster, shouldn't four eggs have a 50% chance of producing two roosters? Confusing. Not that you're not correct. It's just that, to a non-mathematician, it seems counterintuitive.
Do these birds typically have a 50/50 shot at female/male? Or does nature slant them toward more females than males?
I wish. Unfortunately, roosters are 50% of eggs hatched. Hatcheries dispose of half the chicks that hatch by sending them down a chute that takes them, alive, to a grinder. Pretty sad for the little rooster chicks.
A few years back when we still had some laying hens we had one in particular that was pretty broody. I would pull several eggs out from under her every day, her's and a few of the other hen's. One day I decide to let her go ahead and set them out so I marked the 9 eggs she was sitting on with a Sharpie. Each day after I would pull any eggs that weren't marked. After 3 weeks they started hatching, 7 of the 9 hatched, I tossed the other 2 after several days and it was apparent they weren't gonna hatch. I was hoping for 4 or 5 new layers, but all 7 of them darn chicks turned out to be roosters.
They did eat well.
I bought four chicks from Tractor Supply a few years ago, and the girl who collected them for me assured me they were guaranteed to have been checked, and were all hen chicks. Three turned out to be roosters and only one was a hen. When I went back to complain, no one knew anything about any guarantee.
Do these birds typically have a 50/50 shot at female/male? Or does nature slant them toward more females than males?
I wish. Unfortunately, roosters are 50% of eggs hatched. Hatcheries dispose of half the chicks that hatch by sending them down a chute that takes them, alive, to a grinder. Pretty sad for the little rooster chicks.
I've seen video of the sorting process. Yeah you don't want to go down the shiny chute.
It's no wonder that, despite massive effort on my part, I could only wrangle a B in undergrad stats, and another B in grad stats. These concepts elude me.
The first thing we need to know, is are they chicken eggs?
Yes. Gallus gallus domesticus. Collected from my last remaining hen who laid them after my one and only rooster was hit by a car. Hens produce fertile eggs for about a month after the rooster dies.
Still waiting on Mathman to show us the proper formula and how it plots out in two dimensions on the X vs Y axis. Is it algebra, or is it Trig?
I know the curve is exponential from a base point of 1/1 in both directions. As you add flips to the count, the odds of them all being the same approaches zero on the X (heads) and on the Y (tails) axis. But I can not remember the name of the curve.
Come on Mathman, help us out.
People who choose to brew up their own storms bitch loudest about the rain.
God bless Texas----------------------- Old 300 I will remain what i am until the day I die- A HUNTER......Sitting Bull Its not how you pick the booger.. but where you put it !! Roger V Hunter
It's discrete. What we're doing is taking a sample of four pulls from a binomial distribution. If we call roosters successes then we're asking what's the probability of one success in four pulls. The formula would be (the number of combinations of one success in four tries) times (the probability of success)^1 times (the probability of failure)^(4-1). Since the prob of success and prob of failure are here both assumed to be 1/2 them the prob product is always (1/2)^4 = 1/16 no matter how many successes we're investigating. The one success in four tries combination is 4!/(1! x (4-1)1) = 4!/3! = 4. So for one rooster in four tries comes out as a 4 x 1/16 = 4/16 = 1/4 chance.
two roosters 4!/(2! x 2!) = 24/4 = 6 ways, so then 6/16 = 3/8 prob
three roosters 4!/(1! x 3!) = 4 ways, so then 4/16 = 1/4 prob (Notice the symmetry, since roosters and hens are assumed equally likely then three of one and one of the other doesn't care which is the one and which are the three.)
four roosters 4!/(0! x 4!) = 1 then 1/16
zero roosters is same as four hens by symmetry so zero roosters prob is also 1/16
So in terms of ordered pairs with (how many roosters, probability) we have
"Allways speak the truth and you will never have to remember what you said before..." Sam Houston Texans, "We say Grace, We Say Mam, If You Don't Like it, We Don't Give a Damn!"
It's discrete. What we're doing is taking a sample of four pulls from a binomial distribution. If we call roosters successes then we're asking what's the probability of one success in four pulls. The formula would be (the number of combinations of one success in four tries) times (the probability of success)^1 times (the probability of failure)^(4-1). Since the prob of success and prob of failure are here both assumed to be 1/2 them the prob product is always (1/2)^4 = 1/16 no matter how many successes we're investigating. The one success in four tries combination is 4!/(1! x (4-1)1) = 4!/3! = 4. So for one rooster in four tries comes out as a 4 x 1/16 = 4/16 = 1/4 chance.
two roosters 4!/(2! x 2!) = 24/4 = 6 ways, so then 6/16 = 3/8 prob
three roosters 4!/(1! x 3!) = 4 ways, so then 4/16 = 1/4 prob (Notice the symmetry, since roosters and hens are assumed equally likely then three of one and one of the other doesn't care which is the one and which are the three.)
four roosters 4!/(0! x 4!) = 1 then 1/16
zero roosters is same as four hens by symmetry so zero roosters prob is also 1/16
So in terms of ordered pairs with (how many roosters, probability) we have
A few years back when we still had some laying hens we had one in particular that was pretty broody. I would pull several eggs out from under her every day, her's and a few of the other hen's. One day I decide to let her go ahead and set them out so I marked the 9 eggs she was sitting on with a Sharpie. Each day after I would pull any eggs that weren't marked. After 3 weeks they started hatching, 7 of the 9 hatched, I tossed the other 2 after several days and it was apparent they weren't gonna hatch. I was hoping for 4 or 5 new layers, but all 7 of them darn chicks turned out to be roosters.
They did eat well.
I bought four chicks from Tractor Supply a few years ago, and the girl who collected them for me assured me they were guaranteed to have been checked, and were all hen chicks. Three turned out to be roosters and only one was a hen. When I went back to complain, no one knew anything about any guarantee.
I think she wanted to be your Tennessee lamb, assuming you would be her Dixie chicken...
-OMotS
"If memory serves fails me..." Quote: ( unnamed) "been prtty deep in the cooler todaay "
To those pushing gender identity........and not capable of statistics.....this is a bullshit question......
Hahahahaha, yeah, what if one of the hens identifies as a rooster?
Progressives are the most open minded, tolerant, and inclusive people on the planet, as long as you agree with everything they say, and do exactly as you're told.
Yep. If you get 3 heads that means you also got 1 tail. The probability of one tail has to equal the probability of one head, therefore 3 heads = 1 tail = 1 head, so the probability of 3 heads =1 head.
The probability of 1 hen is the same as 1 rooster. If either of those happen, regardless of the arrangement, the other 3 are the other sex. Can't see how anyone can argue otherwise, unless I'm missing something.
Progressives are the most open minded, tolerant, and inclusive people on the planet, as long as you agree with everything they say, and do exactly as you're told.
Progressives are the most open minded, tolerant, and inclusive people on the planet, as long as you agree with everything they say, and do exactly as you're told.
The probability of 1 hen is the same as 1 rooster. If either of those happen, regardless of the arrangement, the other 3 are the other sex. Can't see how anyone can argue otherwise, unless I'm missing something.
That's good for when on a given egg the probability of a rooster is the same as the probability of a hen. If it's not 50/50 then while there are the same number of ways to get one rooster as there are for one hen in four tries, but the probability is skewed.
The probability of 1 hen is the same as 1 rooster. If either of those happen, regardless of the arrangement, the other 3 are the other sex. Can't see how anyone can argue otherwise, unless I'm missing something.
That's good for when on a given egg the probability of a rooster is the same as the probability of a hen. If it's not 50/50 then while there are the same number of ways to get one rooster as there are for one hen in four tries, but the probability is skewed.
For the farm folks here, is the chance of getting a hen or rooster the same? Serious question, I don't know.
Progressives are the most open minded, tolerant, and inclusive people on the planet, as long as you agree with everything they say, and do exactly as you're told.
I have never heard other than 50/50 odds on chickens.
In some species, semen is selected for gender before artificial insemination. The breeders are very successful selecting for males when breeding cattle for bull farms.
People who choose to brew up their own storms bitch loudest about the rain.
Two eggs, same two outcomes for the first egg if the other egg is male.... (MM MF) or female.... (FM FF) hence four equally possible outcome
MM MF
FM FF
Three eggs, same four outcomes for the original two eggs if the third egg is a male, and the same four possible outcomes for the original two eggs if the third eggs is a female.... hence eight equally possible outcomes
MMM MMF MFM MFF
FMM FMF FFM FFF
Four eggs, same eight possible outcomes for the original three eggs if the fourth egg is a male, same eight possible outcomes for those three eggs if the fourth egg is a female. Hence sixteen equally possible outcomes.
MMMM MMMF MMFM MMFF MFMM MFMF MFFM MFFF
FMMM FMMF FMFM FMFF FFMM FFMF FFFM FFFF
One out of sixteen probability of all male MMMM (1/16=0.0625. 6.25%). Same one out of sixteen probability of all females FFFF.
The other combinations yer gonna have to count for yourselves.
"...if the gentlemen of Virginia shall send us a dozen of their sons, we would take great care in their education, instruct them in all we know, and make men of them." Canasatego 1744
1/(2x2x2x2) for the odds of all roosters, 1/(2x2x2) for 3 roosters or hens, etc. It's been quite a while since I took a course that covered statistics/probability material, but I think it's right. 1/16 of a chance for all roosters or hens.
You didn't use logic or reason to get into this opinion, I cannot use logic or reason to get you out of it.
You cannot over estimate the unimportance of nearly everything. John Maxwell
Never have permutations and combinations and the binomial theorem received so much interest as they have in this thread! I might assign this to my students as a good reason to study random draws.
The difference between genius and stupidity is that genius has its limits.- Albert Einstein
Never have permutations and combinations and the binomial theorem received so much interest as they have in this thread! I might assign this to my students as a good reason to study random draws.
Never have permutations and combinations and the binomial theorem received so much interest as they have in this thread! I might assign this to my students as a good reason to study random draws.
Never have permutations and combinations and the binomial theorem received so much interest as they have in this thread! I might assign this to my students as a good reason to study random draws.
Big scrabble score post.
At least we have identified a teacher looking out side the approved curriculum. I guess that means someone is going to label you a racist... Just ignore them... And keep teaching..
-OMotS
"If memory serves fails me..." Quote: ( unnamed) "been prtty deep in the cooler todaay "
As usual, you demonstrate you are a pathetic mentally deranged cyber troll. Been several decades since I took statistics. Where is your homework? Crickets.................
Originally Posted by antelope_sniper
The correct answer had already been provided so I saw no reason to duplicate the work. I find it interesting how different you response is from that of Persian Dog:
Originally Posted by persiandog
you are right . my answer was wrong.
My response to iron bender, admitting I was wrong proves “beyond a shadow of a doubt”, you are a lying troll for using persiandog’s response in an attempt to say I wasn’t man enough to admit I was wrong. I was right from the beginning concerning the 16 denominator, but changed later to an 8, which was wrong too. I was over thinking it, and was tired. The 8 was wrong, but I did have the possible outcomes correct. That was the easy part. Math was my favorite subject all throughout my education. I always made “A”s the majority of the time.
Originally Posted by ElkSlayer91
Originally Posted by ironbender
Nope
Probability HAS to reduce to get more roosters. 4 roosters is not MORE likely than 2 or 3.
I believe you are right. I just need to work the formula out.
Now, you did not quote any other person in this thread to inform them they were wrong, which proves you are nothing but a mentally sick human being, a pathetic mass of cells who waste’s their life away on a computer as a cyber stalking troll.
I came to this thread to have some fun, and it was already extremely late that night, way past midnight if I remember. I was waiting for some more reports to come out on another site concerning the Covid-19 fraud. Lots of evidence is pouring out right now proving all of these shots create permanent blood clots, and there is a high probability millions will die eventually from taking these illegal / unsafe / unproven / deadly / shots.
You trolls on this site are some of the most mentally sick pieces of human excrement I’ve ever come across in my life.
You said I was “wrong as usual”, yet you can not refute one single post of mine with scientific evidence to prove I’m wrong concerning CV19 or any other posting I’ve made on this site, which concretely cements the fact you are a piece of human garbage, and nothing more.
I will not answer any reply you make, because I’m not going to turn TRH’s thread into a pizzing match with scum like you. I’ve made my point with concrete evidence.
Thanks for taking the time, mathman, to lay it out for those following this.
"He is far from Stupid"
”person, who happens to have an above-average level of intelligence”