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.

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

1/16 - 6.25 - 1 rooster
2/16 - 12.5% - 2 roosters
3/16 - 18.75% - 3 roosters
4/16 - 25% - 4 roosters

Elkslayer, something is wrong here.

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.

HHHH HTHH THHH HTHT

HHHT HTTH TTHH THTH

HHTT HHTH TTTH THHT

HTTT TTTT TTHT THTT

4 roostrrs (heads) 1/16
3 heads 4/16
2 heads 6/16 = 3/8
1 head 4/16

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.

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.

Huh?
Same probability of 3 as for 1?
Nope.

No that part is right. 1 T and 3 H is as likely as 3 T and 1 H.

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)

Nope

Probability HAS to reduce to get more roosters. 4 roosters is not MORE likely than 2 or 3.

I hate chicken.

1/16 - 6.25% - 4 rooster
2/16 - 12.5% - 3 roosters
3/16 - 18.75% - 2 roosters
4/16 - 25% - 1 roosters

I believe you are right. I just need to work the formula out.

That looks mo betta!

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

Bingo, or Yahtzee

To those pushing gender identity........and not capable of statistics.....this is a bullshit question......

Gonna bump this for when mathman shows up...

Ballistik is on it though...

Not quite.

What kinda roosters?

You getting into cock fightin? 😂

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).

So many different answers. Seems like an even spit would be a 50/50 chance, no?

You're forgetting there are more ways (for one example) to get three hens and one rooster than there are to get one rooster.

Mathman, what's the answer?

Ah yes, you are correct! I left that one out.

What Milwroad wrote with the addition of a 1/16 chance of zero roosters.

Authoritative answer, please. So far, many of the posts don't seem right. Seems like there should be a 50/50 chance of two roosters out of four eggs.

Okay. Thanks.

I'm hoping for one rooster, and three hens, in case anyone is wondering.

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.

Do these birds typically have a 50/50 shot at female/male? Or does nature slant them toward more females than males?

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.

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 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've seen video of the sorting process. Yeah you don't want to go down the shiny chute.

On a particular set of four eggs it's a 1/16 chance.

In the space of all sets of four eggs it's a 4/16 chance.

RHHH, HRHH, HHRH, HHHR are your one rooster, three hens arrangements out of

HHHH
HHHR
HHRH
HHRR
HRHH
HRHR
HRRH
HRRR
RHHH
RHHR
RHRH
RHRR
RRHH
RRHR
RRRH
RRRR

Geeze.....glad mathman showed up.

Its so clear now....

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?

LOL. Exactly.

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.

yep..

Math problems like these are why chickens and eggs are so popular at the supermarket.

LOL. What? You will have to explain that one.

4R 0H / 0R 4H
3R 1H / 1R 3H
2R 2H / 2R 2H
1R 3H / 3R 1H

RRRR
HHHH
RRRH
RHHH
RRHH
RRHH
RHHH
RRRH

Eight possibilities.

1/8 = 12.5% chance for 4 roosters.

This thread needs rooster fightin music.

Per usual, you're wrong.

If they are liberals, what are the chances one of them identifies as a hamster?

G23

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.

But why did the chicken cross the road?

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

(0, 1/16), (1, 4/16), (2, 6/16), (3, 4/16), (4, 1/16)

If we sum over the probabilities we have 1/16 + 4/16 + 6/16 + 4/16 + 1/16 = 16/16 = 1, the whole probability enchilada.

LOL !

To avoid all the statistical mathematics I suppose

Thanks, very succinct.

I think she wanted to be your Tennessee lamb, assuming you would be her Dixie chicken...

Hahahahaha, yeah, what if one of the hens identifies as a rooster?

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.

4 roosters (heads) 1/16
3 heads 4/16
2 heads 6/16 = 3/8
1 head 4/16
0 heads 1/16

1 in 16 = 0 heads 4 tails
4 in 16 = 1 heads 3 tails
6 in 16 = 2 heads 2 tails
4 in 16 = 3 heads 1 tails
1 in 16 = 4 heads 0 tails
------
16 in 16

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.

Classic.

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.

50/50 -- they are either hens or roosters -- there are no other choices, unless of course you get some LGBTQ 123 chickens.

For the farm folks here, is the chance of getting a hen or rooster the same? Serious question, I don't know.

They're either hens or roosters, true. But the probability that a given egg will produce a rooster may not be 50%.

I don't know what the accepted single egg probability is.

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.

What are the events being labeled independent?

i was wrong , ignore my answer.

The p=0.5 isn't what makes the eggs independent, though they are.

What about double yokers?

you are right . my answer was wrong.

For some foul reason!!

As usual, you demonstrate you are a pathetic mentally deranged cyber troll.

Been several decades since I took statistics.

Where is your homework?

Crickets.................

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:

For the visual learners....

One egg, two equally possible outcomes.

MF

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.

What was the first response?

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.

You're welcome. LOL.

What do you teach?

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..

One half to the fourth.

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.



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.

What are the odds of that happening?