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 "
Television and radio are most effective when people question little and think even less.
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.
