A Math Question

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triviawayne
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A Math Question

Post by triviawayne »

My wife decided to buy chickens. She is hoping she doesn't end up with a rooster.

She said the lot of birds the store gets their supply from claims 80% are hens.

Since she bought 4 birds, I think the odds are 625:1 that one of them is a rooster.

Am I correct?
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Re: A Math Question

Post by Volante »

triviawayne wrote:
Tue Jun 30, 2020 3:09 pm
My wife decided to buy chickens. She is hoping she doesn't end up with a rooster.

She said the lot of birds the store gets their supply from claims 80% are hens.

Since she bought 4 birds, I think the odds are 625:1 that one of them is a rooster.

Am I correct?
Trying to dust the cobwebs off this part of my brain before working with numbers, but I'm concerned about your phrasing "that one of them is a rooster"
Is that the problem or "that at least one of them is a rooster" You could get up to four.

Maybe work the other way....
The chance of getting 4 hens, 80%*80%*80%*80% = 41%, so you have a 59% chance of getting at least one rooster?

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Re: A Math Question

Post by MattKnowles »

Are they African or European chickens?
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triviawayne
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Re: A Math Question

Post by triviawayne »

Volante wrote:
Tue Jun 30, 2020 3:19 pm
triviawayne wrote:
Tue Jun 30, 2020 3:09 pm
My wife decided to buy chickens. She is hoping she doesn't end up with a rooster.

She said the lot of birds the store gets their supply from claims 80% are hens.

Since she bought 4 birds, I think the odds are 625:1 that one of them is a rooster.

Am I correct?
Trying to dust the cobwebs off this part of my brain before working with numbers, but I'm concerned about your phrasing "that one of them is a rooster"
Is that the problem or "that at least one of them is a rooster" You could get up to four.

Maybe work the other way....
The chance of getting 4 hens, 80%*80%*80%*80% = 41%, so you have a 59% chance of getting at least one rooster?
For the first part - we're hoping none of them are roosters.

For the second part...that was one of the ways I thought of it too, but then it just didn't feel like I did the math correct?

This is where statistics make me just think "HUH????????"

I guess what I'm looking for is some hope. What we want are ZERO roosters. What are the chances we have ZERO roosters?
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triviawayne
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Re: A Math Question

Post by triviawayne »

MattKnowles wrote:
Tue Jun 30, 2020 4:02 pm
Are they African or European chickens?
yes
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Re: A Math Question

Post by badgerfellow »

Wouldn't the individual probability of each subsequent fowl not being a rooster change though, and thus the overall probability of not picking a rooster be dependent on number of fowl? Example: If there are 20 hens/roosters to begin, there are 4 roosters and 16 hens. Say you've picked a hen to begin; that one is now out of the pool to select from for round 2. Next fowl now has a 15/19 probability of not being a rooster. Your probability of not picking a rooster in 4 tries is now (16/20)*(15/19)*(14/18)*(13/17), or about 37.6%. Bump up the available flock to 100 birds, and the probability increases to 40.3%.

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Volante
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Re: A Math Question

Post by Volante »

badgerfellow wrote:
Tue Jun 30, 2020 5:00 pm
Wouldn't the individual probability of each subsequent fowl not being a rooster change though, and thus the overall probability of not picking a rooster be dependent on number of fowl? Example: If there are 20 hens/roosters to begin, there are 4 roosters and 16 hens. Say you've picked a hen to begin; that one is now out of the pool to select from for round 2. Next fowl now has a 15/19 probability of not being a rooster. Your probability of not picking a rooster in 4 tries is now (16/20)*(15/19)*(14/18)*(13/17), or about 37.6%. Bump up the available flock to 100 birds, and the probability increases to 40.3%.
"Yes" but the store themselves have stated the probability is already 80/20 so they are probably maintaining a large enough stock that removing a chick keeps the ratio at effectively 80/20. You're getting four out of the high hundreds maybe even thousands.

Course, if wayne wants to name names so we can research the size of this fowl operation...

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Re: A Math Question

Post by talkingaway »

triviawayne wrote:
Tue Jun 30, 2020 3:09 pm
My wife decided to buy chickens. She is hoping she doesn't end up with a rooster.

She said the lot of birds the store gets their supply from claims 80% are hens.

Since she bought 4 birds, I think the odds are 625:1 that one of them is a rooster.

Am I correct?
Embarrassing moment: The city boy here had to look up just to make extra sure that hen/rooster were female/male chickens.

But no, you're off. You computed the odds of getting "Oops, all roosters!".

If 4/5 are hens, then the chance that you get 4 hens are (4/5)^4 = 0.4096, or about 41%. So, 100% - 41% = 59% of the time, you get at least one rooster.

I'm assuming that we're picking hens out of an infinitely large vat of chickens so that "with replacement vs without replacement" doesn't matter. The problem is slightly different if you're picking 4 chickens out of a set of 100 chickens, with 80 hens...but it's probably close. It's VERY different if you're selecting 4 chickens from a set of 8 hens + 2 roosters.

I'm also assuming that we're picking independently. That means that when you pick a hen or a rooster for your first pick, your next pick is still 80% hens - it's not like picking a hen will make a rooster come to the top of the vat to valiantly try to rescue his mate.

For the rest of the odds, we can use the binomial theorem:

(0.8h + 0.2r)^4 = 0.4096h^4 + 0.4096h^3r + 0.1536h^2r^2 + 0.0256hr^3 + 0.0016r^4

Note that your odds of getting all hens is the SAME as the odds of getting 3 hens and a rooster. Why? Well, when you pick HHHR in any order vs HHHH, swapping in an R will decrease your odds by (80 / 20) = 4. But you can also rearrange HHHR to HHRH, HRHH, RHHH, which multiplies your odds by 4. The two factors basically cancel each other out.

As another example, for 2 roosters and 2 hens, your odds of getting them in the exact order HHRR is (4/5)(4/5)(1/5)(1/5) = 16/625. But you can move those two hens into any position - HRHR, HRRH, RHHR, RHRH, RRHH also work. So you multiply by 6. That gives 96/625 = 0.1536, or 15.36%.

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Re: A Math Question

Post by Woof »

The percentage of outcomes in which you get no rooster, assuming as you do that the ratio is effectively invariant, is (4/5)^4= 41%. The percentage of one-rooster outcomes is (4/5)^3*1/5*4= 41% also. The remaining 18% of outcomes are divided up into 2, 3 and 4 rooster scenarios.

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Re: A Math Question

Post by RandyG »

Woof wrote:
Tue Jun 30, 2020 5:39 pm
The percentage of outcomes in which you get no rooster, assuming as you do that the ratio is effectively invariant, is (4/5)^4= 41%. The percentage of one-rooster outcomes is (4/5)^3*1/5*4= 41% also. The remaining 18% of outcomes are divided up into 2, 3 and 4 rooster scenarios.
Would the same logic apply to ghoti?
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Re: A Math Question

Post by Rackme32 »

Are they laden or unladen?

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Re: A Math Question

Post by Woof »

RandyG wrote:
Tue Jun 30, 2020 6:09 pm
Woof wrote:
Tue Jun 30, 2020 5:39 pm
The percentage of outcomes in which you get no rooster, assuming as you do that the ratio is effectively invariant, is (4/5)^4= 41%. The percentage of one-rooster outcomes is (4/5)^3*1/5*4= 41% also. The remaining 18% of outcomes are divided up into 2, 3 and 4 rooster scenarios.
Would the same logic apply to ghoti?
Go ghoti :mrgreen:

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Re: A Math Question

Post by John Boy »

Volante wrote:
Tue Jun 30, 2020 3:19 pm
triviawayne wrote:
Tue Jun 30, 2020 3:09 pm
My wife decided to buy chickens. She is hoping she doesn't end up with a rooster.

She said the lot of birds the store gets their supply from claims 80% are hens.

Since she bought 4 birds, I think the odds are 625:1 that one of them is a rooster.

Am I correct?
Trying to dust the cobwebs off this part of my brain before working with numbers, but I'm concerned about your phrasing "that one of them is a rooster"
Is that the problem or "that at least one of them is a rooster" You could get up to four.

Maybe work the other way....
The chance of getting 4 hens, 80%*80%*80%*80% = 41%, so you have a 59% chance of getting at least one rooster?
(Within minor rounding error) this is the correct answer and exactly the way to solve this problem. Well done.

Now, would you like to field our next question and try for the washer-dryer?

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Re: A Math Question

Post by John Boy »

MattKnowles wrote:
Tue Jun 30, 2020 4:02 pm
Are they African or European chickens?
You, sir, listened to too many episodes of "Car Talk." :mrgreen:

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Re: A Math Question

Post by AFRET CMS »

I was going to make some remark about a sole rooster in the batch ending up a hen-picked husband, but couldn't find a way to make it work.
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Re: A Math Question

Post by davey »

Who cares about the math...I want to know why one can't sex a chicken?...Or is trying bound to get somebody in trouble these days...?... ;)

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Re: A Math Question

Post by alietr »

davey wrote:
Wed Jul 01, 2020 3:09 pm
Who cares about the math...I want to know why one can't sex a chicken?...Or is trying bound to get somebody in trouble these days...?... ;)
Actually, chicken sexing is a big thing. You don't want cockerels (which grow up to be roosters since they can't lay eggs), so common practice is to take the days old cockerels and put them through a shredder (yes, it's as awful as it sounds). This practice is already on the way out, but the ability to determine the sex in the egg would eliminate the need to do so.

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Re: A Math Question

Post by triviawayne »

I really want to thank everyone for jumping in with the math.

My initial thought was either 41% or 59% by doing the 80*80*80*80 thing...but couldn't figure out which was the number I was looking for (we want to see zero roosters).

So far, it's looking like all hens...but time will tell.
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Re: A Math Question

Post by Volante »

alietr wrote:
Wed Jul 01, 2020 3:16 pm
davey wrote:
Wed Jul 01, 2020 3:09 pm
Who cares about the math...I want to know why one can't sex a chicken?...Or is trying bound to get somebody in trouble these days...?... ;)
Actually, chicken sexing is a big thing. You don't want cockerels (which grow up to be roosters since they can't lay eggs), so common practice is to take the days old cockerels and put them through a shredder (yes, it's as awful as it sounds). This practice is already on the way out, but the ability to determine the sex in the egg would eliminate the need to do so.
There's an episode of American Dad that used this as a B-plot that I desperately wanted to reference in video form...but I couldn't find squat.
( https://en.wikipedia.org/wiki/Pulling_Double_Booty )

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Re: A Math Question

Post by talkingaway »

davey wrote:
Wed Jul 01, 2020 3:09 pm
Who cares about the math...I want to know why one can't sex a chicken?...Or is trying bound to get somebody in trouble these days...?... ;)
She wasn't a chicken (well, she sort of was), but we couldn't sex my parrot. Well, until we had her for about 10 years and she laid an unfertilized egg. Their important organs are all on the inside. Apparently, you can do it either via laparoscopy under anesthesia, or you can get a DNA test on feathers, and I'm not sure if that was commercially available in the late 90s, before she laid her egg.

Never really thought about it before, but she did have surgery to remove a benign tumor on her back when she was fairly young - at least 4 or 5 years before she laid the egg. I wonder if we could have done the laparoscopy then...although it probably would have been expensive, and just a "huh, interesting" kind of fact.

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