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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.

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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.


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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.

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Originally Posted by BALLISTIK
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.

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


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Originally Posted by ElkSlayer91
Originally Posted by BALLISTIK
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.

1/16 - 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.


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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


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Originally Posted by Idaho_Shooter
Originally Posted by ElkSlayer91
Originally Posted by BALLISTIK
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.

1/16 - 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.

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.


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Originally Posted by BALLISTIK
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

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Originally Posted by Milwroad
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

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


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Originally Posted by ironbender
Originally Posted by Milwroad
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

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.


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Originally Posted by ironbender
Originally Posted by BALLISTIK
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.

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Nope

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

Last edited by ironbender; 04/15/21.

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I hate chicken.


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Originally Posted by ironbender
Nope

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

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.

Last edited by ElkSlayer91; 04/15/21.

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That looks mo betta!


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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


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Originally Posted by BALLISTIK
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.


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Whatever you said...everyone knows you are a lying jerk.

That's a bold assertion. Point out where you think I lied.

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

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Originally Posted by Daverageguy
I hate chicken.


Gonna bump this for when mathman shows up... laugh

Ballistik is on it though...

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Not quite.

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