# Blue eyes and green eyes

## 16 posts in this topic

Posted · Report post

Let's say that in a certain country, there are two types of people- blue-eyed and green-eyed.
Let's assume that eye color are governed by a single gene, which can be in the dominant form (B) or recessive form (g). This means that people in this country have the genotype BB (blue-eyed), Bg (blue-eyed), or gg (green-eyed). The relative frequency of these genotypes in the population is BB (50%), Bg (25%), and gg( 25%).
Judy (blue-eyed) is married to Jack (blue-eyed). Judy's parents are both blue-eyed; one of Jack's parents is blue-eyed while the other is green-eyed. Judy and Jack's first child has blue eyes.
1) Given the above, what is the probability that Judy's father has the dominant genotype BB?
2) What is the chance that Judy and Jack's second child is green-eyed?
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Posted · Report post

I'm wondering how half the population came to be BB, or how that distribution is sustainable.

That is, given two parents with those genotype probabilities, does even the first generation of children sustain that distribution?

Regardless, what do we know immediately?

1. Because of Jack's green Mom and his own blue eyes,
then regardless of Jack's Dad's type, Jack is type Bg.
2. Corollary: we know only about Jack's Dad's type that it is not gg.
But that has no bearing on the questions.

3. The type of Child1 is a red herring, I think.
Because Jack is not BB, and Judy is not necessarily BB, then C1 could have been green: s/he just happens to be Blue.
Also, each child has the same type probability. Or is has Mr. Bayes already entered the room?

Question 1: What do we know about Judy's Dad? Indirectly ( because Judy is not green-eyed) we know it is not the case that both he and Judy's Mom are gg. I suspect that makes JD's BB type greater than 50%. By how much? Well, in the normal course of events, Judy could have gg with 25% probability, and that probability has been removed. So she's BB (2/3) or Bg (1/3). What parental genotypes produce that distribution? Seems that eliminating gg-gg makes it a lot more probable than 1/3 that she is Bg. So I will leave this unanswered until I can think more about it.

Question 2: Because Jack is Bg, also because Child1 is blue, his kids' shots at being green-eyed are no greater than 50%. Can we say Judy's BB probability is now 2/3 (gg having been eliminated)? If so, then children's gg probability are all (1/3)(1/2) = 1/6. I'll go with that answer.

.

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Posted · Report post

1) 24/35

2) 3/52

And I am not sure the population proportions of 50%, 25%, 25% can be maintained for many generations.
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Posted (edited) · Report post

I took a little time to verify my guess.

Let's construct a proportions table of different union types and their offspring.

Couple = proportion => Offspring(proportion)
1) BB-BB = 1/4 ==> BB(1/4)
2) BB-Bg = 1/4 ==> BB(1/8) + Bg(1/8)
3) BB-gg = 1/4 ==> Bg(1/4)
4) Bg-gg = 1/8 ==> Bg(1/16) + gg(1/16)
5) Bg-Bg = 1/16 => BB(1/64) + Bg(1/32) + gg(1/64)
6) gg-gg = 1/16 => gg(1/16)

Judy's parents could be couples (1), (2), or (5), where both parents are blue eyed. The odds that blue-eyed Judy is from one of those families are (1):(2):(5) = 16:16:3. Where all families (1) have BB father, half of the families (2) have BB father, and none of families (3) have BB father. Therefore, the answer to question 1 is (16+8)/35 = 24/35.

Since one of Jack's parents is green eyed, Jack is Bg type.
From the above table, Judy's genotype odds of BB:Bg = 5:2.
Proportions table for Jack and Judy is as following:

Jack-Judy = probability => Offspring(proportion)
A) Bg-BB = 5/7 => BB(5/14) + Bg(5/14)
B) Bg-Bg = 2/7 => BB(1/14) + Bg(2/14) + gg(1/14)

Thus the odds for Jack-Judy blue eyed kid's family type (A):(B) = 10:3. Meaning, given one blue eyed child, the chance of Jack-Judy union being of type Bg-Bg is 3/13 -- the only type of Jack-Judy family that could produce a green eyed child at the probability of 1/4. Hence, the second child has the green eyes at the probability of (3/13)*(1/4) = 3/52.
Of course, the above relies on assumption of Judy's faithfulness.

Somehow, to me this problem seems similar to Bonanova's “Give monkey enough rope.” (Some serious untangling to perform one must.)

Edited by Prime
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Posted · Report post

I'm wondering how half the population came to be BB, or how that distribution is sustainable.

That is, given two parents with those genotype probabilities, does even the first generation of children sustain that distribution?

Regardless, what do we know immediately?

• Because of Jack's green Mom and his own blue eyes,

then regardless of Jack's Dad's type, Jack is type Bg.

• Corollary: we know only about Jack's Dad's type that it is not gg.

But that has no bearing on the questions.

• The type of Child1 is a red herring, I think.

Because Jack is not BB, and Judy is not necessarily BB, then C1 could have been green: s/he just happens to be Blue.

Also, each child has the same type probability. Or is has Mr. Bayes already entered the room?

Question 1: What do we know about Judy's Dad? Indirectly ( because Judy is not green-eyed) we know it is not the case that both he and Judy's Mom are gg. I suspect that makes JD's BB type greater than 50%. By how much? Well, in the normal course of events, Judy could have gg with 25% probability, and that probability has been removed. So she's BB (2/3) or Bg (1/3). What parental genotypes produce that distribution? Seems that eliminating gg-gg makes it a lot more probable than 1/3 that she is Bg. So I will leave this unanswered until I can think more about it.

Question 2: Because Jack is Bg, also because Child1 is blue, his kids' shots at being green-eyed are no greater than 50%. Can we say Judy's BB probability is now 2/3 (gg having been eliminated)? If so, then children's gg probability are all (1/3)(1/2) = 1/6. I'll go with that answer..

I'm afraid the case of Child 1 is not a red herring. Reverend Bayes has already entered the room since we gained some additional information by knowing the phenotype of Child 1. For instance, if child 1 is green-eyed, intuition says that would vastly affects the answers to 1 and 2; therefore the fact that the child is blue-eyed should also give some information.

I took a little time to verify my guess.

Let's construct a proportions table of different union types and their offspring.

Couple = proportion => Offspring(proportion)

1) BB-BB = 1/4 ==> BB(1/4)

2) BB-Bg = 1/4 ==> BB(1/8) + Bg(1/8)

3) BB-gg = 1/4 ==> Bg(1/4)

4) Bg-gg = 1/8 ==> Bg(1/16) + gg(1/16)

5) Bg-Bg = 1/16 => BB(1/64) + Bg(1/32) + gg(1/64)

6) gg-gg = 1/16 => gg(1/16)

Judy's parents could be couples (1), (2), or (5), where both parents are blue eyed. The odds that blue-eyed Judy is from one of those families are (1):(2):(5) = 16:16:3. Where all families (1) have BB father, half of the families (2) have BB father, and none of families (3) have BB father. Therefore, the answer to question 1 is (16+8)/35 = 24/35.

Since one of Jack's parents is green eyed, Jack is Bg type.

From the above table, Judy's genotype odds of BB:Bg = 5:2.

Proportions table for Jack and Judy is as following:

Jack-Judy = probability => Offspring(proportion)

A) Bg-BB = 5/7 => BB(5/14) + Bg(5/14)

B) Bg-Bg = 2/7 => BB(1/14) + Bg(2/14) + gg(1/14)

Thus the odds for Jack-Judy blue eyed kid's family type (A):(B) = 10:3. Meaning, given one blue eyed child, the chance of Jack-Judy union being of type Bg-Bg is 3/13 -- the only type of Jack-Judy family that could produce a green eyed child at the probability of 1/4. Hence, the second child has the green eyes at the probability of (3/13)*(1/4) = 3/52.

Of course, the above relies on assumption of Judy's faithfulness.

The analysis above does not use the phenotype of Child 1 (blue-eyed) as part of the prior information. As per the discussion in bonanova's post, by knowing the phenotype of the child, we gained some additional knowledge about the probabilities of interest. That knowledge should be incorporated in the computation of 1) and 2).

Somehow, to me this problem seems similar to Bonanova's “Give monkey enough rope.” (Some serious untangling to perform one must.)

Hey, I resent that =). I spend some time trying to make the wording clear and lucid. There is no way this puzzle can match the sheer genius of grammatical obfuscation in 'Give monkey enough rope'.

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Posted · Report post

Current relative frequencies of genotypes are not stable.

They will eventually stabilize at values:

BB = 25/64

Bg = 30/64

gg = 9/64

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Posted (edited) · Report post

I took a little time to verify my guess.

Let's construct a proportions table of different union types and their offspring.

Couple = proportion => Offspring(proportion)

1) BB-BB = 1/4 ==> BB(1/4)

2) BB-Bg = 1/4 ==> BB(1/8) + Bg(1/8)

3) BB-gg = 1/4 ==> Bg(1/4)

4) Bg-gg = 1/8 ==> Bg(1/16) + gg(1/16)

5) Bg-Bg = 1/16 => BB(1/64) + Bg(1/32) + gg(1/64)

6) gg-gg = 1/16 => gg(1/16)

Judy's parents could be couples (1), (2), or (5), where both parents are blue eyed. The odds that blue-eyed Judy is from one of those families are (1):(2):(5) = 16:16:3. Where all families (1) have BB father, half of the families (2) have BB father, and none of families (3) have BB father. Therefore, the answer to question 1 is (16+8)/35 = 24/35.

Since one of Jack's parents is green eyed, Jack is Bg type.

From the above table, Judy's genotype odds of BB:Bg = 5:2.

Proportions table for Jack and Judy is as following:

Jack-Judy = probability => Offspring(proportion)

A) Bg-BB = 5/7 => BB(5/14) + Bg(5/14)

B) Bg-Bg = 2/7 => BB(1/14) + Bg(2/14) + gg(1/14)

Thus the odds for Jack-Judy blue eyed kid's family type (A):(B) = 10:3. Meaning, given one blue eyed child, the chance of Jack-Judy union being of type Bg-Bg is 3/13 -- the only type of Jack-Judy family that could produce a green eyed child at the probability of 1/4. Hence, the second child has the green eyes at the probability of (3/13)*(1/4) = 3/52.

Of course, the above relies on assumption of Judy's faithfulness.

The analysis above does not use the phenotype of Child 1 (blue-eyed) as part of the prior information. As per the discussion in bonanova's post, by knowing the phenotype of the child, we gained some additional knowledge about the probabilities of interest. That knowledge should be incorporated in the computation of 1) and 2).

Somehow, to me this problem seems similar to Bonanova's “Give monkey enough rope.” (Some serious untangling to perform one must.)

Hey, I resent that =). I spend some time trying to make the wording clear and lucid. There is no way this puzzle can match the sheer genius of grammatical obfuscation in 'Give monkey enough rope'.

I did not mean to imply any grammatical obfuscation in this puzzle. I meant probability obfuscation with backward refrences, like first child.

My analysis uses the first blue eyed child as prior information (highlighted in red here). That's how I got probability 3/13 rather than 2/7 for Judy's genotype Bg.

Edited by Prime
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Posted · Report post

1) 65/72

2) 3/52

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Posted · Report post

I took a little time to verify my guess.

Let's construct a proportions table of different union types and their offspring.

Couple = proportion => Offspring(proportion)

1) BB-BB = 1/4 ==> BB(1/4)

2) BB-Bg = 1/4 ==> BB(1/8) + Bg(1/8)

3) BB-gg = 1/4 ==> Bg(1/4)

4) Bg-gg = 1/8 ==> Bg(1/16) + gg(1/16)

5) Bg-Bg = 1/16 => BB(1/64) + Bg(1/32) + gg(1/64)

6) gg-gg = 1/16 => gg(1/16)

Judy's parents could be couples (1), (2), or (5), where both parents are blue eyed. The odds that blue-eyed Judy is from one of those families are (1):(2):(5) = 16:16:3. Where all families (1) have BB father, half of the families (2) have BB father, and none of families (3) have BB father. Therefore, the answer to question 1 is (16+8)/35 = 24/35.

Since one of Jack's parents is green eyed, Jack is Bg type.

From the above table, Judy's genotype odds of BB:Bg = 5:2.

Proportions table for Jack and Judy is as following:

Jack-Judy = probability => Offspring(proportion)

A) Bg-BB = 5/7 => BB(5/14) + Bg(5/14)

B) Bg-Bg = 2/7 => BB(1/14) + Bg(2/14) + gg(1/14)

Thus the odds for Jack-Judy blue eyed kid's family type (A):(B) = 10:3. Meaning, given one blue eyed child, the chance of Jack-Judy union being of type Bg-Bg is 3/13 -- the only type of Jack-Judy family that could produce a green eyed child at the probability of 1/4. Hence, the second child has the green eyes at the probability of (3/13)*(1/4) = 3/52.

Of course, the above relies on assumption of Judy's faithfulness.

I did not mean to imply any grammatical obfuscation in this puzzle. I meant probability obfuscation with backward refrences, like first child.

My analysis uses the first blue eyed child as prior information (highlighted in red here). That's how I got probability 3/13 rather than 2/7 for Judy's genotype Bg.

I was specifically mentioning that the work for problem 1 (see part highlighted in blue) does not use Child 1's color as prior information. As bonanova said, Reverend Bayes has already entered the room. The answer for part 2 is correct, and does use all the available information.

1) 65/72

2) 3/52

I get the same answer for part 2, but different answer for part 1. I could be wrong, and I frequently am. Could you describe your reasoning for part 1 so that I could follow the logic more closely?

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Posted · Report post

Current relative frequencies of genotypes are not stable.

They will eventually stabilize at values:

BB = 25/64

Bg = 30/64

gg = 9/64

Yes, that's true. That's an interesting nugget indeed.

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Posted (edited) · Report post

I suppose, blue eyed child also must play part in figuring out question 1. I only took it into account for figuring out question two, so my previous answer is wrong.

Since blue eyed child reduces Judy's probability of Bg by 1/4, the odds of Judy's parents are as following:

(BB-BB):(BB-Bg):(Bg-Bg) = 32:28:5 (From the table in post #4).
Where all BB-BB families and half of BB-Bg have BB type father.
Therefore:
1) 46/65 Judy's father is BB.
2) 3/52 Judy and Jack's second child has green eyes.
Edited by Prime
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Posted · Report post

I suppose, blue eyed child also must play part in figuring out question 1. I only took it into account for figuring out question two, so my previous answer is wrong.

Since blue eyed child reduces Judy's probability of Bg by 1/4, the odds of Judy's parents are as following:

(BB-BB):(BB-Bg):(Bg-Bg) = 32:28:5 (From the table in post #4).

Where all BB-BB families and half of BB-Bg have BB type father.

Therefore:

1) 46/65 Judy's father is BB.

2) 3/52 Judy and Jack's second child has green eyes.

Bingo.

Somehow, to me this problem seems similar to Bonanova's “Give monkey enough rope.” (Some serious untangling to perform one must.)

Hey, I resent that =). I spend some time trying to make the wording clear and lucid. There is no way this puzzle can match the sheer genius of grammatical obfuscation in 'Give monkey enough rope'.

I did not mean to imply any grammatical obfuscation in this puzzle. I meant probability obfuscation with backward refrences, like first child.

In that case, thank you, I aim to please. Nothing but the best probability obfuscation for my fellow Brainden citizens.

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Posted (edited) · Report post

I made a drawing that shows all possibilities for a blue eyed couple (M=Mother and F=Father) to have a daughter (J=Judy) and a grandChild ( C ) of this daughter with a father of type Bg. Each rectangle represents specific genotype of Father, Mother, Judy and Child. Genotype of the Father can be read on horizontal axis, genotype of Mother is written on vertical axis, genotype of Judy and Child can be read inside of the rectangle. For example rectangle labelled "J:Bg;C:BB" says that Judy is Bg and Child is BB. Area of each rectangle divided by area of the whole square equals probability of the event. Black-hatched rectangles indicate cases that should be excluded due to green eyes of Judy or Child. Blue rectangles indicate cases when Father is of type BB. To answer the question #1 it is sufficient to divide (blue area) by (the area of the whole square minus black-hatched area).

Here is a link to the image:

http://imageshack.us/photo/my-images/545/blueeyesl.jpg/

I'm also trying to attach it, since It looks like I'm not allowed to upload images (why?).

Edited by witzar
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Posted · Report post

I made a drawing that shows all possibilities for a blue eyed couple (M=Mother and F=Father) to have a daughter (J=Judy) and a grandChild ( C ) of this daughter with a father of type Bg. Each rectangle represents specific genotype of Father, Mother, Judy and Child. Genotype of the Father can be read on horizontal axis, genotype of Mother is written on vertical axis, genotype of Judy and Child can be read inside of the rectangle. For example rectangle labelled "J:Bg;C:BB" says that Judy is Bg and Child is BB. Area of each rectangle divided by area of the whole square equals probability of the event. Black-hatched rectangles indicate cases that should be excluded due to green eyes of Judy or Child. Blue rectangles indicate cases when Father is of type BB. To answer the question #1 it is sufficient to divide (blue area) by (the area of the whole square minus black-hatched area).

Here is a link to the image:

http://imageshack.us/photo/my-images/545/blueeyesl.jpg/

I'm also trying to attach it, since It looks like I'm not allowed to upload images (why?).

The logic is impeccable, and the image above looks right with the relative areas. However, I think there might be some arithmetic mistakes in the calculations, since 65/72 is about 90%, and the (blue areas) divided by (the area of the whole square minus black-hatched area) is nearer to 70% than it is to 90%.

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Posted · Report post

I made a drawing that shows all possibilities for a blue eyed couple (M=Mother and F=Father) to have a daughter (J=Judy) and a grandChild ( C ) of this daughter with a father of type Bg. Each rectangle represents specific genotype of Father, Mother, Judy and Child. Genotype of the Father can be read on horizontal axis, genotype of Mother is written on vertical axis, genotype of Judy and Child can be read inside of the rectangle. For example rectangle labelled "J:Bg;C:BB" says that Judy is Bg and Child is BB. Area of each rectangle divided by area of the whole square equals probability of the event. Black-hatched rectangles indicate cases that should be excluded due to green eyes of Judy or Child. Blue rectangles indicate cases when Father is of type BB. To answer the question #1 it is sufficient to divide (blue area) by (the area of the whole square minus black-hatched area).

Here is a link to the image:

http://imageshack.us/photo/my-images/545/blueeyesl.jpg/

I'm also trying to attach it, since It looks like I'm not allowed to upload images (why?).

The logic is impeccable, and the image above looks right with the relative areas. However, I think there might be some arithmetic mistakes in the calculations, since 65/72 is about 90%, and the (blue areas) divided by (the area of the whole square minus black-hatched area) is nearer to 70% than it is to 90%.

(Blue area) equals 46/72 and (the area of the whole square minus black-hatched area) equals 65/72, so my method gives same result as Prime gave: 46/65. There was an arithmetic mistake in my calculations.

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Posted · Report post

Current relative frequencies of genotypes are not stable.

They will eventually stabilize at values:

BB = 25/64

Bg = 30/64

gg = 9/64

There are few interesting questions herein. (Perhaps, people who study gentics and population have some ready-made formulas.)

Won't these stable population proportions (BB:Bg:gg = 25:30:9) be reached in the very next generation?

If the initial population distribution was different, we would get a different proportion for stable population. Does that stable proportion always happen in the very first offspring generation? Why?

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