Best Answer Prime, 01 April 2013 - 01:03 AM

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.

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Started by bushindo, Mar 29 2013 07:28 PM

Best Answer Prime, 01 April 2013 - 01:03 AM

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.

Spoiler for probability

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15 replies to this topic

Posted 29 March 2013 - 07:28 PM

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?

Posted 30 March 2013 - 06:23 AM

Spoiler for First thoughts

.
*Vidi vici veni.*

Posted 30 March 2013 - 07:39 AM

Spoiler for my guess

Past prime, actually.

Posted 31 March 2013 - 02:41 AM

I took a little time to verify my guess.

Spoiler for Long solution

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

**Edited by Prime, 31 March 2013 - 02:48 AM.**

Past prime, actually.

Posted 31 March 2013 - 03:10 PM

Spoiler for First thoughts

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?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.

- 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 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..

Spoiler for comment

I took a little time to verify my guess.

Spoiler for Long solution

Spoiler for Comment

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'.

Posted 31 March 2013 - 06:16 PM

Spoiler for population stability

Posted 31 March 2013 - 06:46 PM

I took a little time to verify my guess.

Spoiler for Long solution

Spoiler for Comment

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, 31 March 2013 - 06:53 PM.**

Past prime, actually.

Posted 31 March 2013 - 09:17 PM

Spoiler for results of my calculations

Posted 31 March 2013 - 11:49 PM

I took a little time to verify my guess.

Spoiler for Long solution

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.

Spoiler for results of my calculations

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?

Posted 31 March 2013 - 11:57 PM

Spoiler for population stability

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

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