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# bushindo

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1. ## Matching values of Die throws pt. 3

I applied witzar's formula to various different N, and I reproduced the same exact table that you gave in here
2. ## Matching values of Die throws pt. 3

Nice compact form!

4. ## Of dogs and men

I think there's an error here (see part highlighted in red above)
5. ## Of dogs and men

bonanova's work checks out to me. I used k-man and TSLF's answers and worked backward to get a distance of about 50.4 feet moved, so I'm casting a vote for bonanova.
6. ## Between six towns

I believe there is a subtle error here
7. ## The Land of Oz

never mind, i explained the wrong part. If it rained today then there is a 50% chance of rain again and 50% it won't If I'm parsing the probability correctly
8. ## A fair coin and an unfair coin

On the contrary, I will, on principle, plug the number in Bayes' formula =)

10. ## Blue eyes and green eyes

Bingo. 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.
11. ## Modifying two standard dice

Fascinating analysis. Thanks for posting this wonderful problem.
12. ## Blue eyes and green eyes

Yes, that's true. That's an interesting nugget indeed.
13. ## Blue eyes and green eyes

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. 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?
14. ## Blue eyes and green eyes

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.. 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'.
15. ## Blue eyes and green eyes

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?
16. ## Modifying two standard dice

Maybe OP meant non-negative? I think I have a proof there is no other solution. Construct the table. Missed the point about positive integers. I agree with bonanova, in that case

18. ## Who gets a free dinner?

Agree. A fair method cannot guaranty any maximum number of tosses. I see your 1 and I will raise you a 0 =) Not acceptable. The winner must be decided by coin toss. My interpretation of the OP is that the coin flip is optional rather than mandatory. The OP says "Pick a person out of n with a sequence of coin toss". I submit that it is possible to construct a sequence of length 0.
19. ## Who gets a free dinner?

I see your 1 and I will raise you a 0 =)
20. ## Who gets a free dinner?

I'm trying to go for the minimum expected number of flips here. This is my best attempt so far Awesome. Same problem though no ceiling. And it seems fair, but not in an obvious way. A proof of fairness would be nice. I don't think this problem can be decided with a fixed ceiling method. That is, I don't think there exist a method that guarantees a maximum number of fixed maximum flips m(N) for every N. See below for reason + proof of fairness
21. ## Who gets a free dinner?

I'm trying to go for the minimum expected number of flips here. This is my best attempt so far

Question
23. ## Four darts

Now, that question 1 is solved, I am inclined to yield the opportunity of solving question 2 to others. I think the proof is fine as it is, but I believe there are a few typos. See below Question for bonanova: about question 1- the probability that the 4th point falls within the triangle,
24. ## For Bushindo - (others try at your own risk)

I think I misinterpreted your position. I don't think bononova and I share the same interpretation of 'uncertain', though. Let me see if we have the following positions correct The way I see it, Bonanova has made his problem statement/position 100% clear: Uncertainty is when Cole's winning chance by shooting in the air = his winning chance by shooting at Alex. The question Bonanova wants us to solve is: For what values of b (accuracy of Bobby) such situation is possible? And solved it we have: Bushindo's view of uncertainty is the same as Bonanova's: Wshoot_air(b,c) = Wshoot_at_Alex(b,c). However, there seems to be some uncertainty as to what exactly Bonanova wants us to find. Prime picks on the usage of the word uncertain. Does not believe there is any uncertainty here at all. (When chances are equal, Cole must shoot Alex, because he hates him more than he hates Bobby.) Also, Prime promotes (unsuccessfully) an alternative strategy whereby Cole shoots himself in the foot with his very first shot. You're right. My view of uncertainty is the same as bonanoba's. It's still too early for me to be senile, *sigh*. Good analysis, though. Your analysis definitely deserves the distinction of Best Answer.
25. ## For Bushindo - (others try at your own risk)

I think I misinterpreted your position. I don't think bononova and I share the same interpretation of 'uncertain', though. Let me see if we have the following positions correct
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