Riddle
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after 4 votes the winner is...
wrichcirw
Voting Style:  Open  Point System:  7 Point  
Started:  1/10/2013  Category:  Entertainment  
Updated:  4 years ago  Status:  Post Voting Period  
Viewed:  2,296 times  Debate No:  29076 
Debate Rounds (2)
Comments (12)
Votes (4)
If my friend does not accept this debate it will be open to the public. The rules are I give you a riddle and you solve it. If you get it right, the voters vote 7 points to you. If you get it wrong or do not know the answer, I win.
There are three clay containers. One of them only pennies from the year 2005 in it, one has only pennies from 1975 in it, and the third has an equal number of each. They are labeled "2005" "1975" and "mixed", however you know that the labels have been switched, so that each container is marked incorrectly. Can you properly label each container by only pulling one coin out from each container? I accept my opponent's sporting challenge.
This is an interesting riddle  I think I have figured it out.  The most important fact I gain from PRO's riddle is that each container is marked incorrectly. Therefore, the key to solving this riddle is the container incorrectly marked "mixed". Whatever coin you pull out of this container, since this container CANNOT be mixed, it is thus the container consisting of solely the type of coin you pulled out. Then the two remaining containers MUST have coins pulled that are contrary to their labels, because AT MOST 2 out of the 3 containers will contain one type of coin  the last container MUST contain ONLY the different coin. To demonstrate why this is so, if the coins pulled from these remaining two containers were to correspond to their label, then one of the containers would be correctly marked, which cannot occur per the rules of this riddle. (note: underlined text refers to example #3, bolded text refers to example #4) Since the two remaining containers will have coins pulled that are contrary to their labels, we can also deduce which container is mixed  it will be the container from which the coin pulled is the same as that of the incorrectly marked "mixed" container. Then, the last container will consist solely of the other type of coin. Example #1 Containers A, B, C A is labeled 2005
B is labeled 1975
C is labeled Mixed.
From A you pull a 1975 coin.
From B you pull a 2005 coin. From C you pull a 2005 coin. Since C is incorrectly labeled mixed, and you pulled a 2005 coin out of it, it CANNOT be mixed, and is thus the 2005 container. Since B is incorrectly labeled 1975 and you pulled a 2005 coin out of it, it CANNOT be the 2005 container, since that is C. Therefore, it MUST be mixed. Since A is incorrectly labeled 2005 and you pulled a 1975 coin out of it, it CANNOT be the mixed container, since that is container B, and is thus the 1975 container. Example #2 Containers A, B, C A is labeled 2005 B is labeled 1975 C is labeled Mixed. From A you pull a 1975 coin.
From B you pull a 2005 coin. From C you pull a 1975 coin. Since C is incorrectly labeled mixed, and you pulled a 1975 coin out of it, it CANNOT be mixed, and is thus the 1975 container. Since A is incorrectly labeled 2005 and you pulled a 1975 coin out of it, it CANNOT be the 1975 container because that is container C, and is thus the mixed container. Since B is incorrectly labeled 1975 and you pulled a 2005 coin out of it, it CANNOT be the mixed container because that is container A. Therefore, it MUST be the 2005 container. Example #3  WRONG scenario (the coins pulled from nonmixed containers correspond to their label) Containers A, B, C A is labeled 2005 B is labeled 1975 C is labeled Mixed. From A you pull a 2005 coin. From B you pull a 1975 coin. From C you pull a 1975 coin.
Since C is incorrectly labeled mixed, and you pulled a 1975 coin out of it, it CANNOT be mixed, and is thus the 1975 container. Since B is incorrectly labeled 1975 and you pulled a 1975 coin out of it, it CANNOT be the 1975 container, since that is C. Therefore, it MUST be the mixed container. Since A is incorrectly labeled 2005 and you pulled a 2005 coin out of it, it CANNOT be the mixed container since that is container B, and is thus the 2005 container. However, this would mean that this container was originally labeled correctly, which is not possible given the rules of this riddle. This example demonstrates why the coins pulled from the remaining two containers must be different than their original labels. Example #4  WRONG SCENARIO (all three coins pulled are the same) Containers A, B, C A is labeled 2005 B is labeled 1975 C is labeled Mixed. From A you pull a 1975 coin. From B you pull a 1975 coin. From C you pull a 1975 coin.
Since C is incorrectly labeled mixed, and you pulled a 1975 coin out of it, it CANNOT be mixed, and is thus the 1975 container. Since B is incorrectly labeled 1975 and you pulled a 1975 coin out of it, it CANNOT be the 1975 container, since that is C. Therefore, it MUST be the mixed container. Since A is incorrectly labeled 2005 and you pulled a 1975 coin out of it, it CANNOT be the mixed container since that is container B, and is thus the 2005 container. However, the 2005 container CANNOT contain ANY 1975 coins. This example demonstrates why you can only pull AT MOST two of the same kind of coin from the three containers.  Examples #1 and #2 are the only two possibilities, because the coins you pull from one of the containers MUST be different from the other two containers, and the coins pulled from the nonmixed containers MUST be different than their labels. Therefore, the only real variation is in the mixed container. I have demonstrated how to correctly label these three incorrectly labeled containers. I have also demonstrated how the coins pulled MUST correspond to a specific logic. Utilizing the logic I've explained and then demonstrated via examples, I have solved this riddle. Give me my seven points please. :) And thanks Azul145 for hosting this debate. :D 

I challenged you to this because I knew you could do it :D. Good job old friend and voters vote con.
Thanks Azul145. I appreciate the mental workout :D
To the audience, give PRO some conduct points lol, he deserves it :) 
4 votes have been placed for this debate. Showing 1 through 4 records.
Vote Placed by Bull_Diesel 4 years ago
Azul145  wrichcirw  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  0  7 
Reasons for voting decision: JFC. I solved the thing and my brain didn't hurt. Then I read con's writeup and my head hurts. Damn.
Vote Placed by tarkovsky 4 years ago
Azul145  wrichcirw  Tied  

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Who had better conduct:      1 point  
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Total points awarded:  0  7 
Reasons for voting decision: Con figured out the riddle.
Vote Placed by likespeace 4 years ago
Azul145  wrichcirw  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  0  7 
Reasons for voting decision: Good show to Con for solving it! Victory concede by Pro. I would honor Con's request for conduct points, except the round one rules state all seven points to Con if he solves it. :)
Vote Placed by DoctorDeku 4 years ago
Azul145  wrichcirw  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  0  7 
Reasons for voting decision: Wow, good job Con 00. I sat here for a good five minutes trying to figure out Pro's riddle before I read your solution. You explained you process of deduction well too, perhaps to much but not to the point that it was overkill.
I agree with you that the table presents the only two choices, but you still need to know which coins were pulled to get the riddle right. Hence my long explanation.
Your table by itself does not solve the riddle, that is all I am saying.
Perhaps the outcome is clear as day to you, but exactly how you reached that outcome is wholly absent from your explanation.
A______B______C
75_____M______05
M_____05______75
These are the only possible orderings of our elements. No need to explain anything, the table of values is more than enough.
For a given arrangement of Jar A, B, and C:
Jar A is labeled '05
Jar B is labeled '75
Jar C is labeled Mixed
If the ordering of said arrangement is incorrect such that each item is incorrectly labeled then the following combinatorial ordering will apply.
For Jars A, B and C:
A B C
75 M 05
M 05 75
These are the only possible correct orderings of the elements.
I solved it. Do I win?