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# Riddle

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wrichcirw
 Voting Style: Open Point System: 7 Point Started: 1/10/2013 Category: Entertainment Updated: 5 years ago Status: Post Voting Period Viewed: 2,611 times Debate No: 29076
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 Pro 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?Report this Argument Con 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 #1Containers 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 #2Containers A, B, CA is labeled 2005B is labeled 1975C 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 non-mixed containers correspond to their label)Containers A, B, CA is labeled 2005B is labeled 1975C 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, CA is labeled 2005B is labeled 1975C 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 non-mixed 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. :DReport this Argument Pro I challenged you to this because I knew you could do it :D. Good job old friend and voters vote con.Report this Argument Con Thanks Azul145. I appreciate the mental workout :D To the audience, give PRO some conduct points lol, he deserves it :) Report this Argument
12 comments have been posted on this debate. Showing 1 through 10 records.
Posted by wrichcirw 5 years ago
Thanks, you too. :D
Posted by tarkovsky 5 years ago
Well I suppose, if you really reaaaaaaaally wanted me to I could enumerate the orderings and say when C= 75, apply 1 or when C = 05 apply two, to indicate the orderings. I just figured it was unnecessary as C is already part of that ordering. Anyway good job.
Posted by wrichcirw 5 years ago
Basically, what I'm saying is that with your table alone, without knowing which coins were pulled, you'd still have a 50/50 chance of getting the riddle wrong.

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.
Posted by wrichcirw 5 years ago
However, the riddle was not "Can you properly label the containers", but "Can you properly label each container by only pulling one coin out from each container?" How can you do so if you do not stipulate which coins were pulled?

Your table by itself does not solve the riddle, that is all I am saying.
Posted by tarkovsky 5 years ago
How I got this table of values could benefit from more explanation but it's not necessary. Again, if I could this would have been perfect time to use latexit but, of course, the comment's text field is quite inferior to the debate's. Much as you don't have to solve some function when you generate a table of values with whatever norm and index suits your inquiries, so too do I not have to explain which coins were pulled. Want to know the outcomes? Refer to the table.
Posted by wrichcirw 5 years ago
Your table does not explain either a) which coins were pulled, or b) the correct labels of the containers.

Perhaps the outcome is clear as day to you, but exactly how you reached that outcome is wholly absent from your explanation.
Posted by tarkovsky 5 years ago
I don't need to elaborate. It's right there, perhaps the stupid text interface in this comment section is what screwed it up.

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.
Posted by wrichcirw 5 years ago
You did not elaborate on which coins were pulled.
Posted by tarkovsky 5 years ago
Did it myself. Was sort of fun. Didn't need the lengthy explanation:

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.
Posted by tarkovsky 5 years ago