adamantius
2006-02-18 09:48:18 |
my puzzle problem
See if you can tackle this one and then i will explain the Problem.
I got this out of a book called, "My Best Puzzles in Logic and Reasoning."
Seventeen Marbles
A bag contains 17 marbles. They are of four different colours: There are 2 of each colour; and of no two colours are there the same number. The colour of which there is the largest number is green. If I draw from the bag enough marbles - but only just enough - to ensure that I have at least 2 of any one colour, and at least 1 of any second colour, I must draw 11.
How many must I draw to ensure that at least one of the marbles is a green one?
Now any one who wants to go ahead and answer this in the forum is quite welcome. I have already attempted it and have seen the books solution.
SPOILER ALERT.
I will now post the issue that I have with this puzzle and the book's solution.
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There are 3 possible arrangements for the number of marbles for each colour.
A) 8 (green) 4 3 2
B) 7 (green) 5 3 2
c) 6 (green) 5 4 2
The book and I agree on the first part of this solution.
So in examining these arrangements I encounter 2 extreme scenarios.
I will explain with arrangement A)
i) it is possible that I draw 4 marbles all of different colours. To get a second marble in any colour I then only need to draw one more. ---5 Marbles.
ii) or I might draw all 8 green marbles, then to get at least one of another colour I need only draw one more. --- 9 marbles.
Therefore the the most marbles I should draw in each case is.
a) 9
b) 8
c) 7
neither of these cases requires me to draw 11 marbles.
At this point I feel the question as it stands is unsolvable.
However the solution gives these responses.
a) 12
b) 11
c) 10
I don't see how these are possible. This is what I need help with. Am I wrong or is the book wrong?
The book then says that b) correct arrangement and states:
"So to make sure of drawing at least one green marble, it is AGAIN necessary to draw 11."
The solution uses the word "Again" so we can assume that the fact that 11 were originally needed is not a typo.
Anyone?
A |
adamantius
2006-02-20 07:41:30 |
Re: my puzzle problem
WOW did I stump everyone?
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Federico Kereki
2006-02-20 08:54:00 |
Re: my puzzle problem
Maybe the problem means "drawing until you have 2 of EACH color", and that fits case (A). |
brianjn
2006-02-20 19:21:36 |
Re: my puzzle problem
Stumped everyone? No! Puzzle posting, and solving, happens elsewhere. |
Gamer
2006-02-20 22:21:23 |
Re: my puzzle problem
I think you mean 1 of each color. That would satisfy both parts of the solution. |
Avin
2006-02-21 15:44:04 |
Re: my puzzle problem
If it were 1 of each color, in all three situations one would need to draw 16 marbles out of the bag, because it's possible for the first 15 to get every marble EXCEPT the two of the least frequent color. Similarly, if the requirement were 2 of each color, in all three situations one would need to draw all 17 marbles.
By the way, I think it's legitimate to post this in the forums as opposed to submitting it to be posted because clearly it's a puzzle that adamantius isn't trying to solve, or posing to us as a puzzle for us to solve; rather it's his attempt to verify that a solution is defective or not to a puzzle that is not in our database and he probably does not intend to submit (particularly since he does not know the right answer). |
brianjn
2006-02-21 19:26:54 |
Re: my puzzle problem
Point taken. I merely glanced at it and assumed it was another person posting puzzles in Forums. |
Gamer
2006-02-21 20:30:06 |
Re: my puzzle problem
The only relationship I see that links 10, 11 and 12 to the numbers is number needed to draw a green one.
There has been some discussion whether a problem that won't pass submission can be posted here. I think this works as it is not asking for the solution but what is wrong with it. |
Avin
2006-02-21 23:10:29 |
Re: my puzzle problem
If the first part of the problem was changed to require drawing at least two marbles of any two colors, then that does seem to fit the numbers given: in the first case, you could draw 8 green, then 1 each of the other three colors ,and the twelveth marble drawn would have to repeat a color, being the second paired color, and so on. |
Vernon Lewis
2006-02-22 00:32:56 |
Re: my puzzle problem
Could this have been submitted as a non-original problem with acknowledgement/credit to the original source? Just wondering for future reference if this situation should arise again - ie doubts about a published solution being correct.
My opinion is that the published solution is correct in this case. |
Tristan
2006-02-22 23:12:24 |
Re: my puzzle problem
Ahem... testing to make sure I remember how to link correctly.
link |
