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Masters of Logic Puzzles III. (stamps)
Posted 06 March 2009 - 08:35 PM
Masters of Logic Puzzles III. (stamps) - Back to the Logic Puzzles
Try this. The grand master takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps:
What are the colors of her stamps, and what is the situation?
Who posted this problem? The actual questions are "What color stamps does B have? What colors have A and C?", not "What are the colors of her stamps, and what is the situation?". The 'situation' is that the second question in the actual problem (please see http://brainden.com/logic-puzzles.htm ) CANNOT BE DETERMINED, not by man, not by computer, not by fish. Whoever posted the problem, please remove the second question.
(The answer to the first question is straight-forward, and hinges on 'C' replying "No". That "No" indicates that whatever else C observes on A and B, She cannot see 2 pairs, therefore at least one of A and B have both green and red stamps on their forehead. Both A and B realize this. A subsequently answers "No" which tells B she cannot have 2 like-colored stamps (otherwise A would have known her colors, which would have been G-R), so B knows She is G-R, 100%.)
Posted 06 March 2009 - 09:02 PM
Some of you people on here are such idiots.
The whole point of the puzzle is you need to work out the situation where
and what B's stamps are.
Why do you get people like Wordblind who start saying "its ambiguous". It is not ambiguous. There is only one solution.
B has 2 different coloured stamps (red and green)
A has a pair of stamps which are the same colour. (red or green).
C has a pair of stamps which are the same colour, but not the same colour as A's pair. (red or green).
It does not matter what colour A or C's pair are, all that matters is the situation. The situation being that both A and C will have the same colour stamps in their pair, and B will have stamps of different colours in B's pair.
Several people have already explained why this is the case. There are no other possible answers if you are being sensible and not just being a fool.
There is NOT only ONE solution, sly. 'A' does NOT have to have a pair (2 of the same color), nor does 'C' HAVE to have a pair. B would know her answer second time around, same pattern of answers, in ANY of the following combos:
A - RG
B - RG
A - RG
B - RG
C - GG (or RR)
A - GG (or RR)
B - RG
C - RG
And your favorite;
A - RR
B - GR
C - GG (or reversing pair colors with A and C, of course)
Let me know if you need this explained.
Posted 09 July 2009 - 06:29 PM
In Fact 2 round is not required at all. Please read this below and tell me why we need second round?
Assuming that all have the fair chance: then here are the combinations:
1. No 2 member will have same colors i.e
A B C
RR RR GG
this type of combination's will not exist's
2. if above is true then below is also true
A B C
RR RG GG
this type of combination's will not exist's too.
in this case since B know's no 2 will be having same colors he will easily identify that he has RG
3. Finaly we have following combinations
A B C
RR RG RG
GG RG RG
RG RR RG
RG GG RG
RG RG RR
RG RG GG
RG RG RG
So in first chance if A say "yes" Group 2 or Group 3 exists.
if A say's "no" means means there exist either group 1 or group 4 combination
So now if Group 4 combination holds true then B will say "no" else if group 1 combination exists then he will say "yes"
Now if both A and B say "no" then C will say "yes" as this has to be group 4 combination:
So in all this is not an fair chance competition infact it's luck.
Edited by Ganesh Sawant, 09 July 2009 - 06:38 PM.
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