[rookie1ja]

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 Grand Master's pocket and the two on his own forehead. He asks them in turn if they know the colors of their own stamps:
A: "No."
B: "No."
C: "No."
A: "No."
B: "Yes."
What color stamps does B have?

Spoiler for Solution:

Masters of Logic Puzzles III. (stamps) - solution
B says: "Suppose I have red-red. A would have said on her second turn: 'I see that B has red-red. If I also have red-red, then all four reds would be used, and C would have realized that she had green-green. But C didn't, so I don't have red-red. Suppose I have green-green. In that case, C would have realized that if she had red-red, I would have seen four reds and I would have answered that I had green-green on my first turn. On the other hand, if she also has green-green [we assume that A can see C; this line is only for completeness], then B would have seen four greens and she would have answered that she had two reds. So C would have realized that, if I have green-green and B has red-red, and if neither of us answered on our first turn, then she must have green-red.
"'But she didn't. So I can't have green-green either, and if I can't have green-green or red-red, then I must have green-red.'
So B continues:
"But she (A) didn't say that she had green-red, so the supposition that I have red-red must be wrong. And as my logic applies to green-green as well, then I must have green-red."
So B had green-red, and we don't know the distribution of the others certainly.
(Actually, it is possible to take the last step first, and deduce that the person who answered YES must have a solution which would work if the greens and reds were switched -- red-green.)

Spoiler for old wording:

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:
A: "No."
B: "No."
C: "No."
A: "No."
B: "Yes."
What are the colors of her stamps, and what is the situation?

[novellabelle]

There is no specification of the two stamps distributed to each person being the same color.

[PookaDot]

That doesn't have to stipulated. If you draw out all the possibilities, this is the only stamp distribution in which one of the three would know without all three knowing. It takes a bit but it's worth it.

[DarkSpin21]

or you can think that the answer would not be able to be determined between red-red and green-green therefore must be red-green

[coolastro1016]

she can as long as the other 2 ladies had 2 of the same colors on there heads and if they had the same colors. So if Lady a and c both had double red then b could say for sure that she had 2 green because all 4 red would be exposed. This is the only way because of the person holding 2 in his hand. If the 2 other ladies have red-Green that leaves 2 left of each color and so the hand can have green green or red red or red green but it's impossable to tell. You would need to know whats in the hand to answer.

we need to know the stamps on the other 2 ladies heads.

[anuraag]

the only possibility is red and green on each

[bkaps44]

we don't need to know the stamps on the other ladies heads. the key here is that it took 'B' a second turn to realize the answer. if both other ladies had the same color, she would've known the answer immediately. since she was able to deduce the colors of the stamps on her head, this is the only situation where she could've done so.

A goes first (with either red-red or green-green) - sees a situation she cannot solve.

B goes next, sees A (with RR or GG) and MUST then see C (with the opposite of A). B cannot know for sure her own stamps.

C goes, sees A (RR or GG) and B (unknown). Since C does not answer, we can infer that B does not have the same color as A. B realizes this, and now knows that since neither A nor C stated their own colors, B must have one of each. B then answers the next time around.

[dsu]

Only for those who haven't figured it out completely yet:

If two or three persons have one of each color (RG), then nobody can say one's own color in any number of tries.

If two people have both same-colored stamps on their head (RR RR or GG GG), the third will have the answer instantly.

Only if one has same-colored and another the opposite but same-colored stamps (RR and GG) on their head can the third person (with different colored or RG) have the answer on her second try.

[Slick_Rick9009]

Quote

Only for those who haven't figured it out completely yet:

If two or three persons have one of each color (RG), then nobody can say one's own color in any number of tries.

If two people have both same-colored stamps on their head (RR RR or GG GG), the third will have the answer instantly.

Only if one has same-colored and another the opposite but same-colored stamps (RR and GG) on their head can the third person (with different colored or RG) have the answer on her second try.

What you said is exaactly how we know that that isn't the case. She DID know the answer and that is how you figure out how she knew.

[Stickguy999]

...umm what? didnt quite understand that one