A moderator takes a set of 8 stamps, 4 red and 4 green, known to three logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those two in the moderator's pocket and the two on his or her own head. He asks them in turn if they know the colors of their own stamps:

1. A: "No"

2. B: "No"

3. C: "No"

4. A: "No"

5. B: "Yes"

What are the colors of B's stamps?

[Taken from Ken's Puzzle of the Day, March 16, 1994.]

Ken's Solution: There are only 19 possible combinations of stamps on heads. They are tabulated below, along with the statement they are eliminated by. Each statement eliminates any unique combinations:

A B C Statement rr rr gg 3 rr rg rg rr rg gg rr gg rr 2 rr gg rg 3 rr gg gg 1 rg rr rg 4 rg rr gg 4 rg rg rr rg rg rg rg rg gg rg gg rr 4 rg gg rg 4 gg rr rr 1 gg rr rg 3 gg rr gg 2 gg rg rr gg rg rg gg gg rr 3In the remaining seven combinations, B has one stamp of each color.

Eric Eilebrecht
provided this insightful solution, which I paraphrase:

The problem asks what colors B's stamps are. They must be one of
each. If they are both red, similar arguments could be used to
say they are both green. Since that sort of argument would lead
to an indeterminate color, B's stamps must be one of each.

[KD: I didn't include this as the final solution, but I have to admit it
uses the general puzzle assumption that "There must be a solution" to
arrive at the correct solution. The problem with it is I could change the
problem statement to read "B's stamps are RR, RG, or GG. What are they?",
and Eric's argument would again lead to RG as the solution.]

Rick Simineo provided the
following solution:

This takes quite a bit of explanation(at least the way I solved it):

There are 21 separate possible ways to affix the stamps.
(Three logicians can each have RR, RG, or GG=27 minus the six which
require more than four Red or Green stamps). Of these, the eight
which contain a matching set of doubles can be eliminated by this
logic:

If A saw B & C both wearing two Red Stamps, knowing that there were
only four Red Stamps to begin with, A should have easily deduced that
he was wearing two Green Stamps in the first round of questioning,
which he did not. The same holds true for inversion of color, and for
the other two logicians as well. This leaves 13 sets.

Of them, four of the ones containingopposite matches can be taken out
by this thought:

If A had GG and B had RR, then C should have thought this: 'If I have
two Red Stamps, then A easily knows what stamps he is wearing (see
above), yet he didn't. The same goes for two Green stamps with B.
I, therefore, must have one of each.' He did not say yes, so this must
not be the case. Similar arguments are made for other colors and for
A. (Please note that these two possibilities for B were not taken out,
B being the only one not to answer No after hearing both of the other
responses.) This leaves nine.

Of those nine, two of them have both A & C with one of either stamp,
and B with a pair. Had one of these situations been the case, B's
logic would have been this:

'If I have a pair, then A should have realized that he does not
have the same pair as me (first argument), since C said No. Also,
A must realize that he does not have a pair opposite mine (second
argument), also because C said no. Therefore, A should have known
that he has one of each, which he obviously didn't realize, having
said No. Therefore, I must not have a pair'
which eliminates those two situations, leaving us with seven.

"What of those seven?" you ask? Doesn't matter. All of them include B with a Stamp of each color, which is precisely the solution we were asked to find.

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