## A Set of Stamps

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?

#### Source: Internet newsgroup rec.puzzles [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        3
```
In 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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