Ultimate Test of 3 Logic Masters

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 her 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? 

'THIS' could be his color combination! 

Earlier logicians had been part of "Spot On The Forehead!" and  "Spot on the Forehead" Sequel Contest

The Wisest Logic Master!

What was the challenge in front of him?

Let's denote red by R and green by G. Then, each can have combination of RR, RG or GG.

So, there are total 27 combinations are possible.

1.  RR RR GG
2.  RR GG RR
3.  GG RR RR

4.  GG GG RR
5.  RR GG GG
6.  GG RR GG

7.  RR RG GG
8.  GG RG RR
9.  RG RR GG
10.RG GG RR
11. RR GG RG
12. GG RR RG

13. RR RG RG
14. GG RG RG
15. RG RR RG
16. RG GG RG
17. RG RG RR
18. RG RG GG

19. RR RR RG
20. GG GG RG
21. RG RR RR
22. RG GG GG
23. RR RG RR
24. GG RG GG

25. RR RR RR
26 .GG GG GG

27. RG RG RG.

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1. Now, obviously (19) to (26) are invalid combinations as those have more than 4 red or green stamps.

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2. In first round, everybody said 'NO' thereby eliminating (1) to (6) combinations. That's because, for example, if C had seen all red (or all green) then he would have known color of own stamps as GG (or RR). Similarly, A and B must not have seen all red or all green.

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3. For (9 - RG RR GG), A would have responded correctly at second round as NO of B had eliminated GG and NO of C had eliminated RR for A in first round. Similarly, (10) is eliminated.

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4. For (11 -  RR GG RG ), C would would have responded correctly immediately after NO of A had eliminated GG and NO of B had eliminated RR for him in first round. With similar logic, (12) also get eliminated.

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5. Remember, B has guessed color of own stamps only in second round of questioning.

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6. For B in (13), his logic would be I can't have RR (total R>4) but GG [ No.(11) - RR GG RG] and RG can be possible. But (11) is eliminated by C's response in first round. That leaves, (13) in contention.

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7. Similarly, if it was (14 - GG RG RG) combination, then B's thought would be - I can't have GG (total G>4) but can have RR as in (12) - GG RR RG which is already eliminated by C's NO response at the end of first round. Hence, I must have RG. That means (14) also remains in contention.

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8. On similar note, (17), (18) remains in contention after A's NO at the start of second round.

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9. If it was (15), then A would have been responded with RG when C's NO in first round eliminates RR (as proved in 2 above) and GG (as proved in 4 above) both. Similarly, (16) is also eliminated after C's NO in first round as above.

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11. For (27), B's logic would be -

"If I had RR then A must had seen RR-RG and had logic - 

"Can't have RR (total R>4); if had GG then C would have answered with RG after I and B said NO in first round itself. Hence, I must tell RG in second round."

Similarly, A's response at the start of second round eliminates GG for me.

Hence, I must have RG combination."

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10. So only possible combinations left where only B can deduce color of own stamps are -

7.  RR RG GG
8.  GG RG RR

13. RR RG RG
14. GG RG RG

17. RG RG RR
18. RG RG GG

27. RG RG RG.

If observed carefully all above, we can conclude that B must have RG color combination of stamps after observing A's and C's stamps as above to correctly answer in the second round.


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"Spot on the Forehead" Sequel Contest

After losing the "Spot on the Forehead" contest, the two defeated Puzzle Masters complained that the winner had made a slight pause before raising his hand, thus derailing their deductive reasoning train of thought. 

And so the Grand Master vowed to set up a truly fair test to reveal the best logician among them.


He showed the three men 5 hats - two white and three black. 

Then he turned off the lights in the room and put a hat on each Puzzle Master's head. After that the old sage hid the remaining two hats, but before he could turn the lights on, one of the Masters, as chance would have it, the winner of the previous contest, announced the color of his hat. 

And he was right once again.

What color was his hat? What could have been his reasoning? 

The winner is wisest for a reason! 

Master of Logic For a Reason!

How the master was challenged?

Let's assume once again A, B and C are those logicians and C has guessed the color of own hat correctly. Here is what he must have thought - 

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"I'm assuming the grand master is conducting this test fairly denying any sort of advantage to any participant.

With that assumption, the grand master can't put 2 white and 1 black hat on heads. In that case, the person having black hat and watching 2 white hats on others' head would know the color of own hat immediately.

For fair play, he can't put 2 blacks and 1 white hat either. That will give unfair advantage to the logicians wearing black hats. Suppose A and B are wearing black and I'm wearing white hat. Now, what A (or B) would be thinking - 

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       " I'm A (or B) and I can see 1 black and 1 white hat (on head of C). If I have white 
         hat on my head then B (or A) would know color of his hat as black as there are 
         only 2 white hats available and those would be on my head and C's head.

         Moreover, 1 black and 2 white hats already eliminated as it's unfair distribution.

         That means I must be wearing black hat."

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That's how the combination of 2 black and 1 white hats also eliminated from fair play.

Hence, all of three must be wearing black hats is only fair distribution giving all of us equal chance of winning and hence I must be wearing black hat only. 

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Note : Here, C is assumed as a winner for only sake of convenience, otherwise either A or B whoever is wisest can be winner.

Spot On The Forehead!

Three Masters of Logic wanted to find out who was the wisest among them. So they turned to their Grand Master, asking to resolve their dispute.

"Easy," the old sage said. "I will blindfold you and paint either red, or blue dot on each man's forehead. When I take your blindfolds off, if you see at least one red dot, raise your hand. The one, who guesses the color of the dot on his forehead first, wins."


And so it was said, and so it was done. 

The Grand Master blindfolded the three contestants and painted red dots on every one. 

When he took their blindfolds off, all three men raised their hands as the rules required, and sat in silence pondering.

Finally, one of them said: "I have a red dot on my forehead."

How did he guess? 

And this is how his logical brain responded! 


Similar kind of puzzles are - 

The Greek Philosophers  

Real Test Of Genius  

Logical Response By Master of Logic!

What was the test?

Let A, B and C be the names of three logicians and C be the logician who correctly guessed the color of dot on forehead. 

Now, this could be the C's logic behind his correct guess - 

"If I had blue dot on my forehead then A and B must had raised hands after looking red dots on the foreheads of other. In case, what A (or B) would have thought? His logic would be -

    "If C is with the blue dot then B (or A) must have raised hand after noticing 
    red dot on my forehead, hence I must have red dot."

So A (or B) would have successfully guessed color of dot on own forehead easily.

But neither A or B not responding that means I must have red dot on my forehead!"  

The logician who guess it correctly could be either A or B not necessarily be C; here it is assumed C is wisest for the sake of convenience.