The question was as follows:
The puzzle has many versions.This one is a little twisted. The riddle is as follows:Solution:
Three persons have 4 black and 4 green papers. Two of them are kept in a drawer and none of them has seen what is . Each person have two of the papers placed at their forehead so that the other two can see the color of the paper but the person himself cannot. Now it was asked one by one if they know about the color of the papers at their forehead. They replied as follows:
A says: No
B says: No
C says: No
A says: No
B says: Yes
How do B know about the color of the papers at his forehead and what are they?
How to go by : If any of them had seen that the other two have the same colors i.e. RRRR and GGGG. Then they must have said Yes at their first turn.
Now as B said he knows the answer, lets assume what he had (Lets go for elimination criteria)
Lets suppose B had either both red or both green. Lets take both green(same logic goes for if he had both red).
Now if A had both red (both green possibility ruled out as stated earlier). And A and B both said "No" , then C can realize that he had RG. But he also said "No". This means the A have RG but again A also said "no" in the second turn. That clearly suggests that the assumption of having B both Red or both Green is wrong.
Hence B have one red and one green.