Paul shows nine colored discs (five
red, two
blue and two
green) to five logicians.
After blindfolding them, Paul picks up the five red discs, sticks one on the forehead of each logician, and hides the other discs elsewhere. After removing the blindfolds, everyone sees the discs (all red) on the others' foreheads but not the one on his own.
After a few minutes, one of the logicians (that reasons a little faster than the others) correctly states the color of his disc. How does he work it out?
I hope there's a shorter solution...
We start from the case RBBGG. The person wearing R will instantly know that his hat is red.
Then we consider RRBBG (same as RRBGG). A R hat guy sees RBBG, so he
knows that if he's wearing a G, the other R hat guy will know that his
hat is R. But if this doesn't happen, he can conclude that his hat is R.
Now we consider RRRGG (same as RRRBB) and RRRBG. A R hat guy sees
RRBG/RRGG, so he knows that if his hat is B or G, from the above case
one of the other 2 R hats should know that his hat is R. If this
doesn't happen, he can conclude that his own hat is R.
Next, we consider RRRRG (same as RRRRB). A R hat guy sees RRRG, so he
knows that if his hat is B or G, from the above case, one of the 3 R
hat guys will realise that his hat is R. If this doesn't happen, then
he can conclude that his own hat is R.
Finally, we consider RRRRR. A R hat guy sees RRRR, so he knows that if
his hat is B or G, from the above case, one of the 4 R hat guys will
have declared that his hat is R. If this doesn't happen for
sufficiently long, then he can conclude that his own hat is R. QED.
I'm assuming, of course, that none of the logicians will be "too slow"
in concluding the above, and give the other logicians a false
intepretation of their actions (or lack thereof).
Long tedious solution
