| Topic Name: |
Brainteaser |
| Message Name: |
Two Questions and the Best Answer |
| Date Posted: |
02/05/2002 |
| In Reply To: |
Qua's was the best answer, but for those who haven't quite got there yet here is a full (and I hope followable) solution. Yes 3 can be blind and work it out.
Here is a list of the possibilities and who would work it out:
123
A) BRR 1 works it out first
B) RRB 2 works it out first
C) BBR 1&2 work it out simultaneously
D) RBB
E) BRB
F) BBB
G) RRB
Assumption: Each logical deductive step takes 15 seconds.
(A) and (B) both require only one logical step [I see both the red hats on other people, so I am wearing a black hat] and so a desperate prisoner will get there quickly (15 seconds) and shout "black".
(C) requires two logical steps [I, and the other seeing prison, can see that the blind prison (3) is wearing a red hat. So me and the blind man must be black+red or red+red. But fifteen seconds has now elapsed and the other seeing guy hasn't said anything. That means me and the blind guy can't both be red, so I must be black.] Now both 1 and 2 will be going through these logical steps at the same time and should both shout out at about 30 seconds that they are wearing black.
Now we come to D,E,F,G. In all of these 1 and 2 see that the blind guy is wearing black and from this neither is able to work out wether they themselves are black or red. Both will remain silent, yet (unlike in C) that silence helps neither of them and the silence continues.
BUT the blind guy has all the time in the world to work out that if there is a long silence then it must be that either D,E,F or G is the case and in all four he is wearing black. Therefore with only the aid of hearing, and the silence of the other two, he can say that he is wearing black and claim his freedom.
Of course things are never so simple, and here is where qua failed to mention two little twists that appear in the tale. As it is told we are only asked to explain how the blind guy knows for certain what colour he is, and this we have done. However a more tricky matter is if we consider what *might* have happened in such a prison and two interesting alternatives appear.
The more dull is that there is a non-conventional prisoner (blind or seeing) who thinks "three black hats, two red, I'll shout black and have a 3/5 chance of freedom".
The second is the intelligent prisoner (blind or seeing) who immediately works out all the above and thinks, "in all scenarios A to G the winning prisoner says that he is wearing black, because a person wearing red can never get to the correct answer first, therefore I will shout black as soon as the prison guard places the hat on my head. I might be wrong, but if I am then I would have lost whatever I did."
Hope that this helps and wasn't too boring.
camstrat |
| Message: |
1) What's with the 15 second time constraint? The original poster didn't say anything about this.
2) Why do you need to list each possibility? The question is asking for a particular scenario, just work through it, like I do below.
The Best Answer:
P1 says no, so he must see either one black and one red or two black. If there is a red, it does not matter who is red.
P2 also says no, so he must also see either one black and one red or two black. But this time, if there is a red, it does matter who is red.
If P3 had been red, then P2 would have know that he was black (otherwise, P1 would have seen two reds and known he was black). So P3 must be black, regardless of whether P1 is red or black.
P3 knows he is black.
As someone else pointed out, this assumes that P2 and P3 are smarter than the average criminable, but allows for P1 to be a typical crack-head inmate.
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