Another Brain Bender

[buttercup]

Now we no longer have the pleasure of one.of.a.kind's delightful threads, here's a logic puzzle to while away the barren hours.

The prison governor calls three favoured prisoners into his office.
He says, "Men, it's my birthday, and I've decided to let you go -- provided you can pass a test in elementary logic."
The governor shows them five discs. Two are black, and three are white. He has the three men face the wall. He pins three of the discs to their backs. Then he conceals the remaining two discs.
In fact, he's pinned the three white discs to the three prisoners' backs.
The governor says, "Any man who tells me whether he is wearing a white disc or a black disc, and can explain to me how he knows, can go free. No conferring."
The governor then has them turn around in front of each other, so that each can see the discs the other two are wearing.

Do the prisoners go free? Which one(s)?

 

[shrek71]

With your stipulation of no conferring amongst the three, none of prisoners are set free. Each prisoner only sees the disc on the back of the other two. He has no way to determine what disc is pinned on his back.

Cheers

 

[Batttman]

the smart one got Away.

 

[DigitallyYours]

This is like the bowler hat question.

If any one of them saw two black discs, he'd know he was wearing a white disc. Since no one sees two black discs, they all remain silent.

If any one of them saw one black disc and one white disc, he'd know he was wearing a white disc. That's because he can't be wearing a black disc since if he was wearing a black disc, one of the others would have deduced that he was wearing a white disc by seeing two black discs. Since no one sees one black, one white, everyone remains silent, and this scenario can be crossed out.

So, if you see two white discs, and no one speaks up, you know you have to be wearing a white disc. You can't be wearing black because then one of the others would be seeing one black and one white and deduced he was wearing white. Since everyone remains silent, we know that's no one case see one black disc and one white disc.

 

[IM469]

Let's start easy ... if you see the two prisoners in front of you wearing the only two black discs - you know must be wearing white and can go free.
If a prisoner is wearing a white and leaves - then you must be wearing a black disc and you can leave.

If you see one prisoner is wearing a black disc and the other a white disc .... if the white disc does not ask to leave immediately (he doesn't see two black discs)- you must be wearing a white disk. With the same logic, if you see two white discs and they both leave claiming their colour - that can only happen if you are wearing a black disc.

If all three of you are wearing white discs (assuming the other prisoners are logical people) - there is a period of time that you realise that this can only occur if you are all wearing white - thus everyone can leave.

Close ???

BTW: While I was writing and thinking - I notice the a similar answer came up when I hit post

 

[shrek71]

This is like the bowler hat question.

So, if you see two white discs, and no one speaks up, you know you have to be wearing a white disc. You can't be wearing black because then one of the others would be seeing one black and one white and deduced he was wearing white. Since everyone remains silent, we know that's no one case see one black disc and one white disc.

Having 1 person see 1 white disc and 1 black disc, still tells him nothing directly about the colour of the disc on his back. He would still have to confer information from the others about their observations, which violates the no conferring rule.

Cheers

 

[oldjones]

Having 1 person see 1 white disc and 1 black disc, still tells him nothing directly about the colour of the disc on his back. He would still have to confer information from the others about their observations, which violates the no conferring rule.

Cheers

If any one of them sees one black and one white, but no one speaks up, then no one saw the other black, that is two blacks at the same time. Which means there can only be one black in the game, and therefore either guy who saw black knows he's wearing white on his back. No conference needed.

There's one Sure Thing, from the getgo: If there are two visible blacks, the guy who sees them knows he's wearing white.

The other Sure Thing is that if everyone's wearing white there's no way to tell what's on your own back by elimination; you need to deduce what the other guys see.

So: Since no one spoke, the first Sure Thing is out, but what if you were wearing black? — you already know no one else is. Then you're betting both the others are as smart as you, and fortunately in the world of logic puzzles that's a given. And we covered that above. Since everyone hears no one speak up, then everyone immediately deduces there are no blacks in the game and everyone instantly shouts, "I'm wearing white" at the same time.

And no one is released by the sadistic warden.

 

[Anynym]

...
and no one speaks up ....

There are actually many reasons why it could be the case that no one speaks up.

Not least among them is that none of the prisoners are gifted logicians - they were, after all, fallible enough to get caught and be imprisoned. And if they were, they MUST assume that at least one of the other two is NOT. Therefore, none can make any deductive choice with enough certainty to speak.

 

[d_jedi]

They all go free.

 

[buttercup]

Here's the official solution.

If I see black + black, I immediately know I'm white, and I call it.

If I see black + white, I cannot tell my own colour directly. But I can determine my colour by observing the behaviour of Whitey (the fellow showing white).

If I'm black, Whitey sees black + black, and Whitey knows directly and immediately that he's white, and calls it.
What does Whitey do if I'm white? Now, Whitey sees black + white. Now, Whitey cannot tell from his own direct observation whether he is black or white, so he does nothing.
But the fact that Whitey does nothing tells me that Whitey does not see black + black, i.e Whitey must see black + white.
The black disc Whitey sees is the one we both see on our comrade, so mine must be white, and I call it.

You say: Ok, if I see black + black, I know for sure I'm white, and I call it.
If I see white + black, I wait for Whitey: if Whitey calls, "I'm white," I know I'm black, and I call it. If Whitey doesn't call, I'm white, and I call it.

But, what if I see white + white? I still don't see how I can possibly tell, then, whether my own disc is white or black?

Answer:
As explained, if my comrades see white + black, although they cannot determine their colour directly, they can do so indirectly, using the above-described method. That is to say: if I see white + black, and Whitey calls, "I'm white", now I know I'm black. If Whitey does not call, now I know I'm white, and I call it.

So, if I see white + white, and my comrades both see white + black, either (or both) of them will call, "I'm white." Their call, if it happened, would tell me I'm black.
If I see white + white, and my comrades both see white + white, no-one will call. The fact of the absence of calls tells me I'm white, and I call it, and I'm gone.

The fact that I call, "I'm white, is of no assistance, directly or indirectly, to my comrades. They will stay where they are unless and until they figure out the same logic I did.