Don't look up answer and if you already know it, please refrain or use spoilers:
You and two others are being kept prisoner. The warden places one disc from a group of three white and two black discs on each of your backs. Each prisoner, including you, is unable to see what color their own disc is. To be released, you are required to give a logical account to the warden of what the color of your own disc is. How do you proceed?
ask the nigga behind me duh
i turn my back and ask the prisoners what colour it is you moron
It's a white collar prison and there is no talking allowed.
How lazy are your minds that you'd try to squirm out of having to think for more than two seconds in answering an optional riddle?
If the other 2 prisoners share the same color, group them together. Otherwise, split them apart. If they’re intelligent, they should get the message and they will perform the same process on you and the other prisoner.
They are all white cuz white supremacy and shiet
im garbage at math but im guessing it's not simply 'the odds are 3/5 that the disc is white'. Im assuming you can see the colors of the discs on the other prisoners' backs? In that case you would have to subtract whatever they had from your probability. If they were one black and one white then it would be 2/3 chances it's white, if they were both white it would be 1/3, and so on.
Is there some other answer? Im always a brainlet with these scenarios.
If the disc in front of me is black, I choose white (because It is statistically more probable). If the disc in front of me is white, I choose black for the aforementioned reason.
okay i'm getting into this logical game, here's what I would do:
I would look the warden in the eyes, see if it dilates and from there I would deduce ... that i'm not a mathematician and that this shit thread of yours should go back to youtube where it intrigues normies who haven't found the answer either and then i would hit my head against the wall till I bled and died
if its white collar than all the discs are white too, qed
Based
Nigga you forgot the MOST important part of this puzzle, mainly that they cannot talk to each other, but you do see their reactions and behavior and are allowed to infer from it what your own disk is. Which btw is just a form of communication, obviously, and the puzzle is thus kinda retarded.
Yeah, i feel that providing that detail is too much of a hint. Brainlets here will still struggle though.
how you could you possibly infer from their nonverbal communication what colour their or anyone else's disc was?
Also, they are wearing black and white outfits. Surely you could look at someone and point at either the black or white stripes on your outfit. This thread is over.
...
That's the riddle.
How low-IQ can we go?
>That's the riddle.
The way the riddle was described in OP is immensely unclear, so you can interact with them as much as you want you just can't talk?
Well the fact that this puzzle is Lacanian should tell you all you need to know, you just assume and make shit up.
>Brainlets
No, the only brainlet is you, for failing to properly lay out the conditions of the puzzle.
Say what you want but I would know my color in less than a minute. Add more parameters if you think my solution isn’t good enough
No interacting. But they are given the same terms and you are able to see if they leave the prison.
>can’t talk to each other
>can’t interact
These are completely different. Which one is it?
fuck this hack
>Jacques Lacan I actually knew. I kind of liked him. We had meetings every once in awhile. But quite frankly I thought he was a total charlatan. He was just posturing for the television cameras in the way many Paris intellectuals do. Why this is influential, I haven't the slightest idea
-Noam Chomsky
Their leaving the prison wouldn't tell you anything, they could have just randomly guessed the right answer, you aren't gaining any information because you already saw what their colour was.
If you see two blacks, you know you have white.
If you see one black and one white, then the white prisoner will leave only if you have the black color.
If you see two whites, and no one does anything, then you also have white.
Clearly you have a white disc, as do the other two.
If you (Tony) have a black disc, Percy can deduce he has a white disc, since if he had a black disc Maria would see 2 black disc and conclude she has a white disc. Since she does not conclude this, Percy has a white disc. The argument is symmetric in Percy and Maria, ergo they both have white discs and so do you (Tony).
Q.E.D.
You haven’t covered the case where both prisoners have white discs and you have black/white.
The riddle clearly states you need to provide a logical argument to the warden, ergo they couldn't have left by guessing randomly.
Their leaving or not is fundamental to solving the riddle and it contains literally all the information to solve it.
I think I did? Since the argument is entirely symmetric in all 3 actors.
I'm pretty sure you solved it. It statistically likely that at least one white disc was broken. If the other prisoners have a black disc, you go white. If they have a white disc, you go black. OP, thoughts?
Not correct. The answer does not involve probability.
>No one leaves and the other 2 discs are white:
You could have black still
>The other's discs are white and black and the one with the white disc leaves:
If he’s judged your inaction, it’s reasonable to conclude that he thinks he is white, and that you are white. If he leaves immediately, then you are black
Isn't it literally impossible to solve this riddle without probability? Wtf is the solution then?
read the thread, you decide based on who leaves. The riddle is impossible if there were infinite black and white discs
The problem with the second-order observation solution to the other two being white is that it's also applicable to the other prisoners.
Over time I would try and use a mechanism of pointing towards a prisoner and closing my eyes to mean black and keep them wide open to mean white. Hopefully they will understand and give my colour as well.
If I couldn't communicate, this would probably be the best bet.
Really ez
The whole problem is being able to deduce what the other see from what you see and how they react to what you see. The "pointing at black and white strip on the prison uniform" and similar answers are nice but they're only dodging the form of the problem, which is perfectly solvable in theory. Here the only assumption is everyone is able to see the other's disk and not too dim about reasoning.
To make things easier: we say that someone who see 2 black disk (and thus know they have a white disk) is lucky.
We say that someone who sees two white disks is full and someone who see one black and one white disk is mixed.
So they are two cases:
1. You or someone else is lucky. Then that person will almost immediately leave and be released. Also in this case you're mixed. You then know that your disk is black, and you can get off.
2. Nobody's lucky. This should be the case if nobody has left after 5 minutes at most (most likely 30s-1 min). This means there is *at most* one black disk among the three of you.
Then there are three possible situations:
a. You're mixed. In this case you can't be black since there is only one black and it's not you. You know for certain that you're white, you can get out.
b. You're black and full. Everybody else is mixed and white. Since the problem is the same for all three of you your mates can do the same reasoning as you, and both of them are in situation a. So the mixed ones will eventually conclude they're not black, and leave. It may take them 5-15 min, 30 min if they're really slow. At that point you know you can only be black, and you leave last.
b. You're white and full, so is everybody. Everybody makes the same reasoning as you, so everybody is waiting for the other to leave. After some long enough time (say one hour) and exchanged look everyone understand they're also white, everyone leaves at the same time.
Again you can't be black and mixed, that would imply there are two black disk and the third of you would be lucky and would have gotten out in 15 min max.
So, the consequence of this analysis is:
1. You see two black disk, you're white, leave now.
2. You see only one black disk. If exactly one of your mate leaves quickly with a smile, you're black, you can get out.
3. You see only one black disk and nobody has left in the first few minutes. If none of your mates is retarded you're white, you can get out.
4. You see two white disks and after 5-15 min your two mates, after exchanging look, have left at about the same time. You're black, you leave right after them.
5. You see two whites disks and nobody has left in quite some time despite exchanging looks and looking repeatedly at each other's disks. All three of you are white, leave at the same time.
In practice it wouldn't always work unless with smart and likeminded people guess, or with repeated exchanges of telling looks.
>expecting Chomsky to understand continental philosophy
Not to defend Lacan, but that's not really telling of anything, Chomsky is a massive philosophical autist.
Just realized you solved it before me and more concisely. Eh, it was good practice.
This requires that the other two prisoners get the first-order inference, but only you get the second-order inference.
>implying chomsky is wrong
If I see two whites, and a prisoner leaves quickly after seeing the inaction of the other prisoner, then I should assume that I am black. If everyone looks at each other for a while, then I am white
couldn't give a rats ass about your dogshit dilemmas OP
modern 21st century life has no dilemmas for a normal human being. FUCK YOUR PHILOSOPHICAL MUSINGS
You've never done this before and have no gauge for what constitutes 'quickly.'
You have to assume the other prisoners reason perfectly, otherwise no strategy is useful. And if they reason perfectly, then their decisions should either be instantaneous, or an instant reaction based on another prisoners lack of immediate action.
Now you're just leaving things to chance again and might as well revert to probability solutions.
Foucault is not even that hard what a hack.
If you have a prisoner that has no idea what’s going on, then yes, you do have to revert to probabilistic thinking
If you can see the other two both have black disks, you know yours is white. If anyone tells the warden their disk is white (someone with a white disk gets out), they must have inferred it from yours being black.
Beats me what you do when there're two or three white disks in play. It looks like you're just expected to see people not move and assume they're stuck rather than stupid. If you can at least see people say "I don't know" (not leave) when the warden questions them, then you get more to work with. With a volunteer answer system instead of turna though, you're fucked.
>pic
>the prisoners also cannot talk to eachother
If it's black they chimpout
>turna
turns*
why are WHITE and BLACK capitalized
We will assume that rounds proceed discretely, and you can see when another prisoner declares their own color. Assume also that prisoners only declare their color when they know it for sure.
The only time I can declare my color without any information other than seeing the color of the other two prisoners is when I see 2 black circles. Then I know that I am white circle. I declare.
If I see another prisoner declare on the first round, then I know I am black circle & declare it. Everyone escapes.
But if no one declares on the first round, I everyone knows that there are less than two black circles among them. Therefore if it's round 2, and I can see one black circle, then I know I am white circle, because otherwise there'd be two black circles among us. I declare so. If I can't see any black circles, I can't determine my own color yet, and don't declare.
So at round 2 there's two cases: either (1) there's actually one black circle (2) all three prisoners are white circle
If (1) is true, the two prisoners with white circles will see the black circle & declare they are white, escaping on round 2. The guy with the black circle will then know he is black and declare on round 3. Everyone escapes.
If (2) is true, no one will declare on round 2. Therefore on round 3, everyone knows they're in case 2 and everyone is white, they declare so, everyone escapes.
What does this have to do with psychoanalysis? This is just a math problem/game theory problem, assuming the people involved are rational. And of course if you can't assume that, or they might just be retarded, the problem is just unsolvable.
Well, it's quite simple: If somebody sees 2 black discs, they would know immediately that they have white on their back.
So, if one of the other prisoners answers immediately, you know for certain that you have white on the back.
If nobody answers right away, everybody sees at least one white disc. That means, if you see one black disc, you know for sure that you have white. Similarly, if somebody else now starts to solve, you know that you have the opposite color of the other remaining guy.
If, for a long time nobody solves, you have to assume that everybody has a white disc and can solve too.
Of course, that is assuming that everybody there is rational enough to play the same way.
If both of them are black, I know I am white and can go up to the warden and be released.
If someone else goes up to the warden, I know I am black.
If nobody goes up to the warden, there are two cases: Three white, or two white and one black. If I see a black and a white, I know it is the latter case and I am white. The other prisoners know this as well. If one of them gets released, I know he saw a black and a white, hence I must be black. If nothing at all happens though, then I am white.