ratsy wrote:

Which means at least one of them definitely wearing a green hat.  b,c or d.

b sees that at least one of a,c,d is  definately wearing green,

and c sees that at least one of a,b,d is definately wearing green. 

agreed.

That means that there are at least 2 green hats out there, between a,b,c. 

wrong - you can't make that assumption. it means there a,b&c all see at least 1 green...nothing more can be certain.

M57 wrote:

weathertop wrote:

my point exactly. if they only guess when there's a 100% certainty then this puzzle is jacked up.

the above is my proof. its a list of the permutations of what they could possibly be wearing. only the blue and red highlights are the 100% certainties. when you consider F-O, there isn't a certainty thus my confusion and thinking the OP is flawed.

You are neglecting to take into account that every time one of them passes, inferences can be made.

the only inference that can be made is that none of the three that pass see 3 blue hats.

Sorry for the delay everyone, I'm on vacation in America's Hat (Canada)

Clarifications

Yes, eventually every prisoner will see the room the same number of times, but you have no idea how often any given prisoner will see the room before all the others.  This means that it's possible that prisoner A could see it 3, 5, X times before prisoner B sees it once.

Every prisoner is watched closely when they are shown the room.  Any indication that they try to provide (dropping a match, scratch on wall, etc) is cleaned up upon exit of the room.

ratsy wrote:

Which means at least one of them definitely wearing a green hat.  b,c or d.

b sees that at least one of a,c,d is  definately wearing green,

and c sees that at least one of a,b,d is definately wearing green. 

That means that there are at least 2 green hats out there, between a,b,c. 

That leaves potentially 2 green or 3 blue hats on the head of guy d.

The only person they have in common that they can all see is Person D.  

You are on the right track. You are correct with your observations, however all but one your above conclusions (sentences with the word 'means') are incorrect.

the only inference that can be made is that none of the three that pass see 3 blue hats.

incorrect.   With each successive Pass, additional inferences can be made.

BorisTheFrugal wrote:

Sorry for the delay everyone, I'm on vacation in America's Hat (Canada)

Clarifications

Yes, eventually every prisoner will see the room the same number of times, but you have no idea how often any given prisoner will see the room before all the others.  This means that it's possible that prisoner A could see it 3, 5, X times before prisoner B sees it once.

Just to be clear.

At some (or just one point in time) every prisoner will have seen the room the exact same number of times,  This condition can only occur once every 23 visits at the very best because the moment one prisoner has visited the room more than every other prisoner, it takes at least 22 more visits for everyone to catch up, and depending on whether or not the warden takes a single prisoner to the room a number of times in the a row, it could take 100s if not 1000s of visits before a condition exists where every prisoner has seen the room the same number of times again.

Is this true?

#4 is wearing a Green hat. The solution is that #4 is ruling out that he is wearing a blue hat. What are the ways that he would be wearing a blue hat?

1. 3 blues and 1 green and #4 is wearing blue. If this were true, then 1,2, or 3 would have seen 3 blues and known he was wearing green.

2. 2 blues and 2 greens and #4 is wearing blue. If this were true, then 2 or 3 would have seen 2 blues (either 2 and 4 or 3 and 4) and known he was wearing green (because if blue then one of the earlier prisoners would have seen 3 blues). So, in this scenario either 2 or 3 would have declared green instead of passing.

3. 1 blue and 3 greens and #4 is wearing blue. Here 1 and 2 will pass because they will see 2 greens and a blue. 3, though, will know he is wearing green because if he was wearing blue then 2 would have known to guess green based on 2 above.

CK66 wrote:

#4 is wearing a Green hat. The solution is that #4 is ruling out that he is wearing a blue hat. What are the ways that he would be wearing a blue hat?

1. 3 blues and 1 green and #4 is wearing blue. If this were true, then 1,2, or 3 would have seen 3 blues and known he was wearing green.

2. 2 blues and 2 greens and #4 is wearing blue. If this were true, then 2 or 3 would have seen 2 blues (either 2 and 4 or 3 and 4) and known he was wearing green (because if blue then one of the earlier prisoners would have seen 3 blues). So, in this scenario either 2 or 3 would have declared green instead of passing.

3. 1 blue and 3 greens and #4 is wearing blue. Here 1 and 2 will pass because they will see 2 greens and a blue. 3, though, will know he is wearing green because if he was wearing blue then 2 would have known to guess green based on 2 above.

#1 i agree with you, that was my assertion too.

#2 i agree that if 2&2 with #4 wearing blue it wouldn't reach him.

#3 i missed that conclusion while working thru the permutations...

i also didn't think that if he was able to rule blue out completely (and that there were fewer blue options to rule out) that would only leave green left. nice job CK!

CK66 wrote:

#4 is wearing a Green hat. The solution is that #4 is ruling out that he is wearing a blue hat. What are the ways that he would be wearing a blue hat?

1. 3 blues and 1 green and #4 is wearing blue. If this were true, then 1,2, or 3 would have seen 3 blues and known he was wearing green.

2. 2 blues and 2 greens and #4 is wearing blue. If this were true, then 2 or 3 would have seen 2 blues (either 2 and 4 or 3 and 4) and known he was wearing green (because if blue then one of the earlier prisoners would have seen 3 blues). So, in this scenario either 2 or 3 would have declared green instead of passing.

3. 1 blue and 3 greens and #4 is wearing blue. Here 1 and 2 will pass because they will see 2 greens and a blue. 3, though, will know he is wearing green because if he was wearing blue then 2 would have known to guess green based on 2 above.

CORRECT!  Congrats CK66.  I wrote a little more detailed explanation in chronological fashion so that folks can see the progression of the logic.

Solution:

There are only 3 blue hats - so at least one of the four prisoners must have been wearing a Green hat. 

Inmate #1 needed to see 3 blue hats on the other inmates in order to be 100% sure that he (#1) was wearing a green hat. However, inmate #1 Passed, therefore he must have seen a green hat on at least one of the other inmates.

Inmate #2 didn’t even need to look at inmate #1.  If he (#2) saw blue hats on BOTH #3 and #4, he would know he was wearing a green hat.  Remember, #1 saw a green hat on at least one of them (#2,#3, and #4), so inmate #2 must have seen a green hat on either #3 or #4 (or possibly on both of them) because he (#2) Passed.

Inmate #3 didn’t even need to look at inmates #1 or #2.  He knew that #2 saw a green hat on either him (#3) OR #4 (or possibly on both of them). Given this knowledge, if #3 saw a blue hat on #4 he would have been 100% sure that he (#3) was wearing a green hat.  However, #3 Passed, so #3 MUST have seen a GREEN hat on #4.

Problem Notes:

There are 8 permutations that would permit the story to unfold the way it did, and interestingly ALL the other permutations (where #4 would be wearing a blue hat) would have enabled an earlier inmate to correctly deduce their hat color (though he would have had to take his blindfold off to do so).  Regardless, one of the inmates was bound to go free with correct play. Kudos to the warden who devised a problem that was guaranteed to lighten the prison population.

I modified this from the original problem I learned as a child, which involved 3 prisoners, 2 black hats and 3 white hats.  I could have just as easily made it have 5 prisoners with 4 tan hats and 5 yellow hats, etc.

M57 wrote:

BorisTheFrugal wrote:

Sorry for the delay everyone, I'm on vacation in America's Hat (Canada)

Clarifications

Yes, eventually every prisoner will see the room the same number of times, but you have no idea how often any given prisoner will see the room before all the others.  This means that it's possible that prisoner A could see it 3, 5, X times before prisoner B sees it once.

Just to be clear.

At some (or just one point in time) every prisoner will have seen the room the exact same number of times,  This condition can only occur once every 23 visits at the very best because the moment one prisoner has visited the room more than every other prisoner, it takes at least 22 more visits for everyone to catch up, and depending on whether or not the warden takes a single prisoner to the room a number of times in the a row, it could take 100s if not 1000s of visits before a condition exists where every prisoner has seen the room the same number of times again.

Is this true?

Technically, yes, that would be potentially true.  But recognize that the point in time that the prisoners are signaling to the warden does not require that all prisoners to have seen the room the same number of times when they signal.  They just have to signal at a point that ALL prisoners have seen it at least once.

Green.

Prisoner 1 passes because he sees at least one green.

Prisoner 2 knows that, so he would know his is green if he sees blue on 3 and 4.  Therefore at least one of 3 and 4 are green.

Prisoner 3 knows that, so he would know his is green if he sees blue on 4. 

Prisoner 4 knows that, so he would know his is green.

Edit: Sorry, I posted this reply without realizing there was another page in the thread already, and the answer had already been posted.

NewlyIdle wrote:

Green.

Prisoner 1 passes because he sees at least one green.

Prisoner 2 knows that, so he would know his is green if he sees blue on 3 and 4.  Therefore at least one of 3 and 4 are green.

Prisoner 3 knows that, so he would know his is green if he sees blue on 4. 

Prisoner 4 knows that, so he would know his is green.

 

Edit: Sorry, I posted this reply without realizing there was another page in the thread already, and the answer had already been posted.

Excellent ..and a good compact answer.

M57 wrote:

BorisTheFrugal wrote:

Sorry for the delay everyone, I'm on vacation in America's Hat (Canada)

Clarifications

Yes, eventually every prisoner will see the room the same number of times, but you have no idea how often any given prisoner will see the room before all the others.  This means that it's possible that prisoner A could see it 3, 5, X times before prisoner B sees it once.

Just to be clear.

At some (or just one point in time) every prisoner will have seen the room the exact same number of times,  This condition can only occur once every 23 visits at the very best because the moment one prisoner has visited the room more than every other prisoner, it takes at least 22 more visits for everyone to catch up, and depending on whether or not the warden takes a single prisoner to the room a number of times in the a row, it could take 100s if not 1000s of visits before a condition exists where every prisoner has seen the room the same number of times again.

Is this true?

The note that every prisoner will eventually (read as: in an infinite number of trials) be in the room the same number of times is just an obfuscated way of saying the prisoners are chosen completely at random.

Given the conditions described in the riddle, if the warden has made 99 selections, you were just as likely to have seen prisoner 1 99 times as you are to have seen prisoner 2 99 times as prisoner 3 99 times, which was just as likely as seeing each of them 33 times. You could have 1000 trials where only prisoners 1 and 2 ever saw the room.

I've seen this riddle logic puzzle before, and it's a great one.

Cramchakle wrote:

The note that every prisoner will eventually (read as: in an infinite number of trials) be in the room the same number of times is just an obfuscated way of saying the prisoners are chosen completely at random.

Given the conditions described in the riddle, if the warden has made 99 selections, you were just as likely to have seen prisoner 1 99 times as you are to have seen prisoner 2 99 times as prisoner 3 99 times, which was just as likely as seeing each of them 33 times. You could have 1000 trials where only prisoners 1 and 2 ever saw the room.

I've seen this riddle logic puzzle before, and it's a great one.

It is a great one, and yes, your conditions above are all accurate.  It's just a means to prove that the prisoners are chosen at random.

I guess I'm asking for a hint here, but is the # 23 significant?  Could the same (or a variation) of the technique be applied for more or less prisoners?

I was thinking that Ozy since a standard problem solving strategy is to make the problem simpler.  So I was thinking what if there were two prisoners?  But I don't see how to get around the idea that if the same prisoner were chosen 4 times in a row, how could he leave the room in a way different from his first time there?

This didn't get me anywhere, but here's another way I was thinking about it.

Since only one switch can be thrown, you can imagine a state transition table like this:

         00

   10        01

         11

Where you can travel either clockwise or counter-clockwise around the circle.   So instead of two levers you could instead imagine it as a single stick that points up/left/down/right, but most be moved in a circle (i.e. not from up to down or left to right).

I was trying to imagine if having some system of rotating clockwise, then counter-clockwise or something might get me in the right direction.   I got no where, but am sharing in hopes of kickstarting someone else's thoughts.

Further along those lines, if you can only move it Clockwise (CW) or Counter-CLockwise (CCW), We can call those 0 & 1.  Then you can have a pattern. For example:

0101010101010...

would mean, first time move CW then move CCW, etc.  Every other time you are undoing your previous time.   But again... I'm not getting anywhere.

FWIW, the "correct" solution does make use of the switches.  But I'm, not at all sure it would work in practice, given reasonable extimates as to how frequently the warden will bring a prisoner to the switch room ("from time to time..."), and the life expectancies of the prisoners (let alone their actual sentences).  I like my solution much better!

Amidon37 wrote:

I was thinking that Ozy since a standard problem solving strategy is to make the problem simpler.  So I was thinking what if there were two prisoners?  But I don't see how to get around the idea that if the same prisoner were chosen 4 times in a row, how could he leave the room in a way different from his first time there?

I can solve it for 2 prisoners. 

4 switch states 00, 01, 10 & 11.

The first time someone is called to the room, they move the switches towards a 00 state.  The second  & subsequent times they move it toward the 11 state.  If you get to the room & see it in 11, you know the other prisoner has been there at least twice.  (need to plan for second visit, because the switches could start in the 11 state).  Since the other guy has been there twice & you have been there at least once, you can say you have all been there.

now how to extrapolate to 23 prisoners.

Or, if it helps, there are 8 possible state changes: (see ozy's diagram above - from 00 to 01, or 00 to 10 etc.)  So if the prisoners could organize themselves onto groups they might be able to something with the state changes.