I think that the explanation using three-hat "cases" makes this puzzle seem more difficult than it really is. I start by just asking what the first prisoner may see. If he sees 2 blue hats, then obviously he has red. Since he does not know what he has, then he sees either 2 reds or a red & a blue. Then on to the second prisoner. If he sees a blue, he will know he must have a red, or else prisoner 1 would know his own color. But since prisoner 2 does not know his color, he cannot be seeing a blue. Ergo, prisoner 3 must have a red hat.
Fun and very very old puzzle. I remember it from at least 30 years ago :) The explanation seem kind of convoluted. In simpler words, P3 would think like this: "If P1 saw 2 blue hats, he would've concluded his hat is red, since there are only 2 blue hats. But he didn't know the answer. So there must be at least 1 red hat on the head of either P2 or mine (P3). Given this, if P2 saw a blue hat on my head, knowing there must be at least 1 red hat between us, he would've known it's on his head. But he didn't know the answer. This means he didn't see a blue hat on my head, therefore I have a red hat"
I had this riddle at a job interview about nine years ago; it was a software developer post. I worked it out and got the job! (Of course, there was more to it than that.) I had forgotten between then and now what the answer was, and I just wrote down a truth table in the same way you did. (I don't know if that's what it would be called in the world of maths but it is in electronics for logic gates.) I arrived at the answer and it was far less stressful!
Much simpler solution: 1) There is only one case where Prisoner 1 knows his own hat: If he sees 2 blue ones in front of him (because there are only 2 total blue hats). Since he doesn't know, that means at least one of the two in front of him is red. 2) Knowing this, if Prisoner 2 sees a blue hat, it means his own would have to be red (since #1 above established there's at least 1 red hat between 2 and 3). Since Prisoner 2 doesn't know, that means Prisoner 3's hat must be red.
Yes. #2 knows he and the front guy can't both wear blue so if the front guy had blue he would know his hat color. Since #2 says he doesn't know, #3 must be wearing red.
My first thought : Let everyone guess and say Red and since there 2 blue hats and 3 Logician, then there would be atleast 1 person with a red hat :) and boom , they are free Edit : Magician => Logician
Solving this after 2 minutes: Assuming that prisoner 3's hat is blue, if prisoner 2's hat is also blue then prisoner 1 would know that their hat was red since there're only 2 blue hats, but they wouldn't, so it gives the only possibly that prisoner 2's hat must be red. And the prisoner 2 would know that and would guess the hat color correctly, but since prisoner 2 also don't know their hat color, the assumption was wrong and therefore prisoner 3's hat must be red.
Solving this without ever stopping the video except right now to write the comment before continuing to the solution: the only way 1 would know his color is if 2 and 3 were both blue, so at least one of them must be red, since 2 doesn't know either, it means that 3 is not blue, otherwise 2 would have known he is the red one
I love how you explain everything so detailed, that even people of English non native language, or people who don't know math that much still understand. 🤝👏
Let's not argue about what kind of justice system gives the warden the right to decide who goes free and who stays in prison , explain to me though how all 3 have proven that they are logicians by just one of them drawing the right conclusion.
For those proposing a simpler/shorter solution. I would argue that this one is better because it shows clearly how to generalize the problem. For those arguing for removing the sequential questioning, I would argue that the reliance on the prisoners estimating how long their peers would take to work it out, whilst it should work, makes for an unsatisfactory problem due to the indefinite time period. If I were the warden I would prefer the stated method as it would be more dramatic 🙂
You don't actually need to ask the prisoners the questions. You just tell them to shout out as soon as they know the colour of their own hat for certain. After a few seconds, prisoner #3, having heard nothing from behind him, should be able to make the inference and shout out "red".
The warden is not entitled to release the prisoners. Regardless of how good logicians they are, only a judge can repeal their sentences, or a parole board. The crooked warden should be in jail with or without all the hats.
If nr. 2 and 3 would both wear a blue hat, nr. 1 would know his own hat color. But nr. 1 does not know. So if 3 has a blue hat, nr 2 would know for sure he has a red hat. But nr. 2 does not know his hat color either. It follows that nr. 3 has a red hat.
EXCELLENT teaching. I think my mom could understand this explanation. Maybe. I've showed her the prisoners and hats riddle like three times. And each time she get a little more mad at me. I'm not going there again.
You missed one important detail in the setup. Now, most of us just assumed that the perfect logicians (or rational people) knew that the others were also rational. But this needs to be stated/assumed in order to make the puzzle work. basically, everyone should be super rational. Rational is when you are perfectly logical. Super rational is when you KNOW everyone else is ALSO rational
Since only one of them needs to correctly identify their hat colour, it seems like an alternative strategy would be to just all answer 'red'. One of them has to be right. And that's probably the strategy that illogical criminals would use as well, since there are more 'red' hats, so the test is not very well-designed by the warden.
That could easily be solved with a rule saying "if any one of you gives a wrong answer, the test is over". In that case, you can either say "I don't know" or give the correct answer.
There's a way better variant of this puzzle. There are 100 prisoners and the hats are assigned to them randomly, and the number of both red and blue hats is random as well. Then they are being asked about the color of their own hat one by one, starting from the prisoner in the back. Only if no more than 1 prisoner guesses wrong, they are let free, otherwise they all loose. However, they are allowed to strategize beforehand. It is solvable, albeit it took me some time to come to a solution first time I heard it :)
Świetna zagadka! 🧠 Film doskonale pokazuje, jak logika i dedukcja mogą prowadzić do rozwiązania nawet najbardziej skomplikowanych problemów. Dziękuję za ten wspaniały materiał, który pobudza do myślenia. Czekam na więcej takich wyzwań! 🌟
Here’s a twist - any one of them should be able to correctly identify the colour of the front prisoner’s hat with just the rules of the game. The logic goes like this - the warden wants to confirm that all three prisoners are logicians, yet only one of them needs to be able to answer. So, in order to test all 3, two of them must be unable to answer with certainty, but the othe two’s logical uncertainty must be enough of a hint for the third to solve the problem using logic. Since everyone has the same information as the person in front, plus that person’s hat colour, the uncertainty of the people in front cannot give extra information, so the uncertainty must pass information forward, therefore, the person at the front must be able to know after the other two have expressed their uncertainty, which brings each of the prisoner up to date with all the information that we had to work with for our solution, without ever seeing a single hat (assuming they’re aware of the number of each colour of hat).
3rd prisoner hat: red 1st prisoner can only know if the two hats in front of him are blue, then his must be red. So if he doesn't know, then either : - one hat is blue and one is red Or - both hats are red (for prisoner 2 & 3) 2nd prisoner deducts the above and realizes the only two possibilities above. He can only 'know' if the hat in front of him is blue, then his has to be red. He won't know, if the hat in front of him is red, then his own can be red too. Last prisoner deducts this, and realizes his own hat is red.
I never saw this one tbf... But I think it's pretty easy. 1) If he doesn't know it's because 2 and 3 have different colours one is red and the other one is blue, because if there was 2 blues he'll know that his hat is red. So the only options is that 2&3 have R&B B&R or R&R 2) If 2 doesn't know, but he knows the 1) steps he knows that if 3) has a blue, he HAD to have red, so, if he doesn't know number 3 has a red hat. 3) Knows that his hat is red.
There's an easier way to think about this that doesn't require explicitly considering 2^3 possibilities: - The only way prisoner 1 would know his hat color is if he were seeing two blue hats, because there are only two blue hats and then he would conclude his is red. He doesn't, so he must see at least one red hat. - Prisoner 2 now knows there is *at least* one red hat between him and prisoner 3, so if prisoner 3 had a blue hat then he would conclude that he must have a red hat. He doesn't, so prisoner 3 can't have a blue hat. - Prisoner 3 now knows he has a red hat.
One of the easiest puzzles of this kind. 1.) If 1 sees 2x blue, 1 says "red" 2.) If 1 says nothing and 2 sees 1x blue, 2 says "red" 3.) If 1 and 2 remain silent, 3 says "red". Reasoning: 1) There are only 2 blue hats 2) Apparently 2 and 3 don't both wear blue. 3) Apparently 3 doesn't wear blue.
Here's an old problem you might find interesting, it requires calculus though "An tiny asteroid is falling from very far away straight towards the sun. It just so happens to be falling at a certain angle such that as it gets closer to the sun, the Earth gets in the way, and the asteroid falls straight down a well exactly on the equator of the Earth. What time must it have been?" . . . . . . . . . . . spoiler 2:30 am is the answer, the trick is realizing Earth's sideways motion around the sun is actually much faster than it's daily rotation rate, so you can simplify to a triangle of vectors, one vector is the sideways orbital motion of the earth, the other vector is the asteroids motion, and the vector connecting them has the same angle that the well would have to be aligned off from midnight pointing exactly opposite the sun, then you convert the angle to a time and it all works out, pretty fun problem.
That reminds me of a much easier one I once came up with. Rounding off numbers, if the day is 24 hours long and the time between full Moons is 28 days, how much faster or slower does the Moon move across the sky than the Sun?
I think your explanation is making this much more complicated. First of all, assuming that prisoner 3 is wearing a blue hat, if 2 also wears blue, 1 would have known that he is wearing red hat and immediately reply positive, but he didn't. This means that either 2 and 3 are both wearing red hats or one blue, one red hat. If 3 was wearing a blue hat, 2 would have known that he was wearing a red hat and immediately reply positive, but he didn't. That means that he is unsure about his own hat colour, which is possible only if 3 is wearing a red hat as there were three of them initially (the same reason 1 would not have been able to answer his hat colour either). That's when 3 knew he is wearing a red hat
Pausing and posting my reasoning: Prisoner 1’s statement can be rephrased as “The other 2 are not both wearing a blue hat”. (If they were, prisoner 1 would know he was wearing a red hat.) Prisoner 2’s statement then says that prisoner 3 is not wearing a blue hat. (He knows they are not both wearing one. If he saw a blue hat, he would know he was wearing a red one.) Prisoner 3 therefore knows he must be wearing a red hat. I think this also shows that if the hats are randomly selected, it’s always possible for the prisoners to win. (Either P1 sees 2 blue hats, and knows he is wearing a red one, or Prisoner 2 sees a blue hat, and knows he’s wearing a red one because P1 didn’t see 2 blue hats, or prisoner 3 knows that P2 didn’t see a blue hat, because of the previous car)
In the first 30 minutes: keep up the good work! In the first day: this is another way I solved the puzzle! After a month: If a prison warden is capable of creating such a good puzzle, he shouldn't be a prison warden After some time: *hate comments*
I would say first that the prisoners trust that each of them truly are masters of logic the warden expects them to be. So we can say that each prisoner has perfect logical ability. I also expect the prisoners to hear each others answers to the questions. If both of the prisoners 2 and 3 would have blue hats the prisoner 1 would know his hat is red. Since Prisoner 1 doesn't know, the two prisoners in front would have either 1 red and 1 blue hat or 2 red hats. Prisoner 2 heard that prisoner 1 doesn't know so he knows that there are two options. If the prisoner 3 in front would have a blue hat prisoner 2 would know his hat is red, but prisoner 2 doesn't know. This gives all the necessary clues to prisoner 3 to know that his hat must be color RED. Because either way the prisoner 2 must have seen a red hat to not know his own hat color.
Logicians: “hey we’re innocent” Warden: “I think you’re innocent too” Logicians: “let us go then” Warden “so I made this game where you have to be able to trust two random strangers judgement, or you’ll all rot in prison. Fun right?” Logicians: “…”
I feel like I have seen this question on this channel before. The answer is red. If 2 and 3 are blue, then 1 will know that his is red, but 1 didn't know, so either 1 saw red-blue or red-red. If 2 saw blue, then 2 would have known that his is red, but he didn't, meaning 3 is red.
After seeing the video, I include myself in the group of other people who said that they solved the problem another way. And this made me wonder which is the better or the easier way to solve this - by creating a table of all variants in your mind and querying it, or by constructing only the possible scenarios for what you see. Probably the former one the better way to present and explain the solution, and the latter one is the way most people would actually use to find the solution.
The simplest way to think about it, in my opinion, is the following. If the first does not know, the next two know they can't both have a blue hat. The second would know they have a red hat if the third has a blue hat. However since they do not know, the third must have a red hat.
From answer 1, 2 + 3 can’t be B,B, because in that case #1 would know his or her hat is R. #2 now knows both 2 and 3 can’t be B. That leaves 2 and 3 as R,R, B,R, or R,B respectively. If #3 was B, # 2 could tell his hat must be R. But, #2 can’t because #3 is R, leaving #2 as R or B. After #2’s answer, #3 knows that R,B, is eliminated as a choice. With either remaining option, R,R, or B,R, #3 knows his or her hat must be R.
Make it 5 logicians in an L shaped corridor and all can determine their hat colour. A more satisfying puzzle IMHO 1 v 2 v 3 > 4 > 5 5 & 3 work as before and 1, 2 & 4 can be derived
In fact, prisoner 3 will always know his hat. Depending on the hat distribution the response to 'do you know your hat' may be : Yes, yes, yes / No, yes, yes / No, no, yes
maybe it's silly and basic, but figuring this out on my own with pen and paper was one of the more rewarding things I've done recently - straight up. that was so much fun. I STRONGLY STRONGLY URGE ANYONE WHO READS THIS TO TRY AND FIGURE IT THEMSELVES! if you just don't think you can, I'm willing to bet all I have that you're smarter than you think. try!!! be well everyone
The deduction isn't that complex (having to go through all possibilities). The *only* reason #1 didn't know his hat's color was because he saw *at least one* red on #2 and #3 (if it were both blue he'd have known his hat's color). Given that, the only reason #2 would not then know his hat's color is if #3 is not waring a blue hat. So, #3 has to be waring red hat at that point.
Luckily, the warden is not a perfect logician, because otherwise he would have realised that only the 3rd prisoner demonstrated himself to be a perfect logician. The other two could have been completely clueless.
The problem is color symmetrical except for the blue hat shortage. If 3 were wearing a blue hat, that would provide more information to 1 and 2. Since neither of them knew, 3 must have a red hat.
if prisoner 2 and 3 had blue hats then prisoner 1 would have answered "yes, I know which color my hat is" because it could only be red. therefore since prisoner 1 did not say that, prisoner 2 and 3 collectively has at least 1 red hat. if prisoner 3 had a blue hat then prisoner 2 would know that they have a red hat, but they don't know this, so prisoner 3 has a red hat.
Solved quickly. Realized number 1 would only know if both hats are blue, so 2 knows that either he or the guy in front of him can’t have blue. If 3 had blue, 2 would know he couldn’t also have blue, so 3 infers he has red.
For those who don't want to read the long comments: There is *at least 1* red hat in P1, P2 (2 blue hats -> P1 is red) < "P2, do you know your hat color? I don't know" > If P3 was blue, P2 would know he was red. > Therefore, P3 is red.
When the warden asks 1, they don't figure it out because there can't be 2 blue hats in front of them. Then 2 realizes they and 3 can't both have blue hats. If 3's hat was blue, 2 would have known that their own is red. So then 3 realizes their hat is red.
I didn't bother listing the possibilities & eliminating them, I just solved it like this: If 2 & 3 had blue, 1 would know that he's red, so there must be at least 1 red hat on 2 or 3. 2 knows this & so if 3 was blue, 2 would know he's red. This means 3 must be red.
*1 say “idk” mean 2 and 3 cant be both blue => only red/blue or red/red available *2 say “idk” mean 3 cant not be blue cause if 3 was blue 2 can guess that his is red *3 now combie those both answer and only left is he is red and 2 is red too
Took me some time, but I came to the same conclusion. Namely that my hat must be red, if they both said that they do not know. I did it mentally, no writing or drawing. It took me some time. More than I expected.
My logic for this one is as follows (spoilers below for those who don't want a solution): First, the only way for the 1st logician to know his hat color is to see two blue hats, so we can conclude that he sees at least one red hat. Second, the 2nd logician then knows that the 1st sees at least one red hat, and if he saw a blue hat, he would know that his hat is red, since he stays silent, we know the 2nd sees one red hat. From there, the 3rd logician knows that the 2nd sees one red hat, and correctly states his hat color.
The 3-line solution... If Prisoner 1 saw two blue hats, he would know he has a red hat (as there are only two blue hats). So prisoners 2 and 3 do not both have blue hats. If prisoner 2 saw a blue hat, he would know he has a red hat. Prisoner 2 therefore saw a red hat, but his own hat could be red or blue. Prisoner 3 knows he has a red hat.
heyy i think im actually getting better at these logic puzzles! i was able to get the answer for this easily and quickly (and ik it’s not a big deal since this was super easy, but i’m still proud since i used to suckkkk at these)
Vey simple. If the 1st prisoner sees 2 blues then he knows his must be red. So, he sees at least one red. Then, if the 2nd prisoner sees a blue then he knows his is red. Therefore, the 3rd prisoner must have a red hat.
prisoner 1 did not see 2 blue hats ahead of him, otherwise his hat color would be red. So he either saw two red or one red and one blue hat ahead of him and was unsure prisoner 2 did not see a blue hat on prisoner 3 because he knows they both cannot have blue hats. If he saw a blue hat his hat had to be red. He saw a red hat on prisoner 3 and was not sure still but prisoner 3 knew this
#1 would only know his hat color if both #2 and #3 were wearing blue. Since #1 doesn't know his hat color it means #2 and #3 can only be wearing both red or one red and one blue. #2 would only know his hat color if #3 was wearing blue. Since #2 doesn't know his hat color it means #3 must be wearing red.
before watching solution: the first would only know he was wearing a red hat if the other 2 were wearing blue hats. Therefore, 2 and 3 could not both be wearing blue hats becuse if they were the 1st would know. the second knows the first knows they are not both wearing blue hats, so if 3 is wearing a blue hat 2 would know if he were wearing a blue hat because 1 would have known that he was wearing a red hat. Therefore 3 must be wearing a red hat.
This is my solution before watching the video's solution: If 1 sees 2 blue hats, he'd know he's red. That leaves the assumption tht he sees 1 blue hat or 2 red hats. With that, if 2 sees a blue hat, then thay meana he's wearing the red one, but since he sees a red hat, he doesn't kmow whether he's wearing a blue or a red pme himself. That leads 3 to conclude that he's wearing the red one.
If 2 and 3 are both wearing blue then 1 knows for sure he’s wearing red which means either 1 or 2 is wearing red (or both are). If 3 is wearing blue then 2 knows that he’s wearing red so therefore 3 knows he’s wearing red for sure.
Wouldn't it be easier if each prisoner just said their hat was red? With three red and two blue hats to share between three prisoners at least one of them is bound to be correct.
Red N1 doesn't know because the 2 others don't have blue and blue. N2 doesn't know because 3 is red. Otherwise if 3 is blue then 2 would be red (because of N1 saying both aren't blue)
Answer is red because, if 1 ould see 2and 3 wearing blue then he will know his hat, but he does not answer , so 2 and 3 are wearing either red and red or blue and red or red and blue Next so from this information, if 3 is wearing blue then he will answer red, but he doesn't answer , so 3 is wearing red which makes 2 to confuse Finally from these informations we come to know that 3 is wearing red hat
Bro the first one doesn't need to look in 7 cases to see what he has if he sees 2 blue hats. He knows instantly he has red because there're only 2 blue hats lol
I think that the explanation using three-hat "cases" makes this puzzle seem more difficult than it really is. I start by just asking what the first prisoner may see. If he sees 2 blue hats, then obviously he has red. Since he does not know what he has, then he sees either 2 reds or a red & a blue. Then on to the second prisoner. If he sees a blue, he will know he must have a red, or else prisoner 1 would know his own color. But since prisoner 2 does not know his color, he cannot be seeing a blue. Ergo, prisoner 3 must have a red hat.
That's exactly my reasoning. Too complicated explanation in the video with the hat table.
What j.r. said.
6:33 YOHOOO! I am so glad they go free, very nice
That's the exact way I was thinking about it. Thank you for writing this one 👏
Exactly. The whole hats table thing really overcomplicates it.
Fun and very very old puzzle. I remember it from at least 30 years ago :)
The explanation seem kind of convoluted. In simpler words, P3 would think like this:
"If P1 saw 2 blue hats, he would've concluded his hat is red, since there are only 2 blue hats. But he didn't know the answer. So there must be at least 1 red hat on the head of either P2 or mine (P3).
Given this, if P2 saw a blue hat on my head, knowing there must be at least 1 red hat between us, he would've known it's on his head. But he didn't know the answer.
This means he didn't see a blue hat on my head, therefore I have a red hat"
I had this riddle at a job interview about nine years ago; it was a software developer post. I worked it out and got the job! (Of course, there was more to it than that.) I had forgotten between then and now what the answer was, and I just wrote down a truth table in the same way you did. (I don't know if that's what it would be called in the world of maths but it is in electronics for logic gates.) I arrived at the answer and it was far less stressful!
Much simpler solution:
1) There is only one case where Prisoner 1 knows his own hat: If he sees 2 blue ones in front of him (because there are only 2 total blue hats). Since he doesn't know, that means at least one of the two in front of him is red.
2) Knowing this, if Prisoner 2 sees a blue hat, it means his own would have to be red (since #1 above established there's at least 1 red hat between 2 and 3). Since Prisoner 2 doesn't know, that means Prisoner 3's hat must be red.
Yes. #2 knows he and the front guy can't both wear blue so if the front guy had blue he would know his hat color. Since #2 says he doesn't know, #3 must be wearing red.
Yep, figured that in about 2s from seeing the question.
Plot twist: they did commit the crime, but did everything in a very logical way.
Because, they, as perfect logicians deduced that they would be assigned to this specific warden with a propensity for logic puzzles.
So at the end, the warden lets 3 evil logicians free!
My first thought : Let everyone guess and say Red and since there 2 blue hats and 3 Logician, then there would be atleast 1 person with a red hat :) and boom , they are free
Edit : Magician => Logician
For real, but I guess they need to explain their choice
@@gtpone916 what he said is an explanation.
Magician?
Exactly he didn’t say you can’t get it wrong
@@daerdevvyl4314 The guy in the video pronounces Logician as Lagician which sounds like Magician at first.
Solving this after 2 minutes: Assuming that prisoner 3's hat is blue, if prisoner 2's hat is also blue then prisoner 1 would know that their hat was red since there're only 2 blue hats, but they wouldn't, so it gives the only possibly that prisoner 2's hat must be red. And the prisoner 2 would know that and would guess the hat color correctly, but since prisoner 2 also don't know their hat color, the assumption was wrong and therefore prisoner 3's hat must be red.
Exactly how I solved it as well
Plot twist: all three were arrested at the gate and found guilty of stealing the hats.
My father told us this riddle in a different way 50 years or more ago. My one brother solved it when he was 12.
Solving this without ever stopping the video except right now to write the comment before continuing to the solution: the only way 1 would know his color is if 2 and 3 were both blue, so at least one of them must be red, since 2 doesn't know either, it means that 3 is not blue, otherwise 2 would have known he is the red one
I love how you explain everything so detailed, that even people of English non native language, or people who don't know math that much still understand. 🤝👏
Let's not argue about what kind of justice system gives the warden the right to decide who goes free and who stays in prison , explain to me though how all 3 have proven that they are logicians by just one of them drawing the right conclusion.
It's a plea bargain.
Wonderful video! Keep up the amazing work!!
For those proposing a simpler/shorter solution. I would argue that this one is better because it shows clearly how to generalize the problem.
For those arguing for removing the sequential questioning, I would argue that the reliance on the prisoners estimating how long their peers would take to work it out, whilst it should work, makes for an unsatisfactory problem due to the indefinite time period. If I were the warden I would prefer the stated method as it would be more dramatic 🙂
You don't actually need to ask the prisoners the questions. You just tell them to shout out as soon as they know the colour of their own hat for certain. After a few seconds, prisoner #3, having heard nothing from behind him, should be able to make the inference and shout out "red".
*yawn*.
That is effectively the same as asking all the prisoners the question. Not speaking up = "I don't know".
Time isn't discrete. You should give them the chance to answer every minute (for example) for your version to work
Yes you watched the older “logician hat video” he did where no one was able to speak, grats.
Prisoner 3 is wrong.
1 & 2 tell him, “Haven’t you heard of dramatic pause?!!!”
The warden is not entitled to release the prisoners. Regardless of how good logicians they are, only a judge can repeal their sentences, or a parole board.
The crooked warden should be in jail with or without all the hats.
Clearly the warden is a logician himself. He's corruptly favoring members of his own group.
If nr. 2 and 3 would both wear a blue hat, nr. 1 would know his own hat color. But nr. 1 does not know.
So if 3 has a blue hat, nr 2 would know for sure he has a red hat. But nr. 2 does not know his hat color either.
It follows that nr. 3 has a red hat.
EXCELLENT teaching. I think my mom could understand this explanation. Maybe. I've showed her the prisoners and hats riddle like three times. And each time she get a little more mad at me. I'm not going there again.
extensive explanation for a simple Problem, but better too much than not enough.
I do enjoy his videos, but his explanations always seem to be long and convoluted, rather than straight to the point.
Finally that nerve wrecking heartbeat is gone. Thank you ♡~
Finally a puzzle I could solve in my head under a minute
You missed one important detail in the setup. Now, most of us just assumed that the perfect logicians (or rational people) knew that the others were also rational. But this needs to be stated/assumed in order to make the puzzle work.
basically, everyone should be super rational. Rational is when you are perfectly logical. Super rational is when you KNOW everyone else is ALSO rational
Since only one of them needs to correctly identify their hat colour, it seems like an alternative strategy would be to just all answer 'red'. One of them has to be right. And that's probably the strategy that illogical criminals would use as well, since there are more 'red' hats, so the test is not very well-designed by the warden.
Yes i also think that....
That could easily be solved with a rule saying "if any one of you gives a wrong answer, the test is over". In that case, you can either say "I don't know" or give the correct answer.
There's a way better variant of this puzzle. There are 100 prisoners and the hats are assigned to them randomly, and the number of both red and blue hats is random as well. Then they are being asked about the color of their own hat one by one, starting from the prisoner in the back. Only if no more than 1 prisoner guesses wrong, they are let free, otherwise they all loose. However, they are allowed to strategize beforehand. It is solvable, albeit it took me some time to come to a solution first time I heard it :)
Świetna zagadka! 🧠 Film doskonale pokazuje, jak logika i dedukcja mogą prowadzić do rozwiązania nawet najbardziej skomplikowanych problemów. Dziękuję za ten wspaniały materiał, który pobudza do myślenia. Czekam na więcej takich wyzwań! 🌟
Here’s a twist - any one of them should be able to correctly identify the colour of the front prisoner’s hat with just the rules of the game.
The logic goes like this - the warden wants to confirm that all three prisoners are logicians, yet only one of them needs to be able to answer. So, in order to test all 3, two of them must be unable to answer with certainty, but the othe two’s logical uncertainty must be enough of a hint for the third to solve the problem using logic. Since everyone has the same information as the person in front, plus that person’s hat colour, the uncertainty of the people in front cannot give extra information, so the uncertainty must pass information forward, therefore, the person at the front must be able to know after the other two have expressed their uncertainty, which brings each of the prisoner up to date with all the information that we had to work with for our solution, without ever seeing a single hat (assuming they’re aware of the number of each colour of hat).
Nice puzzle, got it but took me a small bit to think about it :)
3rd prisoner hat: red
1st prisoner can only know if the two hats in front of him are blue, then his must be red.
So if he doesn't know, then either :
- one hat is blue and one is red
Or
- both hats are red
(for prisoner 2 & 3)
2nd prisoner deducts the above and realizes the only two possibilities above.
He can only 'know' if the hat in front of him is blue, then his has to be red. He won't know, if the hat in front of him is red, then his own can be red too.
Last prisoner deducts this, and realizes his own hat is red.
I never saw this one tbf... But I think it's pretty easy.
1) If he doesn't know it's because 2 and 3 have different colours one is red and the other one is blue, because if there was 2 blues he'll know that his hat is red. So the only options is that 2&3 have R&B B&R or R&R
2) If 2 doesn't know, but he knows the 1) steps he knows that if 3) has a blue, he HAD to have red, so, if he doesn't know number 3 has a red hat.
3) Knows that his hat is red.
There's an easier way to think about this that doesn't require explicitly considering 2^3 possibilities:
- The only way prisoner 1 would know his hat color is if he were seeing two blue hats, because there are only two blue hats and then he would conclude his is red. He doesn't, so he must see at least one red hat.
- Prisoner 2 now knows there is *at least* one red hat between him and prisoner 3, so if prisoner 3 had a blue hat then he would conclude that he must have a red hat. He doesn't, so prisoner 3 can't have a blue hat.
- Prisoner 3 now knows he has a red hat.
I love hat puzzles.
One of the easiest puzzles of this kind.
1.) If 1 sees 2x blue, 1 says "red"
2.) If 1 says nothing and 2 sees 1x blue, 2 says "red"
3.) If 1 and 2 remain silent, 3 says "red".
Reasoning:
1) There are only 2 blue hats
2) Apparently 2 and 3 don't both wear blue.
3) Apparently 3 doesn't wear blue.
Here's an old problem you might find interesting, it requires calculus though "An tiny asteroid is falling from very far away straight towards the sun. It just so happens to be falling at a certain angle such that as it gets closer to the sun, the Earth gets in the way, and the asteroid falls straight down a well exactly on the equator of the Earth. What time must it have been?"
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spoiler 2:30 am is the answer, the trick is realizing Earth's sideways motion around the sun is actually much faster than it's daily rotation rate, so you can simplify to a triangle of vectors, one vector is the sideways orbital motion of the earth, the other vector is the asteroids motion, and the vector connecting them has the same angle that the well would have to be aligned off from midnight pointing exactly opposite the sun, then you convert the angle to a time and it all works out, pretty fun problem.
That reminds me of a much easier one I once came up with. Rounding off numbers, if the day is 24 hours long and the time between full Moons is 28 days, how much faster or slower does the Moon move across the sky than the Sun?
I think your explanation is making this much more complicated.
First of all, assuming that prisoner 3 is wearing a blue hat, if 2 also wears blue, 1 would have known that he is wearing red hat and immediately reply positive, but he didn't. This means that either 2 and 3 are both wearing red hats or one blue, one red hat. If 3 was wearing a blue hat, 2 would have known that he was wearing a red hat and immediately reply positive, but he didn't. That means that he is unsure about his own hat colour, which is possible only if 3 is wearing a red hat as there were three of them initially (the same reason 1 would not have been able to answer his hat colour either). That's when 3 knew he is wearing a red hat
Excellent explanation
You could have two Red Hats and one Ubuntu.
keep up the good work
There's no such thing as a wrongly imprisoned logician. They're acutely dangerous.
Logicians won WW2 by cracking Enigma, but okay.
@@EaglePickingdidn’t logicians also create enigma on the other side?😂
@@dopplr.1204 Good point. I guess logicians are indeed evil.
Pausing and posting my reasoning:
Prisoner 1’s statement can be rephrased as “The other 2 are not both wearing a blue hat”. (If they were, prisoner 1 would know he was wearing a red hat.)
Prisoner 2’s statement then says that prisoner 3 is not wearing a blue hat. (He knows they are not both wearing one. If he saw a blue hat, he would know he was wearing a red one.)
Prisoner 3 therefore knows he must be wearing a red hat.
I think this also shows that if the hats are randomly selected, it’s always possible for the prisoners to win.
(Either P1 sees 2 blue hats, and knows he is wearing a red one, or Prisoner 2 sees a blue hat, and knows he’s wearing a red one because P1 didn’t see 2 blue hats, or prisoner 3 knows that P2 didn’t see a blue hat, because of the previous car)
I got this one thanks to having done the one with four logicians with one behind a wall.
Finally I solve one of these puzzles without overthinking...
In the first 30 minutes: keep up the good work!
In the first day: this is another way I solved the puzzle!
After a month: If a prison warden is capable of creating such a good puzzle, he shouldn't be a prison warden
After some time: *hate comments*
Quite simple this one! Pleased with myself for getting it under a minute
I would say first that the prisoners trust that each of them truly are masters of logic the warden expects them to be.
So we can say that each prisoner has perfect logical ability.
I also expect the prisoners to hear each others answers to the questions.
If both of the prisoners 2 and 3 would have blue hats the prisoner 1 would know his hat is red.
Since Prisoner 1 doesn't know, the two prisoners in front would have either 1 red and 1 blue hat or 2 red hats.
Prisoner 2 heard that prisoner 1 doesn't know so he knows that there are two options.
If the prisoner 3 in front would have a blue hat prisoner 2 would know his hat is red, but prisoner 2 doesn't know.
This gives all the necessary clues to prisoner 3 to know that his hat must be color RED.
Because either way the prisoner 2 must have seen a red hat to not know his own hat color.
Logicians: “hey we’re innocent”
Warden: “I think you’re innocent too”
Logicians: “let us go then”
Warden “so I made this game where you have to be able to trust two random strangers judgement, or you’ll all rot in prison. Fun right?”
Logicians: “…”
Great puzzle and good solution🤩
I feel like I have seen this question on this channel before. The answer is red. If 2 and 3 are blue, then 1 will know that his is red, but 1 didn't know, so either 1 saw red-blue or red-red. If 2 saw blue, then 2 would have known that his is red, but he didn't, meaning 3 is red.
After seeing the video, I include myself in the group of other people who said that they solved the problem another way. And this made me wonder which is the better or the easier way to solve this - by creating a table of all variants in your mind and querying it, or by constructing only the possible scenarios for what you see. Probably the former one the better way to present and explain the solution, and the latter one is the way most people would actually use to find the solution.
The simplest way to think about it, in my opinion, is the following. If the first does not know, the next two know they can't both have a blue hat. The second would know they have a red hat if the third has a blue hat. However since they do not know, the third must have a red hat.
Why did you draw a wall? The puzzle must be incomplete
From answer 1, 2 + 3 can’t be B,B, because in that case #1 would know his or her hat is R. #2 now knows both 2 and 3 can’t be B. That leaves
2 and 3 as R,R, B,R, or R,B respectively. If #3 was B, # 2 could tell his hat must be R. But, #2 can’t because #3 is R, leaving #2 as R or B. After #2’s answer, #3 knows that R,B, is eliminated as a choice. With either remaining option, R,R, or B,R, #3 knows his or her hat must be R.
Make it 5 logicians in an L shaped corridor and all can determine their hat colour. A more satisfying puzzle IMHO
1
v
2
v
3 > 4 > 5
5 & 3 work as before and 1, 2 & 4 can be derived
In fact, prisoner 3 will always know his hat.
Depending on the hat distribution the response to 'do you know your hat' may be : Yes, yes, yes / No, yes, yes / No, no, yes
maybe it's silly and basic, but figuring this out on my own with pen and paper was one of the more rewarding things I've done recently - straight up. that was so much fun. I STRONGLY STRONGLY URGE ANYONE WHO READS THIS TO TRY AND FIGURE IT THEMSELVES! if you just don't think you can, I'm willing to bet all I have that you're smarter than you think. try!!! be well everyone
The deduction isn't that complex (having to go through all possibilities). The *only* reason #1 didn't know his hat's color was because he saw *at least one* red on #2 and #3 (if it were both blue he'd have known his hat's color). Given that, the only reason #2 would not then know his hat's color is if #3 is not waring a blue hat. So, #3 has to be waring red hat at that point.
Luckily, the warden is not a perfect logician, because otherwise he would have realised that only the 3rd prisoner demonstrated himself to be a perfect logician. The other two could have been completely clueless.
The problem is color symmetrical except for the blue hat shortage. If 3 were wearing a blue hat, that would provide more information to 1 and 2. Since neither of them knew, 3 must have a red hat.
if prisoner 2 and 3 had blue hats then prisoner 1 would have answered "yes, I know which color my hat is" because it could only be red.
therefore since prisoner 1 did not say that, prisoner 2 and 3 collectively has at least 1 red hat.
if prisoner 3 had a blue hat then prisoner 2 would know that they have a red hat, but they don't know this, so prisoner 3 has a red hat.
Solved quickly. Realized number 1 would only know if both hats are blue, so 2 knows that either he or the guy in front of him can’t have blue. If 3 had blue, 2 would know he couldn’t also have blue, so 3 infers he has red.
For those who don't want to read the long comments:
There is *at least 1* red hat in P1, P2 (2 blue hats -> P1 is red)
< "P2, do you know your hat color? I don't know"
> If P3 was blue, P2 would know he was red.
> Therefore, P3 is red.
When the warden asks 1, they don't figure it out because there can't be 2 blue hats in front of them.
Then 2 realizes they and 3 can't both have blue hats. If 3's hat was blue, 2 would have known that their own is red.
So then 3 realizes their hat is red.
I didn't bother listing the possibilities & eliminating them, I just solved it like this: If 2 & 3 had blue, 1 would know that he's red, so there must be at least 1 red hat on 2 or 3. 2 knows this & so if 3 was blue, 2 would know he's red. This means 3 must be red.
This is unnecessarily harder than it's alternative, the 2 black, 2 white hat puzzle
الحاجات اللي بحب اسهر عليها ❤
*1 say “idk” mean 2 and 3 cant be both blue => only red/blue or red/red available
*2 say “idk” mean 3 cant not be blue cause if 3 was blue 2 can guess that his is red
*3 now combie those both answer and only left is he is red and 2 is red too
Took me some time, but I came to the same conclusion. Namely that my hat must be red, if they both said that they do not know. I did it mentally, no writing or drawing. It took me some time. More than I expected.
Damn, it's always the puzzle maniac death row prisoners solving the puzzle posed by the apuzzle maniac warden
My logic for this one is as follows (spoilers below for those who don't want a solution):
First, the only way for the 1st logician to know his hat color is to see two blue hats, so we can conclude that he sees at least one red hat.
Second, the 2nd logician then knows that the 1st sees at least one red hat, and if he saw a blue hat, he would know that his hat is red, since he stays silent, we know the 2nd sees one red hat.
From there, the 3rd logician knows that the 2nd sees one red hat, and correctly states his hat color.
The 3-line solution...
If Prisoner 1 saw two blue hats, he would know he has a red hat (as there are only two blue hats). So prisoners 2 and 3 do not both have blue hats.
If prisoner 2 saw a blue hat, he would know he has a red hat. Prisoner 2 therefore saw a red hat, but his own hat could be red or blue.
Prisoner 3 knows he has a red hat.
No. The warden just playing with them.
After this game they have put back to their torture chambers to teach the good manners.
heyy i think im actually getting better at these logic puzzles! i was able to get the answer for this easily and quickly (and ik it’s not a big deal since this was super easy, but i’m still proud since i used to suckkkk at these)
Vey simple. If the 1st prisoner sees 2 blues then he knows his must be red. So, he sees at least one red. Then, if the 2nd prisoner sees a blue then he knows his is red. Therefore, the 3rd prisoner must have a red hat.
prisoner 1 did not see 2 blue hats ahead of him, otherwise his hat color would be red. So he either saw two red or one red and one blue hat ahead of him and was unsure
prisoner 2 did not see a blue hat on prisoner 3 because he knows they both cannot have blue hats. If he saw a blue hat his hat had to be red. He saw a red hat on prisoner 3 and was not sure still but prisoner 3 knew this
Though my favorite puzzle about prisonera is about 100 green eyed prisoners. It breaks my mind every time!
I can't save them sorry.
*Later that day in a spaceship after peeling of the human face mask*
Mission accomplished commander!
#1 would only know his hat color if both #2 and #3 were wearing blue.
Since #1 doesn't know his hat color it means #2 and #3 can only be wearing both red or one red and one blue.
#2 would only know his hat color if #3 was wearing blue.
Since #2 doesn't know his hat color it means #3 must be wearing red.
there is a further answer: The warden says "I lied! Solitary confinement for all of you!"
What would happen if we start by three red hat and the next part u consider pri-3 has red on his head?? Will prisoner 3 get red hat??
Secret ending: Prisoner 1 says that he doesn't know, and then shouts the hat color of the other 2 prisoners, freeing them all.
I like how the wall is in the puzzle just for aesthetic purposes.
I suspect that's connected to the assorted variations of the puzzle that involve another prisoner standing on the other side of the wall
@@sirpetethegreat Indeed.
Magic of permutation &Combination .😢
before watching solution:
the first would only know he was wearing a red hat if the other 2 were wearing blue hats. Therefore, 2 and 3 could not both be wearing blue hats becuse if they were the 1st would know.
the second knows the first knows they are not both wearing blue hats, so if 3 is wearing a blue hat 2 would know if he were wearing a blue hat because 1 would have known that he was wearing a red hat. Therefore 3 must be wearing a red hat.
Why does it always seem to be prisoners and wardens? I long for a puzzle with some other kind of framing device
This is my solution before watching the video's solution:
If 1 sees 2 blue hats, he'd know he's red. That leaves the assumption tht he sees 1 blue hat or 2 red hats.
With that, if 2 sees a blue hat, then thay meana he's wearing the red one, but since he sees a red hat, he doesn't kmow whether he's wearing a blue or a red pme himself.
That leads 3 to conclude that he's wearing the red one.
If 2 and 3 are both wearing blue then 1 knows for sure he’s wearing red which means either 1 or 2 is wearing red (or both are). If 3 is wearing blue then 2 knows that he’s wearing red so therefore 3 knows he’s wearing red for sure.
Anyone else confused when the solution gave the possibility of 3 blue hats when there were only 2?
Wouldn't it be easier if each prisoner just said their hat was red? With three red and two blue hats to share between three prisoners at least one of them is bound to be correct.
What happened to the outro music?
You'd have to be a true logician not to see what's right in front of you: namely, the brim of the hat you're wearing.
Why do you reupload your puzzles now?
Red
N1 doesn't know because the 2 others don't have blue and blue.
N2 doesn't know because 3 is red. Otherwise if 3 is blue then 2 would be red (because of N1 saying both aren't blue)
Solved it in 5 seconds from just the thumbnail
Then how did you know the question?
I solved it last week.
I solved it before the question was even written
@GeezSus How? How do you know the question then?
@@maxhagenauer24 It's obvious
Idk how i did this , i think because of watching this channel's every video?
Answer is red because, if 1 ould see 2and 3 wearing blue then he will know his hat, but he does not answer , so 2 and 3 are wearing either red and red or blue and red or red and blue
Next so from this information, if 3 is wearing blue then he will answer red, but he doesn't answer , so 3 is wearing red which makes 2 to confuse
Finally from these informations we come to know that 3 is wearing red hat
They only proved that number 3 was a logician, not all 3.
And that is the story how the three evil logicians were set free to torment the entire world.😁
When do we pause the video..?
This should be in thd next Saw movie
Bro the first one doesn't need to look in 7 cases to see what he has if he sees 2 blue hats. He knows instantly he has red because there're only 2 blue hats lol
I got it right!