If Logician 1 sees two hats the same colour then he knows his hat is the opposite colour and speaks out. After a while of not speaking out Logician 2 realises that 1 must be looking at two different colours and therefore his hat is the opposite colour to the one in front of him and he speaks out. Basically it's the same, but without the need for the logic table.
Absolutely; this was very easy to figure out without complicating the problem with all the possibilities. Once #1 doesn’t immediately answer, #2 knows his hat is not the same as #3.
number 2 never got the message. He's trained in logic alone, and ends up never knowing if number 1 doens't answer because he didn't figure it out, doesn't want to, or any other reason
Three logicians walk into a bar. The bartender asks them "Would you three like something to drink?" The first logician says "All three of us? I don't know." The second logician says "I don't know either." The third says "Yes, we would."
@@Blade.5786nah, your Version is worded badly. There is no way for them to logically deduce whether he *can* get them a beer. Might be considered nitpicking but seems like a relevant distinction in a logic joke which is mainly based around wording. This would be a better way to phrase it: Do you all want a beer? Idk Idk Yes
@@SH0907 Actually there is, the logical deduction would be that he can (considering the fact that he's the bartender). Using logic even further, this is information the bartender himself is aware of, and therefore wouldn't be asking if he wasn't implying something else (whether they *wanted* a beer). Logic is ingrained in our everyday speech, so it's not exactly wrong to use it in a logic joke.
@@antijokecommenter5003 1's answer indicates 1 wants a beer but does not know about 2 and 3. 2 and 3 then know 1 wants a beer. 2's answer indicates 2 wants a beer but still does not know about 3. 3 then knows 1 and 2 want a beer. Since 3 wants a beer, 3 now knows all want a beer.
You don't really need to enumerate all 6 possibilities in this case. If the 2nd and 3rd person wears the same color, then the 1st person can immediately answer. If they wear different color, then the 1st will stay silent. And that silence becomes information to the 2nd person, who knows the color of the 3rd and, hence, also knows his. I think this concept is called common knowledge
That would've been a far more interesting puzzle. In the case of 2&3 having same colour A, 1 will shout colour B. 2 can then conclude he is wearing colour A (because 1 sees two same hats). 3 can conclude he is wearing colour A on same premise. 4 knows he's wearing colour B by process of elimination. In the case of 2&3 having different colours (A&B, respectively), 1 will stay quiet and 2 will conclude he is wearing A. 1 still doesn't know if he's A or B, and stays quiet. 3 can now conclude this is a "different colour scenario", and knows he is B (opposite of 2). I see no way for 1&4 to figure out what they're wearing from here.
Once 2 has spoken, 3 knows their hat is red (if it had been the same as 2, 1 would have spoken first; so, knowing now that 2 is blue, 3 knows theirs must be the opposite). But here's where it stalls - no-one can see either 1's or 4's hat. They all know they're opposites, but not which way round they are. It could be resolved if 3 were allowed to turn round when they'd got their answer, but that's kinda kludgy. It would be nice if we could find a way for 4 to be the one that resolves it - say, if they can see 1's and only 1's hat. Anyway. Three logicians walk into a bar, and the server says "three beers, is it?" The first logician says "I don't know", and the second logician says "I don't know", and the third logician says "yes, please".
It could be expressed more simply. If #1 sees two of the same color hat, he says he has the other one. If #1 says nothing, then #2 knows he and the guy in front of him have different color hats, so he says the opposite of whatever color he sees on the guy in front of him.
From the problem statement, it is not clear that the first person can see the next two hats. It is really important to know this information for problem solving
@@pauldzim actually, the chance is 2/3. because the second guy knows the guy in front has red, so there are 2 blue and 1 red left. And if you compound the fact that the chance of the 4 prisoners being logicians is > 0, it means that the chance that the prisoners would go free is actually more than 2/3
@@mynthics no, I am not talking about the chance for the left guy, I am talking about the overall chance for everyone to guess correctly. Nr 2 and 3 are a given, so 2 people will already have passed. The 50/50 is for the remaining 2
2:20 you're overthinking it. You don't need combinatorics or tables. If 1 sees two red or two blue he'll know his own color with a couple seconds' reflection. Once 2 has given him that time, and not heard an answer, 2 knows he differs from 3. MORE GENERALLY, 2 knows that his hat plus everything he sees before him won't yield an answer. As long as the puzzle would have let 1 solve the problem had 2 one color, 2 knows he has the other color. For instance: 10,000 people are on the hill, with the hats split 50/50. 1 sees 4999/4999; if he saw 5000 of either he'd know his own hat. 2 sees 4998/4999 and knows his own hat is the 4998 color.
That was... an unnecessarily long explanation. As other people pointed out, if they know there are 2 red hats and 2 blue hats (which is the case according to the sources in description), logician 1 will have to remain silent, as seeing different hat colors on 2 and 3 means he could have either color. Then logician 2 notices 1's silence, and thinks, "If 3 and I had the same color, 1 would immediately know his own hat color. As this was not the case, 3 and I have different hat colors, meaning that since 3 has a red hat, my own hat is blue".
For bonus points, by the same logic 3 immediately knows his own hat color as soon as 2 calls out. I don't think there's any way for 1 or 4 to know their own color, though.
I think firmly stating that they can't communicate in any way was a little confusing here. If we had clarified that there are clock ticks, and on each tick a logician can answer (and all other logicians take note of when no-one else answers), then it would've made it more clear that the passage of time could be used to communicate whether a logician does/doesn't know the answer. It's a pattern used in other hat style logic puzzles so it's not too far out of reach to imagine, but it's worth clarifying. e.g.: the blue-eyed people on an island logic puzzle uses the concept of "days" to let the puzzler understand that the passage of time is divided into distinct rounds, which can tell agents at which point other agents were uncertain of the answer.
Exactly. I don't need the fluff story. Just clearly state the rules. I had to look it up somewhere else to understand that they get turns to answer or decline and it immediately became clear that the first logician not answering is the information the second one needs to know his hat is different color than the third.
The clock ticks thing would be more relevant if there were several logical steps involved instead of just one. The first logician, given no clock ticks, would answer instantly if he saw two same coloured hats, as the first colour is trivial in that case. Since they don't answer instantly, the second logician needs to only give it like ten seconds to know their own colour with certainty.
Let's assume that "they are not allowed to talk to each other" means that they can't communicate to each other at all. Then the only way that #2 would know that #1 was uncertain was via the fact that the warden has not already set everyone free. The problem is therefore better posed if there are time constraints.
@@ashleybroughton7713 Even if they couldn't hear each other answer, Logician 2 would still realize they haven't been set free after a few seconds and then figure out what's going on. Even if that weren't the case, there's a second solution here. The logicians were given the information that the Warden specifically set up this scenario to test if they were logicians and not criminals. For this scenario to even be a logic puzzle, logicians 2 and 3 have to have different colored hats since any criminal in the 1 spot could figure out his hat color if he sees two of the same colored hats in front of him. With that information alone, Logician 2 should be able to instantly say he has a blue hat, even if he never gets any information from Logician 1.
"Let's assume that "they are not allowed to talk to each other" means that they can't communicate to each other at all." - My point exactly. Many of these logical problems seem to rely on ignoring the initial conditions, not actually working out the logic.
I misheard the instructions at first and try figuring out how they could all state the color of the hat for certain. I’m glad I listened to the instructions again.
Why bother with the 6 possibilities? Irrelevant. Solution starts at 3:28 and ends very quick, L1 sees two colors so he won't speak, which meants L2 is blue.
Your solution works perfectly, but it is unnecessarily complicated. The beginning with excluding 3 and 4 is fine. But then 1 and 2 can forget about any combinations. If 1 saw two blue hats, he'd know his hat is red, and vice versa if he saw two red hats, he'd know his hat is blue. Since he's not saying anything, 2 realizes 1 must see both colours on his and 3's head and concludes his hat is red if 3's is blue and blue if 3's is red.
3 does not even have to wait. He knows that he has different color than 2, for same reason 2 knows. So when 2 says his hat color, 3 immediately knows his color. Then 1 and 4 have no way to determine their color, as no one sees them. Which is a bit disappointing.
Even better: If the distribution of hats is arbitrary, logicians 2 and 3 are nevertheless always sure to find out their hat colours. The other two only find out if they wear the same colour (and 2 and 3 wear the other colour). Plus, logician 3 doesn't even have to know the order of people behind him. He only has to wait for someone to call out a colour and will then be sure he wears the opposite colour.
@@yafriendceko That's the neat thing here, 3 doesn't need to know. All they know is that the first one to call out a colour (provided they only do so once they're 100% certain, which is generally assumed anyway) cannot wear the same colour as 3 does.
You should include a timeframe for when they answer to make the puzzle more clear something like every 5 minutes the warden will ask for an answer so that the time between isn’t arbitrary
5 seconds would be enough for person 1 to realize that person 2 and 3 have different colors. After those 5 seconds, person 2 can say that his color is the opposite of the one in front of him.
@@Bob94390 true it may be enough but if you’re a perfect logician and you aren’t given a timeframe for answering and you know you have to wait for the second answer slot how do you know how long it will take for the first answer slot to go unanswered. Maybe the person in spot 1 is also a perfect logician but they take a little more time than you think they would and end up with a wrong answer. The time slots make the puzzle cleaner for us and the fake computer people the puzzle is interested in.
Any logicians should be able to figure out “I see both red hats so I must be wearing blue” almost immediately. Unless Person 1 is daydreaming or something, it shouldn’t take more than a few seconds for Person 2 to be certain that Person 1 doesn’t have enough information to know. If anything, you just need to add the assumption that everyone’s actually paying attention.
No time-frame needed. No. 1 would know the answer instantly (sees two hats of same colour) or never. By your reasoning, we have to also account for the fact that no.2 could be an a**hole who wants to be in prison and never speaks up (although he knows the answer).
@@DreadX10 in this configuration yes but what if it’s a different configuration and no1 takes an extra second before answering. No2 has suddenly trapped them all in jail. The whole point of these puzzles is that they’re guaranteed with perfect logic but not giving them a set time to answer can lead to mistakes. If no2 solved it slightly faster than no1 and then answered and the configuration was different that would be it. With an answering cadence problem solved.
I got asked a version of this in an interview about 10 years ago. I remember when the question was posed to me I thought there couldn't possibly be an answer. But the interviewer asked me to try working it anyway. I spent an agonizing amount of time on it, building a logic table. Eventually I figured it out and was very happy! I didn't get the job, but I was still pleased with myself for answering this puzzle correctly in a nerve wracking situation.
That question got asked in Honkai Star Rail too xD In short: After some waiting, 1 stays silent, meaning 1 and 2 aren't the same color (he'd know his color otherwise) 2 understands that now and knows his color must be blue, since 3 has red. So 2 will be able to say his hat color with certainty
An interesting variation of a this kind of problem is this: 3 logicians wanted to figure out which of them is smartest, so they went to the king. The king devised a test in which they were to sit in a circle and close their eyes. They were told that a hat, either red or blue, will be placed on each of them. They were also told that there are 2 blue and 3 red hats in total. When they open their eyes, they are supposed to figure out which hat colour they are wearing. Each will only see the hats of the two logicians in front of them. The first to shout out their hat colour is the smartest. While their eyes were closed, the king placed a red hat on each of them. When they opened their eyes, after a short time, one of them shouted that they were wearing a red hat. He was declared the smartest. The question is, how did he think?
Not seen that variation, but it's cute. 1. If any of them were looking at two blue hats, they would instantly know they were wearing a red hat, but no-one answers so it's not two blue hats. 2. Having figured this out, if someone saw one blue hat, they would instantly know their hat was red, but no-one answers meaning there isn't one blue hat. 3. The only remaining possibility is all 3 hats are red, and so it's just down to who is fastest to answer red.
@@manudude02 The logic is nice, but you start to get a difficulty with how to define 'instantly + instantly'. How long should one of them wait to determine that they are now in the second consecutive silence and not the first one? Because of this difficulty they cannot simply rush to be the fastest to say 'red' the second they reason by this method that it could logically be the answer, just in case one or both the others are still in the first of the two consecutive silences.
The king asks the first time and silence, so it excludes 2 blue hats. The king asks second time and silence again, so it excludes one blue hat. When the king asks for the third time, they all know they're wearing red hats, because that's the only remaining possibility, and the fastest to speak wins.
The test needs to be fair, so all of them must be wearing the red hat. Therefore the first to realise that answered the question. I know this doesn't employ logic but common sense but I feel the given task isn't suited for it
This logic was quite simple. I didn't use your method exactly but somewhat similar to your answer. As logician 1 is confused, means 2 will understand that the hat of 2 and 3 are of different color. So he sees 3 has red color and concludes 2 himself is blue.
Oh Presh, you do love an overcomplicated solution. This takes exactly two very easy-to-follow steps-no formulae, no tables. • If #1 saw the same color on #2 and #3, he would shout that he had the other color. • He doesn't shout, so #2 knows he is wearing a different color than #3 and shouts out his hat color. DONE. Why would you start with the two who have least information?!
What an overcomplicated explanation for such a simple solution! I'm neither a logician nor a mathematician, yet I figured it out in less time than it took to explain it.
Simple explanation (spoilers!): If L1 saw two hats of the same color, he'd know that his hat was the other color. Since L1 does not know his hat color, L2 knows that his own hat must be the color that he does not see on L3.
Yes. The problem is whem the two hats in front of him are different. Therefore. When number 1 doesnt know the colour. Number 2 will now know that the colurs of 2 amd 3 are different. Cos as you said if they are the same then 1 would just give the amswer. But they are different and somce number 2 can see that number 3s hat is red. He now knows taht his being different means his hat is blue. Thats the point.
This video made the question more complex Simple explanation:- If Logician 2 and 3 have the same colour hats, Logician 1 knows that his hat is the opposite colour as he knows there are only 2 of them. If Logician 1 doesn't know the answer, Logician 2 knows that him and Logician 3 have a different coloured hat. So, Logician 2 says the opposite of Logician 3's hat colour. In all the 6 possible arrangements, either No.1 or No.2 can say the right answer
Solved it in under a minute, but, then again, I had "help"; I already knew a similar problem (the Three Chinese Philosophers riddle, in which a queen paints either a red or a green dot on each of three philosophers' foreheads and challenges them to find out, without conferring, whether their own dot is red or green). To be perfectly honest, I thought Presh's explanation was a bit over-complicated; he could have been a bit more succinct and to the point.
I seem to remember a harder version of this puzzle where all 4 men had to guess the color of their hats. As soon as 2 guesses, number 3 immediately knows the color of his hat as well. I don't remember how 1 and 4 got their color but I think it had to do with the wording of the puzzle.
@@drewsworthh ah ok. I remember a prisoner scenario where everyone has to guess their color in order to be freed. But it had some twist iirc like the guy behind the wall was actually on the very left behind another wall and could only glimpse the tip of the last person's hat.
The difficult version is 100 people (could have been any number) lining up. There are 2 colours but the proportion of hats is not given. Everyone takes turn guessing starting from the back. They're allowed time to strategize. They can only make a guess but not signalling using other things like volume. The optimal strategy ensures everyone except the one at the back can deduce the colour correctly.
Must admit, I didn't even think about the probability. Just that 2 would know that 1's inability to determine what he's wearing must mean that he's seeing a blue hat and a red hat and the fact that 2 can see a red hat means he must be wearing a blue hat. Thanks for posting as I can now explain to my wife that I am fully entitled to a celebratory glass of wine.
I've seen one similar to this on TED-Ed, though that was with more people, and a random distribution of hats. In this case, if the top person sees two red hats, he knows he has blue, and thus they win, same thing if he sees two blue hats, he knows he has red. However, if he sees a red and a blue, he has a 50/50 chance, and thus hesitates to answer. This then tells the next person in line that he sees a hat of each color, since he doesn't automatically know his own hat color. So he looks at the hat in front of him, and instantly knows his own hat color, being the opposite of the person in front of him, thus they win. The third and fourth people don't even need to answer.
This is a simple problem of pure logic, and by that I mean SIMPLE, and LOGIC. There's no need to introduce any mathematics whatsoever. Sometimes you really can reason without formulas.
Number 1 remains silent since he sees 1 blue and 1 red hat, leaving two options open. Number 2 (logician as he is) realizes that when number 1 remains silent, number 1 must face 1 blue and 1 red hat. Since number 2 can see number 3 is wearing a red hat, he knows for sure he is wearing blue himself.
You made that explanation so much more complicated than it needed to be! All you needed to say was that there were only two hats of each color, and thus if logician one saw two hats of the same color he would know that he was looking at all the hats of that color and thus his hat must be the other color, and logician 2knows this, so if he doesn’t hear logician one state the color of his hat, then he and logician three must be wearing two different color hats and thus his hat is the color he *doesnt* see in font of him. (And fwiw, once logician three hears logician two shout out that he’s wearing a blue hat, he would know that his own hat must be red, not that he would have any reason to care at that point.)
You don't really need the first step (2 eliminating half the possibilities). 1's silence immediately means that 2 and 3 have different colours, so upon noting 1's silence 2 can announce his colour as the opposite of the one he sees on 3. The question, though, is how long should 2 wait to be sure that 1 doesn't see two hats the same colour rather than, say, is still fumbling with his glasses?
If hats on 2 and 3 are the same color 2 can expect 1 to speak, but it 1 does not speak for a time, 2 can assume that his hat is of the opposite color to 1 whose hat he can see and can speak.
I love "a group of logicians" problems. I thought your solution was overly complex though -- you simply have to realize that when 1 doesn't speak, 2 realizes that means that he and 3 have different colored hats, and since 3 has a red hat, his must be blue. (If 2 and 3 had the same color, 1 would know their hat color immediately).
Pretty simple actually. The first man would speak if he saw two identical hats in front of him. Since he doesn't, it should become clear to the others after a few minutes that positions 2 and 3 contain one red and one blue hat. The second man would then know this, see a red hat in front of him, and state that his hat is blue.
At timestamp 0:46 the warden says they are not allowed to talk to each other. So right at 3:17 when #1 talks, the warden should have said in the voice of Bill Paxton from Aliens, "Game over, man", and they should have been marched back into jail.
Exactly. Every logician should only find out if another logician guessed right after taking a guess himself/herself. Or more generally, no communication, since silence is communication as well, it communicates that 2 and 3 have different hats. That'd make the problem unsolvable though, so the flaw had to be intentional.
The riddle can be solved by just Logician 1, 2 and 3 interacting with each other. Logician 4 is pretty much just a distraction for the solver. It would be the same riddle with 3 guys, 3 hats, where 2 is the same color and one is different.
Interesting, heard this logic problem on a car talk radio program many years ago. Like a good humorous story nice to see this make the rounds again. Thanks for sharing.
If 2 and 3 had the same color, 1 would announce they’re wearing the opposite color. 2 could hear that announcement but it isn’t made, despite allowing one whole second for a perfect logician to make it. 2 then knows that they’re wearing the opposite of 3, i.e. blue. 4 is a little annoyed to be a background prop just to wear an unseen hat, but is glad to go free.
I didn't iterate the possible arrangements but just worked out that in general, Logician 4 can see 2 logicians and if they have different colored hats, he cannot be certain of the color of his own hat. So, given Logician 1's uncertainty, Logician 2 knows his hat must be the opposite color to that which he can see on Logician 3, hence he announces he is wearing a blue hat.
Your channel is awesome but this is the worst explanation of a problem ever. Much more simply stated as “If position 1 sees two hats the same, he calls out other colour, but if silent, position 2 calls out the opposite of what he sees 3 wearing”.
You must not have had many things explained to you if you think this explanation that takes one unnecessary detour is "the worst explanation of a problem ever"
m1 - nice m2 - well done m3 - good work m4 - don't even have to be here autor - slowpoke another variant of this question that teacher asked us at school some1 have 5 hats (2 blue and 3 red) and he will put 3 of them on 3 ppl and place them in circle in front of each other, if some1 says his hat color they win, other rules is the same
I overheard the fact that only one prisoner needs to answer correctly.. So I tried to solve this for like ages until I watched again. But then I thought the solution is to easy if 1 can see 2 and 3. What I'm trying to say is that I spend an unreasonable amount of time trying to solve this puzzle.😂
I've always enjoyed puzzles, particularly logic puzzles. I first heard this one around 25 years ago from a guy who worked in the carpark of the company where I worked. When he told me, I thought about it for a few minutes before realising the answer. As I told him he stated that he already knew the answer but didn't know the reason why, which I was able to explain.
The flaw is that 2 can logically never be sure that 1 isn't going to speak up. He can't know how long it is reasonable to wait before assuming 1 isn't going to speak up - that element is social, not logical, as I'm sure any autistic person will agree. If the warden explicitly asked each person one at a time then this would be resolved.
Since they're not allowed to communicate, this timing strategy does also kind of rely on chance. My take was that the logician 1 must say "I'm 50% certain that...".
Hmmm but you said they can't communicate with each other at all. Person 2 can only guess their own hat is blue because they were awaiting for the shout of Person 1 and didn't hear it, that is a form of communication. If the one who knew their hat color had to whisper it to the warden instead of shouting, there would be no solution, and no communication like the rules stated.
I’ll go one further. Even with the stipulation that this counts as communication (which it absolutely does, you nailed it), the puzzle is trivially soluble. The illustration shows that the hats are brimmed, and the narration makes no claim that the men cannot see their own hats. So, an acceptable answer is: All 4 men can immediately state with certainty the color of their own hat, because they can see their own hats.
The assumption is that all four prisoners know how many of each hat is in play and that each credits the other with being perfectly capable and infallible logicians. One can think of the problem as having rounds. Each prisoner enters his response on each round. After each round all the prisoners know each of the other prisoners' responses. The responses are PASS, RED, BLUE. Number 2 can see that number 3's hat is red. If his own (number 2's) hat were red, then number 1 would instantly know that his hat is blue. Number 1's silence convinces number 2 that his own had must be blue. No one else can tell what color their own hat is. Until number 2 speaks, number 3 knows for sure that he and number 2 have different colors. On round one all prisoners pass. On round two, number 2 can infer from number 1's pass that he, prisoner 2 has a blue hat, so the responses are: PASS, BLUE, PASS, PASS On round three: RED, BLUE, PASS, PASS. No further progress is possible, but at this point 1 and 2 know all four colors, and everyone knows 1 and 2.
No, it's not an assumption. The 4 are logicians and friends, so they know this info. And the Warden told them it's 2 red hats and 2 blue hats, so all is known, no assumptions required.
Actually, round three (assuming the answers are not given aloud in order) would be: pass, blue, red, pass. After Number 2 is able to deduce that he is wearing a blue hat, Number 3 knows that Number 1 saw a blue hat on Number 2 and a red hat on him (Number 3). If each person responds aloud and in order of their number, then Number 3 would announce the color of his hat in round 2.
I agree with the idea but the results are wrong Round 1: Pass, Pass, Pass, Pass Round 2: Pass, Blue, Pass, Pass Round 3: Pass, Blue, Red, Pass That's it, it's not possible for person 1 and 4 to guess their colors so it's 50/50 chance.
You give the rules that they cannot communicate to each other, but then says #2 knows their hat because they have the information that #1 see 2 different colors. But without communicating he cannot know that …
#2 knows he has blue because if logician 1 saw 2 red hats he would say that his is blue so they can be free.. So by not being free already he assumes #1 doesn't see 2 red hats but mixed colors. That's why he says mine is blue
An easy puzzle after a long time. But i feel that you unnecessarily elongated the solution by giving all possibilities. Just say that 1st person didn't shout his colour which give information to 2nd one that his colour is different than 3rd one.
Simple answer: - Logician 2 knows that Logician 1 can see 2 and 3. - And therefore if 2 and 3 share the same color, Logician 1 will know his color. - Since Logician 1 doesn’t say anything, this only means 2 and 3 have different color. - Logician 2 can see red on Logician 3 ==> he has a blue hat
Idk if I would say this is strictly a logic puzzle, since the crucial step in it is an inductive inference. 1 taking a long time to answer does not strictly entail that he is stumped, since he could have for example, just fallen asleep, or struck a deal with one of the guards, or maybe has a grudge against the other logicians. However improbable these scenarios might be, they prevent this from being a logic puzzle strictly, since it relies on the assumption that the first logician would take a long time to answer only if he didn’t have enough information to complete the puzzle. Still though a very interesting puzzle, and enjoyable video
I used to present this to my undergrad students years ago, plus three different cute variations of this riddle (three guys in a circle seeing each other, then once again but one of them blind, then many logicians on a train)… Will you present those advanced versions too, in another video? 👍
Exactly. Further, the diagram with the problem shows #2 is a position where he is completely blocking #1’s view of #3. So if anything, the initial presentation makes clear that #1 CANNOT see #3.
Is the entire hat in that color, or just the top of it? If it's the whole hat, couldn't one of them just look up at the interior (the underside of the brim), and say the color?
Yes, and if you lifted your pencil off the paper you could place a mark at the endpoint without ever having to navigate the maze. But that would rather defeat the point, wouldn't it?
you must specify all the conditions. here is why: I thought that ligicians are not allowed to talk (otherwise how would one logician "eliminate" the wrong possibilities so that the rest 3 see the remaining possibilities), but at 3:21 "logician 1 says ...", so it is allowed to talk afterall. then if it is allowed to talk why don't just logician nr.1 says outload the colours of hats for logicians nr.2 and nr.3 so that logicians nr.2 and nr.3 could say outloud the colours of the hats they wear and be freed?
in other words: 1. logician nr.1 says: "hey, logician nr.2, your hat is of X colour" 2. logician nr.2 says: "hey warden, my hat is of colour X, so you must free us". I still don't get it. at 0:47 you say "they are not allowed to talk to each other", but HOW/WHO draws all 6 possibilities and then eliminates them?????
That's just slightly careless commentary. Presh should have said "...logician 1 THINKS, I don't know..." I don't think Presh meant to imply that Logician 1 says anything out loud. Logician 1 doesn't need to talk. Their silence is enough to tell Logician 2 the answer.
I don't know how you came up with the six different options of hat arrangement. The deductive reasoning is beautiful this must be a superpower and I feel like if I watch this video 100 times I still wouldn't get it. Because I can see the patterns laid out I can use deductive reasoning could you imagine doing this with your mind like a photographic memory. I am fascinated by analytical thinking is that what it feels like?
I know a different yet similar logic puzzle: Three Logicians are behind another. They each have a red or blue hat. Logician 1 can see both other hats, Logician 2 can see one and Logician 3 can see no hat. They are told, that the capturer has 2 red hats and 3 blue hats. The not used hats were hidden. After a long while Logician 3 says with certainty his hat-color. Why and which color is it? If Logician 1 could see two red hats, he could assume, that he has a blue hat. But as he says nothing, he can't see two red hats. Therefore, if Logician 2 could see a red hat, he could assume, he has a blue hat, because otherwise Logician 1 would say his color. Therefore, Logician 2 can't see a red hat, so Logician 3 knows, that he has a blue hat.
- i cant think of any way for 3 or 4 to know their colour - if 1 sees that 2 and 3 have the same hat colour, he knows his colour is the opposite one - if 1 cannot know as 2 and 3 have different colours, 2 will know that his colour is the opposite of 3 as 1 is unable to solve his own colour, freeing them all
question: Can Person #1 see person 3 and 2? Even if person one is on a step higher then 2, I can't imagine he could see 3 unless they aren't directly in a straight line or the distance they are standing apart is not the same as they are shown on screen. If yes (Person 1 can see 2 and 3) then this is a clear case logical deduction via process of elimination.
What I find interesting is we use a combination formula, 4C2, to determine a permutation ( the ordered number of ways they can be arranged with the hats).
I thought of it this way: if #1 sees blue blue or red red on #2 and #3 he instantly knows he is the other color. So, when #2 sees #1 hesitate, he knows that #2 and #3 are either blue red or red blue. He can see #3 has a red hat, so he yells out "my hat is blue". MUCH easier to think of it this way. And at this point, #3 would know all of this, and HE could state "my hat is red". But I think that's as far as it could go, #1 and #4 would still not be able to state their hat colors. So if THAT had been the problem (ALL four have to give their colors), I think they're stuck in prison (unless they abandon logic and go for a 50/50 guess). EDIT - looks like a lot of commenters had the same solution method that I did. :-)
Nice. Thanks for detailing the different possibilities.
ปีที่แล้ว +1
That's an overcomplicated solution. You don't need possibilites. If #1 sees 2 same color hats on #2 and # 3's head, he can tell his hat's color. If he doesn't tell anything, #2 knows that he and #3 wear different color hats. Because he sees a red hat he knows he wears a blue one.
I think it is nCr. There is another function nPr and these are two key statistical functions (based on factorials). Something to do with how many options for different situations (I can't explain it further here).
3:29 this can't be solved. it has got a big problem. The #2 logician can't know if the first is not answering the question because the two hats in front of him are of the opposite colors. The #2 actually could think that maybe the #1 is thinking, so the #2 doesn't know how much time he should wait. The other parts of the reasoning are correct, but this one is just not logical..you can't say "since the first logical is not saying anything, the second should know his hat Is blue. That's because the #2 can't know how much he should wait because he doesn't know how much it takes for the First to understand the solution.
You added way too much unrequired logic to this with all the possibilities. It's simply that logician 1 doesn't see two red hats or they would know they had blue. Therefore logician 1's silence tells two that they have blue.
I got as far as working out the probabilities and how it would be simple if 2 and 3 had the same colour. I didn't think of the final step. Very clever.
When I first saw this logic puzzle in the late 70’s, the three prisoners (Never called logicians in the original.) were on the same level, 1, could only see 2’s hat, not 3’s. How is that older and much harder version worked out
This explanation is way more complicated than it needs to be. The quickest way to describe the solution is the following. Logician #2 sees a red hat in front of him. He knows that if his hat was red that #1 would see two red hats and #1 would immediately deduce that his hat was blue, because there can only be 2 red hats ( which didn't happen ). Thus, #2 is able to definitely deduce that he is wearing a blue hat. If I was in #2's positions, I'm definitely not considering all the position configurations. I'm just thinking of what would happen if the guy behind my saw 2 red hats.
If Logician 1 sees two hats the same colour then he knows his hat is the opposite colour and speaks out. After a while of not speaking out Logician 2 realises that 1 must be looking at two different colours and therefore his hat is the opposite colour to the one in front of him and he speaks out. Basically it's the same, but without the need for the logic table.
Yea the table just made things more complicates
Agreed, no need to work through all permutations.
Absolutely; this was very easy to figure out without complicating the problem with all the possibilities. Once #1 doesn’t immediately answer, #2 knows his hat is not the same as #3.
This is how I solved it too.
In seconds nice thinking 👍
Number 1 isn't confused, he's a perfect logician, he's impatiently waiting for 2 to hurry up and get the message.
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number 2 never got the message. He's trained in logic alone, and ends up never knowing if number 1 doens't answer because he didn't figure it out, doesn't want to, or any other reason
But the weren't supposed to talk to each other.
@@larrydene6379they didnt talk
they're friends and in jail. 2 knows 1 is trying.@@shiinondogewalker2809
Three logicians walk into a bar. The bartender asks them "Would you three like something to drink?" The first logician says "All three of us? I don't know." The second logician says "I don't know either." The third says "Yes, we would."
The version I read went:
"Can I get you all a beer?"
"I don't know"
"I don't know"
"Yes"
I prefer it more than yours tbh, less obvious
@@Blade.5786nah, your Version is worded badly. There is no way for them to logically deduce whether he *can* get them a beer. Might be considered nitpicking but seems like a relevant distinction in a logic joke which is mainly based around wording. This would be a better way to phrase it:
Do you all want a beer?
Idk
Idk
Yes
i wont get it
@@SH0907 Actually there is, the logical deduction would be that he can (considering the fact that he's the bartender). Using logic even further, this is information the bartender himself is aware of, and therefore wouldn't be asking if he wasn't implying something else (whether they *wanted* a beer). Logic is ingrained in our everyday speech, so it's not exactly wrong to use it in a logic joke.
@@antijokecommenter5003
1's answer indicates 1 wants a beer but does not know about 2 and 3.
2 and 3 then know 1 wants a beer.
2's answer indicates 2 wants a beer but still does not know about 3.
3 then knows 1 and 2 want a beer.
Since 3 wants a beer, 3 now knows all want a beer.
You don't really need to enumerate all 6 possibilities in this case. If the 2nd and 3rd person wears the same color, then the 1st person can immediately answer. If they wear different color, then the 1st will stay silent. And that silence becomes information to the 2nd person, who knows the color of the 3rd and, hence, also knows his. I think this concept is called common knowledge
That is my thinking also.
This is how I worked it out as well 👍
came here to write this as well
Even more - after the second man claim his hat, third one can do it as well
exactly
...I feel like you've made this more complicated than it had to be.
Video had to be long enough to be monetized
@@Nako3 isnt it min 8 minutes for monetization tho
@@randomgamer-te8op Im not sure
@@Nako3then why are writting such a bs?!
complicated? how???
What a convoluted way to explain a simple solution.
Exactly my thoughts
I misheard the rules. I thought all four had to state the color of their own hat for them to be set free, and I just couldn't see that happening.
Same. In that case only 2 and 3 would know their color.
me too
That would've been a far more interesting puzzle.
In the case of 2&3 having same colour A, 1 will shout colour B. 2 can then conclude he is wearing colour A (because 1 sees two same hats). 3 can conclude he is wearing colour A on same premise. 4 knows he's wearing colour B by process of elimination.
In the case of 2&3 having different colours (A&B, respectively), 1 will stay quiet and 2 will conclude he is wearing A. 1 still doesn't know if he's A or B, and stays quiet. 3 can now conclude this is a "different colour scenario", and knows he is B (opposite of 2). I see no way for 1&4 to figure out what they're wearing from here.
@@BlurbFish Yep, they'd have a 50/50 chance, which is what they all start with anyway...
Once 2 has spoken, 3 knows their hat is red (if it had been the same as 2, 1 would have spoken first; so, knowing now that 2 is blue, 3 knows theirs must be the opposite). But here's where it stalls - no-one can see either 1's or 4's hat. They all know they're opposites, but not which way round they are. It could be resolved if 3 were allowed to turn round when they'd got their answer, but that's kinda kludgy. It would be nice if we could find a way for 4 to be the one that resolves it - say, if they can see 1's and only 1's hat.
Anyway. Three logicians walk into a bar, and the server says "three beers, is it?" The first logician says "I don't know", and the second logician says "I don't know", and the third logician says "yes, please".
It could be expressed more simply. If #1 sees two of the same color hat, he says he has the other one. If #1 says nothing, then #2 knows he and the guy in front of him have different color hats, so he says the opposite of whatever color he sees on the guy in front of him.
Unfortunately not long enough an explanation to make this video profitable
Still not 100% sure. Number one could be bad at his job.
yes this also works!! there are so many other possibilities to solve this puzzle.
Part of the rules of the puzzle is that all four people are PERFECT logicians.@@mattwatson7510
Absolutely. The author again is supposedly making all three friends equally capable, with the same skills.. that's not true.
You made this so much more convoluted than it needed to be
The obvious answer is to look at the brim of your own hat
😂
Should have more likes. XD I saw this puzzle before with simple conical hats, but didn't think about why they need to be, haha.
From the problem statement, it is not clear that the first person can see the next two hats. It is really important to know this information for problem solving
The warden is the real true logician.
Not so! The warden missed the possibility that the four prisoners are criminal logicians.
No! The logicians had a 50/50 chance of going free even if they just guessed. That's terrible wardening.
@@pauldzim actually, the chance is 2/3. because the second guy knows the guy in front has red, so there are 2 blue and 1 red left. And if you compound the fact that the chance of the 4 prisoners being logicians is > 0, it means that the chance that the prisoners would go free is actually more than 2/3
@@Simqerthe chance is in fact 1/2, as the first person can see 1 red and 1 blue hat which means there’s a 50/50 chance for person 1 to get correct
@@mynthics no, I am not talking about the chance for the left guy, I am talking about the overall chance for everyone to guess correctly. Nr 2 and 3 are a given, so 2 people will already have passed. The 50/50 is for the remaining 2
2:20 you're overthinking it. You don't need combinatorics or tables. If 1 sees two red or two blue he'll know his own color with a couple seconds' reflection. Once 2 has given him that time, and not heard an answer, 2 knows he differs from 3. MORE GENERALLY, 2 knows that his hat plus everything he sees before him won't yield an answer. As long as the puzzle would have let 1 solve the problem had 2 one color, 2 knows he has the other color. For instance: 10,000 people are on the hill, with the hats split 50/50. 1 sees 4999/4999; if he saw 5000 of either he'd know his own hat. 2 sees 4998/4999 and knows his own hat is the 4998 color.
That was... an unnecessarily long explanation. As other people pointed out, if they know there are 2 red hats and 2 blue hats (which is the case according to the sources in description), logician 1 will have to remain silent, as seeing different hat colors on 2 and 3 means he could have either color. Then logician 2 notices 1's silence, and thinks, "If 3 and I had the same color, 1 would immediately know his own hat color. As this was not the case, 3 and I have different hat colors, meaning that since 3 has a red hat, my own hat is blue".
Same reasoning. Glad I do not have to write all that !
i think it's not the first time he presents this puzzle either.
For bonus points, by the same logic 3 immediately knows his own hat color as soon as 2 calls out.
I don't think there's any way for 1 or 4 to know their own color, though.
One more of them could know, as at some point either 1 or 4 would have to guess their hat colour and right/wrong this would inform the other person.
im not sure how 3 would know their colour, as 3 can only figure out his hat is different colour to 2 but 3 cant see 2 to know the exact colour
you've brought in a time dimension into the problem, and also the rule of not talking to one another is broken if silence is considered communication
I think firmly stating that they can't communicate in any way was a little confusing here. If we had clarified that there are clock ticks, and on each tick a logician can answer (and all other logicians take note of when no-one else answers), then it would've made it more clear that the passage of time could be used to communicate whether a logician does/doesn't know the answer. It's a pattern used in other hat style logic puzzles so it's not too far out of reach to imagine, but it's worth clarifying.
e.g.: the blue-eyed people on an island logic puzzle uses the concept of "days" to let the puzzler understand that the passage of time is divided into distinct rounds, which can tell agents at which point other agents were uncertain of the answer.
Exactly. I don't need the fluff story. Just clearly state the rules. I had to look it up somewhere else to understand that they get turns to answer or decline and it immediately became clear that the first logician not answering is the information the second one needs to know his hat is different color than the third.
The clock ticks thing would be more relevant if there were several logical steps involved instead of just one. The first logician, given no clock ticks, would answer instantly if he saw two same coloured hats, as the first colour is trivial in that case. Since they don't answer instantly, the second logician needs to only give it like ten seconds to know their own colour with certainty.
Yup. Actually makes the whole thing illegitimate. Because right now 2 can't be absolutely certain that 1 is done thinking.
Let's assume that "they are not allowed to talk to each other" means that they can't communicate to each other at all. Then the only way that #2 would know that #1 was uncertain was via the fact that the warden has not already set everyone free. The problem is therefore better posed if there are time constraints.
I think they'd notice being set free
This is what I thought if they can't talk to each other then how are they able to communicate halfway through??
@@ashleybroughton7713 Even if they couldn't hear each other answer, Logician 2 would still realize they haven't been set free after a few seconds and then figure out what's going on.
Even if that weren't the case, there's a second solution here. The logicians were given the information that the Warden specifically set up this scenario to test if they were logicians and not criminals. For this scenario to even be a logic puzzle, logicians 2 and 3 have to have different colored hats since any criminal in the 1 spot could figure out his hat color if he sees two of the same colored hats in front of him. With that information alone, Logician 2 should be able to instantly say he has a blue hat, even if he never gets any information from Logician 1.
"Let's assume that "they are not allowed to talk to each other" means that they can't communicate to each other at all."
- My point exactly.
Many of these logical problems seem to rely on ignoring the initial conditions, not actually working out the logic.
I misheard the instructions at first and try figuring out how they could all state the color of the hat for certain. I’m glad I listened to the instructions again.
Why bother with the 6 possibilities? Irrelevant. Solution starts at 3:28 and ends very quick, L1 sees two colors so he won't speak, which meants L2 is blue.
Your solution works perfectly, but it is unnecessarily complicated. The beginning with excluding 3 and 4 is fine. But then 1 and 2 can forget about any combinations. If 1 saw two blue hats, he'd know his hat is red, and vice versa if he saw two red hats, he'd know his hat is blue. Since he's not saying anything, 2 realizes 1 must see both colours on his and 3's head and concludes his hat is red if 3's is blue and blue if 3's is red.
Surprisingly, 3 is the next person that can guess his hat, after 2 shouted blue then if 3 is blue, 1 would shout right away, so 3 knows his hat is red
3 does not even have to wait. He knows that he has different color than 2, for same reason 2 knows. So when 2 says his hat color, 3 immediately knows his color.
Then 1 and 4 have no way to determine their color, as no one sees them. Which is a bit disappointing.
Even better: If the distribution of hats is arbitrary, logicians 2 and 3 are nevertheless always sure to find out their hat colours. The other two only find out if they wear the same colour (and 2 and 3 wear the other colour). Plus, logician 3 doesn't even have to know the order of people behind him. He only has to wait for someone to call out a colour and will then be sure he wears the opposite colour.
@@yafriendceko That's the neat thing here, 3 doesn't need to know. All they know is that the first one to call out a colour (provided they only do so once they're 100% certain, which is generally assumed anyway) cannot wear the same colour as 3 does.
You should include a timeframe for when they answer to make the puzzle more clear something like every 5 minutes the warden will ask for an answer so that the time between isn’t arbitrary
5 seconds would be enough for person 1 to realize that person 2 and 3 have different colors. After those 5 seconds, person 2 can say that his color is the opposite of the one in front of him.
@@Bob94390 true it may be enough but if you’re a perfect logician and you aren’t given a timeframe for answering and you know you have to wait for the second answer slot how do you know how long it will take for the first answer slot to go unanswered. Maybe the person in spot 1 is also a perfect logician but they take a little more time than you think they would and end up with a wrong answer. The time slots make the puzzle cleaner for us and the fake computer people the puzzle is interested in.
Any logicians should be able to figure out “I see both red hats so I must be wearing blue” almost immediately. Unless Person 1 is daydreaming or something, it shouldn’t take more than a few seconds for Person 2 to be certain that Person 1 doesn’t have enough information to know. If anything, you just need to add the assumption that everyone’s actually paying attention.
No time-frame needed. No. 1 would know the answer instantly (sees two hats of same colour) or never.
By your reasoning, we have to also account for the fact that no.2 could be an a**hole who wants to be in prison and never speaks up (although he knows the answer).
@@DreadX10 in this configuration yes but what if it’s a different configuration and no1 takes an extra second before answering. No2 has suddenly trapped them all in jail. The whole point of these puzzles is that they’re guaranteed with perfect logic but not giving them a set time to answer can lead to mistakes. If no2 solved it slightly faster than no1 and then answered and the configuration was different that would be it. With an answering cadence problem solved.
I got asked a version of this in an interview about 10 years ago. I remember when the question was posed to me I thought there couldn't possibly be an answer. But the interviewer asked me to try working it anyway. I spent an agonizing amount of time on it, building a logic table. Eventually I figured it out and was very happy! I didn't get the job, but I was still pleased with myself for answering this puzzle correctly in a nerve wracking situation.
I’ve interviewed a lot of people and I think question like these are MUCH harder in a stressed situation, so kudos to you
That question got asked in Honkai Star Rail too xD
In short:
After some waiting, 1 stays silent, meaning 1 and 2 aren't the same color (he'd know his color otherwise)
2 understands that now and knows his color must be blue, since 3 has red.
So 2 will be able to say his hat color with certainty
An interesting variation of a this kind of problem is this: 3 logicians wanted to figure out which of them is smartest, so they went to the king. The king devised a test in which they were to sit in a circle and close their eyes. They were told that a hat, either red or blue, will be placed on each of them. They were also told that there are 2 blue and 3 red hats in total. When they open their eyes, they are supposed to figure out which hat colour they are wearing. Each will only see the hats of the two logicians in front of them. The first to shout out their hat colour is the smartest. While their eyes were closed, the king placed a red hat on each of them. When they opened their eyes, after a short time, one of them shouted that they were wearing a red hat. He was declared the smartest. The question is, how did he think?
Not seen that variation, but it's cute.
1. If any of them were looking at two blue hats, they would instantly know they were wearing a red hat, but no-one answers so it's not two blue hats.
2. Having figured this out, if someone saw one blue hat, they would instantly know their hat was red, but no-one answers meaning there isn't one blue hat.
3. The only remaining possibility is all 3 hats are red, and so it's just down to who is fastest to answer red.
@@manudude02 The logic is nice, but you start to get a difficulty with how to define 'instantly + instantly'. How long should one of them wait to determine that they are now in the second consecutive silence and not the first one? Because of this difficulty they cannot simply rush to be the fastest to say 'red' the second they reason by this method that it could logically be the answer, just in case one or both the others are still in the first of the two consecutive silences.
The king asks the first time and silence, so it excludes 2 blue hats.
The king asks second time and silence again, so it excludes one blue hat.
When the king asks for the third time, they all know they're wearing red hats, because that's the only remaining possibility, and the fastest to speak wins.
The test needs to be fair, so all of them must be wearing the red hat. Therefore the first to realise that answered the question. I know this doesn't employ logic but common sense but I feel the given task isn't suited for it
@@manudude02there were 2 blue hats😐 so if one blue hat was in front of me then how would I assume I m wearing red hat? It can still be blue hat?
This logic was quite simple. I didn't use your method exactly but somewhat similar to your answer. As logician 1 is confused, means 2 will understand that the hat of 2 and 3 are of different color. So he sees 3 has red color and concludes 2 himself is blue.
these guys cannot communicate with eachother.
Proceeds to make them communicate.
When?
???
Oh Presh, you do love an overcomplicated solution. This takes exactly two very easy-to-follow steps-no formulae, no tables. • If #1 saw the same color on #2 and #3, he would shout that he had the other color. • He doesn't shout, so #2 knows he is wearing a different color than #3 and shouts out his hat color. DONE. Why would you start with the two who have least information?!
What an overcomplicated explanation for such a simple solution! I'm neither a logician nor a mathematician, yet I figured it out in less time than it took to explain it.
_"After a while after #2 doesn't answer"_ is kind of a dishonest modification to the rules.
Simple explanation (spoilers!):
If L1 saw two hats of the same color, he'd know that his hat was the other color. Since L1 does not know his hat color, L2 knows that his own hat must be the color that he does not see on L3.
Yes. The problem is whem the two hats in front of him are different.
Therefore. When number 1 doesnt know the colour. Number 2 will now know that the colurs of 2 amd 3 are different.
Cos as you said if they are the same then 1 would just give the amswer.
But they are different and somce number 2 can see that number 3s hat is red. He now knows taht his being different means his hat is blue.
Thats the point.
This video made the question more complex
Simple explanation:-
If Logician 2 and 3 have the same colour hats, Logician 1 knows that his hat is the opposite colour as he knows there are only 2 of them. If Logician 1 doesn't know the answer, Logician 2 knows that him and Logician 3 have a different coloured hat. So, Logician 2 says the opposite of Logician 3's hat colour. In all the 6 possible arrangements, either No.1 or No.2 can say the right answer
Solved it in under a minute, but, then again, I had "help"; I already knew a similar problem (the Three Chinese Philosophers riddle, in which a queen paints either a red or a green dot on each of three philosophers' foreheads and challenges them to find out, without conferring, whether their own dot is red or green). To be perfectly honest, I thought Presh's explanation was a bit over-complicated; he could have been a bit more succinct and to the point.
Excellent video! I went through this a few times to fully understand the logic (ok, I'm slow), but now I do and I think this is terrific!
I seem to remember a harder version of this puzzle where all 4 men had to guess the color of their hats. As soon as 2 guesses, number 3 immediately knows the color of his hat as well. I don't remember how 1 and 4 got their color but I think it had to do with the wording of the puzzle.
Yeah, I was trying to figure out the same. Like, what'd be different if 1 and 4 had their colors switched?
@@couchpotatoe91nothing because in order for them all to get out only one person has to know their hat color
@@drewsworthh ah ok. I remember a prisoner scenario where everyone has to guess their color in order to be freed. But it had some twist iirc like the guy behind the wall was actually on the very left behind another wall and could only glimpse the tip of the last person's hat.
The difficult version is 100 people (could have been any number) lining up. There are 2 colours but the proportion of hats is not given. Everyone takes turn guessing starting from the back. They're allowed time to strategize. They can only make a guess but not signalling using other things like volume. The optimal strategy ensures everyone except the one at the back can deduce the colour correctly.
Must admit, I didn't even think about the probability. Just that 2 would know that 1's inability to determine what he's wearing must mean that he's seeing a blue hat and a red hat and the fact that 2 can see a red hat means he must be wearing a blue hat. Thanks for posting as I can now explain to my wife that I am fully entitled to a celebratory glass of wine.
I've seen one similar to this on TED-Ed, though that was with more people, and a random distribution of hats. In this case, if the top person sees two red hats, he knows he has blue, and thus they win, same thing if he sees two blue hats, he knows he has red. However, if he sees a red and a blue, he has a 50/50 chance, and thus hesitates to answer. This then tells the next person in line that he sees a hat of each color, since he doesn't automatically know his own hat color. So he looks at the hat in front of him, and instantly knows his own hat color, being the opposite of the person in front of him, thus they win. The third and fourth people don't even need to answer.
This is a simple problem of pure logic, and by that I mean SIMPLE, and LOGIC. There's no need to introduce any mathematics whatsoever. Sometimes you really can reason without formulas.
Number 1 remains silent since he sees 1 blue and 1 red hat, leaving two options open. Number 2 (logician as he is) realizes that when number 1 remains silent, number 1 must face 1 blue and 1 red hat. Since number 2 can see number 3 is wearing a red hat, he knows for sure he is wearing blue himself.
Understood bitzh
except when it comes down to number 1 and number 4 still not knowing which color hat they are wearing.
You made that explanation so much more complicated than it needed to be! All you needed to say was that there were only two hats of each color, and thus if logician one saw two hats of the same color he would know that he was looking at all the hats of that color and thus his hat must be the other color, and logician 2knows this, so if he doesn’t hear logician one state the color of his hat, then he and logician three must be wearing two different color hats and thus his hat is the color he *doesnt* see in font of him. (And fwiw, once logician three hears logician two shout out that he’s wearing a blue hat, he would know that his own hat must be red, not that he would have any reason to care at that point.)
This was a very very easy puzzle. I’ve never solved one of these before, but this one was extremely easy
But the problem is the temporal nature of it..
How long should #2 wait for #1 to not give an answer?
You don't really need the first step (2 eliminating half the possibilities). 1's silence immediately means that 2 and 3 have different colours, so upon noting 1's silence 2 can announce his colour as the opposite of the one he sees on 3. The question, though, is how long should 2 wait to be sure that 1 doesn't see two hats the same colour rather than, say, is still fumbling with his glasses?
Good catch
If hats on 2 and 3 are the same color 2 can expect 1 to speak, but it 1 does not speak for a time, 2 can assume that his hat is of the opposite color to 1 whose hat he can see and can speak.
I love "a group of logicians" problems. I thought your solution was overly complex though -- you simply have to realize that when 1 doesn't speak, 2 realizes that means that he and 3 have different colored hats, and since 3 has a red hat, his must be blue. (If 2 and 3 had the same color, 1 would know their hat color immediately).
Pretty simple actually. The first man would speak if he saw two identical hats in front of him. Since he doesn't, it should become clear to the others after a few minutes that positions 2 and 3 contain one red and one blue hat. The second man would then know this, see a red hat in front of him, and state that his hat is blue.
At timestamp 0:46 the warden says they are not allowed to talk to each other. So right at 3:17 when #1 talks, the warden should have said in the voice of Bill Paxton from Aliens, "Game over, man", and they should have been marched back into jail.
Exactly. Every logician should only find out if another logician guessed right after taking a guess himself/herself. Or more generally, no communication, since silence is communication as well, it communicates that 2 and 3 have different hats. That'd make the problem unsolvable though, so the flaw had to be intentional.
Logician 1-4 realize that their eyes are in fact green, and they are all let free.
This is a very easy puzzle. But it was explained the very complicated way. That’s kinda the opposite of what was supposed to happen
no it's very hard i couldn't solve it after an hour
The riddle can be solved by just Logician 1, 2 and 3 interacting with each other. Logician 4 is pretty much just a distraction for the solver. It would be the same riddle with 3 guys, 3 hats, where 2 is the same color and one is different.
Interesting, heard this logic problem on a car talk radio program many years ago. Like a good humorous story nice to see this make the rounds again. Thanks for sharing.
If 2 and 3 had the same color, 1 would announce they’re wearing the opposite color. 2 could hear that announcement but it isn’t made, despite allowing one whole second for a perfect logician to make it. 2 then knows that they’re wearing the opposite of 3, i.e. blue. 4 is a little annoyed to be a background prop just to wear an unseen hat, but is glad to go free.
I didn't iterate the possible arrangements but just worked out that in general, Logician 4 can see 2 logicians and if they have different colored hats, he cannot be certain of the color of his own hat. So, given Logician 1's uncertainty, Logician 2 knows his hat must be the opposite color to that which he can see on Logician 3, hence he announces he is wearing a blue hat.
Logician 4 doesn't see any other, just like logician 3...
@@Azarathification Yeah, I realized after that when I said Logician 4, I meant 1 but now this is getting complicated to explain.
Plot twist: Logician 4 is watching them solve this with the warden
Your channel is awesome but this is the worst explanation of a problem ever. Much more simply stated as “If position 1 sees two hats the same, he calls out other colour, but if silent, position 2 calls out the opposite of what he sees 3 wearing”.
You must not have had many things explained to you if you think this explanation that takes one unnecessary detour is "the worst explanation of a problem ever"
@@noahblack914its still a lot longer than needed, we didnt really need math or anything, this comment is way easier to explain and to visualise
m1 - nice
m2 - well done
m3 - good work
m4 - don't even have to be here
autor - slowpoke
another variant of this question that teacher asked us at school
some1 have 5 hats (2 blue and 3 red) and he will put 3 of them on 3 ppl and place them in circle in front of each other, if some1 says his hat color they win, other rules is the same
I overheard the fact that only one prisoner needs to answer correctly.. So I tried to solve this for like ages until I watched again. But then I thought the solution is to easy if 1 can see 2 and 3. What I'm trying to say is that I spend an unreasonable amount of time trying to solve this puzzle.😂
I've always enjoyed puzzles, particularly logic puzzles. I first heard this one around 25 years ago from a guy who worked in the carpark of the company where I worked. When he told me, I thought about it for a few minutes before realising the answer. As I told him he stated that he already knew the answer but didn't know the reason why, which I was able to explain.
The flaw is that 2 can logically never be sure that 1 isn't going to speak up. He can't know how long it is reasonable to wait before assuming 1 isn't going to speak up - that element is social, not logical, as I'm sure any autistic person will agree. If the warden explicitly asked each person one at a time then this would be resolved.
I believe it was the condition in original riddle that he asks them few times.
Since they're not allowed to communicate, this timing strategy does also kind of rely on chance. My take was that the logician 1 must say "I'm 50% certain that...".
Hmmm but you said they can't communicate with each other at all. Person 2 can only guess their own hat is blue because they were awaiting for the shout of Person 1 and didn't hear it, that is a form of communication. If the one who knew their hat color had to whisper it to the warden instead of shouting, there would be no solution, and no communication like the rules stated.
I’ll go one further. Even with the stipulation that this counts as communication (which it absolutely does, you nailed it), the puzzle is trivially soluble. The illustration shows that the hats are brimmed, and the narration makes no claim that the men cannot see their own hats.
So, an acceptable answer is: All 4 men can immediately state with certainty the color of their own hat, because they can see their own hats.
@@jaarnealahah fair enough! guess you could say they just had to _look up_ the answer
No good. They may only look ahead, not up. And no cheating by whispering either. Only pure logic.@@zewzit
Actually number 2 know that he has to have a hat of the opposite colour to number 3 or number 1 has no puzzle to solve.
if the hats were all one color throughout the whole hat, all they'd have to do is look up at the brim of the hat to see what color they are wearing.
The assumption is that all four prisoners know how many of each hat is in play and that each credits the other with being perfectly capable and infallible logicians.
One can think of the problem as having rounds. Each prisoner enters his response on each round. After each round all the prisoners know each of the other prisoners' responses. The responses are PASS, RED, BLUE.
Number 2 can see that number 3's hat is red. If his own (number 2's) hat were red, then number 1 would instantly know that his hat is blue. Number 1's silence convinces number 2 that his own had must be blue. No one else can tell what color their own hat is. Until number 2 speaks, number 3 knows for sure that he and number 2 have different colors.
On round one all prisoners pass.
On round two, number 2 can infer from number 1's pass that he, prisoner 2 has a blue hat, so the responses are:
PASS, BLUE, PASS, PASS
On round three:
RED, BLUE, PASS, PASS.
No further progress is possible, but at this point 1 and 2 know all four colors, and everyone knows 1 and 2.
how you commented 2 days ago video is published few minutes ago only????????
@@uditisgaming5872 What are you, a logician?😂
No, it's not an assumption.
The 4 are logicians and friends, so they know this info.
And the Warden told them it's 2 red hats and 2 blue hats, so all is known, no assumptions required.
Actually, round three (assuming the answers are not given aloud in order) would be: pass, blue, red, pass. After Number 2 is able to deduce that he is wearing a blue hat, Number 3 knows that Number 1 saw a blue hat on Number 2 and a red hat on him (Number 3). If each person responds aloud and in order of their number, then Number 3 would announce the color of his hat in round 2.
I agree with the idea but the results are wrong
Round 1: Pass, Pass, Pass, Pass
Round 2: Pass, Blue, Pass, Pass
Round 3: Pass, Blue, Red, Pass
That's it, it's not possible for person 1 and 4 to guess their colors so it's 50/50 chance.
This is so incredibly easy. It was so easy in fact, that I questioned if it really only took more 4 seconds to figure out.
Just like the honesty of thumbing up your own comment.
You give the rules that they cannot communicate to each other, but then says #2 knows their hat because they have the information that #1 see 2 different colors. But without communicating he cannot know that …
Actaully number 2 can logically conclude he must have a different colour to number 3 otherwise number 1 can solve the puzzle.
They cannot communicate with each other, except for stating for sure the color of their own hat or staying silent.
#2 knows he has blue because if logician 1 saw 2 red hats he would say that his is blue so they can be free.. So by not being free already he assumes #1 doesn't see 2 red hats but mixed colors. That's why he says mine is blue
3:44 no communication between 1 and 2 then how to remove blue,red,red,blue ?
An easy puzzle after a long time.
But i feel that you unnecessarily elongated the solution by giving all possibilities.
Just say that 1st person didn't shout his colour which give information to 2nd one that his colour is different than 3rd one.
No need for that logic tell us hats 2 and 3 must be different colours or 1 can solve the puzzle.
Simple answer:
- Logician 2 knows that Logician 1 can see 2 and 3.
- And therefore if 2 and 3 share the same color, Logician 1 will know his color.
- Since Logician 1 doesn’t say anything, this only means 2 and 3 have different color.
- Logician 2 can see red on Logician 3 ==> he has a blue hat
I find these logic problems so ridiculous sometimes. Why take a simple thing and make a math problem out of it? Thats madness😅
Idk if I would say this is strictly a logic puzzle, since the crucial step in it is an inductive inference.
1 taking a long time to answer does not strictly entail that he is stumped, since he could have for example, just fallen asleep, or struck a deal with one of the guards, or maybe has a grudge against the other logicians. However improbable these scenarios might be, they prevent this from being a logic puzzle strictly, since it relies on the assumption that the first logician would take a long time to answer only if he didn’t have enough information to complete the puzzle.
Still though a very interesting puzzle, and enjoyable video
I used to present this to my undergrad students years ago, plus three different cute variations of this riddle (three guys in a circle seeing each other, then once again but one of them blind, then many logicians on a train)… Will you present those advanced versions too, in another video? 👍
Excellent, I was searching this puzzle since long.
it wasn't made clear in the initial rules if # 1 could see # 2 & # 3 or just # 2.
Exactly. Further, the diagram with the problem shows #2 is a position where he is completely blocking #1’s view of #3. So if anything, the initial presentation makes clear that #1 CANNOT see #3.
What throws people off is that they think that No. 1 must be the one to answer as he has _the most information_ ...!
how was the factorial formula derived?
3:30 how? They can't communicate to each other right 0:46? How did he know and become so sure?
Is the entire hat in that color, or just the top of it? If it's the whole hat, couldn't one of them just look up at the interior (the underside of the brim), and say the color?
Yes, and if you lifted your pencil off the paper you could place a mark at the endpoint without ever having to navigate the maze. But that would rather defeat the point, wouldn't it?
you must specify all the conditions.
here is why:
I thought that ligicians are not allowed to talk (otherwise how would one logician "eliminate" the wrong possibilities so that the rest 3 see the remaining possibilities), but at 3:21 "logician 1 says ...", so it is allowed to talk afterall. then if it is allowed to talk why don't just logician nr.1 says outload the colours of hats for logicians nr.2 and nr.3 so that logicians nr.2 and nr.3 could say outloud the colours of the hats they wear and be freed?
in other words:
1. logician nr.1 says: "hey, logician nr.2, your hat is of X colour"
2. logician nr.2 says: "hey warden, my hat is of colour X, so you must free us".
I still don't get it. at 0:47 you say "they are not allowed to talk to each other", but HOW/WHO draws all 6 possibilities and then eliminates them?????
That's just slightly careless commentary. Presh should have said "...logician 1 THINKS, I don't know..."
I don't think Presh meant to imply that Logician 1 says anything out loud.
Logician 1 doesn't need to talk. Their silence is enough to tell Logician 2 the answer.
@@gavindeane3670 You are correct. I rushed into writing the comments :)
I don't know how you came up with the six different options of hat arrangement. The deductive reasoning is beautiful this must be a superpower and I feel like if I watch this video 100 times I still wouldn't get it. Because I can see the patterns laid out I can use deductive reasoning could you imagine doing this with your mind like a photographic memory. I am fascinated by analytical thinking is that what it feels like?
when silence is an option :
logicians : we bout to end this warden's whole career
I know a different yet similar logic puzzle:
Three Logicians are behind another. They each have a red or blue hat. Logician 1 can see both other hats, Logician 2 can see one and Logician 3 can see no hat. They are told, that the capturer has 2 red hats and 3 blue hats. The not used hats were hidden. After a long while Logician 3 says with certainty his hat-color. Why and which color is it?
If Logician 1 could see two red hats, he could assume, that he has a blue hat. But as he says nothing, he can't see two red hats.
Therefore, if Logician 2 could see a red hat, he could assume, he has a blue hat, because otherwise Logician 1 would say his color.
Therefore, Logician 2 can't see a red hat, so Logician 3 knows, that he has a blue hat.
I feel like "logicians" 3 and 4 in everyday life situations 😀
I am happy every time i was able to solve a problem on this nice channel.
- i cant think of any way for 3 or 4 to know their colour
- if 1 sees that 2 and 3 have the same hat colour, he knows his colour is the opposite one
- if 1 cannot know as 2 and 3 have different colours, 2 will know that his colour is the opposite of 3 as 1 is unable to solve his own colour, freeing them all
3 could figure out their hat color after 2 responds. 1 and 4 won't be able to figure out their own hat colors as they don't have enough information.
Took me about 10 seconds. Wish I could make a living off of puzzles like this. Alas, no joy.
question: Can Person #1 see person 3 and 2? Even if person one is on a step higher then 2, I can't imagine he could see 3 unless they aren't directly in a straight line or the distance they are standing apart is not the same as they are shown on screen. If yes (Person 1 can see 2 and 3) then this is a clear case logical deduction via process of elimination.
What I find interesting is we use a combination formula, 4C2, to determine a permutation ( the ordered number of ways they can be arranged with the hats).
haha best comment!
You said they are not allowed to talk to each other and no cheating so theres no way the 2nd could get notified of the 1st confusion...
I thought of it this way: if #1 sees blue blue or red red on #2 and #3 he instantly knows he is the other color. So, when #2 sees #1 hesitate, he knows that #2 and #3 are either blue red or red blue. He can see #3 has a red hat, so he yells out "my hat is blue". MUCH easier to think of it this way.
And at this point, #3 would know all of this, and HE could state "my hat is red". But I think that's as far as it could go, #1 and #4 would still not be able to state their hat colors. So if THAT had been the problem (ALL four have to give their colors), I think they're stuck in prison (unless they abandon logic and go for a 50/50 guess).
EDIT - looks like a lot of commenters had the same solution method that I did. :-)
Nice. Thanks for detailing the different possibilities.
That's an overcomplicated solution.
You don't need possibilites.
If #1 sees 2 same color hats on #2 and # 3's head, he can tell his hat's color.
If he doesn't tell anything, #2 knows that he and #3 wear different color hats. Because he sees a red hat he knows he wears a blue one.
What is that Choose thing at the begining of the solving process?
I think it is nCr. There is another function nPr and these are two key statistical functions (based on factorials). Something to do with how many options for different situations (I can't explain it further here).
@@asphaltpilgrim ah ok that's probably the only thing they didn't teach on my calculus classes back in the day 😅 Cheers 🙏
3:29 this can't be solved. it has got a big problem. The #2 logician can't know if the first is not answering the question because the two hats in front of him are of the opposite colors. The #2 actually could think that maybe the #1 is thinking, so the #2 doesn't know how much time he should wait. The other parts of the reasoning are correct, but this one is just not logical..you can't say "since the first logical is not saying anything, the second should know his hat Is blue. That's because the #2 can't know how much he should wait because he doesn't know how much it takes for the First to understand the solution.
Jp here again, Thank you MindYourDecision :)
I actually have this riddle in an old book, you cannot trick me today you wizard!
You added way too much unrequired logic to this with all the possibilities. It's simply that logician 1 doesn't see two red hats or they would know they had blue. Therefore logician 1's silence tells two that they have blue.
I was presented with this problem during an interview. I totally biffed the answer, but they still hired me!
I got as far as working out the probabilities and how it would be simple if 2 and 3 had the same colour. I didn't think of the final step. Very clever.
When I first saw this logic puzzle in the late 70’s, the three prisoners (Never called logicians in the original.) were on the same level, 1, could only see 2’s hat, not 3’s. How is that older and much harder version worked out
This explanation is way more complicated than it needs to be.
The quickest way to describe the solution is the following.
Logician #2 sees a red hat in front of him. He knows that if his hat was red that #1 would see two red hats and #1 would immediately deduce that his hat was blue, because there can only be 2 red hats ( which didn't happen ). Thus, #2 is able to definitely deduce that he is wearing a blue hat.
If I was in #2's positions, I'm definitely not considering all the position configurations. I'm just thinking of what would happen if the guy behind my saw 2 red hats.