In all honestly I thought the hat one was a trick, and that prisoner 3 could see his reflection in the wall lol The last one I also thought was a trick, but I wasn't paying close attention. I figured if you cut the deck in half (like cutting a card in half) that would be the easiest solution lol. maybe next time I see a video with riddles, I'll pause it and think about it lol
There is a problem with the last riddle. It assumes the 13 gave up cards in the deck are distributed RANDOMLY. that's what the riddle says. But the answer implies that the 13 face up cards are distributed EVENLY. Huge error in the last riddle
@@roberjohnsmith no, no it does not. Every possible outcome would work. If the small pile has zero flipping them all would make that 13 just like the big pile, same goes for every other outcome
Got 4 of 5 - I ran out of patience (a virtue in most of life's problem-solvings) on the last one. Really liked the solution and will remember the logic from now on. Well, I'm 75-years-old and no longer one of the sharper knives in the kitchen. Ironically, now I can't remember where I left the sharpener.
0 Then it's good. If age is an indicator for how many riddles you could solve. He needs to be 94 years old to answer all 5. You on the other hand would be able to answer all 5 at the age of 35. But of course in reality that's not how it works. But keep trying!
@@koolwond4264 if you first take the sheep to the other side then the wolf will be alone with the other sheep and eat it, and thats what we're trying to avoid, no?
Then man4 would lose his chance to be free. He must be the last as he has no comparison to think what hat he wore. 1,3 will know once they discover the pattern of what man 2 is thinking.
The objective is to make the same number of cards face up, it it can be 1 and 1 or 26 and 26, it doesnt have to use the original 13 face up cards and split them equally.
I got 4 out of 5. The only one I didn’t get was the wolf and sheep one lol. Felt really dumb after the explanation so had to make up for it with 4 and 5
@@randomperson-cq6lb in the interview the interviewer asked me a similar question and I gave the correct answer quickly 😁 Hence it was one of the key moments of that interview
The hat riddle is based on assumption that prisoner 1 is silent. Suppose he sees that Prisoner 2 and 3 are wearing the same coloured hat, then he would be the first person to shout. Prisoner 2 is able to interpret only on the silence of Prisoner 1 and not otherwise
Loved all of these! The only thing that bothered me was the question wording of riddle 4. I don't believe it's fair to ask the question "who will be the FIRST....person to definitely know his hat color and shout it out?" The answer to that question can be either person1 and person2. - If the middle two men have same colored hats (blk or white), then person1 will be the FIRST PERSON to definitely know his hat color and yell it out - if the middle two men have different colored hats, then person2 will be the FIRST PERSON to definitely know his hat color and yell it out. He'll know it via first person's silence.
@@artugert I have to agree with Dy here, I also have a problem with the way the riddle is set up in this video. We are told that the prisoners' places are as shown in the picture, they do not specify that the hats also match the picture. The hats are introduced afterwards and the wording makes the hat positions seem arbitrary to the problem. In my mind when I heard it, it seemed like the hats could have been arranged any way. Still a good riddle, I would have just added the specification that the hats are placed as shown in the picture.
@@philipchervenkov7933 I agree. I googled the riddle and found it asked in several other publications, and in those problems the hats were said to be assigned randomly (we, the riddle solvers don't know who has what hat either). This video's version of the riddle was worded poorly, as the answer is not "2"... it's "1 or 2" depending on who has what hats.
He lost his arm due to someone hating him so much for his riddles he forced the guy to go to war, and he lost his arm due to a sea king while saving Monkey D. Luffy.
For the last one, i was thinking of tearing the cards in two. Then create two piles from the halves of each other. There will surely be two piles of cards facing up
Ditto, I didn't think of the correct answer because I made the mistake of assuming either two equal piles - instead I allowed the definition of card to not require them to be whole undamaged playing cards - hence tear them in half! There could be a follow up riddle where you don't know how many are face down to start with and then this would be the answer.
Sometimes it's easier to reduce the #5 riddle to smaller numbers. Try to solve it with 4 cards and just 1 flipped over. Then start thinking but only assume what is feasible (the two decks can be of different size and you can flip cards even being blind). It's easy to see that putting aside one of the four cards and flipping it you will get the answer. Most common mistakes are to assume what you shouldn't (two decks of same size and not being able to flip cards).
You are right, it is way easier if you break it down to something simple like 4 cards. And I made the stupid mistake assuming the piles had to be the same size...
52 cards: If you decide to consider card flips (I didn't), then start with 1 card on the first pile and flip it. Then you end up with either 13/1 or 12/0. With a 2 card pile that is flipped, you get either 13/2, 12/1, or 11/0. With a 3 card pile: 13/3, 12/2, 11/1, 10/0. with a 4 card pile: 13/4, 12/3, 11/2, 10/1, or 9/0. The pattern (eg, the difference in number of face up cards between the two piles = 13 minus the number of cards in the small pile) quickly suggests to try.. With a 13 card pile: 13/13, 12/12, 11/11, ... or 0/0. Voila, 13 card pile flipped leads to the same number of up cards as in the other 39 card pile in all scenarios.
I "solved" the first two, but only because I've seen them before. I think being able to at least understand the explanation is satisfactory. They're interesting and change up your method of thinking.
Febha Mathew if you take the first 13 cards and all of them are face down, that means the second pile have all 13 face up cards. Then by flipping the first 13 card pile over, you will always have the same amount of face up cards. Max having 2 piles with 13 cards each pile face up.
I didn't solve the last one because I didn't give myself enough time to think, I actually knew it would involve flipping one of the piles because I know a magic trick that works on the exact same principle, I just failed to realize both piles didn't need to be the same size. I have seen and solved variations of all the other problems before, so they were all pretty easy to me. I never got why people struggle so much with the prisoners and hats one. Just put yourself in the minds of each prisoner, and the solution should be very obvious.
Exactly the same for me. I immediately remembered the cards puzzle and knew that you had to flip part of the cards over, but it only worked with 13 and 39 cards, not two stacks of 26 cards…
Riddle #5: For those of you experiencing difficulty with this particular riddle, here's a hint: Don't focus on finding the 13 actual cards that are face-up in the deck.
I solved this tough puzzle took me more than 2 hours and ended up using math instead. In my opinion, the premises provided felt incomplete, should at least add one that could serve as a hint.
SPOILER 1. Insight: You can´t divide 13 face-up cards into two. 2. Insight: Flipping some cards is the only way to change the number of face-up cards in total. 3. Insight: If you flip a stack with uneven number of cards in total, it will change the number of face-up cards within from even to uneven or vice versa. 4. Conclusion: By flipping an uneven amount of cards, the overall number of face-up cards will be even. 5. Insight: As you don´t have any more information as stated initially, the turned cards must be one of the two stacks afterwards, any mixing up with the remaining leads to an unsolvable situation with no chance to divide the face-ups in two halves by any counting or blind sorting of the cards. This one is crucial, but also obvious. 6. Insight: For sure it doesn´t matter where from within the original stack you take the cards to turn. You may take the top ones, or pick them one by one out of the deck, doesn´t matter. 7. Conclusion: It has to be the number of cards to flip and build the 2nd stack that matters only. 8. Heuristic: Flipping all or half of the deck doesn´t work (don´t even think about it, you have to flip an uneven number, see 3.). The only other given count is "13". And now start thinking, the steps before were just "technical thoughts". 9. Solving: Turning a stack of 13 cards with x face-up cards will lead to a stack witch 13-x face-up cards. Because this stack had x face-ups before turning, the remaining stack of 39 has to contain 13-x face-ups, too.
@@JaydenLawson no, that wouldn't work. imagine the piles one has 5 of the 13 flipped cards, the pile three has the other 8 of the 13 and the piles two and four don't have any flipped cards. you flip one and three, so one has 8 and three has 5. you put them together and they have 13 flipped cards. two and four don't have any flipped cards, so your theory is wrong
@ANIMAL!!! I mean the video pretty much explains it out the entire answer for you. In short Prisoner 2 can infer that his hat is the opposite of Prisoner 3 because the Prisoner 1 behind him is NOT saying anything. Prisoner 1 doesn't know what he's wearing only because 2 and 3 have different colors. If 2 and 3 DID have the same colors then Prisoner 1 could easily just say the opposite color, boom done. However because Prisoner 1 is NOT saying anything, Prisoner 2 can infer that they have opposite colors and can just state the opposite color of Prisoner 3
@ANIMAL!!! Lol, Cheers for figuring it out. Yes Prisoner 2 has to make an assumption that Prisoner 1 is not saying anything because he sees P2 P3 have different colors. However P1 could easily just be goofing around or what not, or dumb. This exercise assumes all Prisoners have sound logical deduction ability for this puzzle to work.
Xzor is asking WHY you would even want the wolf in the first place if you knew it wanted to eat your sheep. And honestly my best guess is that the wolf is worth a lot of money or it's meat tastes good.
There is a problem with prisoners riddle in 3:59 it says “who is DEFINITELY THE FIRST ONE to find out”. Here is the problem: prisoner 2 is guaranteed to find the answer, but he may be the first one or the second one to do that. In fact the video itself explains the situation in which prisoner 1 is the FIRST one to solve the problem. Therefore, I think the problem should be changed to this: “who will definitely find the color of his hat in any situation (no matter being the first or second person to do that). Am I right?
I was thinking about that too , if 2nd and 3rd are wearing same hats , 1st will know his hat's color. But I guess the question demanded the same pattern of distribution that was shown in the picture. There are 6 ways in which hats can be distributed. But only the distribution shown was to be considered.
This is true. Also, the last riddle was flawed greatly. And there is no way to solve it the way it was asked. In the last riddle, it was implied in the question that 13 cards were face up random throughout the deck. But the answer implied they were distributed EVENLY throughout the deck. If it was truly random then I could take off 13 of the first cards of the deck and get no face up cards.
Amir Mansoor Kamali Sarvestani no, you’re not. The riddle states that the 4 criminals are lined up as in the picture. White/black/white/black. That means the answer is always number 2. Yes, they explain the other situation only to clarify the answer but I guess people are just overall confused when you try to clarify things.
Hi C1. In the video it never says that the order of the colors is fixed. I guessed they meant that the colors are only and only in white, black, white and black order. But it never ever explicitly says that the order should be as the graphics shows. I watched it again and again. He lists a number of rules but never says that the order is fixed to white, black, white, black. As a result, every one can assume that this riddle is about finding the answer with any possible order of colors.
The last one kind of needs the caveat that "the piles don't have to be equal amounts of cards" I know it wasn't stated, but that's why I was getting stumped :P
I looked back at it and honestly I think his diagram is what made me get that mental block. He shows two perfectly distributed piles in the animated explanation.
Milad Zavar The second riddle is better explained visually rather than with words, but lets see. (Also the video said "No maths needed, only logic" but im kinda stupid and cant explain it with words so ill try to explain it mathematically) For the sake of simplicity, lets say the area of one super grass patch is 2. It can now be represented by 2^d = A, where "d" is the number of days which had passed and "A" is the area in meters. [ e.g if 3 days had passed d = 3. Input it into the formula to get 8 meters squared (2^3 is 2x2x2 = 8) ] Since the garden took 10 days to be completely covered, the area of the garden is 2^10, or 1024 meters squared. If there were 2 patches, it would take 9 days because they would both grow up an area of 2^9, or 512 meters squared. Since there is 2 patches, it would be 512 + 512, which equals to 1024, equal to the total area of the garden, therefore taking only 1 day less than one patch would.
@@simoncordero384 your answer is very good but just to be precise: The size of one patch is not 2 but 1. As at the first day when he buys the super grass no day has passed the size is 2^0 = 1 and on the second day (one day passed) the size is 2^1 = 2 which is the starting point when buying two super grass but as i said - the explanation was very good :)
Split the deck into 2 random piles. Place them both sideways (lean them against each other or prop them up against something). Now you have 2 piles with the same amount of face up cards = 0. A real life genius > math genius 😎
If the first 13 cards are face-up cards, then the result ends up with no any face-up cards, but the answer is still logically correct. Thanks for the interesting riddles.
I think in the fourth puzzle, the correct answer cannot be given because prisoners one and two have the same chance of winning and shouting first. If the second and third prisoners had two white or two black hats, then the first prisoner would be sure what his hat is. If, on the other hand, the second and third prisoners had black and white or white and black hats, then the second prisoner would be sure what hat he has, according to the solution in the film. It depends on the fate that the prisoner wins, so the correct prisoner number cannot be given. It is possible that when solving this task, you only need to consider the example of setting the hats that was in the video. However, it is not explicitly stated that this is the case. It is known that the example can be used to imagine the place of the event and an example of an arrangement of hats, which may also take a different order. So it sticks to my opinion.
I came to the exact same conclusion. They can't shout at the same time... The second guy has to wait for the first one's reaction... Either it's silence (the second guy will shout first) or the first guy shouted first and the second guy can only be second. This means this riddle is in fact flawed or as you said some assumptions were not explained clearly in the video.
It wasn't obvious to me that you could "create" dozens of new face-up cards by flipping the deck, or that there was a maximum of 2 piles allowed. My solution was 26 2-card piles (a pile being 2 or more cards). I am giving myself credit for the win 🙂.
I figured out that they didn't have to be evenly split and that you could flip the cards but still couldn't figure out how many cards I needed to split and flip in order to win... Trust me you would have still lost with this information 🤣 I solved the other 4 riddles with relative ease btw...
On the last one, I cut the deck in half making 2 piles of 52 half-cards each, both with the same number of face up and face down cards. Please wear eye protection and thick gloves if you copy this method.
@@sfiarif5622 It has to do with Exponential Growth. So basically if you start with 1 patch of grass that doubles each day, in 10 days it will have the equivalent of 1024 patches of grass (1*2*2*2*2*2*2*2*2*2*2). If you instead start with 2 patches of grass, that each double each day, and your asking how many days will it take for you have 1024 patches it would be Day Zero (day you bought the grass):2 Day 1: 4 Day 2: 8 Day 3: 16 Day 4: 32 Day 5: 64 Day 6: 128 Day 7: 256 Day 8: 512 Day 9: 1024 Therefore it would take 9 days for the 2 patches to fill the same amount of space. (And it would take 8 days for 4 patches to cover it, and 7 days for 8 patches to cover it, and so on and so forth)
I don't think it's a great riddle, you need to specify that you're allowed to flip any cards. My solution was to just make 4 piles with 3 existing out of 1 card only which is - without specifying what is forbidden and what not - totally correct. Or as someone said, cut the piles in half, totally correct answer. If you are allowed to flip, you are allowed to cut or anything else, either.
how the heck can you fail the first and get the forth? fives a hard one though, also got to give you credit for getting the forth that was a bit of a meta one
I think the first is harder than the second. Those numbers just hella confused me. I got the fourth one because I’ve heard similar riddles before. All about “what information does silence give to this person?”.
In the riddle 4....if man2 and man3 are given same colour then man 1 will be the one who shouts first....so, we can't finalize that man 2 will shout out first...coz man1 and 2 have same probability
I got all 5. The first 4 puzzles were not difficult. For the fifth one, when you divide the deck into two stacks, if there are A face-up cards in the first stack then there are (13-A) face-up cards in the second stack, so if the stack has a number of cards equal to C, there are C-(13-A) face-down cards. If C=13 then the face down cards are 13-13+A = A, so if you flip the stack of 13 cards over you have two stacks with the same number of face-up cards. Thank you for the awesome brain teasers!
I am still lost. If there are more face up cards in the top half of the deck, how can it give you the same answer as if there are more face up cards in the _bottom_ half of the deck?
@@TheNoiseySpectator if by "top half" you mean the first 13 cards, the amount of face up cards in that stack has to be 13-A, where A is the amount of face up cards in the "bottom stack" of 39 cards. It doesn't matter which one is greater, A or 13-A, the amount of face down cards in the "top stack" has to be 13-A, equal to the amount of face up cards in the "bottom stack".
the answer to the first riddle is wrong if it takes 5min to make 5 t-shirts then the speed of making the t-shirts of these machines is 5t-shirt/5min=1t-shirt/1min so it 1t-shirt per min so it would take 1 min for 100 machines to make 100 t-shirts the five-minute answer would make sense if the machine would only produce in a bash of 5 and it would give us the t-shirt only after 5 min but it was never implied anywhere so it's safe to assume that it could produce 1 t-shirt at a time. and if we calculate the number that 100 machines produce in 5 min it would be 100*5=500 which is 5 times bigger
I got them all. The last one took me a while because I was stuck thinking about doing something with 4 equal piles of 13 😬. 4 x 13 equaling 52 feels so unintuitive, so I thought the trick was in there for at least 5 minutes 🤔
I missed the last one. I assumed the two decks had to be the same size and thought I needed to flip some cards around (to make even face-up cards), but never got to a solution, obviously.
Logical Criticism you don’t need to separate anything, aside from 13 random cards. They didn’t explain it Super clearly, and it works best if you realize this simple fact: No matter what, you won’t know how MANY face up cards are in EITHER pile. But by taking out 13 randos and flipping them, you can now say “These two piled have the same number of face up cards. I have no idea how many, I ONLY KNOW THAT THEY’RE EQUAL Imagine the possibilities: if all face up cards were in your pile of 13, and then you flipped it over, then BOTH DECKS WOULD HAVE 0 faceup cards What if there were 1 in your pile, and 12 in the other? Well, after you flipped your pile over, that 1 is upside down, and the other 12 are now face up. Thus BOTH STACKS HAVE 12 fave up cards This works out for every possible scenario. Try it out
On the last one, I misunderstood the assignment. I thought the goal was to create two packs of 26 cards each AND the same number of face-up cards (which has no solution). However, while looking for that non-existent solution, I did flip 13 cards (but it only worked sometimes, of course).🙂
Yeah that last one totally hosed me. I thought it was a trick question since you cant have 2 even stacks of 13. He didn’t say you couldn’t flip cards tho, so in the end it WAS a trick question, just a way more complex one! Thats a good one tho.. I’d be pretty impressed if someone came up with that answer on their own. Never having heard anything like it..
Sad, but the author did not properly explain puzzle 4. In order for the answer to be definitely Man 2, another fact has to be added, like "at first no one says anything. Than, one man shouts out the answer, and he is right. Which man was it" Otherwise, either Man 1 or Man 2 could be the right answer, depending on the hat distribution, as author actually explains in his answer!
Except the answer is not Man 2 but no man or only Man 1 if he is smart , he call out the correct answer after waiting a while even if he has the answer worked out by seeing two white hats or two black hats. The will force Man to call out the wrong answer. Then Man 1 calls out the answer if he has it. BTW The 5th puzzle is very good.
TheEvilJade the answer is man 2, because they’re lined up as in the picture with alternating colours. Since man 1 can only see 2 hats of different colours, he is not certain what colour hat he is wearing as he cannot see man 4. Since these are all perfectly intelligent prisoners, and thus perfect logicians, man 2 realizes that because man 1 did not immediately call out a colour, he must have seen a white and black hat, and since man 2 can see a white hat in front of him, he instantly knows his hat is black. If you read the instructions, it says “as in the picture” which includes the had orders.
For the wolf one I just said: Just take the wolf first and then start taking the sheep. The riddle never said the sheep were supposed to be safe after crossing the river.
Why did you not mention that the criminals were aware of their position in relation to the other criminals. I didn't know that man 2 was aware that there was another criminal behind him. I only knew that he saw one man in front of him, but I didn't know that he knew the location of the other criminals. The riddle was impossible to solve without this crucial information.
Agreed and also he should have also mentioned that the hats were distributed to the criminals as in the picture and not randomly. At least for me there was too little time to read it in the end and then we needed also remember the positioning when we read that. I started first thinking that the hats were distributed randomly and was wondering that naturally it depends what color hats prisoner number 2 and 3 has because if they have same color, then prisoner 1 would know and shout first, but if they have different color, then prisoner 2 knows his hat color as in the riddle.
Fun puzzles! I have a different solution for #5: create 15 piles (it doesn’t matter how many in each). At least two of them will have no face-up cards. (You didn’t say we needed to know which two piles had the same number of cards!)
It was explained well. What did you think of doing that they didnt say you couldnt? Also cutting the cards or something like that is obviously not allowed
Ellie H. It's probably because you didn't think properly about the first one (since it was labeled as easy) after that you started thinking more carefully and got the other ones right
yea I don't know how my brother did that Edit: probably helps that I play strategy games like supcom, civ and other 4X strategy games 2nd edit: actually all of the first three are things you calculate when playing strategy games
Some research found that people use different part of the brain to process easy and difficult question. (something like that) one question being asks differently will make people's answer different. Maybe when you see 5 5 5, 100 100 __, then you want to put 100 in the blank.
First 4 puzzles were a breeze. Last one, I was considering unequal decks & Flipping, but I was trying with 25 & 27 & then trying out possibilities to make sure. To help in solving, I'll summarise - Flipping is obvious since it's not said there has to be same original cards + we're cutting in 2 decks, so the sum within these 2 decks must be an even no. Also - Odd + Odd = Even O + E or E + O = Odd E + E = E Thus we need an E + E or O + O possibility to split 2 decks equally. You can see that IFF Odd(1) = Odd (2) & Even (1) = Even (2), THEN sum will be even & they'll be split into 2 equal halves. Also even if one of the decks has an odd no of cards face-up, flipping means it becomes an even no (eg - 7 flipped becomes 13-7 = 6, if we took a deck of 13 & flipped it over, other deck has 13-7 = 6, similarly - 4 flipped becomes 9, other deck has 13-4 = 9. Nevertheless, the solution is easy, just need to know the trick. It helps to get the Basics right & use logic. Yes, 13 isn't divisible by 2 & no, we don't end up with 6.5 in each.
Hadn't I for some reason assumed in the last riddle that both of the piles should be the same size, I'd have gotten them all without pausing the video.
Please explain me riddle 4. For me the answer is : it depends on the ways hats are distributed. If 2 and 3 have same color : 1 shouts, then all the other shouts at the same time (assuming they have same IQ). If 2 and 3 have different colors : 2 shouts first, then all the other
The Hat Riddle, I solved it differently, Man 1 speaks first yells white. Man 1 sees 2 different hat colors, He knows he either has white or black, If he yells black, he knows man 2 sees white hat in front which leaves him with 2 possibilities and gets them nowhere. If man 1 yells white, he knows man 2 sees man 3 with a white hat thus he can state he is wearing a black hat
The thing is, the riddle specifically asks who will DEFINITELY shout their hat colour first. Number 1 has a 50% chance of shouting his hat colour depending on the colours of the hats in front of him, whereas Number 2 is guaranteed to know his hat colour regardless of the circumstances. That's why the answer is Number 2
Thank you for not having a thumbnail like: *99%* of people *CANT DO THESE*
Yodernote Boy ;
Right said...
Yeah and these were actually hard unlike those other clibaiting ones designed to make u feel better about yourslef
i didnt do only last one, all other were easy
Facts
I got all of them right after the video finished.
Same
In all honestly I thought the hat one was a trick, and that prisoner 3 could see his reflection in the wall lol
The last one I also thought was a trick, but I wasn't paying close attention. I figured if you cut the deck in half (like cutting a card in half) that would be the easiest solution lol.
maybe next time I see a video with riddles, I'll pause it and think about it lol
The last one I failed. I believed that there is something fishy. But the actual answer is really elegant and satisfiing.
Петър Петров Haha yeah that one was interesting. I had misunderstood and thought each pile had to be 26 cards which locked me into that thinking
There is a problem with the last riddle. It assumes the 13 gave up cards in the deck are distributed RANDOMLY. that's what the riddle says.
But the answer implies that the 13 face up cards are distributed EVENLY.
Huge error in the last riddle
@@roberjohnsmith no, no it does not. Every possible outcome would work. If the small pile has zero flipping them all would make that 13 just like the big pile, same goes for every other outcome
Robert Percival No it's not stupid.
yeah, I half expected "you make two piles of zero cards (and leave the deck as it is)" to be the answer.
Got 4 of 5 - I ran out of patience (a virtue in most of life's problem-solvings) on the last one. Really liked the solution and will remember the logic from now on. Well, I'm 75-years-old and no longer one of the sharper knives in the kitchen. Ironically, now I can't remember where I left the sharpener.
Haha you seem sharp to me!
hostarepairman I’m 14 and I only got 2 lol
0 Then it's good.
If age is an indicator for how many riddles you could solve. He needs to be 94 years old to answer all 5. You on the other hand would be able to answer all 5 at the age of 35.
But of course in reality that's not how it works. But keep trying!
Are you a pencil🤔
I got them all however it wasnt mainly from my own logic but from watching many other riddles and using the logic from those to be able to do these
Wolf and sheep is literally the easiest one
Yes...1st 3 are quite easy...i got the 4th's answer but not with quite a reasonable explanation
The first one was obviously the easiest, but the wolf one wasnt hard at all. I got stuck on the prisoners one, though.
@@koolwond4264 if you first take the sheep to the other side then the wolf will be alone with the other sheep and eat it, and thats what we're trying to avoid, no?
@@alicia-hd2cs that's true
@@koolwond4264 My IQ dropped after reading your comment...
I thought the piles of cards had to contain the same number of cards :/
I thought I could cut the pack in half
Too
I didn;t think that, I knew that was the trick but it didn;t cross my mind to flip those freacking cards :)))
There’s a way to do that as well @Crabbi5
I kiiiiinda thought of something similar (something that wasn't thr target), but I would still probably fail this one
The one with the prisoners is crazy. It's obviously man 4: he just shouts whatever he wants and legs it; he's already over the damn wall!
Agree 100% they are playing a game in order to gain their freedom. 3:09
So what's the use to play this game if the guy is already free?
Love that
Lmao this is gold
They are in different rooms of buildings. But nice joke still.
Then man4 would lose his chance to be free. He must be the last as he has no comparison to think what hat he wore. 1,3 will know once they discover the pattern of what man 2 is thinking.
I can’t do the last one because I don’t actually have a friend with only one arm.
BIG BRAIN
I thought the last question was a trick question since 13 cant be divided by 2.
@SHREYA RAMAN I just thought about cutting al the card in to 2 and then you have two decks with the same amount of face up cards
well, 6.5
The objective is to make the same number of cards face up, it it can be 1 and 1 or 26 and 26, it doesnt have to use the original 13 face up cards and split them equally.
That's what is was thinking off and the answer was hussh
Same here... Got stumped by the last one the others i could solve
I only solved the wolf and sheep one as it is quite practical. The last one I didn't even understand after the explanation 😂
All face-up=13; Takeout card=13; takeout face-up=?; leftover face-up=13-? ;flipped takeout then face-up=13-?; whatever ? it is, we are sure 13-?=13-?
With the last one because they picked 13 (an odd number) I realised I'd have to double it somehow to make an equal pile. Took me ages.
First take out your agent Smith, because you don't understand after the explanation.
Same with me
I got 4 out of 5. The only one I didn’t get was the wolf and sheep one lol. Felt really dumb after the explanation so had to make up for it with 4 and 5
4:55 why doesn’t the guy in blue have an arm?
I noticed that too. I think it was a way to employ amputee cartoons.
More weird he puts cards with 13 flipped against.
Thats the hardest riddle
Technically, he does.
I've passed my colorblind tests but that guy definitely wears green, not blue
The prisoners hat problem helped me to get a job
Thanks
How
@@randomperson-cq6lb in the interview the interviewer asked me a similar question and I gave the correct answer quickly 😁
Hence it was one of the key moments of that interview
The hat riddle is based on assumption that prisoner 1 is silent. Suppose he sees that Prisoner 2 and 3 are wearing the same coloured hat, then he would be the first person to shout. Prisoner 2 is able to interpret only on the silence of Prisoner 1 and not otherwise
Yeah it’s a bitch getting into mi6
@@random.videos007 ahhaa good JOB
did u get it
THe biggest riddle here is what happened to that guy's arm
He was handicapped
The wolf ate in the last boat ride. That guy was also used in the last riddle to show that he couldn't rip the cards in half.
Lol omg.
Ahahahhahahahahha
Perhaps "supergrass" has contributed to my inability to solve any of these riddles lol
I was on some supergrass just now and did quite well until the last one.
Facts
I am on Super Skywalker Kush and I did all of the riddles. But I'm good at maths, so.. EZ
I'm smart af but I failed all of the riddles
i only got that one after the explanation AND a good walk...
I already failed the first one 😂
haha yep, it seems simple but is actually quite tricky ;)
ok riddle channel the answer for 2nd question is 5 days sense 2×5 is 10
Riddle Channel I failed all the questions. But I sloved the last one. ;D
XCEL Music yeah that's what I tough
Vanessa Mellark same
Loved all of these!
The only thing that bothered me was the question wording of riddle 4. I don't believe it's fair to ask the question "who will be the FIRST....person to definitely know his hat color and shout it out?"
The answer to that question can be either person1 and person2.
- If the middle two men have same colored hats (blk or white), then person1 will be the FIRST PERSON to definitely know his hat color and yell it out
- if the middle two men have different colored hats, then person2 will be the FIRST PERSON to definitely know his hat color and yell it out. He'll know it via first person's silence.
The actual hat colors were shown on screen. Man 2 and 3 did not have the same color.
@@artugert I have to agree with Dy here, I also have a problem with the way the riddle is set up in this video. We are told that the prisoners' places are as shown in the picture, they do not specify that the hats also match the picture. The hats are introduced afterwards and the wording makes the hat positions seem arbitrary to the problem. In my mind when I heard it, it seemed like the hats could have been arranged any way. Still a good riddle, I would have just added the specification that the hats are placed as shown in the picture.
@@philipchervenkov7933 I agree. I googled the riddle and found it asked in several other publications, and in those problems the hats were said to be assigned randomly (we, the riddle solvers don't know who has what hat either).
This video's version of the riddle was worded poorly, as the answer is not "2"... it's "1 or 2" depending on who has what hats.
Absolutely agree with Dy
The same is the case with me. Absolutely agree with @dy
4:57 where is his arm?
He lost it in the war
LOL
I don't even know
shanks
He lost his arm due to someone hating him so much for his riddles he forced the guy to go to war, and he lost his arm due to a sea king while saving Monkey D. Luffy.
For the last one, i was thinking of tearing the cards in two. Then create two piles from the halves of each other. There will surely be two piles of cards facing up
I thought that was the answer also
Ditto, I didn't think of the correct answer because I made the mistake of assuming either two equal piles - instead I allowed the definition of card to not require them to be whole undamaged playing cards - hence tear them in half! There could be a follow up riddle where you don't know how many are face down to start with and then this would be the answer.
@@manticore5733 Agreed. Sometimes in life, destroying things is the solution. Lol
Sometimes it's easier to reduce the #5 riddle to smaller numbers. Try to solve it with 4 cards and just 1 flipped over. Then start thinking but only assume what is feasible (the two decks can be of different size and you can flip cards even being blind). It's easy to see that putting aside one of the four cards and flipping it you will get the answer. Most common mistakes are to assume what you shouldn't (two decks of same size and not being able to flip cards).
You are right, it is way easier if you break it down to something simple like 4 cards. And I made the stupid mistake assuming the piles had to be the same size...
exactly the same
52 cards: If you decide to consider card flips (I didn't), then start with 1 card on the first pile and flip it. Then you end up with either 13/1 or 12/0.
With a 2 card pile that is flipped, you get either 13/2, 12/1, or 11/0.
With a 3 card pile: 13/3, 12/2, 11/1, 10/0.
with a 4 card pile: 13/4, 12/3, 11/2, 10/1, or 9/0.
The pattern (eg, the difference in number of face up cards between the two piles = 13 minus the number of cards in the small pile) quickly suggests to try..
With a 13 card pile: 13/13, 12/12, 11/11, ... or 0/0.
Voila, 13 card pile flipped leads to the same number of up cards as in the other 39 card pile in all scenarios.
You can't just assume anything.
I "solved" the first two, but only because I've seen them before. I think being able to at least understand the explanation is satisfactory. They're interesting and change up your method of thinking.
The last riddle was actually great
Jeankrikri i don’t get it
Ask any card mechanic!
Febha Mathew if you take the first 13 cards and all of them are face down, that means the second pile have all 13 face up cards. Then by flipping the first 13 card pile over, you will always have the same amount of face up cards. Max having 2 piles with 13 cards each pile face up.
I wanted to cut all cards in half to solve the puzzle, but love the real solution.
Ikr
I didn't solve the last one because I didn't give myself enough time to think, I actually knew it would involve flipping one of the piles because I know a magic trick that works on the exact same principle, I just failed to realize both piles didn't need to be the same size.
I have seen and solved variations of all the other problems before, so they were all pretty easy to me. I never got why people struggle so much with the prisoners and hats one. Just put yourself in the minds of each prisoner, and the solution should be very obvious.
Exactly the same for me. I immediately remembered the cards puzzle and knew that you had to flip part of the cards over, but it only worked with 13 and 39 cards, not two stacks of 26 cards…
I’m not sure how this relates to any of our comments, nor how this would make any difference.
5:17 he clearly showed 2 piles with same size... so...
It's obvious that you have to flip some cards, don't make excuses. But yes, most of the people here aren't that smart.
EAVDR lol dumbass
Don't consume Peter's super grass before trying to solve the riddles!
Clemens Kindermann it's weed... i knew it
Riddle #5: For those of you experiencing difficulty with this particular riddle, here's a hint:
Don't focus on finding the 13 actual cards that are face-up in the deck.
I solved this tough puzzle took me more than 2 hours and ended up using math instead. In my opinion, the premises provided felt incomplete, should at least add one that could serve as a hint.
Yea, I used maths too...
The next hint is that you're allowed to flip cards (and you have to since it's an odd number of cards facing up).
SPOILER
1. Insight: You can´t divide 13 face-up cards into two.
2. Insight: Flipping some cards is the only way to change the number of face-up cards in total.
3. Insight: If you flip a stack with uneven number of cards in total, it will change the number of face-up cards within from even to uneven or vice versa.
4. Conclusion: By flipping an uneven amount of cards, the overall number of face-up cards will be even.
5. Insight: As you don´t have any more information as stated initially, the turned cards must be one of the two stacks afterwards, any mixing up with the remaining leads to an unsolvable situation with no chance to divide the face-ups in two halves by any counting or blind sorting of the cards. This one is crucial, but also obvious.
6. Insight: For sure it doesn´t matter where from within the original stack you take the cards to turn. You may take the top ones, or pick them one by one out of the deck, doesn´t matter.
7. Conclusion: It has to be the number of cards to flip and build the 2nd stack that matters only.
8. Heuristic: Flipping all or half of the deck doesn´t work (don´t even think about it, you have to flip an uneven number, see 3.). The only other given count is "13". And now start thinking, the steps before were just "technical thoughts".
9. Solving: Turning a stack of 13 cards with x face-up cards will lead to a stack witch 13-x face-up cards. Because this stack had x face-ups before turning, the remaining stack of 39 has to contain 13-x face-ups, too.
Last riddle: Rip all 52 cards in half to create 2 equal piles.
It will need two arms, and he only have one.
Or four piles of 13, flipping piles one and three, then adding together piles one and three, and two and four. Boom
@@JaydenLawson no, that wouldn't work. imagine the piles one has 5 of the 13 flipped cards, the pile three has the other 8 of the 13 and the piles two and four don't have any flipped cards. you flip one and three, so one has 8 and three has 5. you put them together and they have 13 flipped cards. two and four don't have any flipped cards, so your theory is wrong
@@baaron_ that would be two solutions in one
I lost a brain cell for thinking too much
But gained a 1000 more! :)
You don't loose brain cells due to thinking but due to lack of it.
@Milad Zavar why are you triggered lol
So how can you live now without brain???
I knew I smelled something burning
I figured out the fourth logic puzzle on my own after a few seconds and I felt so boss. Thank you brain for making me feel good for the day.
@ANIMAL!!! I mean the video pretty much explains it out the entire answer for you. In short Prisoner 2 can infer that his hat is the opposite of Prisoner 3 because the Prisoner 1 behind him is NOT saying anything.
Prisoner 1 doesn't know what he's wearing only because 2 and 3 have different colors. If 2 and 3 DID have the same colors then Prisoner 1 could easily just say the opposite color, boom done. However because Prisoner 1 is NOT saying anything, Prisoner 2 can infer that they have opposite colors and can just state the opposite color of Prisoner 3
@ANIMAL!!! Lol, Cheers for figuring it out. Yes Prisoner 2 has to make an assumption that Prisoner 1 is not saying anything because he sees P2 P3 have different colors. However P1 could easily just be goofing around or what not, or dumb.
This exercise assumes all Prisoners have sound logical deduction ability for this puzzle to work.
@ANIMAL!!! he can see the 3 ones hat
@ANIMAL!!! yes he need to assume that prisioner 1 has a brain
tricky questions, good riddle
i dont understand about no 3
is no 3 doesn't make sense?
Why would you want to bring the wolf if it's going to eat your sheep ?
as long as the man is with the wolf the wolf wont eat the sheep
Xzor is asking WHY you would even want the wolf in the first place if you knew it wanted to eat your sheep. And honestly my best guess is that the wolf is worth a lot of money or it's meat tastes good.
he needs that wolf as a fuck buddy
so it doesn't eat his wife
because if you have a wolf then everyone knows you're tough.
...Yep. Still know nothing.
YouOnlyLiveTwice You'll always know nothing, Jon Snow.
Stop trying to pretend like you know that you know nothing, Jon Snow. You know nothing.
YouOnlyLiveTwice You are on the right track 😆
You know at least one thing, winter is coming...
who cut the deck in half with a knife? two stacks, with as an added bonus, 13 face up cards on each. You didn't specify the cards should be whole.
hahaha thats some real out of the box thinking :D :D
nukevapes i said the same thing
hi five!
I fucking thought the same thing bro! AHAHAHAHAHA
he also didnt say you could only make 2 decks
There is a problem with prisoners riddle in 3:59 it says “who is DEFINITELY THE FIRST ONE to find out”. Here is the problem: prisoner 2 is guaranteed to find the answer, but he may be the first one or the second one to do that. In fact the video itself explains the situation in which prisoner 1 is the FIRST one to solve the problem. Therefore, I think the problem should be changed to this: “who will definitely find the color of his hat in any situation (no matter being the first or second person to do that). Am I right?
I was thinking about that too , if 2nd and 3rd are wearing same hats , 1st will know his hat's color.
But I guess the question demanded the same pattern of distribution that was shown in the picture.
There are 6 ways in which hats can be distributed. But only the distribution shown was to be considered.
This is true. Also, the last riddle was flawed greatly. And there is no way to solve it the way it was asked.
In the last riddle, it was implied in the question that 13 cards were face up random throughout the deck. But the answer implied they were distributed EVENLY throughout the deck. If it was truly random then I could take off 13 of the first cards of the deck and get no face up cards.
Amir Mansoor Kamali Sarvestani no, you’re not. The riddle states that the 4 criminals are lined up as in the picture. White/black/white/black. That means the answer is always number 2.
Yes, they explain the other situation only to clarify the answer but I guess people are just overall confused when you try to clarify things.
Hi C1. In the video it never says that the order of the colors is fixed. I guessed they meant that the colors are only and only in white, black, white and black order. But it never ever explicitly says that the order should be as the graphics shows. I watched it again and again. He lists a number of rules but never says that the order is fixed to white, black, white, black. As a result, every one can assume that this riddle is about finding the answer with any possible order of colors.
@@roberjohnsmith So if you flip the cards of the second pile with the 13 cards you would have 13 cards facing up.
Riddle 4: After hearing #2 shout black, #3 should know that his hat color is white.
The funniest thing is #1 and #4 would never know the color of their hats
@@TheOne-jm6tg number one is the first to know his own color if number 2 and 3 have the same color.
@@TheOne-jm6tg unless 2 and 3 had the same colour!
The last one kind of needs the caveat that "the piles don't have to be equal amounts of cards" I know it wasn't stated, but that's why I was getting stumped :P
I looked back at it and honestly I think his diagram is what made me get that mental block. He shows two perfectly distributed piles in the animated explanation.
There’s a way to do this riddle with two even piles
Created your own mental block with phantom rules. I did the same.
Well I knew it wasn't necesary but I still didn't get the answer
That omission was deliberate.
I was able to solve all of them except for the last one, that really got me. Great riddle
Lol i wasnt able to solve any of the first but I got the last one, probably just lucky
Milad Zavar The second riddle is better explained visually rather than with words, but lets see. (Also the video said "No maths needed, only logic" but im kinda stupid and cant explain it with words so ill try to explain it mathematically)
For the sake of simplicity, lets say the area of one super grass patch is 2. It can now be represented by 2^d = A, where "d" is the number of days which had passed and "A" is the area in meters. [ e.g if 3 days had passed d = 3. Input it into the formula to get 8 meters squared (2^3 is 2x2x2 = 8) ]
Since the garden took 10 days to be completely covered, the area of the garden is 2^10, or 1024 meters squared. If there were 2 patches, it would take 9 days because they would both grow up an area of 2^9, or 512 meters squared. Since there is 2 patches, it would be 512 + 512, which equals to 1024, equal to the total area of the garden, therefore taking only 1 day less than one patch would.
@@simoncordero384 your answer is very good but just to be precise:
The size of one patch is not 2 but 1. As at the first day when he buys the super grass no day has passed the size is 2^0 = 1 and on the second day (one day passed) the size is 2^1 = 2 which is the starting point when buying two super grass
but as i said - the explanation was very good :)
Spanching Ah, for the first day I just made it 2^1, second day 2^2 and so on so its a bit simpler :3
@@simoncordero384 for the solution it does not matter, just thought that on day one no day has passed :)
Finally a video that has actually HARD riddles
I got 4/5.
I'm happy. :)
dont lie to yourself.
alexander martinez ikr
Elsanne J it's possible don't be jealous because they were able to something you weren't.
right so don't lie to yourself Elsanne J
alexander martinez I replied to the wrong person. I was talking about you two above.
Split the deck into 2 random piles. Place them both sideways (lean them against each other or prop them up against something). Now you have 2 piles with the same amount of face up cards = 0.
A real life genius > math genius 😎
Your video was astonishing
I am subscribing !!!
Got ‘em all. I guess the time I spend watching riddle videos on UA-cam isn’t “wasted” after all 😂
I have watched over 100 riddles on yt and it helps me think out of the box
If the first 13 cards are face-up cards, then the result ends up with no any face-up cards, but the answer is still logically correct. Thanks for the interesting riddles.
I think in the fourth puzzle, the correct answer cannot be given because prisoners one and two have the same chance of winning and shouting first. If the second and third prisoners had two white or two black hats, then the first prisoner would be sure what his hat is. If, on the other hand, the second and third prisoners had black and white or white and black hats, then the second prisoner would be sure what hat he has, according to the solution in the film. It depends on the fate that the prisoner wins, so the correct prisoner number cannot be given.
It is possible that when solving this task, you only need to consider the example of setting the hats that was in the video. However, it is not explicitly stated that this is the case. It is known that the example can be used to imagine the place of the event and an example of an arrangement of hats, which may also take a different order. So it sticks to my opinion.
I came to the exact same conclusion. They can't shout at the same time... The second guy has to wait for the first one's reaction... Either it's silence (the second guy will shout first) or the first guy shouted first and the second guy can only be second. This means this riddle is in fact flawed or as you said some assumptions were not explained clearly in the video.
Also, “silence” , is not an objective metric.
I think the riddle was about the situation on the image, with 2 and 3 wearing different hats
It's not a randomized trial - it's the situation as pictured, so #1 can't make a conclusion
They could have made that clearer
It wasn't obvious to me that you could "create" dozens of new face-up cards by flipping the deck, or that there was a maximum of 2 piles allowed. My solution was 26 2-card piles (a pile being 2 or more cards). I am giving myself credit for the win 🙂.
I figured out that they didn't have to be evenly split and that you could flip the cards but still couldn't figure out how many cards I needed to split and flip in order to win... Trust me you would have still lost with this information 🤣
I solved the other 4 riddles with relative ease btw...
It says create 2 piles, though
The last one ist great. Almost like a computer does subtraction by bit flipping.
wow I have once not been dissapointed by youtube. thanks for the last 2 really great riddles
The way he said SUPERGRASS was so funny 😂😂
Failed all of them lol
On the last one, I cut the deck in half making 2 piles of 52 half-cards each, both with the same number of face up and face down cards. Please wear eye protection and thick gloves if you copy this method.
on the super grass how the hell and how come the grass cover a half of garden on 9 days i dont get it help me..
@@sfiarif5622 It has to do with Exponential Growth. So basically if you start with 1 patch of grass that doubles each day, in 10 days it will have the equivalent of 1024 patches of grass (1*2*2*2*2*2*2*2*2*2*2). If you instead start with 2 patches of grass, that each double each day, and your asking how many days will it take for you have 1024 patches it would be
Day Zero (day you bought the grass):2
Day 1: 4
Day 2: 8
Day 3: 16
Day 4: 32
Day 5: 64
Day 6: 128
Day 7: 256
Day 8: 512
Day 9: 1024
Therefore it would take 9 days for the 2 patches to fill the same amount of space. (And it would take 8 days for 4 patches to cover it, and 7 days for 8 patches to cover it, and so on and so forth)
@@chrisk1039 still dont get it. how the hell 2 patches cover the whole garden in 1 day. sry im so dumb
@@sfiarif5622 1 day LESS. So 9 days :D
@@sfiarif5622 It's not in 1 day, it's in 9 days. 1 Day less than the single patch would do it
But, steel is heavier than feathers.
they're both a kilogram
@@crystallogician get the pun bruh
@@aryankhullar7101 he might be referring to a video
What's going on
@@demise.8748 the Irish guy
That card one was extremely difficult compared to the others, but great riddles
I don't think it's a great riddle, you need to specify that you're allowed to flip any cards. My solution was to just make 4 piles with 3 existing out of 1 card only which is - without specifying what is forbidden and what not - totally correct.
Or as someone said, cut the piles in half, totally correct answer. If you are allowed to flip, you are allowed to cut or anything else, either.
"You are blindfolded and can't see anything"
"Also, you only have one arm"
With no head
I got 1,2,3 & 4
1,2 and 3. Those 2 last riddles were difficult.
Lolykiller90 Mozombite I got none right
André to be honest I've seen the one with the wolf before so kind of cheated on that one
same
André same here
i solved all, exept the last one :/
Thanks for wasting your time typing this.
dodo
You are welcome!👍
Lmao🤣🤣🤣🤣🤣
@@crinklyten2152 r/iamverysmart
What the fuck is happening in this comment section lmao
I failed number 1 and 2
I got correct number 3 and 4 I didn't even bother with number 5
holy shit same thing lmao even 5
Same. lol
same
how the heck can you fail the first and get the forth?
fives a hard one though, also got to give you credit for getting the forth that was a bit of a meta one
I think the first is harder than the second. Those numbers just hella confused me. I got the fourth one because I’ve heard similar riddles before. All about “what information does silence give to this person?”.
Wow, what an riddle man, I need more to solve.
In the riddle 4....if man2 and man3 are given same colour then man 1 will be the one who shouts first....so, we can't finalize that man 2 will shout out first...coz man1 and 2 have same probability
Yes, but they aren't look at the picture
@@reetajaiswara8410 That picture is a case given at random dude.....
@@saiavinash1200 na, it's given for our reference cuz I've seen it in the final pages of a notebook (those amazing fact pages).
Chill bro....we both have different perspectives.lets stop this discussion here😂😂😂.
By the way nice meeting you..
@@saiavinash1200 👏👍
I love how he kept showing the pause button. If you assume we know these hard ass riddles then we probably know what a pause button looks like.
When did he assume that?
He shows a big-ass pause button that block the riddle hints
Gio nathaniel once u pause... press the screen one more time to remove pause button from screen
Did get the last one. Thanks for sharing, it's really a great one.
I got all 5. The first 4 puzzles were not difficult. For the fifth one, when you divide the deck into two stacks, if there are A face-up cards in the first stack then there are (13-A) face-up cards in the second stack, so if the stack has a number of cards equal to C, there are C-(13-A) face-down cards. If C=13 then the face down cards are 13-13+A = A, so if you flip the stack of 13 cards over you have two stacks with the same number of face-up cards. Thank you for the awesome brain teasers!
I am still lost.
If there are more face up cards in the top half of the deck, how can it give you the same answer as if there are more face up cards in the _bottom_ half of the deck?
@@TheNoiseySpectator if by "top half" you mean the first 13 cards, the amount of face up cards in that stack has to be 13-A, where A is the amount of face up cards in the "bottom stack" of 39 cards. It doesn't matter which one is greater, A or 13-A, the amount of face down cards in the "top stack" has to be 13-A, equal to the amount of face up cards in the "bottom stack".
İ just failed the first one xd
How did you fail the first one? Did you get tricked? And if so what did you think was the answer
@@alex2005z I just didnt think at all.
@@ACE-qb1ux that's not failing, that's just not doing it
People who watch TedEd: that's not even a challenge
I failed the first one!!
How could u failed if he literally said 500 saw machine 500 shirt 5 mins
the answer to the first riddle is wrong
if it takes 5min to make 5 t-shirts then the speed of making the t-shirts of these machines is 5t-shirt/5min=1t-shirt/1min
so it 1t-shirt per min
so it would take 1 min for 100 machines to make 100 t-shirts
the five-minute answer would make sense if the machine would only produce in a bash of 5 and it would give us the t-shirt only after 5 min but it was never implied anywhere so it's safe to assume that it could produce 1 t-shirt at a time.
and if we calculate the number that 100 machines produce in 5 min it would be 100*5=500 which is 5 times bigger
Something caught my ear.
How could the prisoners shout if they were told not to speak?
I assume that they are ONLY allowed to scream a color, and they are not allowed to discuss.
SPEAK TO EACH OTHER
@@beyblader11111 and they can only say the color if they're sure about it, they can't guess
I got them all. The last one took me a while because I was stuck thinking about doing something with 4 equal piles of 13 😬. 4 x 13 equaling 52 feels so unintuitive, so I thought the trick was in there for at least 5 minutes 🤔
5/5!
I missed the last one. I assumed the two decks had to be the same size and thought I needed to flip some cards around (to make even face-up cards), but never got to a solution, obviously.
Look at the comment section
Full of geniuses bravo fellas!!!
Yeah, yet no one asked how in the hell you separate face-up cards from the deck blindfolded.
Logical Criticism you don’t need to separate anything, aside from 13 random cards. They didn’t explain it Super clearly, and it works best if you realize this simple fact:
No matter what, you won’t know how MANY face up cards are in EITHER pile. But by taking out 13 randos and flipping them, you can now say “These two piled have the same number of face up cards. I have no idea how many, I ONLY KNOW THAT THEY’RE EQUAL
Imagine the possibilities: if all face up cards were in your pile of 13, and then you flipped it over, then BOTH DECKS WOULD HAVE 0 faceup cards
What if there were 1 in your pile, and 12 in the other? Well, after you flipped your pile over, that 1 is upside down, and the other 12 are now face up. Thus BOTH STACKS HAVE 12 fave up cards
This works out for every possible scenario. Try it out
All of these were difficult lol. I didn’t get any
Yeah me too . Feel like a dumbass now.
I only got the first one xd
i got 2 the first and third
4 out of 5. Last one was a toughie... Fun tho!
On the last one, I misunderstood the assignment. I thought the goal was to create two packs of 26 cards each AND the same number of face-up cards (which has no solution). However, while looking for that non-existent solution, I did flip 13 cards (but it only worked sometimes, of course).🙂
Same here
I can't believe I solved the harder one but failed at the easy, medium and very hard!
Yeah that last one totally hosed me. I thought it was a trick question since you cant have 2 even stacks of 13. He didn’t say you couldn’t flip cards tho, so in the end it WAS a trick question, just a way more complex one! Thats a good one tho.. I’d be pretty impressed if someone came up with that answer on their own. Never having heard anything like it..
Last one got me
I thought the two piles must be the same cards number 😂😂
Thanks, Now I KNOW that im dumb...
Ah don't worry I'm dumb too :)
Surely I'm am too I didn't get not one lol ;)
With you buddy
Sad, but the author did not properly explain puzzle 4. In order for the answer to be definitely Man 2, another fact has to be added, like "at first no one says anything. Than, one man shouts out the answer, and he is right. Which man was it"
Otherwise, either Man 1 or Man 2 could be the right answer, depending on the hat distribution, as author actually explains in his answer!
I guess "lined up as in the picture" is too hard to understand lol
Except the answer is not Man 2 but no man or only Man 1 if he is smart , he call out the correct answer after waiting a while even if he has the answer worked out by seeing two white hats or two black hats. The will force Man to call out the wrong answer. Then Man 1 calls out the answer if he has it. BTW The 5th puzzle is very good.
TheEvilJade the answer is man 2, because they’re lined up as in the picture with alternating colours. Since man 1 can only see 2 hats of different colours, he is not certain what colour hat he is wearing as he cannot see man 4. Since these are all perfectly intelligent prisoners, and thus perfect logicians, man 2 realizes that because man 1 did not immediately call out a colour, he must have seen a white and black hat, and since man 2 can see a white hat in front of him, he instantly knows his hat is black. If you read the instructions, it says “as in the picture” which includes the had orders.
@@ianthomas1560 Relooked and where exactly is "as in the picture". There 4 written instruction , in no case they define what man 1- 4 is wearing.
@@TheEvilJade look at the fucking picture
First 4 I got easily. Last one totally stumped - clever answer.
Last one was the easier to me.
For the wolf one I just said:
Just take the wolf first and then start taking the sheep. The riddle never said the sheep were supposed to be safe after crossing the river.
Jurgen Klopp... Is that you?
Why did you not mention that the criminals were aware of their position in relation to the other criminals. I didn't know that man 2 was aware that there was another criminal behind him. I only knew that he saw one man in front of him, but I didn't know that he knew the location of the other criminals. The riddle was impossible to solve without this crucial information.
Agreed
Agreed and also he should have also mentioned that the hats were distributed to the criminals as in the picture and not randomly. At least for me there was too little time to read it in the end and then we needed also remember the positioning when we read that. I started first thinking that the hats were distributed randomly and was wondering that naturally it depends what color hats prisoner number 2 and 3 has because if they have same color, then prisoner 1 would know and shout first, but if they have different color, then prisoner 2 knows his hat color as in the riddle.
I solved it...Nothing wrong with info stupids
Exactly... what if the second man was given a black hat also? It’s all 50/50
but the criminals have eyes
The third one is only difficult if you've never come across a riddle of this sort, but pretty much everyone has
I never have, but it was really self explanatory. I just realized that you will always have to keep the wolf with you. It was really obvious
last one is great, the rest is so easy... thanks for the entertainment!
I missed the first 2 but then got the last 3... what is life?!?
Maybe you've seen the last 3 before 😂😂😂
Same.
me too but i also got the 5th
I think 40% of your brain is missing
no you didn't
Fun puzzles! I have a different solution for #5: create 15 piles (it doesn’t matter how many in each). At least two of them will have no face-up cards. (You didn’t say we needed to know which two piles had the same number of cards!)
But he said you must use all the cards in those two piles, your solution would go against the rules..
I feel like I could’ve maybe got the last one if it was explained a little bit better
Riddles leave out some information to make them tricky
It was explained well. What did you think of doing that they didnt say you couldnt? Also cutting the cards or something like that is obviously not allowed
So good!
I got all except the last one. I'm happy with my result!🤓
4th one was tougher than 5th one...... 1,2,3,5
I got all of them right EXCEPT the FIRST one! How can i even do that?
Ellie H.
idk but me too
Ellie H. It's probably because you didn't think properly about the first one (since it was labeled as easy) after that you started thinking more carefully and got the other ones right
It was a warm-up round it doesn't count.
yea I don't know how my brother did that
Edit: probably helps that I play strategy games like supcom, civ and other 4X strategy games
2nd edit: actually all of the first three are things you calculate when playing strategy games
Some research found that people use different part of the brain to process easy and difficult question. (something like that)
one question being asks differently will make people's answer different.
Maybe when you see 5 5 5, 100 100 __, then you want to put 100 in the blank.
First 4 puzzles were a breeze.
Last one, I was considering unequal decks & Flipping, but I was trying with 25 & 27 & then trying out possibilities to make sure.
To help in solving, I'll summarise -
Flipping is obvious since it's not said there has to be same original cards + we're cutting in 2 decks, so the sum within these 2 decks must be an even no.
Also -
Odd + Odd = Even
O + E or E + O = Odd
E + E = E
Thus we need an E + E or O + O possibility to split 2 decks equally. You can see that IFF Odd(1) = Odd (2) & Even (1) = Even (2), THEN sum will be even & they'll be split into 2 equal halves.
Also even if one of the decks has an odd no of cards face-up, flipping means it becomes an even no (eg - 7 flipped becomes 13-7 = 6, if we took a deck of 13 & flipped it over, other deck has 13-7 = 6, similarly -
4 flipped becomes 9, other deck has 13-4 = 9.
Nevertheless, the solution is easy, just need to know the trick. It helps to get the Basics right & use logic.
Yes, 13 isn't divisible by 2 & no, we don't end up with 6.5 in each.
the mad scientist can do it all
This was difficult and fun!
These were awesome riddles...
I just couldn't solve 5th and 2nd...
😅😅
Same
the second cannot be 9 days, it's too stupid.
@Max Gulyaev yes i realized It too , 9 days Is correct
I solved all of them, but I almost gave up on the last one because thought the stacks had to have an equal number of cards.
Same here. Solved it once i realised they did not need to be the same pile size.
Hadn't I for some reason assumed in the last riddle that both of the piles should be the same size, I'd have gotten them all without pausing the video.
same here
Lol no one cares
Yeah, that was made unclear, I had as solution to create 4 stacks of 13
I missed 2 was trying to count like it was a math problem but got the rest except for 5 thought the same way lel
I had the same problem, but really, you would've gotten them all, without pausing! You had me until you modified your success.
Got all but the last one. Quite happy with the fact I figured out the hats riddle!
only got none XD
me 3
me 4
me 5
me 6
... I'm Mr. Me6!
geratrics s me 7
Please explain me riddle 4.
For me the answer is : it depends on the ways hats are distributed.
If 2 and 3 have same color : 1 shouts, then all the other shouts at the same time (assuming they have same IQ).
If 2 and 3 have different colors : 2 shouts first, then all the other
Martin Bayot You are 100% right. The solution you said here, is correct. It does depend on hat distribution and your logic is good! Cheers!
The Hat Riddle, I solved it differently, Man 1 speaks first yells white.
Man 1 sees 2 different hat colors, He knows he either has white or black, If he yells black, he knows man 2 sees white hat in front which leaves him with 2 possibilities and gets them nowhere. If man 1 yells white, he knows man 2 sees man 3 with a white hat thus he can state he is wearing a black hat
The thing is, the riddle specifically asks who will DEFINITELY shout their hat colour first. Number 1 has a 50% chance of shouting his hat colour depending on the colours of the hats in front of him, whereas Number 2 is guaranteed to know his hat colour regardless of the circumstances. That's why the answer is Number 2
@@MAJIN_MAGIC see above
Josue Garcia that’s not solving it.
The guy's missing arm is a big riddle to me
Awesome all the riddles blow up my mind and change my thought process. I have great time while solving them
Riddle 3: Cross with sheep in boat and wolf swimming alongside.
1 got the first 3 right and now I feel like should buy a cookie.