3 Riddles Popular on Logic | Can you solve it? 

The hardest part of solving these riddles is understanding the explanation of what you are trying to solve.

took 15 mins...but i solved the 2nd riddle!!!!! YESSSSS!! Finally one video where i solved something

3rd one is formulated really Bad. The prisoner who can look in all boxes is he one of the 100 prisoners? So only 99 should know the number of the Box with their name in it. because the prisoner who can look in all boxes has a 100% Chance

So this "Anna" wakes up in the morning and cannot find her socks. Wtf is there no light in her room? Or even a window?

I got the last one correct but the first 2 I failed EDIT: THANKS SO MUCH FOR THE LIKES GUYS😃😃😃😎😎

I don't agree with your solution to #3. Yes, the first prisoner can ensure that no loop exceeds fifty boxes. So there will be two or more loops. That doesn't guarantee that any prisoner will find his own name. To do that he'd have to work the correct loop, and he has no hint on where to find it.

The biggest question is, how is her room pitch dark in the morning when she overslept?

I think Anna is a doctor because i can't understand her writing

Ok, The first on my me laugh so hard, it was just so stupid and I know it's a riddle but still like who does that

On the last riddle Prisoner: LETS RIOT

I confirm the solution is perfectly viable. The ones who disagree simply have to reflect on the riddle a bit more and understand it. The most frequent argument is that even if the first prisoner will guarantee there's no loop of more than 50 names, it's not guaranteed each and every prisoner will pick the loop with his name in it, thus finding it. This argument is fully invalid, as it is guaranteed he will pick the right loop simply because the box with his name always leads to the box of his number (thus they're in the same loop) and every prisoner starts looking into the box of his number.

The third one was easy, I got stuck on the freaking socks

Why not: • Every prisoner gets his number • First prisoner swaps names that Prisoner 2 gets his name in box 2. Prisoner 3 gets his name in box 3 100% wins

On the first one i just said take all the socks...😂

Prisoner Riddle: The first one puts all names into Box 1. Done.

6:15 "If there's a loop that is over 50 steps long, there can only be 1" I don't understand. A loop can even be 99 steps long, say if each box contained the name of prisoner "n+1." If we are to proceed with the strategy, many prisoners will miss their names because the can only open until the 50th box. In fact, I believe almost all of them will miss their names. EDIT: Figured out how to close 100 step loop into 2 loops: The 100th box is supposed to contain what should be in the 1st box. Just switch the contents of the 100th box and the 50th box.

first 2 are too ez, but the third, iyes they are smart, but they're not good memorizers, so they cant memorize the names, cant the first one just walk in and align the names in alphabatical order and then let the prisoners go in, in alphabatical order too?

the first riddle was a no shit

It was amazing.Specially the prisoner loop .Just amazing!

Michael:Jeff is the knave Julien:Michael is the knight Jeff: My name is jeff

The first one is somehow incomplete. Yes you took 3 socks and for sure two are matching colours. Still, in the dark you can't tell which ones, so you can't wear them before lighting the room. You should have stated clearly that the riddle was only grabbing a matching pair with 100% sure, not actually put them on.

I only got the 2 one right

Riddle 3: Another way to solve it that makes it easier for every prisoner except the first one: The first prisoner memorizes the order of the prisoners entering, and each prisoner memorizes their own number. The first prisoner puts the second person's name in the second box, third in the third box, etc. The prisoners open their own box and immediately leave.

Dude I have found another strategy... When they discuss the strategy they tell the first prisoner that is aloud to switch all boxes as he wants to put every prisoners name in alphabetic order....the alphabet is like 21 letters long sooo every prisoner can open 50 boxes and the possibility to find his name is tooo high

Riddle 3: The first guy puts the names in alphabetical order.

I knew the second one was jullien!!

1:41 MY NAME IS JEFF XD!!?>>>

Julien may be the spy but because Jeff CLAIMED to be the spy he can be held as an accompanist

I got number 2 and number 3.

YES I GOT THEM ALL...I'm way too happy about that aren't I

Riddle 3- Easiest way As 1st person can change any box and can look them He will plan with other mates that he will arrange everyone name from 1-100 acc to alphabet sequence (like dictionary). As everyone know there name they will able to know the box Number with there name as already discussed with eachother . Quite easy

2nd is the easiest

I got the 2nd one right but before hand my first thought was to get the king to execute all 3 of them 😂

The third one wasn't as hard as I thought it would be

Anna wakes up from a nap and just like "OH SH*T I NEED TO BE AT AN INTERVIEW"

I don't get what you are trying to say for the last one ?

2riddle is easy he said there is a spy BETWEEN his men and there is only three so it got to be the second one

The wording of the last riddle gives you way more possibilities to solve it. The first person can just arrange all the names in the same order the prisoners will enter the room so they just look in the box numbered with their entering position. Or if the first person can't memorize that many names in order, the first prisoner can just organize the names alphabetically and the others just have to remember how many names come before them. And so on... If they can set the order of the prisoners and change the order of the names in the boxes there is really no need to do this with a sorting algorythm. They don't even have to be abled to open 50 boxes, after the first persone opening one box each would be enough.

In riddle 3: In the explanation or solution there was no hint that the first prisoner has a limit of swapping names. So it could be very simple to solve just by swapping the names of the prisoners until they are in the same order as the order they enter. This way only the first prisoner would open all the boxes reorder the names in them and than all of the other prisoners would solve the riddle by opening only the box corresponding to their order number. They could even order themselves alphabetically so that the first prisoner doesn't even need to remember their names, but only order the names in the boxes alphabetically. The other prisoners would not be required to remember anything else than their number in the order.

100 prisoners, one guard. Just bust out already.

the last riddle is stupid.. lol are there cameras or something.. is there no rule in changing the boxes themselves... lol if not.. the first guy should be someone who knows everyone's name.. and put the boxes themselves in the order the prisoners came up with.. id say best to put them in alphebitcal order lol... the numbers are irrelevant here.. just like they were in the sock riddle... lol how are the numbers on the boxes.. can't you just swap the numbers and organize the boxes in alphabetical order.. so the prisoners are guaranteed to know it's their box no matter what lol... haha all this lies on number one prisoner lol he can't mess this up or everyone is executed

3:47 so if they couldnt communicate how did they come up with a plan EXPOSE HIM!!!!!!!

About that last one riddle - there is one much better solution. Everything starts with mutual agreement, a week before the game. Every prisoner will have his unique number. For example - prisoner John Smith will have the number "6". When the game begins, some prisoner goes to the boxes, opens them all and finds the name "John Smith". After that, it's enough to knock on the lid of that box for six times. His friend John Smith will understand that in exactly THIS box is his name. This ''knock-guy'' prisoner will knock on the other boxes in this way, sending the relevant signals to other prisoners each according to the number of knocks. In this way, everyone will know where to go and which box to open because everyone has his own unique number that is the same as the number of knocks, and that ''knock-guy'' knows them. And even if the prisoners were wrong to see it, just go to the area where it was knocked - each of them has 50 attemps.

If the prisoners were so smart they wouldn't have got caught

4:38 he spelled two wrong (to)

i got 1 and 2 but 3 was too hard

If i had a week to talk with 99 other smart prisoners i think we would just break out of prison

Never heard the swap variant of riddle 3. Very cool!

1: fail 2: fail 3: literally the hardest one..... I got it 🤦‍♂️

I don't get number 3?

Got first two. No way I'm getting the third. Having hard time even after the explanation lol

That “easy” one was still difficult lol

With the third, why can't the first prisoner check who's next, find their name and swap it with box 2? Dunno id i'm missing something but it says he can check in all of them and swap two around, this should work. (Hopefully)

I'm high asf Wutttttttt?????😰

Broken Lamp? that's simple, turn on the light that's built in with the house

Bro, the answer to the first one is "EFF THIS! IT DOESNT MATTER IF YOU EFFING WEAR MIX-MATCHED SOCKS!"

Riddle 1: what if there's 5 black socks in a row

why the fuck would you say your the spy?!?!?!??!?!

I think Anna's lamp was made in China And doesn't she has a tubelight, bulb or something. What an amazing house 🏡

Did nobody else think about the fact that on the last riddle the first.prisoner could just open every box find his name, and leave it that way seeing as how they could but didn't have to close the boxes again?

I thought the prisoners had planned their order so couldn't the first person change the boxes by order, so third person would pick box three? I probably am missing something and sound really dumb

What if there's 3 socks in a row that are all the same color? Not really thought out I see

If there was no prisoner No.1 who can limit the largest possible loop to 50, the third riddle would not have 100% efficiency, but only about 30%. This way, it is ensured that each prisoner's number will begin at the end of the loop which is 50 boxes or less.

Riddle 3: If prisoner 1 is allowed to open up all the boxes and change the names the way he wants, why doesn't he just rearrange them in an alphabetical order (family name)? The day before, they line up alphabetically and see e.g. "Krustchow is number 11" etc. If the first prisoner then rearranges the cards, every prisone besides the first one has to open only one box. I foresaw your solution, but the problem is, you cannot learn 99 names in one night. More importantly, 100 prisoners will unlikely know each 99 prisoner in just one night. One might fail to remember a name which will result in failure for the team. Go on, rant on me for being so picky. It will work either by remembering the names or by rearranging the cards alphabetically.

my name is jeff

How did i get the hardest one right, but the other 2 wrong??

The first prisoner can't switch more than two names from the boxes, but there is no rule preventing him from rearranging the boxes themselves. He can just put the boxes in the order the prisoners come in starting from the box containing the name of the first prisoner and they will all just need to open one box to find their names. Think outside the box, lol.

The 1st riddle actually became a test in my school XD on the combinatorics chapter, pigeon hole principle section. The 2nd riddle is actually easy. It's only hard if you refuse to do it in the first place XD My way of thinking: If Jeff were the spy, then the other two are liars - won't work since both of them can't be the knight. Since Jeff told the lie and is not the spy, then Jeff is the knave. Julien's statement revealed that he can't be the knight therefore he must be the spy. For the 3rd riddle... Thanks, minutephysics.

You don't know how to tell these riddles my friend. No disrespect to you, I swear I don't mean any disrespect but I've heard them before and you're not doing a good job saying them. People can get confused because you can not articulate the riddles correctly.

i sorry but i dont get it!

"What's in the box?!!!" (Brad Pitt quote from the movie Seven)...look it up.

Love this channel so much.

How come I got riddles two and three right but not riddle one

The first one you can turn on the light or get a flash light

I love riddles so I subscribed

I know how to answer #3 with a 99% chance of success. Basically, have the prisoners line up in alphabet order. Then each prisoner will count what place they are in. Once the first prisoner goes in to look at the 100 boxes, he just places them all in alphabetical order. From there, the prisoners just need to remember what place they belonged to. The only way this can fail is if one prisoner forgets their place number.

Pshh, I knew all of these immediately, so easy. somebody get me a thrid grade tutor fast

Too easy (all of them)

Why Dosent all of the prisoners open one Box?

1) Alternative: Black socks are pretty much thicker than most white socks, let alone how she should at least identify which is lighter and darker. 2) The guy in the mid. 3) If the first pri. Can arrange it, they should just arrange it chonologically, different from yours though..

1st is wrong

final riddle: who ever is first to go in puts all names in one box, then they just find their name from there

if people want a longer explanation of the last riddle, check out minutephysics's video on it ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-C5-I0bAuEUE.html

1 wrong 2 right 3 right

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for riddle two, I think a confession would be good enough for the king to kill you.

I have better solution for this riddle first one can arrange it in the form of alphibitic form souppose, if someone name starts with A he should numbered at starting few boxes and next B and so on. by this others 99 can open 50 boxes in which they can open according to their first alphabet so by this they don't need 50 boxes to open they can easily find by just 25

Answer to riddle 1: Turn the light on or use you're phones light and take two pairs.

julien is the spy, Jeff is a knave , michale is a knight

damn, I chose correctly on the second on because you said 'there is a spy between his men' and Julien was in the middle..

Which software u use for ur editing

Holy shit I solved all three. The third was easier for me then I thought it would be.

I believe the problem with #3 is how does Prisoner#1 know everyones cell# and name by memory? (you might say "extremely intelligent") which leads to the question many people have wondered, why tf can't prisoner 1 just remember the names in each box? i'm sure prisoner#1 isnt allowed multiple days, nonetheless hours to go through all loops starting with #1 ect ect which like stated above, once opened box 1 he would have to know the corresponding Cell# with the name in the box. Which if he can remember 100 names and each of their corresponding cell#s, then why couldn't he just remember the names in each box... I suppose the only workaround for this, is if the piece of paper also INCLUDES the Cell# with their names, but that wasn't stated within the riddle.. EXTRA: even if the paper had both the Name and Cell# for each prisoner, the question still remains how long does he have to go through all 100 and choose the 2 swaps..? given enough time sure it could be done but ONLY if the cell# was given as well because if its not, then that would mean he remembered everyones cell# as well as their names for each cell, and idk about anyone else, but remembering 100 names is hard enough but adding in a 1-100 number to go along with those 100 names...not happening, besides in prison they probably don't use their real names all the time. oh well.

I always know the middle and hard ones but always get stuck in the easy ones😂

The knight, knave, and spy riddle is actually taught in a Discrete Mathematics class... It sucks

Dude, I can't believe I got number three! Though, tbh, I missed the first one. I only got the third and second one right

There's a strategy that GUARANTEES they will escape. Oh wait it's only a >30% chance.

He can't just swap any two name in a loop to cut it in half ... if you swap two success numbers you merely reduce the loop by one. Box 1 (Name 3) -> Box 3 (Name 8) -> Box 8 (Name 72) -> Box 72 (Name 4) -> Box 4 (Name 10) -> Box 10 (Name 1) If you swap The Name in Box 8 with Box 72 ... you get a single loop Box 72. But the loop goes Box 1 (Name 3) -> Box 3 (Name 8) -> Box 8 (Name 4) -> Box 4 (Name 10) -> Box 10 (Name 1) Now if you swap Box 3 with Box 4 Box 1 (Name 3) -> Box 3 (Name 10) -> Box 10 (Name 1) And Box 4 (Name 8) -> Box 8 (Name 72) -> Box 72 (Name 4) Or if you swap Box 72 with Box 10 Box 1 (Name 3) -> Box 3 (Name 8) -> Box 8 (Name 72) -> Box 72 (Name 1) And Box 4 (Name 10) -> Box 10 (Name 4) So the idea just swapping two names in the sequence will "cut the loop in half" is wrong.

2nd answers Michael:knight Julien:spy Jeff:the LiEr

The third one makes no sense because the warden said only the 1st prisoner could look through more than one box.