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
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
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.
There is the thing. if the first is in a smaller loop, that is inproblable, all die. there is no chance of full success. It can be a chance of all go througt or all die. Even if the first on identify that there is a loop bigger than 50. He is alredy dead. Becouse he doesn't know wich box replace.
No light in their room, no job, getting up late for an interview, not having their clothes already laid out. If a person can't get their stuff together that would be their best option.
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.
I agree with you... I think the person posting the riddles is suffering from brain scramble on #3... I started reading through the comments to see if someone agreed with me ... If we give this thread a thumbs up, others will find it faster ...
The prisoner will always be in their own loop. It is necessarily true that their start box and name box must be in the same loop, since their name box is the one that comes before their start box (thus forming the loop)
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.
Killua Zoldyck and if you want to nit pick about capitalisation, I'll nit pick about your lack of punctuation. Shouldn't it be: You didn't need to capitalise " the "?
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?
mR Dub and Paul Smith , no there arent 100 letters in the laphabet, but if 2 ppl have the same letter, then u order them according to their seccond letter's alphabetical order, and in case the second is also the same, u do the same process with the third letter, and so on. example: if there are 3 people names: nelson, nijel, jeffery first is jeffery then is nelson, because letter (e) comes before the letter (i) in the alphabets and last is nijel
Tabirca Radu ah ok, makes sense, can u explain the solution though? i dont see how doing what he explained garuantees in all of them finding there names.
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
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
one prisoner writes the 100 names on 100pieces of paper and give each other a paper with is name. they each chose a number from 1 to 100. Prisoner k removes and eat the original paper from box k and replaces it with the paper with his name written on.
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
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.
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.
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..
+Sweet Niya, He said you can't communicate in cells, however the warden gave them some time (Break actually to make a Strategy ) Which they did , i mean you can't like a prisoner go to pick his number (Name then come back and tell the next cell "Hey.. i got no.1 , try picking the other one" That what he meant
I don't know if it wasn't communicated well or what, but even if someone can swap 2 names at he very beginning, there's still the chance for failure. There are effectively 99 boxes and 99 people each only only get 50 tries at once, since only the first person can swap names & look at all boxes. All it takes is for 1 person not finding their own name. Maybe I'm just misunderstanding the solution or something
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
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.
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)
only 1 prisoner can look in all 100 boxes and cqn only switch 2 names. the rest can only open 50 boxes and have to leave the boxes exactly how they were.
or since they numbered and he/ she memorized numbers with names. he would just put them into numbers... idk if ui make sense, but you are allowed to switch as many times
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.
JIGYASA SETHI WTF IS WRONG WITH YOU? Get tf out of here, if this guy is a TH-camr and is trying to entertain people he should be prepared for criticism. Go hide in ur safe space you snowflake
You really did not get that one didn't you? That's not how it works. Since there are only two colors and if you get two socks that are diff colors then the possibility of getting the same color of either one of the previous two socks is 100% since you used up the 1/2 probability. You're using a paradox to the riddle saying that you can get three of the same color when in the riddle it starts off of grabbing only *two* socks.....and by obvious means, if the two socks are both the same color then what is the point of getting a third sock in the first place-disregarding if it might be the same color or not.
IEatWhenI'mBored it still doesn't matter like explained before as long as you grab 3 socks your good 1. If you grab white then white your already good 2. If you grave black then black your already good 3. If you grave black then white or white then black guess what when you grave the third it has to be either white or black now you have a pair for sure hence why yo u wanna just grave 3 it's a paradox by grabbing 3 your 100% to have a pair in your hand with the info the riddle gives us we are assuming she's grabbing these socks n putting them on in car where she can see but she wants to be sure when she make it to where ever she can see she want to have a pair
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.
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?
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.
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.
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 1- got it right easily, but i didnt undestand what u said how to do it afterwards xD riddle 2- i guessed because i didnt wanna spend my time on it lol- i got it right (wow) riddle 3- that was pretty easy tbh... it took me about 20 seconds
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
![[Live สด] การออกรางวัลสลากกินแบ่งรัฐบาล งวดวันที่ 16 พฤศจิกายน 2567](http://i.ytimg.com/vi/wPd7VFfWHIA/mqdefault.jpg)
took 15 mins...but i solved the 2nd riddle!!!!! YESSSSS!! Finally one video where i solved something
th-cam.com/video/kDKLMv1jKEs-/w-d-xo.html try this new challenge if u r really genius..
I solve every riddle without pausing
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 true
No they are not allowed to communicate
The hardest part of solving these riddles is understanding the explanation of what you are trying to solve.
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?
God,
if she lives in a basement, where's the stairs or at least the door out? i mean how could she find her clothes or the socks, IN THE FUCKING DARK.
lol and she can just crab like a couple socks and go outside in the sunlight to see!
okay... just say, she's blind. feeling better now?
Then she wouldn't know which one of the three is incorrect, since she can't see,
On the last riddle
Prisoner: LETS RIOT
I got the last one correct but the first 2 I failed
EDIT: THANKS SO MUCH FOR THE LIKES GUYS😃😃😃😎😎
Sounds like you got a weird brain ;)
Riddle Channel yep
Same with me !!
VKarl exactly
Yash Londhe glad I'm not the only one
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
Jakub Mroczek you can swap 2 names only once
switch the boxes not the names
To_5BG- Казармата на Крийпърите he was talking about if it would be like that
Jakub Mroczek that’s what I thought I was totally lost on that last one
Me too
I think Anna is a doctor because i can't understand her writing
lmao
;)
The biggest question is, how is her room pitch dark in the morning when she overslept?
Because she usually gets up at 3. happy now?
Or maybe she sleeps in the basement with no Windows. That could be a reason as to why it is still dark
She lives above the arctic circle so uses blackout curtains
3rd Shift
I don't get what you are trying to say for the last one ?
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.
There is the thing. if the first is in a smaller loop, that is inproblable, all die. there is no chance of full success. It can be a chance of all go througt or all die. Even if the first on identify that there is a loop bigger than 50. He is alredy dead. Becouse he doesn't know wich box replace.
On the first one i just said take all the socks...😂
No light in their room, no job, getting up late for an interview, not having their clothes already laid out. If a person can't get their stuff together that would be their best option.
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.
Exactly right. I'm disappointed no one else has realized this yet.
I agree with you... I think the person posting the riddles is suffering from brain scramble on #3...
I started reading through the comments to see if someone agreed with me ... If we give this thread a thumbs up, others will find it faster ...
yea exactly, wtf? you can't gurantee that everyone will find their name in under 50 boxes by cutting 1 loop
The prisoner will always be in their own loop. It is necessarily true that their start box and name box must be in the same loop, since their name box is the one that comes before their start box (thus forming the loop)
They are always in their loop because they start with their corresponding number.
The third one was easy, I got stuck on the freaking socks
I said 4
_ TheBigPig _ same
Yup ikr
_ TheBigPig _ mine was the other way around
Same!!!
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.
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
When I've heard the sock riddles it's always with at least 6 or more colored socks, not just black and white lol.
Carter learn how to use grammar bitch
Carter and general spelling you capitalized the for no fucking reason
Killua Zoldyck and if you want to nit pick about capitalisation, I'll nit pick about your lack of punctuation.
Shouldn't it be:
You didn't need to capitalise " the "?
Killua Zoldyck It should be:
'Learn how to use grammar bitch.'
'And general spelling; you capitalized 'the' for no fucking reason.'
Prisoner Riddle:
The first one puts all names into Box 1.
Done.
Hahaha
Riddle 3: The first guy puts the names in alphabetical order.
you can only switch 2 names, once.
Lin J 4:30 read the bottom rule
Now realise that he made a typo and the prisoner could only switch 'two names', not 'to names'
Bodelicious that what I thought too
but they didnt say anything about switching the boxes
why the fuck would you say your the spy?!?!?!??!?!
people like you make me want to shoot children.
Meaty one 😂😂😂😂😂
It's a riddle, it's not real life
Yay een nl
Ik ben een pedofiel he would need to memorize the order of the 100 prisoners
It was amazing.Specially the prisoner loop .Just amazing!
I got number 2 and number 3.
good job :)
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?
AFK able the first can only swap two names once.
AFK able there isnt 100 letters in the alphabet
But not all names are ONE LETTER LONG DINGUS
mR Dub and Paul Smith , no there arent 100 letters in the laphabet, but if 2 ppl have the same letter, then u order them according to their seccond letter's alphabetical order, and in case the second is also the same, u do the same process with the third letter, and so on.
example: if there are 3 people names: nelson, nijel, jeffery
first is jeffery
then is nelson, because letter (e) comes before the letter (i) in the alphabets
and last is nijel
Tabirca Radu ah ok, makes sense, can u explain the solution though? i dont see how doing what he explained garuantees in all of them finding there names.
I only got the 2 one right
mackayla campbell BAAAAAAD CHANNEL number one was wrong number 3 was long
Number 1 was correct. You only need to grab 1 more sock than there are colors to guarantee at least 2 will match. Number 3 was confusing.
Michael:Jeff is the knave
Julien:Michael is the knight
Jeff: My name is jeff
Never heard the swap variant of riddle 3. Very cool!
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
Any two boxes
Not every box
the first riddle was a no shit
1: fail
2: fail
3: literally the hardest one..... I got it 🤦♂️
Anna wakes up from a nap and just like "OH SH*T I NEED TO BE AT AN INTERVIEW"
I got all correct and I'm 10
2nd is the easiest
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
There is a typo in the statement of the rules. The first prisoner can swap any *two* names. No more.
one prisoner writes the 100 names on 100pieces of paper and give each other a paper with is name. they each chose a number from 1 to 100. Prisoner k removes and eat the original paper from box k and replaces it with the paper with his name written on.
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
Bro it was mentioned that he can only open the boxes and have to leave em just the way they were
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.
Michael Knight.... I see what you did there. :D
1:41 MY NAME IS JEFF XD!!?>>>
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
Family's Kids Nice one.
What funny is I heard his name was Julian and my first thought was "That's the name a damn spy would have." and by Jove I was right...
Tomiho
I knew the second one was jullien!!
Yeah, that's wrong, bud.
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.
Exactly that’s actually the solution
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..
4:38 he spelled two wrong (to)
Riddle 1: what if there's 5 black socks in a row
Then obviously you’ll have a pair of black socks?
3:47 so if they couldnt communicate how did they come up with a plan EXPOSE HIM!!!!!!!
+Sweet Niya, He said you can't communicate in cells, however the warden gave them some time (Break actually to make a Strategy ) Which they did , i mean you can't like a prisoner go to pick his number (Name then come back and tell the next cell "Hey.. i got no.1 , try picking the other one" That what he meant
I don't know if it wasn't communicated well or what, but even if someone can swap 2 names at he very beginning, there's still the chance for failure. There are effectively 99 boxes and 99 people each only only get 50 tries at once, since only the first person can swap names & look at all boxes. All it takes is for 1 person not finding their own name.
Maybe I'm just misunderstanding the solution or something
Julien may be the spy but because Jeff CLAIMED to be the spy he can be held as an accompanist
Got first two. No way I'm getting the third. Having hard time even after the explanation lol
i got 1 and 2 but 3 was too hard
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
You are correct. I think he misspoke in regards to the rules of that riddle...
He can switch 2 boxes only once
I'm high asf
Wutttttttt?????😰
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.
That “easy” one was still difficult lol
I don't get number 3?
shadow nova neither did i
Think its fake :)
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 mean, he could but all that does is leads number 2 to getting his name. It doesnt bring any security to prisoners 3-100
only 1 prisoner can look in all 100 boxes and cqn only switch 2 names. the rest can only open 50 boxes and have to leave the boxes exactly how they were.
or since they numbered and he/ she memorized numbers with names. he would just put them into numbers... idk if ui make sense, but you are allowed to switch as many times
Mr Neos48 THE FIRST PRISONER JUST DOESNT CHANGE ANY NAMES BC IT WOULD BE STUPID TO ENDANGER HIMSELF
So does everyone need to get their names or is it like an every man for himself thing where if u get ur name u live?
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.
Aaron Monroe I don't think that is really nice
JIGYASA SETHI yeah that is not
Sujatha Ramakrishnan I know. He's just being a bully. It doesn't matter if he talks funny or is not good. At least he's trying!
JIGYASA SETHI WTF IS WRONG WITH YOU? Get tf out of here, if this guy is a TH-camr and is trying to entertain people he should be prepared for criticism. Go hide in ur safe space you snowflake
JIGYASA SETHI I’m not trying to be “mean” but I just get sick of this political correctness now. Merry Christmas and God bless you!
Yes I got all of them!
damn, I chose correctly on the second on because you said 'there is a spy between his men' and Julien was in the middle..
What if there's 3 socks in a row that are all the same color? Not really thought out I see
You really did not get that one didn't you? That's not how it works. Since there are only two colors and if you get two socks that are diff colors then the possibility of getting the same color of either one of the previous two socks is 100% since you used up the 1/2 probability. You're using a paradox to the riddle saying that you can get three of the same color when in the riddle it starts off of grabbing only *two* socks.....and by obvious means, if the two socks are both the same color then what is the point of getting a third sock in the first place-disregarding if it might be the same color or not.
rivetRose No I got it, I just wrote a comment saying that it doesn't make sense, that's sarcasm if you didn't get that either
Christian Maund he doesnt
What if you're not certain that you got 2 different colored socks at the start?
IEatWhenI'mBored it still doesn't matter like explained before as long as you grab 3 socks your good
1. If you grab white then white your already good 2. If you grave black then black your already good 3. If you grave black then white or white then black guess what when you grave the third it has to be either white or black now you have a pair for sure hence why yo u wanna just grave 3 it's a paradox by grabbing 3 your 100% to have a pair in your hand with the info the riddle gives us we are assuming she's grabbing these socks n putting them on in car where she can see but she wants to be sure when she make it to where ever she can see she want to have a pair
i sorry but i dont get it!
my name is jeff
Shane Kim ugly fat kid
are you the same jeff of coc?
i am abig fan of coc
What does my name is jeff have to do with riddles, ugly and you stupid
“Mayday mayday. We are sinking....”.Guy who made zis video: “wat are you sinking about?”
LOL i just took off my socks XD
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
Oh really, smartypants? What if you don't have one or the electricity went out? Do you really think you're smart?
If you were listening clearly it said that the girl's lamp was broken so she has no light.
I was right on the 2nd one WOOWHOO ._.
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.
Awesome!
I love riddles so I subscribed
1 wrong
2 right
3 right
LOL!!! I got all riddles but the easiest.
Third was interesting, first two were easy, not need to pause video even. Got all three of them anyway :-)
The third one makes no sense because the warden said only the 1st prisoner could look through more than one box.
Love this channel so much.
I am the spy, and they even spelled my name right!
final riddle: who ever is first to go in puts all names in one box, then they just find their name from there
i fist thought the riddles were on Logic, the rapper
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?
Holy shit I solved all three. The third was easier for me then I thought it would be.
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 last riddle was the easiest XD! I didnt get the rest!
I got all 3 right :)
I got the last one but not the other once😂
Bro, the answer to the first one is "EFF THIS! IT DOESNT MATTER IF YOU EFFING WEAR MIX-MATCHED SOCKS!"
My brain broke and left me through my ear
If the prisoners were so smart they wouldn't have got caught
"What's in the box?!!!" (Brad Pitt quote from the movie Seven)...look it up.
So difficult #3 😂
But the final riddle you must know all 100 prisoner name
If i had a week to talk with 99 other smart prisoners i think we would just break out of prison
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.
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 1- got it right easily, but i didnt undestand what u said how to do it afterwards xD
riddle 2- i guessed because i didnt wanna spend my time on it lol- i got it right (wow)
riddle 3- that was pretty easy tbh... it took me about 20 seconds
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
What a smart and deficut questions
riddle 3: wouldn't the warden see the order that everybody is going in and get suspicious
Broken Lamp? that's simple, turn on the light that's built in with the house
they live in a shed.
100 prisoners, one guard. Just bust out already.
Awesome
Hey !!
MY NAME IS JEFF!
julien is the spy, Jeff is a knave , michale is a knight
OMG OMG he ANSWERD, you are the BEST
OMG I ACTUALLY GOT ALL OF THEM RIGHT WOOOO
i didnt know 1 or 2 but figured out 3 so easily