I am glad that the smaller version of the game was demonstrated. It helped my pre-breakfast mind to comprehend the larger version's properties and why it works.
One of the 21 prisoners will count how many times the left switch has been turned on. So when he enters the room, if it's on he switches it off. The left switch serves a counting purpose. The right switch is used by prisoners to get rid of their move when the left switch's position doesn't meet the required conditions: The other prisoners don't touch the left switch unless it's off. And if it is, they turn it on; but maximum 2 times. If a prisoner already turned the left switch on 2 times, he doesn't touch it anymore and only uses the right switch as "garbage moves". That way, the counter only needs to keep track of how many times he reset the left switch. When that count reaches 40, it means everyone affected the switch at least once and the counter can safely say he's the last man to switch it.
Also why would the inmates flip switch A twice? That makes no sense. All that needs to be counted is one trip per person, so why would you not limit each of them to flipping switch A once? And they can clearly communicate amongst themselves (also, realistically, it's not like the guards can stop them from talking), so why would they not just flip any switch, then tell the counter to count them when they come out? Or maybe find an easier way to solve the problem than this bullshit?
I think the rules of that riddle are clear: no communication between prisoners once it starts. You're just allowed to plan ahead. Also, for the "why you have to do it twice" part: 7:51
The only problem with this riddle is that you could theoretically never get released because it is statistically possible--however improbable--that at least one of the inmates never gets randomly selected...or that the counter person stops getting randomly chosen before everyone has entered the room at least once.
Good puzzle, but too much confusion. Proper rewording of the puzzle: A prison has a room with a lightswitch and a broken ceiling light. The switch can be turned on and off, but it does nothing. The evil warden decides to torture four prisoners with a tough game. He tells them he'll set them all free if any of them can correctly announce that all of them have visited the room during the game. He shows them the room and says a guard will take them to the room individually many times, in random order, and each time will let them flip the switch if desired, and then take them back to lockup. They won't know whether the switch is on or off when the game starts. None will ever see or hear the others during the game, but now all of them must talk about how to use the switch-flipping to determine when one of them should announce that they've all visited the room. What should they do? Answer: Have one of them be "Mr. Counter," and only he will ever turn the switch off. All others either flip it on or do nothing. Whenever one of them is brought to the room and sees the switch off, he flips it on, until he's done so twice. After that, he leaves it alone. Whenever Mr. Counter is brought to the room and sees it on, he counts it and flips it off. When his count reaches six, he announces that they've all been to the room. Why that strategy works: There are 3 prisoners that Mr. Counter needs to be sure have been to the room, and each will flip the switch on 2 times. 3 x 2 = 6 ons that Mr. Counter needs to see. Each time Mr. Counter sees it on, he flips it off. Since he's the only one who ever flips it off, he knows that whenever he sees it on, a prisoner has flipped it after he had flipped it off (or he's seeing it on the very first time he's there because it happened to start in the on position). The sixth time he sees an on, he knows either all three have flipped it on twice, or it started on and they've flipped it on five times. Either way, all three have been there, and so has he, so he announces that they've all been there.
Aaarggh I think I'm really close but the one thing that is throwing me is not knowing the starting positions. What I have is that one of the switches is flipped every time by all of the prisoners except one. One of the prisoners (chosen before this starts) flips the other switch the first time he enters the room. The next person will see the other switch is flipped, and flips it back. When the chosen prisoner returns he counts 1 and flips that same 'important' switch. The next person who sees it in this state who hasn't already returned it to its original position flips it back. By repeating this, the chosen person can count up to 20 and ensure that everyone else has been in the room at least once. However this relies on knowing the starting positions and I can't see a way around unknown starting positions Gonna keep thinking on it but I don't know how much patience I have left!
I'm usually the riddle/puzzle bringer for our yearly man weekend, a riddle is proposed at the beginning of the trip and worked on the whole time. This will be in my arsenal this weekend.
Dude, I figured that out in like 10 minutes. Too easy! Just kidding I gave up after 10 minutes I just wanted to be "that guy in the comments" for once!
I wish it was made clear that the switches can be given names A and B. Could also be that you get put into the room blindfolded and can't tell which switch is the right or left one if you don't know from which side to look at the two switches. Without that info my solution was to just never proclaim anything and technically survive until you die of natural causes. The guards or whatever even state that the inmates will never see each other again, implying the problem is unsolvable and that they will never be released. The problem of watching these at home is that you can't ask for clarification on the rules so the wording has to be perfect the first time.
Just go in and the first time youre in there leave a hair in the corner and flip a switch. When theres 20 hairs in the corner and you havent left your hair your the last one.
the fact that he used words like 'either', 'last' and never implied that 21st prisoner was a current count lead me to believe that there was in fact only one or two prisoners, with a riddle the delivery needs to be exact because any information is seen as valuable, if the riddle is delivered oddly speculated fractures and it becomes possible to find multiple correct answers, in turn rendering the riddle in-valid
Great solution. Still, seems like there's another solution. The tactic "no one ever proclaims" also provides the stated winning strategy: "...make sure you survive". The prisoners survive - they survive in a never-ending life of solitary confinement and switch -flipping, but the winning condition is fulfilled. Since the prison is going to be closed, this can't really go on forever, but neither do humans - plus, plans to close Guantanamo Bay prison have historically not gone too rapidly. (Okay, here "survive" was just a poor word choice by Brian (or by Nick the puzzle writer), but on first hearing, one can read easily it as intentionally specific).
So the counter has to visit the room at least 40 times? Wouldn't that take a very long time chosen at random from pool of 21? And wouldn't his declaration possibly come some time after everyone had visited? So with death on the line why not just flip switches for a couple months then declare? I'm probably missing something.
Wait... there's a major problem with this riddle. If the guards are trying to decide whether to execute or free all the prisoners, there's no way they'd use a system that could go forever. What if nobody ever proclaims that everybody has been in? Will the guards just wait forever?
The answer can also be something far more simple. Either way the person who says they are the last are going to be the last, so no matter what they are correct and everyone must be let go. Either they legit are the last ones and everyone would go free, or they say they are the last and the game stops and everyone is executed, but that would make them the last person, so technically the guards would have to let them go.
I never heard how they'd designate which switch is A or which is B? Are they labeled? If you've never been in the room you couldn't say the switch on the left is A and to the right is B. They may be configured any number of ways.
They should leave a piece of clothing only for the first time they were selected except one. Therefore when the 21st prisoners sees 20 pieces of clothing he will proclaim he's the last one to go.
Options: 1) You could leave some sort of marking in the room, maybe a piece of clothing, or etch something into the wall. 2) Every inmate was in that room before, as it was the room you were all in last night or is a common room, so the first person can proclaim everyone was there already. 3) Since the riddle did not say you have to proclaim it the first instance every inmate entered the room, you could wait an infinite amount of time, and you'd would know for sure that when time reaches infinite, every inmate has gone into the room. 4) "Leave the fate in your hands." So does that mean I'm the last one in the room? 5) "Men, this is the last night either of you will see each other" So there is only two of you? 6) "You are assigned as the last inmate" So there is only one of you?
7) You could kill all the other inmates in the last night phase, then you would know that when you enter the room, all the remaining inmates have entered the room.
Great puzzle. Got me stumped, but hear me out: I don't think the non-counters have to flip switch A twice. They can just do this: if A is on, leave it alone; if A is off, flip it on, but only do it once. The counter is the only one allowed to flip switch A off. This way, the counter will only have to count to 21 (rather than twice that). He doesn't have to count himself because he'd know if he's been in the room or not, but in case switch A started out as "on", he should count one extra, making the total count 21. Seems like a faster solution. Anyone find any fault with it?
Except he doesn't know whether or not the switch started ON or if someone has entered the room before him. By adding one, if the switch started OFF, and someone entered before the counter and turned it ON, the game would never end, as everyone has flipped the switch once but the count remains at 20 and they have no way of knowing if the last person just hasn't been called yet. The concept of flipping it twice each is adding that last degree of certainty, where if the switch starts ON, one prisoner will only interact once, but it is guaranteed that everyone has done so.
so the answer is that it's never explicitly stated that the final prisoner must declare that he's the final one on his first visit into the room, just that someone at some point must declare that all 21 prisoners have been in the room, and it's possible that all prisoners have been into the room, but none of them need to make a declaration yet because there is no penalty for being without declaration, the only penalty is if the declaration is wrong?
The key condition is that "everyone HAS entered the room and switched the switch." It doesn't matter if they all HAVE went in once and this was your first time entering the room and NOT declare they all have entered. They will keep going until someone mess up by declaring too early OR until someone declare after they know for sure that everyone has participated.
FASCINATING puzzle ! I spent several nights working on it (better way to fall asleep than counting sheep if you have insomnia !) and I think there's a simpler (and more logical) solution - or at least easier to explain! The key missing information ( SCAM !!!) is that the prisoners will be individually taken into the room AS MANY TIMES AS IT TAKES until one finally proclaims "all the 21 prisoners have been in the switch room" . And based on the fact that there are 21 prisoners and each must go into the room randomly at least as often as the "Mr Counter" prisoner does (which is minimum 21 times randomly), that's at least 21 x 21 = 441 times on average that prisoners must go into the room = much more than one night's work ! OK ....... 1. One prisoner is " Mr Counter" as per the official solution 2. It's also agreed that one switch (e.g left hand one ) is the counter switch as per the official solution 3. The other (right hand) switch is a "junk" switch which has zero meaning (can be toggled at random just because one switch must be toggled - if for no other reason) 4. Only Mr Counter can turn the counter switch from OFF to ON 5 All other prisoners can only turn the counter switch from ON to OFF and ONLY ONCE (thus if they have already turned the left hand counter switch from ON to OFF they just toggle the right hand junk switch if /when they go back to the room again and again) 6. Every time Mr Counter sees the counter switch in OFF he knows one more prisoner has been in the room for the first time. . He then turns the counter switch back to ON and waits for the next time it is turned OFF to count "one more prisoner" . 7. However, the initial state of the counter switch is a problem if it is initially in the ON state which can obviously lead Mr Counter to miss one prisoner. The solution to this is for every prisoner to turn the counter switch on TWICE. 20 x 2 = 40, so if Mr Counter reaches 39 times ON he knows for certain that all 20 other prisoners have been in the room (and so the trial requires - on average - 21 x 21 x 2 times in the room = 884 x Based on this solution it would be simpler to just state the puzzle as having one switch and the prisoners have the option to toggle - OR NOT . But that would be a big clue to the answer, so by having two switches it makes the puzzle more difficult. The second switch is a red herring (and so is the number of prisoners = 21) ! SCAM / SCAM/ SCAM .... Brian I "hate" = love you for setting such nasty puzzles !!!!!! _____________________ Having listened to the video a dozen times I now understand that both Brian and I are saying the same thing in different ways so I was right ! Here's the same puzzle and (easily explained) solution from the internet (except that it does not correctly deal with the A switch being initially in the ON position, as far as I can see) Question: A prison has 23 prisoners in 23 different cells. The prisoners have no way to communicate with each other in any way from their cells. There is another room, the rec room, that has two switches on the wall (A and B). The switches have on and off positions but they start in an unknown position. Prisoners are randomly taken to and from the rec room one at a time. They must change the position of only one of the two switches each time they go to the room. At any point a prisoner can yell out, "Every prisoner has been here!" If the prisoner is correct that all of the prisoners have visited the rec room, then they all go free. If they aren't correct then they are all executed. Before they start they are given one planning session during which they can discuss a method to win the game. What method can they use to ensure they all go free? Answer Here are the rules they can use to ensure they will all go free eventually: The prisoners will choose one 'leader' and everybody else will be a follower. If you are a follower: If switch A is in the on position, toggle switch B. If switch A is off, you have not toggled switch A yet, and you have seen switch on during a previous visit; then toggle switch A. Otherwise toggle switch B. If you are a leader: If switch A is off, turn it on. If switch A is on, turn it off. If you did not turn switch A on during their previous visit, increment the count of prisoners. Once the leader increments the count to 23 they can yell, "Every prisoner has been here!" and all of them will be released.
im at 7:08 and now have 'worked out' the answer. although sounds like you dont need to be the first one after everyone else has already been in. which also means, that you could just say: no one ever say everyone else has been in. then the guards will be stuck there doing this forever edit: yep, sounds like that little clue was enough. also what was with different audio and video at 7:32?
What if it took more than a day for the counter to enter the room and be able to flip the A switch 40 times? If it's closing the next day, wouldn't everyone be executed?
Guards: "you must predict for certain that everyone has been in the room before. You have one chance to come up with plan." Me: "Hey, can you show us the room with the switches while we plan?" Guard: "um, i guess." Everyone goes to the room. Me: "I can say for certain that we have all been to this room." Everyone goes free.
I don't get it What if it's the same two prisoners, one of which being the counter guy and one just a random prisoner. Once the counter guy gets to 21 he will think he's been the last prisoner but he actually hasn't. Can someone please explain this to me????
Let's run through an example with 3 prisoners. (1 counter and 2 others) So say this is the console A》 On. Off B》 The counter is only worried about two things. 1:reseting A to the off position. 2: keeping track of the number of times he does so.(he is counting to TWICE the number of prisoners excluding himself.) All other prisoners have one objective 1: turn on switch A TWICE and only twice. So we start in some arbitrary position. Let's say. 《A On. Off B》 A prisoner is chosen at random. Let's assume it's the counter! So the counter sees A is on and therefore resets it. A》 On. Off B》 His count raises to 1 (now notice the counter has a count of 1 even though no other prisoners have preceded him. This is why the count needs to be doubled in order to account for the possible + 1 the starting position can inflict on his count.) So the next prisoners chosen. Let's call him prisoner 2. He sees A is off and so switches on for his first time. 《A On. Off B》 Counter is next again and resets A another time. Bringing his count to 2 A》 On. Off B》 Next is prisoner 2 again. He turns on a for his second and last time. 《A On. Off B》 So next is Prisoner 3 he sees A is in the on position and therefore switches B(the trash switch) 《A On. Off 《B Next up is the counter again. He resets A bring his count to 3 A》 On. Off 《B Next is Prisoner 2. He has already switched A the maximum number of times therefore he switches B A》 On. Off B》 Next is prisoner 3 and he switches A for his first time. 《A On. Off B》 Next is counter and he resets A. A》 On. Off B》 bringing his count to 4 and ensuring everyone has indeed gone!
Isaac Long each peisoner only flips the first switch on twice after they have flipped it on twice, they flip the second switch on their turn. The counter doesn't cares about the second switch.
I've heard a similar riddle before, but you get out much sooner if you just wait enough days such there is 99% that everyone must have been in the room and then claim it. Small risk, but faster to get out.
Why are the 40 counts necessary? Everyone but the counter flips the agreed switch exactily once to on (or off, whatever is agreed upon), and the counter ignores the first state (counting himself as 1, the first known prisoner in the room) After 21 counts, he's sure. Why 40?
If the switch starts in the on position, then after 21 counts only 20 people have visited the room. At 40, 19 of the people have gone twice, and the 20th has gone at least once.
But it depends on how frequently the guards bring you in the room. in mean there will be 21 visits for each of the 21 inmates. If the guards pick one person per week. This strategy would take 8.45 years to come to an end. But when the counter reach half the count at 22, there is only one chance out of 2^21 that someone was left behind. Would you really expect every to spend 4.2 years more in prison to avoid the single chance in 2 million to die ?
Small problem, the guards could have withheld one prisoner allowing only 20 prisoners so prisoner 21 could have made a mistake. The riddle only shows that prisoner 21 has been in the room twice in a row.
The prisoners better hope that they get taken to the switches often, because they have to go in there almost 1000 times (45 each) with this strategy, mostly just waiting on the counter. If it only happens about once per day, then it would take almost 3 years for them to escape. I'm sure that one of the inmates would crack long before the strategy is over.
I've just realized this is the same visual setup as the 6-digit miracle multiplication (same shirt, same shwood, same venue, same approximate time), but with different guests. Huh.
ACTUALLY... hold the fort a dang minute!! I just thought that the guards themselves could add shoes to the pile, thereby messing that plan up... but by the same token, the guards could flick the switches themselves between bringing in the inmates, and that would mess the real answer up too. I guess there's no solution to this puzzle after all.
I had a guess at 4:00, which turned out to be wrong, but it is so much easier than the actual answer. If it is your first time in the room, and you are wearing two shoes, you count the number of shoes in the corner of the room. If it is twenty, you declare that you are the last of the prisoners to enter the room. If it is less than twenty, you take off a shoe, put it in the corner, flick a switch and leave the room. If you come in the room wearing one shoe, you just flick a switch because you have to. Nobody said that the ONLY thing you could do in this room was flick the switch!!
he rambled off that expenation so fast, that my brain nearly exploded. 3 persons and beer explenation makes more sence.... you just have to reach 4 minimum, either way because if on = count = 1 person 1 2x = 3 person 4 1x= 4 every one went in at least once person 4 2x = 5 or if off. person 1 2x = 2 person 2 2x = 4
The problem is that I thought that if you were the last one to enter, you'd have to immediately state that everyone has been in there. That info was missing.
So the key is there is no time limit and you don't need to announce all people have been in the room at the earliest point. So the missing info is that this could take twice as long as you think.
i think a even simpler way to do it is if you go in the room and leave a single item behind they can count the items that are in the room Note: they said a empty room so theres nothing in there to begin with so this is a possibly solution
A problem with the video was that Brian didn't correct them until a bit later on that saying "last person" is totally erroneous, as the win condition is just correctly stating that everyone has been in the room before at least once. A/the problem with this puzzle is that it gives no time limit. Considering that it's closing up there would need to be some sort of time limit, yet this strategy not guaranteed work with a time limit if the selection was random. Making the "win condition" more clear would have also been more appropriate, since clearly people were misunderstanding the puzzle, which is never a good thing. In other words, it should be explained that the guards will be ensuring that everyone does get selected multiple times before the time limit is up (and obviously the prisoners won't be informed of the time limits's duration since then they could just wait until the end), because that is _required_ for the strategy to work. In other words this is either a somewhat unfair puzzle or else a bad analogy. A better analogy might be that you're trapped in some sort of eternal prison and need to get back to reality/life. This way the solution is one that is not only certain but which gets out with an optimal amount of time lost.
I thought it would be easiest for everybody to leave the same article of clothing in the room, and once 20 other people have gone, you'll know you're the 21st since there's 20 other articles in the room.
I didn't give up. Just never answer and u all live. THe question was how to make sure you all survive. So that's a possibility a lonely one but w/e. It works
You could just say you're the last because you are. What you say ends the cycle you are always you last one until the next comes in. So if you say you are last you are not wrong and everybody is set free
Game ends when someone proclaims. Goals is for all of them to survive right, not for all of them to escape. So solution: Keep playing forever. But yeah, I don't think the initial condition matters. You just don't keep track of if it's on or off, just if it's different. Everyone keeps flipping B until the counter comes in, changes A. Next guy comes in, if it's his first time, he switches B. If not, he sees A has changed so switches it back. The rest will not notice A has been changed. Counter comes in again, sees A is changed so counts 1 and changes A again. Not sure if this is faster than the solution in the video (I think it is, because everyone only needs to switch it once), but at least it is not dependent on the initial conditions. Although I guess this becomes a problem if the counter is the first person to enter the room, because nobody would know he's been in there, while he would just be waiting for someone to notice him.
wait but how do you tell the counter of that entire plan if you cant communicate with him and how do you get that plan to the entire rest of the prison???
if the goal is to answer correctly if you were the last person or not, shouldn't all the 21 prisoners simply agree in saying, by the moment they got picked, "i'm not the last one". Because independent who is chosen first, he can say that he was not the last one and he will be right.
the number of prisoners is actually not important. you could have 3 prisoners, like in their end example, or you could have 300! you only need 1 Counter, and the rest to follow the plan. its a very small misdirection in that people will think an odd number like 21 has to be important, or youd just use a normal number like 5 or 10.
The statement "Men this is the last night either of you will ever see each other again" implies that, while you may be the 21st prisoner in the prison, there are only two prisoners currently in the prison. At that point a designated person can remember the state of the switches after they leave the room, and when you enter the room again they'll know if the other prisoner has been in there. I waited and saw the answer before submitting my comment, and I am going to leave the rest of it because I call shenanigans on the wording.
but the counter guy in the strategy would need to enter the room so many times and it would take too much time guards wouldnt give them infinite time though the prison is shutting down remember?
For those still confused: one person is recording the data. They send him a message by flipping the switch to on, he then flips the switch to off so a new message can be sent. When he has received enough messages saying "I've been here" he knows everyone has been there
What made it hard for me is I thought you had to proclaim when everyone had been in the room at least once when actually you can wait for everyone to have been in room twice.
3 mins in and I'm already confused. He said you're the 21st prisoner, not that there are 20 others still there right now. He said "either" which has to mean there are only two prisoners. The way the kept talking seems to mean those things are not true. He also didn't say what position the switches started in.
What if the prisoner who keeps counting the numbers doesn't enter the cell when more than 2 or more prisoners enter for the first and last time, and after that the sequence ends? Am I wrong at something or the riddle is wrong?
You cannot flip a switch that is on on, or a switch that is off off, so only one of these is counted. The one that originally turned the witch on. If neither of these prisoners turned the switch on, they still consider that they have two times left of flipping the switch on. Only the counter is permitted to flip the switch off.
okay but if the prisoners have no communication with each other how could they dedicate one person to a certain role? no one could talk about what they need to do as individuals.
I had an alternative answer: the question was... what strategy should be chosen to make sure you SURVIVE & the only condition where the prisoners are killed are if someone makes a false statement of everyone having been on the room, therefore, if you all agree to never make the statement, you will survive; you won't be set free either, but you will survive... Ps* The question should be rephrased as: how do you make sure you are set free
"Either of you"? So there's two inmates. You're the 21st inmate, but the first 19 are gone. I don't know why I'm doing this, 'cause if I'm right, I easily could have cheated. But if I'm wrong, *either* you flubbed the reading of the riddle by saying "either," *or* I get to learn I'm actually a moron who doesn't know what "either" means. Which'd be good to know. So. If there are just two people: go in; flip a switch. Remember the positions. When you go in again, if they're different than how you left them, you declare for freedom. Also, I hate those kind of gotcha riddles, which rely on misdirection; on hiding a key piece of information amongst useless fluff. But that is a perfect mirror of magic. Magic's just honest about what it is. If you did flub the reading by saying "either," I've no clue. And I can't be arsed to think too hard on it, on the assumption the riddle was messed up, 'cause then no part of it's reliable. EDIT: So, yeah, you said "Men, this is the last night either of you will ever see each other." Webster defines "either" as: "being the one and the other of two." Which... I honestly don't follow, but the point is it means there are *two* prisoners. So nyah, I'm right. But I'm also glad that wasn't it, 'cause it's actually a pretty cool solution. Cheers. Even if you done *goofed* telling it.
i knew this one heard it before everyone used light switch 1 then one person is designed to flip light switch 2 after 42 time light switch 2 has been flicked then all of them have been in that room
Your plan is flawed sir... if, the counter is randomly chosen first and then never gets chosen again, then no one ever gets out... improbable but statistically possible
only if you assume a finite run-time of the experiment... if you run it infinitely long the the counter MUST get in there as often as everyone else that's how probabilities work sure the inmates wont live infinitely long but that is not the question...
This riddle seems flawed. Are the prisoners allowed to keep entering indefinitely forever, even after all 21 have entered the room? So maybe they have all entered the room multiple times... and eventually the "counter" enters and declares everyone has entered? If so, that wasn't stated clearly, and this is a very boring riddle.
there still is a catch, u can mess up if the last person counts 39 and assumes the initial condition was on, while it was actually off, then then r all dead !!!
I'm now reading the comments, and (at the very least) TheZomiBeBeast and MemeLord_YaBish agreed with me! I then stopped reading the comments because there are too many of them. I have to go out now, but my bedroom light is on. I guess I'll just flip the switch and leave the room... hey....?!
The answer confused me even more than the original riddle did.
Screw all of that, I'd just use my skills from The Modern Rogue to break out everybody.
Wooops, thanks for catching that.
same, GET THE SPEARS
Right? That seems too hard....prison break is in order!
OK that news-anchorman bit was perfect
I am glad that the smaller version of the game was demonstrated. It helped my pre-breakfast mind to comprehend the larger version's properties and why it works.
This is a good computer science riddle.
Thinking in binary and memory states.
9:18 Brian : "And in your mind you're thinking?" Me: "Which beer belongs to who again?"
Brian's the only RUclipsr that makes the sponsored advertisements funny and watchable (and he's educated in memes lol)
I still don't get it even after the video...
One of the 21 prisoners will count how many times the left switch has been turned on. So when he enters the room, if it's on he switches it off.
The left switch serves a counting purpose.
The right switch is used by prisoners to get rid of their move when the left switch's position doesn't meet the required conditions:
The other prisoners don't touch the left switch unless it's off. And if it is, they turn it on; but maximum 2 times.
If a prisoner already turned the left switch on 2 times, he doesn't touch it anymore and only uses the right switch as "garbage moves".
That way, the counter only needs to keep track of how many times he reset the left switch. When that count reaches 40, it means everyone affected the switch at least once and the counter can safely say he's the last man to switch it.
spi rale: i understood your rundown a hell of a lot better than brians :)
everyone have to do it twice and ONLY twice. that way when its been done 40 times, he knows everyone have had atleast one go at it.
Also why would the inmates flip switch A twice? That makes no sense. All that needs to be counted is one trip per person, so why would you not limit each of them to flipping switch A once?
And they can clearly communicate amongst themselves (also, realistically, it's not like the guards can stop them from talking), so why would they not just flip any switch, then tell the counter to count them when they come out? Or maybe find an easier way to solve the problem than this bullshit?
I think the rules of that riddle are clear: no communication between prisoners once it starts. You're just allowed to plan ahead.
Also, for the "why you have to do it twice" part: 7:51
The only problem with this riddle is that you could theoretically never get released because it is statistically possible--however improbable--that at least one of the inmates never gets randomly selected...or that the counter person stops getting randomly chosen before everyone has entered the room at least once.
+Allan Kennedy why is that a problem?
They keep playing forever then. At least nobody is executed.
Actually you have a 20 in 21 chance of not being picked every time they pick, slim chance of not being picked ever but every time 20 in 21 chance.
Actually, over n selections, the chance of not getting released approaches 0 as n approaches infinity.
UnderdoneElm77 hopefully n approaches infinity before you die of old age lol.... numbers are infinite, life on earth is not
Good puzzle, but too much confusion. Proper rewording of the puzzle:
A prison has a room with a lightswitch and a broken ceiling light. The switch can be turned on and off, but it does nothing. The evil warden decides to torture four prisoners with a tough game. He tells them he'll set them all free if any of them can correctly announce that all of them have visited the room during the game. He shows them the room and says a guard will take them to the room individually many times, in random order, and each time will let them flip the switch if desired, and then take them back to lockup. They won't know whether the switch is on or off when the game starts. None will ever see or hear the others during the game, but now all of them must talk about how to use the switch-flipping to determine when one of them should announce that they've all visited the room. What should they do?
Answer: Have one of them be "Mr. Counter," and only he will ever turn the switch off. All others either flip it on or do nothing. Whenever one of them is brought to the room and sees the switch off, he flips it on, until he's done so twice. After that, he leaves it alone. Whenever Mr. Counter is brought to the room and sees it on, he counts it and flips it off. When his count reaches six, he announces that they've all been to the room.
Why that strategy works:
There are 3 prisoners that Mr. Counter needs to be sure have been to the room, and each will flip the switch on 2 times. 3 x 2 = 6 ons that Mr. Counter needs to see. Each time Mr. Counter sees it on, he flips it off. Since he's the only one who ever flips it off, he knows that whenever he sees it on, a prisoner has flipped it after he had flipped it off (or he's seeing it on the very first time he's there because it happened to start in the on position). The sixth time he sees an on, he knows either all three have flipped it on twice, or it started on and they've flipped it on five times. Either way, all three have been there, and so has he, so he announces that they've all been there.
Aaarggh I think I'm really close but the one thing that is throwing me is not knowing the starting positions.
What I have is that one of the switches is flipped every time by all of the prisoners except one. One of the prisoners (chosen before this starts) flips the other switch the first time he enters the room. The next person will see the other switch is flipped, and flips it back. When the chosen prisoner returns he counts 1 and flips that same 'important' switch. The next person who sees it in this state who hasn't already returned it to its original position flips it back. By repeating this, the chosen person can count up to 20 and ensure that everyone else has been in the room at least once. However this relies on knowing the starting positions and I can't see a way around unknown starting positions
Gonna keep thinking on it but I don't know how much patience I have left!
If it closes in a day just wait till 11:59 and then declare you're the last prisoner... after that the prison is closed anyway
I'm usually the riddle/puzzle bringer for our yearly man weekend, a riddle is proposed at the beginning of the trip and worked on the whole time. This will be in my arsenal this weekend.
Dude, I figured that out in like 10 minutes. Too easy!
Just kidding I gave up after 10 minutes I just wanted to be "that guy in the comments" for once!
I wish it was made clear that the switches can be given names A and B. Could also be that you get put into the room blindfolded and can't tell which switch is the right or left one if you don't know from which side to look at the two switches. Without that info my solution was to just never proclaim anything and technically survive until you die of natural causes. The guards or whatever even state that the inmates will never see each other again, implying the problem is unsolvable and that they will never be released.
The problem of watching these at home is that you can't ask for clarification on the rules so the wording has to be perfect the first time.
Just go in and the first time youre in there leave a hair in the corner and flip a switch. When theres 20 hairs in the corner and you havent left your hair your the last one.
the fact that he used words like 'either', 'last' and never implied that 21st prisoner was a current count lead me to believe that there was in fact only one or two prisoners, with a riddle the delivery needs to be exact because any information is seen as valuable, if the riddle is delivered oddly speculated fractures and it becomes possible to find multiple correct answers, in turn rendering the riddle in-valid
I thought the same thing when he said "either", which was going to make it waaay easier.
Great solution. Still, seems like there's another solution. The tactic "no one ever proclaims" also provides the stated winning strategy: "...make sure you survive". The prisoners survive - they survive in a never-ending life of solitary confinement and switch -flipping, but the winning condition is fulfilled. Since the prison is going to be closed, this can't really go on forever, but neither do humans - plus, plans to close Guantanamo Bay prison have historically not gone too rapidly. (Okay, here "survive" was just a poor word choice by Brian (or by Nick the puzzle writer), but on first hearing, one can read easily it as intentionally specific).
This seemed complex once you revealed it, it when you showed the example with the beers it made sense. Great puzzle!
Awesome!
The bit that threw me, is I didn't get that if no one ever proclaims anything, they will just continue playing the game forever.
I’m so confuzzled by this
But Brain I have class in the morning, I can't rack my brain on this one all night
Brain I have class in the morning ^^
Stefan Reich You can see how tired was, didn't catch that
Damn it...I glanced and thought I saw Paul F. Tompkins...but it was just JuRY
That meme though, love you Brian. Awesome adds as usual.
I love those Riddles a lot. There are so many good ones out there.
So the counter has to visit the room at least 40 times? Wouldn't that take a very long time chosen at random from pool of 21? And wouldn't his declaration possibly come some time after everyone had visited? So with death on the line why not just flip switches for a couple months then declare? I'm probably missing something.
Wait... there's a major problem with this riddle. If the guards are trying to decide whether to execute or free all the prisoners, there's no way they'd use a system that could go forever. What if nobody ever proclaims that everybody has been in? Will the guards just wait forever?
how can someone that talented end up being prisoner
criminal mastermind
The answer can also be something far more simple. Either way the person who says they are the last are going to be the last, so no matter what they are correct and everyone must be let go. Either they legit are the last ones and everyone would go free, or they say they are the last and the game stops and everyone is executed, but that would make them the last person, so technically the guards would have to let them go.
If you say 'I am the last prison' technically, at that moment, that prisoner *is* the latest prisoner at that time.
The goal isn't to make a true statement, it is to make the statement when the conditions have been met.
1psplegend In other words don't go full smartass on the guards or you are screwed. 😉
I never heard how they'd designate which switch is A or which is B? Are they labeled? If you've never been in the room you couldn't say the switch on the left is A and to the right is B. They may be configured any number of ways.
Really enjoying the new RUclips speed playback settings
The explanation of both the riddle and the solution could've been better, but it's still a really cool riddle!
They should leave a piece of clothing only for the first time they were selected except one. Therefore when the 21st prisoners sees 20 pieces of clothing he will proclaim he's the last one to go.
Not even close
Options:
1) You could leave some sort of marking in the room, maybe a piece of clothing, or etch something into the wall.
2) Every inmate was in that room before, as it was the room you were all in last night or is a common room, so the first person can proclaim everyone was there already.
3) Since the riddle did not say you have to proclaim it the first instance every inmate entered the room, you could wait an infinite amount of time, and you'd would know for sure that when time reaches infinite, every inmate has gone into the room.
4) "Leave the fate in your hands." So does that mean I'm the last one in the room?
5) "Men, this is the last night either of you will see each other" So there is only two of you?
6) "You are assigned as the last inmate" So there is only one of you?
7) You could kill all the other inmates in the last night phase, then you would know that when you enter the room, all the remaining inmates have entered the room.
Great puzzle. Got me stumped, but hear me out: I don't think the non-counters have to flip switch A twice. They can just do this: if A is on, leave it alone; if A is off, flip it on, but only do it once. The counter is the only one allowed to flip switch A off. This way, the counter will only have to count to 21 (rather than twice that). He doesn't have to count himself because he'd know if he's been in the room or not, but in case switch A started out as "on", he should count one extra, making the total count 21. Seems like a faster solution. Anyone find any fault with it?
By counting one more. The switch starting as ON is the same as there being an extra person who turned it on.
Except he doesn't know whether or not the switch started ON or if someone has entered the room before him. By adding one, if the switch started OFF, and someone entered before the counter and turned it ON, the game would never end, as everyone has flipped the switch once but the count remains at 20 and they have no way of knowing if the last person just hasn't been called yet. The concept of flipping it twice each is adding that last degree of certainty, where if the switch starts ON, one prisoner will only interact once, but it is guaranteed that everyone has done so.
Oh snap. You're totally right. I'm convinced. Thanks for clarifying.
so the answer is that it's never explicitly stated that the final prisoner must declare that he's the final one on his first visit into the room, just that someone at some point must declare that all 21 prisoners have been in the room, and it's possible that all prisoners have been into the room, but none of them need to make a declaration yet because there is no penalty for being without declaration, the only penalty is if the declaration is wrong?
that guy rocks that moustache
The key condition is that "everyone HAS entered the room and switched the switch." It doesn't matter if they all HAVE went in once and this was your first time entering the room and NOT declare they all have entered. They will keep going until someone mess up by declaring too early OR until someone declare after they know for sure that everyone has participated.
FASCINATING puzzle !
I spent several nights working on it (better way to fall asleep than counting sheep if you have insomnia !) and I think there's a simpler (and more logical) solution - or at least easier to explain!
The key missing information ( SCAM !!!) is that the prisoners will be individually taken into the room AS MANY TIMES AS IT TAKES until one finally proclaims "all the 21 prisoners have been in the switch room" . And based on the fact that there are 21 prisoners and each must go into the room randomly at least as often as the "Mr Counter" prisoner does (which is minimum 21 times randomly), that's at least 21 x 21 = 441 times on average that prisoners must go into the room = much more than one night's work !
OK .......
1. One prisoner is " Mr Counter" as per the official solution
2. It's also agreed that one switch (e.g left hand one ) is the counter switch as per the official solution
3. The other (right hand) switch is a "junk" switch which has zero meaning (can be toggled at random just because one switch must be toggled - if for no other reason)
4. Only Mr Counter can turn the counter switch from OFF to ON
5 All other prisoners can only turn the counter switch from ON to OFF and ONLY ONCE (thus if they have already turned the left hand counter switch from ON to OFF they just toggle the right hand junk switch if /when they go back to the room again and again)
6. Every time Mr Counter sees the counter switch in OFF he knows one more prisoner has been in the room for the first time. . He then turns the counter switch back to ON and waits for the next time it is turned OFF to count "one more prisoner" .
7. However, the initial state of the counter switch is a problem if it is initially in the ON state which can obviously lead Mr Counter to miss one prisoner. The solution to this is for every prisoner to turn the counter switch on TWICE. 20 x 2 = 40, so if Mr Counter reaches 39 times ON he knows for certain that all 20 other prisoners have been in the room (and so the trial requires - on average - 21 x 21 x 2 times in the room = 884 x
Based on this solution it would be simpler to just state the puzzle as having one switch and the prisoners have the option to toggle - OR NOT .
But that would be a big clue to the answer, so by having two switches it makes the puzzle more difficult.
The second switch is a red herring (and so is the number of prisoners = 21) !
SCAM / SCAM/ SCAM .... Brian I "hate" = love you for setting such nasty puzzles !!!!!!
_____________________
Having listened to the video a dozen times I now understand that both Brian and I are saying the same thing in different ways so I was right !
Here's the same puzzle and (easily explained) solution from the internet (except that it does not correctly deal with the A switch being initially in the ON position, as far as I can see)
Question: A prison has 23 prisoners in 23 different cells. The prisoners have no way to communicate with each other in any way from their cells. There is another room, the rec room, that has two switches on the wall (A and B). The switches have on and off positions but they start in an unknown position.
Prisoners are randomly taken to and from the rec room one at a time. They must change the position of only one of the two switches each time they go to the room. At any point a prisoner can yell out, "Every prisoner has been here!" If the prisoner is correct that all of the prisoners have visited the rec room, then they all go free. If they aren't correct then they are all executed.
Before they start they are given one planning session during which they can discuss a method to win the game. What method can they use to ensure they all go free?
Answer
Here are the rules they can use to ensure they will all go free eventually:
The prisoners will choose one 'leader' and everybody else will be a follower. If you are a follower:
If switch A is in the on position, toggle switch B.
If switch A is off, you have not toggled switch A yet, and you have seen switch on during a previous visit; then toggle switch A. Otherwise toggle switch B.
If you are a leader:
If switch A is off, turn it on.
If switch A is on, turn it off. If you did not turn switch A on during their previous visit, increment the count of prisoners.
Once the leader increments the count to 23 they can yell, "Every prisoner has been here!" and all of them will be released.
im at 7:08 and now have 'worked out' the answer. although sounds like you dont need to be the first one after everyone else has already been in. which also means, that you could just say: no one ever say everyone else has been in. then the guards will be stuck there doing this forever
edit: yep, sounds like that little clue was enough. also what was with different audio and video at 7:32?
What if it took more than a day for the counter to enter the room and be able to flip the A switch 40 times? If it's closing the next day, wouldn't everyone be executed?
Guards: "you must predict for certain that everyone has been in the room before. You have one chance to come up with plan." Me: "Hey, can you show us the room with the switches while we plan?" Guard: "um, i guess." Everyone goes to the room. Me: "I can say for certain that we have all been to this room." Everyone goes free.
My brain!!! It Hurts.... I`m pretty sure that this puzzle broke my brain...
I don't get it
What if it's the same two prisoners, one of which being the counter guy and one just a random prisoner. Once the counter guy gets to 21 he will think he's been the last prisoner but he actually hasn't.
Can someone please explain this to me????
Let's run through an example with 3 prisoners. (1 counter and 2 others)
So say this is the console
A》
On. Off
B》
The counter is only worried about two things.
1:reseting A to the off position.
2: keeping track of the number of times he does so.(he is counting to TWICE the number of prisoners excluding himself.)
All other prisoners have one objective
1: turn on switch A TWICE and only twice.
So we start in some arbitrary position. Let's say.
《A
On. Off
B》
A prisoner is chosen at random. Let's assume it's the counter!
So the counter sees A is on and therefore resets it.
A》
On. Off
B》
His count raises to 1
(now notice the counter has a count of 1 even though no other prisoners have preceded him. This is why the count needs to be doubled in order to account for the possible + 1 the starting position can inflict on his count.)
So the next prisoners chosen. Let's call him prisoner 2. He sees A is off and so switches on for his first time.
《A
On. Off
B》
Counter is next again and resets A another time. Bringing his count to 2
A》
On. Off
B》
Next is prisoner 2 again. He turns on a for his second and last time.
《A
On. Off
B》
So next is Prisoner 3 he sees A is in the on position and therefore switches B(the trash switch)
《A
On. Off
《B
Next up is the counter again. He resets A bring his count to 3
A》
On. Off
《B
Next is Prisoner 2. He has already switched A the maximum number of times therefore he switches B
A》
On. Off
B》
Next is prisoner 3 and he switches A for his first time.
《A
On. Off
B》
Next is counter and he resets A.
A》
On. Off
B》
bringing his count to 4 and ensuring everyone has indeed gone!
Isaac Long each peisoner only flips the first switch on twice after they have flipped it on twice, they flip the second switch on their turn. The counter doesn't cares about the second switch.
I've heard a similar riddle before, but you get out much sooner if you just wait enough days such there is 99% that everyone must have been in the room and then claim it. Small risk, but faster to get out.
Why are the 40 counts necessary? Everyone but the counter flips the agreed switch exactily once to on (or off, whatever is agreed upon), and the counter ignores the first state (counting himself as 1, the first known prisoner in the room) After 21 counts, he's sure. Why 40?
If the switch starts in the on position, then after 21 counts only 20 people have visited the room. At 40, 19 of the people have gone twice, and the 20th has gone at least once.
But it depends on how frequently the guards bring you in the room. in mean there will be 21 visits for each of the 21 inmates. If the guards pick one person per week. This strategy would take 8.45 years to come to an end. But when the counter reach half the count at 22, there is only one chance out of 2^21 that someone was left behind. Would you really expect every to spend 4.2 years more in prison to avoid the single chance in 2 million to die ?
Small problem, the guards could have withheld one prisoner allowing only 20 prisoners so prisoner 21 could have made a mistake. The riddle only shows that prisoner 21 has been in the room twice in a row.
watch again.
I got so close, but I got hung up on the starting position - I didn't think to make each prisoner flip the counter twice instead of just once.
The prisoners better hope that they get taken to the switches often, because they have to go in there almost 1000 times (45 each) with this strategy, mostly just waiting on the counter. If it only happens about once per day, then it would take almost 3 years for them to escape. I'm sure that one of the inmates would crack long before the strategy is over.
I've just realized this is the same visual setup as the 6-digit miracle multiplication (same shirt, same shwood, same venue, same approximate time), but with different guests. Huh.
ACTUALLY... hold the fort a dang minute!! I just thought that the guards themselves could add shoes to the pile, thereby messing that plan up... but by the same token, the guards could flick the switches themselves between bringing in the inmates, and that would mess the real answer up too. I guess there's no solution to this puzzle after all.
Cool puzzle but how do the prisoners know how to pick a counter when no communication is allowed?
I got it. I have heard a variation of that riddle. It is easy.
i've heard the riddle before (and solved it after a couple of hours). great riddle.
This is one of those riddles where I can never remember the solution no matter how many times I come across it.
I had a guess at 4:00, which turned out to be wrong, but it is so much easier than the actual answer.
If it is your first time in the room, and you are wearing two shoes, you count the number of shoes in the corner of the room. If it is twenty, you declare that you are the last of the prisoners to enter the room. If it is less than twenty, you take off a shoe, put it in the corner, flick a switch and leave the room. If you come in the room wearing one shoe, you just flick a switch because you have to. Nobody said that the ONLY thing you could do in this room was flick the switch!!
I still like my answer. If no one ever proclaims they're the last guy, there's no loss condition or win condition and it continues ad infinitum.
he rambled off that expenation so fast, that my brain nearly exploded.
3 persons and beer explenation makes more sence.... you just have to reach 4 minimum, either way
because if on = count = 1
person 1 2x = 3
person 4 1x= 4 every one went in at least once
person 4 2x = 5
or if off.
person 1 2x = 2
person 2 2x = 4
The not-knowing-the-original-position is the bit that threw me. :P
The problem is that I thought that if you were the last one to enter, you'd have to immediately state that everyone has been in there. That info was missing.
Figured it out after first hint. Another hint is that not all the prisoners have to be following the same rules.
So the key is there is no time limit and you don't need to announce all people have been in the room at the earliest point.
So the missing info is that this could take twice as long as you think.
i think a even simpler way to do it is if you go in the room and leave a single item behind they can count the items that are in the room Note: they said a empty room so theres nothing in there to begin with so this is a possibly solution
A problem with the video was that Brian didn't correct them until a bit later on that saying "last person" is totally erroneous, as the win condition is just correctly stating that everyone has been in the room before at least once.
A/the problem with this puzzle is that it gives no time limit. Considering that it's closing up there would need to be some sort of time limit, yet this strategy not guaranteed work with a time limit if the selection was random. Making the "win condition" more clear would have also been more appropriate, since clearly people were misunderstanding the puzzle, which is never a good thing.
In other words, it should be explained that the guards will be ensuring that everyone does get selected multiple times before the time limit is up (and obviously the prisoners won't be informed of the time limits's duration since then they could just wait until the end), because that is _required_ for the strategy to work.
In other words this is either a somewhat unfair puzzle or else a bad analogy. A better analogy might be that you're trapped in some sort of eternal prison and need to get back to reality/life. This way the solution is one that is not only certain but which gets out with an optimal amount of time lost.
I thought it would be easiest for everybody to leave the same article of clothing in the room, and once 20 other people have gone, you'll know you're the 21st since there's 20 other articles in the room.
-"This week on Scam School,- "Every week on Scam School, a puzzle from one of you guys that broke my brain!"
I didn't give up. Just never answer and u all live. THe question was how to make sure you all survive. So that's a possibility a lonely one but w/e. It works
You could just say you're the last because you are. What you say ends the cycle you are always you last one until the next comes in. So if you say you are last you are not wrong and everybody is set free
Game ends when someone proclaims. Goals is for all of them to survive right, not for all of them to escape. So solution: Keep playing forever. But yeah, I don't think the initial condition matters. You just don't keep track of if it's on or off, just if it's different. Everyone keeps flipping B until the counter comes in, changes A. Next guy comes in, if it's his first time, he switches B. If not, he sees A has changed so switches it back. The rest will not notice A has been changed. Counter comes in again, sees A is changed so counts 1 and changes A again. Not sure if this is faster than the solution in the video (I think it is, because everyone only needs to switch it once), but at least it is not dependent on the initial conditions. Although I guess this becomes a problem if the counter is the first person to enter the room, because nobody would know he's been in there, while he would just be waiting for someone to notice him.
Loving this channel
wait but how do you tell the counter of that entire plan if you cant communicate with him and how do you get that plan to the entire rest of the prison???
But what happens if the counter only enters the room one time?
if the goal is to answer correctly if you were the last person or not, shouldn't all the 21 prisoners simply agree in saying, by the moment they got picked, "i'm not the last one". Because independent who is chosen first, he can say that he was not the last one and he will be right.
the number of prisoners is actually not important. you could have 3 prisoners, like in their end example, or you could have 300! you only need 1 Counter, and the rest to follow the plan.
its a very small misdirection in that people will think an odd number like 21 has to be important, or youd just use a normal number like 5 or 10.
I thought that was a nice touch, too.
Not sure I'm following. What if the counter goes in 1st or 5th or 10th and never goes back in? Who keeps count then? I think this solution is flawed.
Never goes back in again*
The statement "Men this is the last night either of you will ever see each other again" implies that, while you may be the 21st prisoner in the prison, there are only two prisoners currently in the prison. At that point a designated person can remember the state of the switches after they leave the room, and when you enter the room again they'll know if the other prisoner has been in there.
I waited and saw the answer before submitting my comment, and I am going to leave the rest of it because I call shenanigans on the wording.
+whotheheckcares oh, jeez. So you're that guy.
How do you ASSIGN anything? He explicitly stated that there is no communication.
*once you're imprisoned. The whole point is to come up with a plan beforehand.
but the counter guy in the strategy would need to enter the room so many times and it would take too much time guards wouldnt give them infinite time though the prison is shutting down remember?
5:13 binary counter? Maybe? You have 1’s and 0’s
they should make a mark when its there turn
Reminds me of the puzzle with the 3 light bulbs but way harder
For those still confused: one person is recording the data. They send him a message by flipping the switch to on, he then flips the switch to off so a new message can be sent. When he has received enough messages saying "I've been here" he knows everyone has been there
What made it hard for me is I thought you had to proclaim when everyone had been in the room at least once when actually you can wait for everyone to have been in room twice.
3 mins in and I'm already confused. He said you're the 21st prisoner, not that there are 20 others still there right now. He said "either" which has to mean there are only two prisoners. The way the kept talking seems to mean those things are not true. He also didn't say what position the switches started in.
What if the prisoner who keeps counting the numbers doesn't enter the cell when more than 2 or more prisoners enter for the first and last time, and after that the sequence ends? Am I wrong at something or the riddle is wrong?
You cannot flip a switch that is on on, or a switch that is off off, so only one of these is counted. The one that originally turned the witch on. If neither of these prisoners turned the switch on, they still consider that they have two times left of flipping the switch on. Only the counter is permitted to flip the switch off.
Vergos Ioannis they continue forever, so eventually he will enter multiple times
oh i've missed the infinite sequence part, thank you for the feedback ;)
okay but if the prisoners have no communication with each other how could they dedicate one person to a certain role? no one could talk about what they need to do as individuals.
I had an alternative answer: the question was... what strategy should be chosen to make sure you SURVIVE & the only condition where the prisoners are killed are if someone makes a false statement of everyone having been on the room, therefore, if you all agree to never make the statement, you will survive; you won't be set free either, but you will survive...
Ps* The question should be rephrased as: how do you make sure you are set free
But you only got a day. So if you don't solve it in that time you'll be executed anyways
You totally have to do something with that news caster character
"Either of you"? So there's two inmates. You're the 21st inmate, but the first 19 are gone. I don't know why I'm doing this, 'cause if I'm right, I easily could have cheated. But if I'm wrong, *either* you flubbed the reading of the riddle by saying "either," *or* I get to learn I'm actually a moron who doesn't know what "either" means. Which'd be good to know.
So. If there are just two people: go in; flip a switch. Remember the positions. When you go in again, if they're different than how you left them, you declare for freedom. Also, I hate those kind of gotcha riddles, which rely on misdirection; on hiding a key piece of information amongst useless fluff. But that is a perfect mirror of magic. Magic's just honest about what it is.
If you did flub the reading by saying "either," I've no clue. And I can't be arsed to think too hard on it, on the assumption the riddle was messed up, 'cause then no part of it's reliable.
EDIT: So, yeah, you said "Men, this is the last night either of you will ever see each other." Webster defines "either" as: "being the one and the other of two." Which... I honestly don't follow, but the point is it means there are *two* prisoners. So nyah, I'm right.
But I'm also glad that wasn't it, 'cause it's actually a pretty cool solution. Cheers. Even if you done *goofed* telling it.
before I saw more than to 2:58 I was thinking every prisoner could drop something in the room. Like Hansel and Gretel
Amazing!
i knew this one
heard it before
everyone used light switch 1
then one person is designed to flip light switch 2
after 42 time light switch 2 has been flicked then all of them have been in that room
Your plan is flawed sir... if, the counter is randomly chosen first and then never gets chosen again, then no one ever gets out... improbable but statistically possible
only if you assume a finite run-time of the experiment... if you run it infinitely long the the counter MUST get in there as often as everyone else that's how probabilities work
sure the inmates wont live infinitely long but that is not the question...
This riddle seems flawed. Are the prisoners allowed to keep entering indefinitely forever, even after all 21 have entered the room? So maybe they have all entered the room multiple times... and eventually the "counter" enters and declares everyone has entered? If so, that wasn't stated clearly, and this is a very boring riddle.
there still is a catch, u can mess up if the last person counts 39 and assumes the initial condition was on, while it was actually off, then then r all dead !!!
I'm now reading the comments, and (at the very least) TheZomiBeBeast and MemeLord_YaBish agreed with me! I then stopped reading the comments because there are too many of them. I have to go out now, but my bedroom light is on. I guess I'll just flip the switch and leave the room... hey....?!
The riddle doesn’t specify that the switches are in a “ left right” row so without knowing how the switches are configured this is impossible.
That ads. Somebody actually need to do a full episode of just scamschool ads.