The Riddle That Seems Impossible Even If You Know The Answer

[@pyguy9915]

Something seems wrong at 9:00 What is the probability of a loop of length 1? (Can't be 1/1) Length 2?

[@joekemp83]

That calculation applies only for n>50.

[@veritasium]

Yes it is 1/1 but you have to treat it as an expected value. On average in any arrangement of 100 slips in boxes you should expect one loop of length 1. But sometimes you won’t get one. Sometimes you’ll get two or three etc. So if you had 1000 random arrangements of 100 numbers, you’d expect to find 1000 loops of length 1 500 loops of length 2 333 loops of length 3 etc. 10 loops of length 100 In the graph we show something slightly different, the probability that a loop of length L is the longest loop. If L>50 it must be the longest loop so the probability it exists and the probability it’s the longest are the same. If L

[@_timelapmaker_9755]

@@veritasium Agreed

[@gc852]

@@veritasium I think one should applies a variation of the fixed point theorem here. i.e., a function mapping a domain to itself has a fixed point.

[@joekemp83]

@@gc852 Only applies to continuous functions.

[@DrDJX]

As somebody that's tried tracking down a CD left in the wrong CD case, I can attest that the loop strategy does indeed work 31% of the time. (The other 69% of the time it turns up weeks later on the kitchen table.)

[@albevanhanoy]

Best comment, 10/10

[@alveolate]

lmaooo what an amazingly real-life example of this! unfortunately... CDs are no longer common enough for most people under a certain age to really get this example :(

[@NZsaltz]

To be fair, it should work 100% time if you don't have a warden forcing you to only open half.

[@R.B.90]

Omg I forgot about that. Lol I swear we all did this for CDs, DVDs, video games. The loop has been right in front of us all along :)

[@theglitch312]

@@NZsaltz Wait? You _don’t_ have somebody execute you if you open more than half of your CD cases?

[@magdasg9571]

Memorizing this just in case I'm ever trapped in a prison with a sadistic mathematical prison warden

[@spyro440]

Well - are you Korean? ;)

[@94XJ]

@@spyro440 you win. 🤣🤣🤣🤣

[@EgoExit]

Lmaooooooo

[@Garvit_M]

The only way you're coming out alive foso is you're the only participate , otherwise you're fucked 😂

[@jackinkc767]

Thanks for the suggestion. Also, bring canned fool.

[@wetbadger2174]

When you factor in the odds of one nerd convincing 99 other convicts to go with this strategy, your chances quickly fall back to zero.

[@guido3721]

Underrated comment

[@ashcheung4325]

Yeah

[@tolep]

You need a nerd with charisma.

[@margaretjones5572]

Hmmmm!!!?! Where theoretical maths meet the "real world" and provide the opportunity to show that the human species' success is tied to cooperation within the species as well as groups within that species...like family units. When we cooperate as a family to follow known solutions to problems we have a 1/3 chance of succeeding. Where as groups who refuse to cooperate towards a goal have an almost zero chance of success!!

[@MarkoMikulicic]

That shows your bias. Thanks to KZbin, Derek and Brilliant, we're heading towards a bright future were all prisoners are going to be nerds. Wait...

[@tfdtfdtfd]

The main point here is that "failing hard" comes with no harsher penalty than "failing little"....hence, you can redistribute your loss function to take advantage of this. Great video, btw!

[@ZamanAristoOrCleon]

Imagine being the first inmate and not finding your number. “Oof, we tried”

[@Andy-lm2zp]

or the last!

[@funnygreenguy]

@@Andy-lm2zp if you were the last one and you failed, then there should've been MANY others also failing.

[@ramuroy6088]

@@funnygreenguy only when you follow the strategy mentioned in the video...

[@funnygreenguy]

@@ramuroy6088 yeah, I automatically assumed it from the original comment 😀

[@Spectre0799]

I like to think that the people constructing the test purposely made no loops of >50, so that the prisoners could go free if they are intelligent enough to be worth returning to society (if they figure out the loop method then they will be given 100% chance)

[@fixed-point]

Interesting corollary: If prisoner #1 (or any other prisoner) finds that his own loop has a length of exactly 50, he immediately knows there's a 100% chance of success.

[@Ryuseigan]

Really?

[@AthAthanasius]

@@Ryuseigan Yes, because then the other 49 on that loop will obviously also succeed, and anyone else is on a loop of *at most* 50 (two loops of 50 is all there is), and thus will also succeed.

[@shinichigojir12]

Because then there would be no other loops that can be greater than 50. Any other loops guaranteed to be lower than 50 so its guaranteed win.

[@deegobooster]

@@Ryuseigan yes really. The largest loop size of the remaining boxes cannot exceed 50, because 50 has already been used in that first loop prisoner #1 found.

[@nitinkriplani5811]

@@Ryuseigan Yes, because there cant be same box on different loops. So a loop of 50 boxes would imply that remaining boxes at max can be 50 so no loop goes above 50 in size, hence 100% success

[@gregsquires6201]

I think the chance of convincing 99 other prisoners that this strategy is their best chance of survival is much lower than 31%.

[@lavarel]

@@pokejinwwi one? Out of 100? That's even more impossible. We're talking inmates here,

[@tvao9010]

Some guy will want to go on his lucky number and a lot of other dumb stuff would happen

[@gobbedy]

at least 7 of them will pray to jesus for the which boxes to open

[@zraakuladann3946]

One will try to escape and get everyone killed.

[@user-sl6gn1ss8p]

@@tvao9010 the play there is to add (lucky number - inmate number) to all boxes and proceed as usual : )

[@sparkythesecretsquirrel4013]

I needed this. I woke up this morning not feeling sufficiently stupid. Problem solved.

[@astronut2463]

😂

[@bscorvin]

My actual concern if this ever somehow became a situation I got myself into is that someone would decide this is stupid and just pick boxes at random

[@miloszforman6270]

_"that someone would decide this is stupid and just pick boxes at random"_ That's frequently a problem in real life. Not so much in the fabricated setup of the video. Persons with a strong determination are usually the ones which make the decisions, and not always they are very smart. Like this example from recent (and many others from less recent) history: Smart mind says: "Don't go to war against xxx, IT WILL NOT WORK." Strong will says: "Of course it will, you're a coward and a traitor." 20 years later, it did not work. Strong will says: "Nobody could have known that, therefore it was the right decision at that time."

[@szymonl4363]

Exactly. Every idiot deciding to go for random nubers roughly halves the chance of winning. Anyway, here is some useless math I spent half an hour on (what am I doing with my life?). Theoretically even with googol or googolplex people (assuming that we just don't worry abut time it would take to open half a googolplex boxes gogolpex times) you would have an over 30% chance of winnig, but when talking abut actual people, not idealized beings, this just doesn't work. Even with the googol people case and (10^-48)%, wich is 0.000000000000000000000000000000000000000000000001% people being idiots and choosing randomly, your chance would be(3/2^(10^50))%, wich i think is way under 1/10^10000, or 1/googol^100. When I plugged this into calculators (even the most precise ones I could find), they straight out gave me 0, witch speaks for itself.

[@Solo_man69]

If that happened in real life to you u would have a better chance just starting a jailbreak

[@Drumaier]

A legit concern.

[@mibsmibs9166]

If one person did it, it would just reduce the odds of sucess by 50%, 15% still isnt bad

[@Bismuth9]

6:35 I like that Derek's "random" numbers were all odd numbers. We have a bias towards perceiving odd numbers as more random than even numbers. Even more so, of the 9 digits in these numbers, only a single one was even.

[@jynx619]

Your observation is not necessarily true. It could just be that Derek randomly picked 5 odd numbers, and this has a probability of 2.8%

[@johnhunter7244]

They are called *odd* for a reason

[@HYPERWATER]

Raoodnm

[@josenobi3022]

@@jynx619 how did you calculate that probability

[@janbacer]

@@josenobi3022 I'd do 1/2⁵ but that's 3.1% 😕

[@reifrei1170]

really sad that these prisoners were so good at math and cooperation, yet still ended up in jail 😢

[@markgunther2502]

White collar criminals no doubt.

[@shardulthakur7362]

Lmao laughed too hard at this

[@shardulthakur7362]

May be all are in for stock market manipulation

[@hollamonm]

They're all just people who went to jail for maximum time for possession of weed in the US and one of them happens to be a math professor who explains how it works to the other 99. (This is literally a thing that occurs in the US and especially among black and latino communities because racism! I'm being sarcastic, but this is true. You're less likely to end up in prison for possession alone if you're white and I hate that that's true still.)

[@Darker7]

Political dissidents: First time being genocided? :Ü™

[@sarahs.6377]

Mentally filing this away next to "the mitochondria is the powerhouse of the cell."

[@michaelgove9349]

As a former professional gambler, the key to understanding this in real-world terms is at 9:48. Every prisoner's individual chance of success is still 50%. The strategy works by making the prisoners' individual chances contingent on each other: linking them together. Imagine a related puzzle - the sadistic warden has been told that the median human heart rate is 75 beats per minute. He devises a game where he measures the 100 prisoners' heart rates. He has two large bins marked "UNDER 75 BPM" and "75 OR OVER", and he places each prisoner's ID tag in the bin corresponding to their measured heart rate. But there are two catches. Firstly, he has privately flipped a coin before the game to decide which will be the winning bin: the prisoners don't know which is which. And secondly, *all 100* prisoners have to win the game for them to be freed. So each prisoner's chance of winning is 50%. And with no collective strategy, the chance of everyone winning is 0.5 to the power 100. Tiny. But the collective strategy is simple: everyone does extremely vigorous exercise immediately before getting tested. Now *everyone's* heart rate is above the median. So although each person's individual chance of winning the game is still 50%, now the collective chance of winning is also 50% - because *everyone is now in the same bin.* Basically, that's how to grok the video's strategy for the boxes problem - in a sense it puts everyone "in the same bin" - and the bin is marked *"Are all the loops shorter than 51 ?".*

[@michaelgove9349]

Notice that the Skyum "loops" strategy works because the room *is the same room for everyone.* And the strategy takes advantage of a particular mathematical property of the way the numbers are sorted into boxes *in that one particular room.* If there were 100 *different* rooms, with the numbers in the 100 boxes sorted randomly in each one, then the chances of *everyone* succeeding would again be huge. Because now all of the 100 outcomes would be statistically independent.

[@yusufahmed2233]

Ok, that was an AWESOME example! Mind = blown 🤯🤯

[@balintvarga5146]

Absolutely stunning example.

[@LiveHappy76]

Did you retire early and because of success in that? One of my brothers learned a roulette betting strategy, something about (a) betting on 1/3 of the numbers each time and (b) setting a predetermined loss limit at which you'll stop betting. You win more often than lose, but do lose. He was kicked out of casinos for using it, even though it breaks no rules. I've been tempted for years to try such things but have always resorted to status quo of work a job for a paycheck....

[@thomasrosebrough9062]

I appreciate this version of the explanation to point out where the "magic number" comes from by reminding us that the criteria for "the bin" is arbitrary

[@inemanja]

As someone that went to prison, I can tell you with 100% confidence, that they got more chances to win by randomly picking boxes (one in 8*1^32), than 100 of them to agree to ANY strategy.

[@flogzer0]

I actually had this happen to me in a Turkish prison. I came up with it on the spot and saved us all.

[@aaron2112]

@Michael Ritsema sure dude. Whatever.

[@Houstonruss]

@@aaron2112 He save hundreds of us! I owe my life to michael for his solution!

[@Brormable]

@@aaron2112 it's true, I was one of the inmates and Michael is a true genius

[@aspiringdiamond]

@@aaron2112 Michael saved my life in that Turkish prison, he isn't lying

[@GuitarGuise]

The way I like to think about the solution: you're no longer betting that each individual prisoner will find their number with a pattern that they choose (arbitrary or intentional), but you're betting on the probability that a pattern (a loop exceeding a length of 50) does not exist in the set of boxes. And that's a static property of the set you're betting on, in contrast to rolling the dice every time on 50 different prisoners. So, in a way, you've already succeeded or failed by choosing the loop strategy, whether you know it or not. Random chance no longer has anything to do with the prisoners' choices (unless they mess up the execution of the loop strategy), but entirely on how the box set is arranged and the loop strategy that the prisoners decide to employ at the start. Fascinating mathematics! Thanks for sharing this!

[@quixoticPrancer]

Yeah exactly. For a video that set out to make a complicated thing understandable, there is room for considerable improvement...

[@brendawilliams8062]

51 would be set of three boxes. : 5.0999011. Or 352319696. You can’t jump 51

[@sonkeschmidt2027]

With the implementation of the strategy you change an element of randomness in the problem. The randomness of the prisoners behaviour. Just like a conductor changes noise into order with the swing of a wooden stick based on mutual understanding of the symbolism of the stick. The power of cooperation, hence the saying a bad plan is better than no plan.

[@enzzz]

@@sonkeschmidt2027 I mean with this one, it's so much about the magnitude of difference if you notice this behind the hood pattern of loops. You could also create a very bad plan where everyone cooperates, like maybe prisoners might think that each should go 1. 1 - 50. 2. 2 - 51. 3. 3 - 52. ... 50. 50-99. 51. 51-100. 52. 52-1. 100. 100-49. thinking this is better than completely random. This is for example the first idea I thought to test if it would have any influence at all, what if they increase number one by one. I'm not sure if this actually will increase odds in any meaningful way. Didn't actually test it out, but I knew it definitely won't give near 1/3 odds. So I would say it's more about noticing this really obscure but effective strategy rather than cooperation really. You of course need cooperation to carry it out.

[@sonkeschmidt2027]

@@enzzz what is this obscure strategy going to do without cooperation? How you going to implement it? Or even find it out? You would need someone to come up with it and convince everyone to follow a strategy with just 30% chance of winning. That is not a good plan. It might be the best but it's not a good one. To get this to happen requires a shitton of cooperation.

[@geraldcollins5684]

I've seen a variation of this some years ago. Didn't remember the way to solve but it came back quickly when you gave the solution.

[@ltmcolen]

you've got to admire these mathematicians for thinking out of the box

[@MusicSounds]

literally this time

[@-Jethro-]

I see what you did there!

[@donc-m4900]

But why where they in jail to begin with 😂

[@abinbaby4044]

Or out of the loop!

[@alveolate]

@@abinbaby4044 actually, into the loop xD

[@charliehorse8686]

If you think the riddle is hard, imagine trying to convince 99 fellow prisoners to follow the plan to the letter.

[@davidjames1684]

No letters, just numbers, ha ha.

[@StabbyJoe135]

@senni bgon you've clearly never been in prison. Or met someone with ADHD.

[@andrzejbozek]

xd

[@christiankrause1594]

Then i have another riddle for you: You are sitting in a restaurant and listening to the neighbors table. You listen to three different woman talking. a) One says: I have two childs, Martin, the older child, just got his driver license. What is the probability for her other child also being a boy? b) Two says: I have two childs, Martin just got his drivers license. What is the probability for her other child also being a boy? c) Three says: I have two childs, Martin just got his drivers license. He was born on a wednesday. What is the probability for her other child also being a boy? SOLUTION: a) 1/2, b) 1/3 c) 13/27 Explanation: b) Not possible is the birth of G/G, possible is B/B, G/B (girl older) and B/G (boy older). Each of the three B/B , G/B and B/G are equal with a probability of 1/3, but only on B/B the other child is a boy, so it is 1/3. a) As Martin is the older child, the question is simple: What is the probability for a new born child being a boy. What is the probability for your own next child being a boy/girl. It is 1/2. The difference is, that Martin is "fixed" in the birthorder being said he is the older one. c) This one is really hard: The more you "fix" one of the child in the birthorder with detail information, the more the probability increases from 1/3 to 1/2 being a boy. In this situation we have to draw: B/B G/B B/G 1234567 1234567 1234567 1oooxooo 1oooxooo 1ooooooo 2oooxooo 2oooxooo 2ooooooo 3oooxooo 3oooxooo 3ooooooo 4xxxxxxx 4oooxooo 4xxxxxxx 5oooxooo 5oooxooo 5ooooooo 6oooxooo 6oooxooo 6ooooooo 7oooxooo 7oooxooo 7ooooooo The x marks all wednesdays and the "o" marks all other weekdays the second child could be born. The sum is 4x7 -1 = 27 possible birthdays. Only all events in the left diagram (2x7 - 1 = 13) marks the events where the other child is also a boy.

[@charliehorse8686]

@@christiankrause1594 Obviously we don't have enough information. We don't know if Martin is right handed, and we don't know if he prefers chocolate or strawberry ice cream. We also don't know if he bites his nails or has a birthmark on his left shoulder.

[@NZ-fo8tp]

This is actually, in my opinion, the least controversial thing he has posted in a while. Good work. This makes alot of sense to me, I would never have thought of it but it works

[@JensPilemandOttesen]

Just by the title I thought Parkers video of the same puzzle.😀

[@ELYESSS]

It's just maths, it's either right or wrong, it can't be controversial.

[@flashstar1234]

Yeah same makes perfect sense to me

[@pirojfmifhghek566]

I'm disappointed, honestly. I was looking forward to more angry ElectroBOOM response videos.

[@CertifiedSlamboy]

@@ELYESSS Erm. Have you heard of the -1/12th video?

[@bartoszporzezinski4842]

Love this! The concept of loops does break the mind slightly, but is very logical once you look at it with a reduced number of boxes. I would never come up with this solution myself, however - eliminating randomness of prisoners' choices in this manner is truly ingenious.

[@wearwolf2500]

I think the key to understanding why you are always in the loop that contains your number is that the slips are all unique. The only number that can complete the loop is your own because that's the only slip that can point back to another box you have already opened. If you open box 3 and find a 5 and then open box 5 to find a 7 you can't open box 7 and find another 5. It will either by 3 to complete the loop or a number you haven't seen already.

[@cameodamaneo]

I think this might be a better explanation than the video

[@happywhale1786]

I think the key to understand is "there are only loops". And loops always go back to your start. --- My first thoughts viewing his proof are: Then why there are always only loops? => Or in other words why only loops can fullfill the requirement for a set of 1 to 1 pair to be all contained? => OR why the 1to1 and one-direction basic structure can only form a line or a loop? oh, this can be easily proven with induction. --- So we have the basic: there can be only lines and loops. --- why there are only loops is because the structure can only form lines and loops. Since lines has slip outside of box which is invalid in our scenario, there can only be loops.

[@Akronox]

For me, it was simpler to consider that the box and the number must be part of the same loop regardless of its length. Any box must be part of a loop that points to itself since all numbers are unique and are just a permutation, the worst case is just having to go through all the boxes. So if you start with the box with your number you know it is part of the correct loop.

[@paulparker1425]

For me, I just tried to break the strategy. How? By imagining that I didn't find my number for the first 99 boxes. That final box MUST complete the loop, there's no possible alternative and that's the WORST case. Every other combination (loop) containing fewer boxes will terminate faster at my number because that's the terminating condition.

[@Akronox]

Actually, it is because of the definition of the loop, all boxes and corresponding numbers must be part of the same loop. Since the number in the box gives you the box that gives the next number and so on and the loop can only end with the original box and the loop has to end.

[@Campfire_Bandit]

It's a small thing but I find Destin's response to your claim (5 seconds of deep thought followed by "teach me") is inspirational. He switched gears so fast from peer to student, it's the kind of attitude I want and he makes it look so simple.

[@-PSJ]

I actually think 5 second of thought is still too short

[@ericlevy3317]

The best teachers never stop being students.

[@RJFerret]

Instead of having separate peer/student relationships, perceive every relationship as potential for learning, we can learn from young children, learn from old wisdom, learn from other perspectives, learn from animals, which is why Destin's reaction appears so simple, because he is always seeking to learn, it's not being in a learning mode, it's just a constant way of living/learning. This will sound trite, but instead of being inspired by it/want to do it, instead just do it, seek out what you might learn from any interaction, there's always something!

[@bruhbutton4520]

Dudes so honest and genuine. I strive to be like him

[@yossam6722]

A true master is an eternal student.

[@joseph-fernando-piano]

A really incredible feature of the loop strategy is comparing how well it works even against random guessing with more chances to open boxes. For example, if each prisoner were allowed to open 99 of the 100 boxes, instead of 50, to find their own number, the total probability of success by randomly guessing is only 0.99^100, or 36.6%! (Whereas the loop strategy gives a comparable chance of success while only opening 50 boxes, and succeeds 99% of the time if you can open 99 boxes) If you were allowed to open 98 of 100 boxes, the chance of winning via random guessing drops to 13.3%, and to below 5% for 97 boxes!

[@jakestory8748]

If this is true thats absolutely awesome. Missed chance to point that out in this video

[@bloxer9563]

Maybe

[@RockyRoad213]

if you open 99 of the 100 boxes, then by elimination you can guarantee which box your number is in

[@justanordinaryguy658]

​@@RockyRoad213well yeah but if you don't actually open it you lose

[@joshjason9460]

What if the evil warden gave an empty box

[@basimqasim7113]

i made a smaller version of this riddle by decreasing the number of participants to 4 and i made artificial intelligence to randomly pick numbers for me with the loop strategy they succeeded 5 times out of 10 with random picking they didn't succeed in 10 trials. it was really fascinating.

[@chavab8753]

I did that also. Please see my posts. It raised a few questions for me. Would be happy if you'd weigh in.

[@GodbornNoven]

Funny how you can solve pretty much the craziest logic riddles with AI

[@hundredpercentjuice]

@@GodbornNoven .... or you could just use a die. Don't be silly.

[@BigPapaMitchell]

12:20 I have a better intuitive explanation: The only way you could start on a chain and not eventually reach itself is if either that chain forms a line with an endpoint, or that chain loops back on itself in the middle. The first one requires a box to have no number in it, which is impossible, and the second requires that two boxes have the same number, which is impossible, meaning that it must be the case it loops back on itself.

[@irakyl]

I was thinking the same thing, his explanation with the link at 11:25 wasn't very helpful. He should have instead hammered on the point that you could never fail to find your number, you will never have to try again and find a different loop. When Dustin said "I feel like it's possible that you start with the loop but don't end up finding your number", Derek should have said "And what would that look like?" at which point you realise pretty quickly that it's impossible, and thus loop length is the only factor that determines your succes. Perhaps Derek could also have spent a little more time analysing the situation before starting his explanation. If the viewers felt like they discovered the properties of the loops themselves it would feel much more intuitive. Edit: if you still feel like this is unintuitive, like this is cheating math somehow, that this shouldn't be possible, consider the following: there is only one 'check' to see if the experiment is a succes or a fail, it's right at the beginning, if the starting configuration has a loop of 51 or higher. Imagine if the box configuration was rondomised for each individual prisoner, then the chance of all 100 of them passing would be astronomically low again, and all is right in the world.

[@MasterHigure]

@@irakyl Derek should've said "Then how would you manage to loop back to the box you started at?" Assuming we all agree beforehand that these are closed loops, I think that's the easiest argument for why you must be in the correct loop.

[@godgige]

@@MasterHigure interestlingly I found that explanation to be perfect for my taste, but it might be that I already understood why is that (the loops) beforehand. Anyways great video!

[@DanielReyes-pr1rd]

Your right everyone is on the chain, but not being on the chain is just a death sentence for the other prisoners, not the one who had just found his number. i dont really know what the point of this is besides just being an exercise. although when your sitting in front of a computer all day you start to come up with creative ways of thinking

[@bumpsy]

I thought that was kinda obvious as soon as Derek presented the strategy. Obviously there cannot be any dead ends when each number is still in one of the boxes. I.e. no empty boxes and no boxes with no number on them, which is obviously both not possible here

[@smartereveryday]

Thanks for teaching me.

[@jhaz89]

You're welcome

[@daivomjoshi56]

Hey destin when is your 2nd part of "Kodak film making" video coming ?

[@Ryanisalive]

There you are!

[@alberteinstein9626]

Thanks , @SmarterEveryDay ,, you just questioned the same questions we had watching and following by you 👍🏻

[@strikerj_]

2:23 "Teach me." is one of the best responses ever

[@brianrussell463]

Destin had my favorite response ever, "teach me!" I love that.

[@hunterbroadnix9609]

I work in irrigation and we actually use closed loops like this! If I have zones numbered from 1-10 (for houses 1-10) but the stations do not follow the order of the houses, we just pick the first wire and move it to the correct order (for example: station 5 is house 1; put station 5 wire into station 1 slot, and station 1 into the next assigned order; repeat until finished). I’ve had 2-3 closed loops and always thought it was fascinating, really neat to see a video that carries into the profession!

[@frogbutts3628]

Do you get executed if you mess it up?

[@1SSJA]

that is so cool

[@Robert08010]

I like this warden. He has reasoned out that if he can turn all his prisoners into math wizards or at least willing to work together and trust one another, he can let them out.

[@KhangNguyen-ij4xh]

Then you get something like Vento Aureo. Basically a group of criminal with a prodigy and are willing to work together

[@Solo_man69]

The wardens watching too much saw and squid game

[@teeemm9456]

The missing part of the strategy is, if the first prisoner fails, they implement plan B and break out.

[@Solo_man69]

@Tee Emm And now probability has been in creased to 100% cus there’s like 1 warden

[@mattsnyder4754]

The real “magic” of this solution is that it ties the probability of success to one single condition (the state of the “loops” in the boxes) instead of a repeated condition (the odds of each prisoner finding the correct number). You’ve immediately removed the exponential scaling of the probability. So even if you choose a sub-optimal method. Any method that is based on a single fixed condition is immediately an improvement.

[@giangtran-to6tb]

ok

[@daburnd]

That is quite certain indeed. But i think vertasium forgot that not everything will form a loop as a set of boxes can end up with a box that has its own number.

[@duitakarbhat]

@@daburnd lol no he didn't forget that. In fact, you start off with that. Hence, in your scenario, you'd find your number in the first attempt.

[@daburnd]

@@duitakarbhat that is only for the only for the prisoner that has the number of the box containing its own number. But not for the x - amount of boxes leading up to that one in the pointer line-up. For example : Box 1 points to box 2, box 2 points to box 3, box 3 points to itself. Hence, not everything needs to be a loop ( it can be a finite non looping set ), unless u put up the constraint that no box can contain its own number. Just handling this case asif it could only be closed loops therefore seems not logical.

[@hO_Oman]

@@daburnd if both boxes 2 and 3 point to box 3, it means they both contain the same number, 3. that's not allowed

[@mattsnyder4754]

So, I think there’s an easier explanation for why you’re guaranteed to eventually circle back to your own box. For a loop to close, you have to pick a slip of a box you’ve already been to (otherwise you’re just continuing on down the loop). But the fact that you’ve already been to a box requires that you’ve already found its slip in a previous box. The one exception to that, is the box you started with. Which, in this case, is your own number.

[@adrian-w3h2c]

Intuitively, the solution was difficult to grasp. But it makes a lot of sense when he explains the math behind the problem. Sometimes our intuition can only take us so far. This video has made me really appreciate the value of math as a problem solving tool in a way that no traditional math class could.

[@gorak9000]

Your "intuition" definitely leads you down the wrong path in probabilities at least 76% of the time. The human brain is just not wired to deal with probabilities intuitively (and also 95% of statistics are completely made up)

[@legendgamer204]

I thought of a nice way to phrase why the strategy works: the strategy works to reduce the amount of variation between prisoners' sets of guesses. This reduces the amount of things that have to go right for a successful outcome, just like flipping 3 coins instead of 10.

[@Takin2000]

It becomes more intuitive when you first find a way to increase your odds _at all_ . A simple way to increase your odds is this simple method: First prisoner picks boxes 1 - 50. Second prisoner picks the boxes 51-100 If the rest picks at random, your probability of winning is higher than everyone picking at random. Why? Well, the first prisoner has a 50% chance of winning. No changes here. However, the second prisoner has a slightly higher chance because: if the first prisoner found his number, that means that numbers 51-100 DONT contain the first prisoners number. For the second prisonder, that is one guaranteed failing option removed. If the first prisoner did NOT fond his number, then it must be in one of the boxes that the second prisoner is checking. _This doesnt matter though because the first prisoner lost already_ Meaning: *decreasing your odds in the case that someone before you lost doesnt decrease the odds of the whole game* . Thats also why the loop strategy works: It trades individual win% to increase collective win% because if one prisoner already lost, then the rest of the prisoners also losing doesnt hurt your odds

[@thomasmaughan4798]

Figuring out this loop structure does not seem like math. I'm not sure what it is, but it isn't something you can pop into a calculator or slide rule. Math easily finds the odds (the permutations) and it is a really big number. Your calculator will probably choke on 100 factorial.

[@aleph_zero1]

Miltersen: "Hey, i made this riddle. Anyone want to try it?" Everyone: "Dang, it is kinda difficult. What is the answer?" Miltersen: "Idk go figure it out"

[@itismethatguy]

Haha yeah His name bro is funny, u could have user that

[@hahanamegobrrr6667]

"This problem is left as an exercise to the reader" is just a fancier way of saying "idk lol"

[@ynemey1243]

@@hahanamegobrrr6667 I've had someone say that to me before. That's how you know they are full of BS.

[@wouterpomp5014]

Imagine coming up with this massively smart idea and still only having 31% chance not to get executed.

[@kevinz8554]

Life do be like that sometimes

[@aarondavis8943]

Imagine trying to explain probability to a bunch of prisoners. I put the actual real-world chance at something around 0.001%

[@Eveer.r]

Imagine knowing this for a fact and no one listens 🤣

[@idiatico]

You'd use up your strategy time trying to convince them it's smart and gives a 31% chance at success then someone will speak louder then you saying "all that work for less than the 50% chance we get picking randomly?" And then everyone dies

[@clover7359]

31% chance of success is a hell of a lot better than effectively 0%.

[@pradyumnachakraborty3262]

This is actually exactly how the turtle and hare algorithm works for arrays .Given an array on N+1 elements with numbers 1 to N, with only 1 element being repeated, you can actually find the repeated element using this concept. Though the algorithm is a little different, the concept is very similar. You can maybe make a video on that. It's a very interesting problem and a beautiful variation of Floyd's algorithm.

[@miloszforman6270]

I wonder how you could find the one duplicate element using Floyd's algorithm. You may find it by chance if you use a random element to start with, but usually it would require to examine the complete loop structure. Or doesn't it?

[@WavyCats]

As a programmer, this reminds me of cyclic sort, and we deal with cycles all the time. Once you explained the strategy, it instantly blew my mind. Very clever solution!

[@rachadelmoutaouaffiq5019]

Linked lists :))

[@codahighland]

@@rachadelmoutaouaffiq5019 The worst kind of list!

[@markingraham4892]

It's a idiotic video. Any system to consistently search boxes would have the same result.

[@codahighland]

@@markingraham4892 Uh... no, that's not true at all. It's true that other systems CAN have equally good results (nothing better) but not ALL systems will. Take, for example, the system of "everyone search the first 50 boxes." That's a consistent way to search, but it would obviously fail.

[@WavyCats]

@@markingraham4892 care to elaborate?

[@jenius00]

I think the most important thing to note about this is the prisoners' choices are no longer random variables. It's the setup of the boxes that is the random variable. If the prisoners' follow their strategy on a good setup they are guaranteed to succeed and on a bad setup they are guaranteed to fail. So it doesn't seem that surprising that they have a much better shot than if they are randomly choosing boxes.

[@HeythemMD]

+

[@markmuller7962]

Yep, in other words the strategy creates a finite system of loops that is predefined as a winning system or a losing system with a 30% - 70% ratio. The strength of the strategy is about that the individual actions doesn't really matter, once the strategy is chosen the entire system of loops 4:45 is automatically determine and it's either a winning system or a losing one (30 - 70)

[@chrishillery]

So it turns it from a game of chance into Candyland.

[@hansschwarz1338]

a deterministic aproach doesnt make something more likely, just for being deterministic. It gets more likely, because the probability of the biggest loop being length 50 or less is higher than the random aproach. if every prisoner could only could choose one box to open the aproach wouldnt impact the result. So the deterministic nature doesnt have a part in it, but rather the finite setup and structure of this riddle in combination with the right strategy.

[@questionedsanity785]

@@hansschwarz1338 This is wrong. It is correct that determinism doesn't matter, but the real reason the loop method is superior is that it coordinates successes among the group. In particular this means it does actually help if you only get to open one box, and in fact this situation makes it clearer why it works. If everyone opens one box they can only succeed if they all choose different boxes, it is therefore clear that you improve your probability of group success as long as you coordinate to never have 2 people open the same box. This means that coordination is the important part, it just happens to be the case that the optimal method of coordination in the many box case is the loop method.

[@neildmoss]

I feel that "Metersen's colleague" should definitely get at _least_ a name check here! So, hat tip to Sven Skyum, reader emeritus at Department of Computer Science, University of Aarhus.

[@ApequH]

* Tips Hat*

[@tim40gabby25]

"Skyum's protocol" sounds a bit like a namecheck...

[@eliaskjrbo8142]

I live in Aarhus!

[@osiris1102]

@@ApequH tips fedora m'skyum

[@neildmoss]

I'd love to know the process by which Skyum arrived at this answer. Years of work in a field where this kind of "loop" structure has been studied already? Flash of inspiration after a night of pizza and cola? Or was it an immediate "well, duh..., isn't it obvious?" savant-level intuitive grasp? That's as fascinating as the original riddle.

[@larkefedifero]

@ 10:44 - As far as the question of not coming back to your own number: One short answer is - remember that YOU (and the prisoners' group) are the one defining these "loops." Hence, by having faith in these loop definitions, you can be *certain* that, given enough chances (as an individual prisoner), you will ALWAYS find your number. The PROBLEM is, that, 69% of the time, *one* of the loops will be 51 numbers long - or longer. So, 69% of the time, because of the unfortunate "luck of the draw," the 100 prisoners are going to die, for certain. They will discover this fate with the first prisoner's number to fall on the "51+ length" loop, as he will use up all of his 50 chances before finding his number. However, the positive flip-side, as seen on the (Z-factor?) diagram above, is that 31% of the time, ALL of the "prisoner-defined" loops are 50 numbers long, or *less.* This guarantees that, following the "prisoner-defined" loop definition system, and loops defined therein, EVERY SINGLE PRISONER will be *guaranteed* to find his number 100% of the time - in 50 chances or fewer. This is what thus also guarantees that, as a group, the prisoners will certainly succeed, 100% OF that 31% of the time! And compared to (1/2)^100, the standard odds of such a thing happening at random, that's really quite an amazing improvement! 😉👍

[@cultofmel]

I was really confused at first on how whatever number you start with is guaranteed to be in your loop, but once I started to type out a comment questioning it I totally realized how it works. In order to finish your loop you have to end up back where you start, and since none of the boxes can be empty, you're guaranteed to be in some sort of loop.

[@TheJakoubecek]

Then you have two boxes which contain “3”… Every number occurs exactly once in all the boxes. If box 1 contains “3” then no other box have “3” you are guaranteed at this point that your loop you are on cannot have box which contains the same number (1 has 3, 3 has to have something else ie 5, then 5 has to have somethine else and so on, until you find box pointing you back to 1)

[@TheJakoubecek]

Try it with something small like 3 or 4 boxes. Whatever you try you will always see only loops. No dead ends. If you find dead end it means there are duplicates and some number is missing (duplicate took its place)

[@YEC999]

@@TheJakoubecek Oh man you are right🤯 i somehow didn't see that Thanks!!

[@nicolass.5849]

In a loop, yes. But why in your loop ? I don't understand it...

[@cultofmel]

@@nicolass.5849 You have to end up back where you start in a loop. And since in this case you start at your own number, one of the boxes in your loop must also contain your number. If none of them did, you would never be able to finish the loop.

[@grys9245]

I paused at 3:44 to try this method out using Python code. I ran the code 100 times, of which 32 runs were successful (every prisoner was able to complete the task under the given conditions). In other words, it achieved a probability of 0.32. Very close!

[@adolforodolfo6929]

Good stuff.

[@Harshil_Uppal]

Bruh how do you even code that?

[@ramsyfpp6418]

@@Harshil_Uppal it would be easier in javascript But it's very easy anyways

[@HEYJO77]

nice

[@Linaiz]

@@Harshil_Uppal Maybe have an array of numbers, size 100. Fill the array randomly with values from 1 to 100. Then, every prisoner searches the array. Every prisoner can have up to 50 attempts. You use the strategy of "index_to_search = array[index_to_search]", where on the first attempt, "index_to_search = prisoner_number". Each search must last less than 50 attempts. If prisoner_number != array[index_to_search] and attempt > 50, the experiment fails.

[@CimolOk-nz5yj]

🎯 Key Takeaways for quick navigation: 00:00 *🧩 Introduction to the 100 Prisoners Riddle* - Introduction of a counterintuitive riddle that challenges perceptions, even after the solution is known. - Mention of widespread confusion and incorrect explanations on the topic. - Setup of the riddle involving 100 prisoners and a unique challenge they face. 02:02 *🤔 The Counterintuitive Solution* - Explanation of how a seemingly impossible probability can be significantly improved with the right strategy. - Introduction of the concept of mathematical strategy to solve the riddle. 03:11 *🔄 The Loop Strategy Explained* - Detailed explanation of the loop strategy for prisoners to increase their survival odds. - Illustration of how this strategy transforms theproblem and its solution. 05:44 *📊 Probability and Loops* - Analysis of how loops determine the success of the strategy. - Introduction to the calculation of success probability based on loop lengths. 09:46 *🤝 Impact of Strategy on Individual Probabilities* - Discussion on how the loop strategy affects individual and collective probabilities. - Clarification of misconceptions regarding loop participation and success rates. 12:43 *🛡️ Countermeasures and Adaptations* - Exploration of scenarios involving external interference with the strategy, and how prisoners can adapt. - Insight into the resilience of the strategy against manipulation. 14:15 *📈 Scaling the Problem* - Examination of the problem's scalability and the consistent success rate across different prisoner numbers. - Explanation of the mathematical limit of the strategy's success rate. 16:44 *💡 Learning from the Riddle* - Conclusion highlighting the educational value of the riddle and its implications. - Promotion of a platform for further learning and problem-solving skills development. Made with HARPA AI

[@miloszforman6270]

Now please let Harpa AI explain how this peculiar 1/n formula works, and where this 31.18% winning chance comes from.

[@testingreadaboutit]

8:25 - That flip to unexpectedly new stuff on the whiteboard was so smooth. Nice job.

[@gaminawulfsdottir3253]

Hah! I was thinking the same thing!

[@chriswasia413]

This riddle definitely seems impossible but the brute force approach confirms. Ran a program that played the scenario 50,000 times and yep... 3,450,000 prisoners died. Survival rate was 31% of the time. Nice work and as always, thanks for teaching us something new Derek!

[@ace10414]

I'm writing my own simulation for this, and I'm not getting the same results. Could I see your code please? I'm thinking I'm doing something wrong and looking to learn.

[@saurabhkumarsingh3986]

Yes please a github link would be appreciated. More to learn

[@muhammadmaazwaseem7452]

Following

[@Liberty46]

@@ace10414 he lied

[@imranq9241]

I did a simulation and got about 29% chance. So seems pretty accurate. I'll share my code if anyone wants to see it

[@billrexhausen4221]

Writing a research paper and still telling the reader to figure it out for him or herself is peak professor energy.

[@manstuckinabox3679]

T h e p r o o f o f t h i s i s t o o m a g n i f i c e n t f o r t h i s b o o k.

[@pieriniedoardo5839]

And people still wonder why math graduates tend to hide in the woods sending bombs to people

[@Khaerulbtg]

Why are the replies gone?

[@irrelevant_noob]

@@Khaerulbtg Happens sometimes. I'm guessing either they were removed individually, or the accounts have been poofed. 🤷‍♂️

[@MK-fg8hi]

It would be really awesome to see the original reviews of that paper (from when is was submitted to publication). I bet those reviewers were also beaming with professor energy 😂😂

[@LuqmanRahamat]

If everyone agreed to only look from box 1-50, then half of them would find their number.

[@miloszforman6270]

Yes, that's about the same result as that of pure random strategy, and it guarantees failure.

[@rtripp25]

This type of content is unbelievable! As an AP Calc and AP Stats teacher, it keeps me motivated to learn more and find more interesting problems for my students to keep them motivated.

[@nejnovejsi]

​​@@ibrahim-sj2cr I dont think in this situation you are supposed to go to box 6. You stil go to your box 1, it is just some other box than before. I have not thought this through, but just from the top of my head, I dont see any issues with this.

[@ibrahim-sj2cr]

@@nejnovejsii yes you are correct...had to get the pen and paper out on his one

[@ibrahim-sj2cr]

@@nejnovejsi all the numbers were changed, that was what i didnt realise

[@ibrahim-sj2cr]

@@nejnovejsi and thanks

[@mdtexeira]

The minute you mentioned loops, it no longer seemed impossible and actually seemed retroactively obvious. Math is so damned cool.

[@micahturpin8042]

Same here. I had the same question that Destin did, but as soon as I realized that, unless the box you start with contains it's own (and hence your) number, you are guaranteed to be on the proper loop. It's only a question of how long that loop is. As soon as I was able to wrap my head around that, it made perfect sense.

[@nomoneyball5423]

I have a different solution that grants greater than a 31% chance. It approaches this as if it were a trick question, and still completely satisfies all the rules/requirements. "Each must leave the room as they found it." They found it with 100 closed boxes. They didn't "find" any boxes open nor did they find any information or clues regarding which numbered slips were in which boxes. On that premise, #1 (beforehand tells all of this strategy), and opens 1/50 (a 50% chance). He then takes ALL numbered slips placed in boxes 1-50 and he stuffs all of them in box #1 (just as he had told the rest of the group beforehand that he would do...Mind you; this is a "trick answer," but according to the rules, NOTHING specifies this is not allowed). Then, prisoner #2 (who was beforehand told of this overall strategy-and understand/assume the rest were as well for the sake of time), then opens boxes 1 (which now has all numbered slips which were originally placed in boxes 1-50) and he opens boxes 51, 52, 53.... (you get the point) to box 99 (50 boxes). As long as HIS number (2) isn't in box 100 (a 1% chance which lowers our overall group chance only to 49%!!!!), everything is still a go. Then, he also places all numbered slips he has now discovered (in boxes 1 & 51-99) all into box 1 (just like the first prisoner did). Now, prisoner #3 has a 100% chance now as he just checks boxes 1 and 100 (which have all the numbered slips in them as the remaining boxes are empty) and then he places the numbered slip located in box 100 into box 1 as well! (So now ALL numbered slips are now located in box 1). Prisoners #4, 5, 6 and so on only have to check box 1 and their chance is 100%, thus keeping our risk level/chance at 49%! All the way to prisoner 100. This way completely plays by the rules by the way according to the information/restrictions that was given us. As we only talked strategy beforehand, only one entered at a time, we did not communicate during, and we each left the room as we found it; with 100 closed boxes that never got moved. Yes, the SLIPS got moved, but we "left the room as we found it," because prisoner #1 FOUND the room with 100 closed boxes in such order. Also matching the approach of a "trick question/trick answer," you could also have prisoners 2-100 arranged tightly around the open doorway as prisoner 1 goes in, and all 2-100 could watch as he opens each box. They could each take notes on what box he opened and what number they witnessed him discover. And so on and so forth. This still allows for them entering 1 at a time as well as not communicating in any way with any of the others. Another way is to say ok the room is sealed but with glass walls. Just the way my mind works and another way of looking at it. These answers give me more peace about it than the one discussed in this video. However, the solution discussed in this video is more fascinating and I'm sure true math experts (I am NOT one I'm just someone who greatly enjoys strategy and basic wisdom/philosophy) will prefer it the way it is in the video for the math aspects!

[@user-bu9xh4sg6v]

But how will the prisoners know the loops to find their numbers? Like which number to open next to get to their loops?

[@gavissing5225]

@@user-bu9xh4sg6v they open what ever number is in the box then open that box see what number is in that then they go to that number and open that box’s d so on until they find there number

[@superkilleryt3764]

@@nomoneyball5423 wait what if the first person cant find his number...

[@tomgray8156]

To be honest, I didn’t struggle with understanding the strategy, or how the loops work. I was confused about the 31%, but after you explained that this all made a lot of sense, and it’s really fun maths.

[@kevinbean3679]

Especially going the opposite direction to shoe that approximately 69% of the time, the prisoners get executed. Rather morbid problem 😅

[@Legendendear]

@@kevinbean3679 69? Nice!

[@AvanaVana]

Same, I knew that it had to be able to be related to some nice little expression. Now I’m trying to figure out how and if the “2” in 1 - ln(2) is related somehow to the ratio of choices per total boxes the prisoners are allowed to make

[@giyanvice]

So out of hundred groups made up of 100 Prisoners, only 31 groups will get all the answers and go home, the rest 69 groups will have to die.

[@nimrod06]

@@AvanaVana It is obvious if you do the integration. int^2n_n 1/x dx = ln (2n) - ln (n) = ln 2, and that upper limit of 2 is the ratio you mentioned.

[@ThatThumbNailedit]

2:45 Basically pick a random box,if the box says a number thats not yours,count the boxes till to that number u go and open it, example: u have number 6, if u opened the 5th box out of 10 boxes and the fifth box shows the number 2,go to the 2nd box and open it,if shows 8 then go to 8th,repeat all that process untill u get ur number,even luckier if u got ur number less than 50.

[@WiiAndii]

Alternative explanation for why you're guaranteed to be on the same loop as your number: Keep in mind each slip only exists once. If you start at the box labeled with your number and follow the loop, the only way you could NOT eventually find your number is if you come across a slip pointing you to a box you've already opened, which of course leads you on the same path you've already been on, trapping you in an infinite sub-loop. But that is impossible, because that box you would be pointed back to was already pointed to by the slip you found just before that, and you can't find that same slip again. The only way you'll be pointed back to a box you already opened is that it's the box you started with, because you went to that one without having found the slip that points to it yet.

[@Vfulncchl]

Ah, he should have said that in the video, that is actually a good explanation

[@Vfulncchl]

So in order to get back to the initial box picked, with your number on it, you have to find the right slip. If you find the slip you are good. So your number HAS to be in the loop that also contains the box with your number. Hence, if the longest chain is 50 or less, everyone will find their number. Damn it actually makes sense now

[@john_hunter_]

That's a good way of thinking about it.

[@filipposchiabel]

@Any Body no, it isn't: every prisoner has a *unique* number, so you can't find 2 boxes that bring you to the same box

[@WiiAndii]

@Any Body The way the riddle is set up this shouldn't be possible. If each slip exists only once, box 23 can't have slip 54, because that slip was already in box 7.

[@rashadisayev]

The best thing about this loop strategy would be if someone finds their number in their last chance of opening a box, they make sure that they will be freed since if one 50 length loop exists, the others can be maximum 50 length

[@lethalwolf7455]

Friend! None of the other comments or the video itself allowed me to understand this. But your comment just did! Just visualized what you said and now I get it. Sincere thanks!

[@Ren-fo4lg]

I’ve read this multiple times and each time I flip flop between understanding and not understanding

[@rashadisayev]

@@lethalwolf7455 Happy to have helped you to understand because I, too, hardly understood it after reading some comments

[@JJforShie1]

@@Ren-fo4lg lol me too

[@yourtinerary]

This comment is 🙌🏻

[@Lindeman08]

10:55. Because the system was explained so well I did not find this confusing at all. You're gonna be on the right loop as long as you pick the box with your number. The only question remaining is how long that loop is.

[@AmxCsifier]

Well, Destin still didn't watch the Veritasium video back then

[@AnthonyGoodley]

@unfaithfulevil 🅥 Your a joke. You have no content!

[@lanceslance2930]

@unfaithfulevil 🅥 that checkmark looks slightly off and then I realized lmao

[@magica3526]

but like also thats a bit of a silly question, as is derrick's answer. The only way to complete the loop is to find the correct slip.

[@dries-pederjanse6249]

For me he has a false title

[@FanTazTiCxD]

Works with 2 prisoners too who can both open 1 box each. If they don't discuss anything before and open one random box each, the probability of success is: 0,5 x 0,5 = 0,25 = 25% If they agree to both open the box labeled their number, then either they will both be wrong (because the guards swapped box 1 and box 2) or both be right if the box numbers match the slips. So a 50% chance

[@Slinky0205]

Imagine spending hours to come up with that strategy and the first prisoner doesn't find their number

[@athletico3548]

the rules are impossible by themselves. One of the rules say: "they must leave the room exactly as they found it". Once you enter the room, the room is not the same as it once was. Since prisoners are allowed to open and close the boxes to look for their number, the room can't be exactly as they found it. Which means, the own problem is a fraud. Another rule says: "If all 100 prisoners find their number "during" their turn in the room, they will all be freed, but if even one fails, they will all be executed." Its a paradox. Its impossible to 100 prisoners to find their number "during" their turn, since they are getting inside the room one after another.

[@SublimeWeasel]

​​@@athletico3548so with the first rule point you explained; you mean that if a prisoner knows the numbers, the math fails? But the prisoners arent allowed to share info with each other. It's still random for each individual. Also, idont really get the second point you made, can you elaborate?

[@SublimeWeasel]

@@athletico3548 how would you rephrase the rules then? Im having a hard time understanding, i do understand the chess lore and heraclitus' river, i dont understand the second point. Maybe its because my main language is not english, but i dont see it :( This is some qualia stuff right here

[@athletico3548]

@@SublimeWeasel rephrasing it: "If each of the 100 prisoners find their number during their turn in the room, they will all be freed, but if even one fails, they will all be executed."

[@pragyanburagohain8751]

​@@athletico3548this just seems like semantic nonsense. Also how does leaving the room exactly as they found it not make sense? Open and close boxes simple. They are closed as before. Seriously what are you trying to say

[@pedroferrari6899]

Had to try it myself, thanks for such thorough explanation! function prisonersFindTheirBoxes(numberOfBoxes) { const boxes = Array.from({ length: numberOfBoxes }, (_, i) => i + 1); const shuffledBoxes = shuffleArray([...boxes]); function shuffleArray(array) { for (let i = array.length - 1; i > 0; i--) { const j = Math.floor(Math.random() * (i + 1)); [array[i], array[j]] = [array[j], array[i]]; } return array; } const map = new Map(); boxes.forEach((value, index) => { map.set(value, shuffledBoxes[index]); }); return { map, boxes: shuffledBoxes, }; }; function getAwayFromPrisonOrDie() { const numberOfBoxes = 100; const numberOfTries = Math.round(numberOfBoxes / 2); const { map, boxes } = prisonersFindTheirBoxes(numberOfBoxes); for (let prisoner = 1; prisoner

[@miloszforman6270]

Well yes, does work. Thanks for posting.

[@jamesbuchanan1913]

Imagine being the prisoner trying to sell this plan: "Now we still have a 70% chance of failure, and it's much to difficult to explain, but you all need to follow my instructions exactly for us to have any chance." Then every dissenter lowers your chance by half agian. At this point, I'm starting to think this time would be better spent coordinating a riot.

[@supersonicgamerguru]

I was very confused for a minute, until i realized you were trying to say "dissenter". Descent is going down, decent is not a verb, dissent is disagreeing.

[@BenjaminRonlund]

Wow you're very smart with a great understanding of spelling and grammar. It's a shame those won't win you any friends or help you communicate effectively with those you manage to piss off.

[@JD-wu5pf]

@@BenjaminRonlund I'm sure going full roid rage in a KZbin comment section is the correct way to make friends?

[@GameFuMaster]

@@BenjaminRonlund too bad your misspellings is actually not going help you communicate effectively.

[@supersonicgamerguru]

@@BenjaminRonlund yep, the problem is definitely with making literally any corrections to anybody ever, not with the people who go "ur dum that's not a word". It's 100% not allowed to comment on spelling or grammar, even when it will help make the original post more clear to future readers and you fully explain what the correct word is and why it's correct. Explaining things never helped anybody, especially those learning an extremely difficult language like English, whether it be as a first language or a second.

[@kristjanbrezovnik6485]

Destin's "Teach me." is possibly the greatest example of doing science right. Or, you know, something to that effect. :D Possibly both most humble and most epic answer ever.

[@terryarmbruster9719]

And not created by them at all. Its a category theory loop. Took this s over 35 years ago. The same problem but worded different as in not prisoners is shown and proven in elementary category theory. Don't need all that stats to prove or show s. All one needs to know is how to categorize the problem and check where the functions associate then if s running link function appears. Lol category theory is like a cannon that swats fly problems like that

[@DevinDTV]

it's not that deep or interesting to want to be taught something lmao

[@JackFate76]

@@DevinDTV But it‘s humble and epic.

[@Dylan_Did_A_Thing]

right? I love it, it gave me chills to hear because I feel like it's such a rare thing these days for people to admit they don't know something, let alone for them to actually ask to learn it

[@Hansulf]

I felt the same

[@user-uo3mm5dg5o]

Bravo, well explained! At 2:51 minutes where the problem is stated, the second line must state for completeness that the boxes are either numbered or will be arranged in a known preset configuration so the prisoners can prearrange using a numbering system that assigns each box a consistent number. I imagined all blank boxes with a number in it randomly placed somewhere in the room (not even adhering to a nice grid pattern) in which case this loop method would not work, nor the pairing that is explained later.

[@tegxi]

Since, like the monty hall problem, this benefits from other ways of explaining, here's another way to think about how every box must form a closed loop: for it to NOT form a closed loop, it'd have to be something like 1 -> 2 -> 3 -> 4 -> 2, where 1 never gets repeated and instead leads to a smaller loop. this, however, indicates 2 slips that point to 2, which isn't possible, since it means no slips point to 1.

[@hishaam5429]

Why can’t u have no slips pointing to 1

[@Mrityunjay7]

@@hishaam5429 There are 100 unique numbers and 100 unique boxes so having 2 numbers that are the same is a contradiction Thus there cannot be 2 boxes pointing to 2 Which further implies that there must be one box pointing to the number 1

[@tegxi]

@@hishaam5429 the rule is that every prisoner has a slip somewhere. the game would be rigged if prisoner 1 had no slip

[@Random-zt4ig]

Your explanation is fantastic. Traveller!

[@reidflemingworldstoughestm1394]

@@hishaam5429 you could have a loop with 99 slips that don't point to 1, but then box 1 would have to contain a slip that pointed to box 1 -- a loop of size one.

[@Astraeul]

Here’s how I thought about the must be on a loop issue, and it might help others. When you look into your numbered box you have started a path which must end. the only way the path ends is by reaching a box you have already opened. This can happen in two ways, you find a box with the starting number or you find a number which points you back to a number you were previously on which would mean you found a number twice in your path, but finding a number twice is not possible as each number only appears once. The only way your path ends is by completing the loop and finding the box with your starting number.

[@softy8088]

This is the key: You can't enter a loop from outside the loop. It's impossible to have "tails" attached to a loop because it would mean two separate labels point to one box. (it would also imply there is a box with nothing pointing to it.) This isn't allowed per the rules of the game.

[@ediartiva]

Or you are unlucky enough to be in the worse-case scenario. (1-100)

[@fortyyearfitness]

Why is that any different than stating at box 1 and ending at box 50? I’m slow cause I’m not understanding the difference. Why is moving around to boxes based on the number in a box any more special…. Now I realize it must be because all the smart people get it, but I just can’t make sense of it…opening a box and reading a number inside, then going to that box seems just as random..

[@mastersKaaP]

@@ediartiva By the rules of the game you're not allowed to keep looking after 50 boxes, but if you ignore the rule and keep on then you will ALWAYS eventually find your number, in the worst case it would be in the very last box that's not opened. If there are no duplicates and no empty boxes there's no way you end the loop without finding your own number.

[@BigRW]

That's actually false. If you search 50 boxes without finding your number, you must leave without completing your loop.

[@ethanhong4153]

I also had the question "what if you're on the wrong loop" My reasoning is the last box in the loop must contain the same number as the first opened box due to the definition of the loop. Otherwise, if the "last box" contains a different numbered slip, it just means that it is not the last box, and the loop is longer. So there are 2 possible outcomes after opening a box: a) it contains the prisoner's number and the loop is completed or b) it does not contain the prisoners number and they must repeat the steps for the next box until the loop is closed. And the loop has to close because there are a finite 100 boxes. If there is a single loop of 100 boxes each with a unique slip, after 99 boxes/slips, the 100th slip is guaranteed to have the number of the first box. It was nice to revisit this riddle!

[@Michael_Aune]

This is a good explanation and helped me understand it

[@Kahadi]

Because its a loop, its impossible to be on the wrong loop as long as you don't screw up picking your boxes.

[@jillianblair1894]

I Aye sea

[@chervilious]

@@voidmayonnaise If the random arrangment has no more than 50 loops (31% chance by the video) then first prisoners has 100% chance to get it right. I disagree with Veritasium saying it's 50%. The probably already determined when the loop is set with this trick

[@sleepingkirby]

exactly what @KahaiMitram said, if you're not on the right loop, it's not a loop. And, by the definition of the all unique numbers being "renumbered", it has to be a loop. The only question is, is the loop shorter than or equal to 50. (edit) the above is still valid, but nm to anything after that. My sleep addled brain did the math wrong. The unique permutation of a loop is (n-1)!, which is the same as n!/n, which, if n=100, is 100!/100.

[@anonymanonym1364]

Really mesmerized! Your videos are just awesome...! They are really exciting and lead me to think deeper!

[@Eindall]

I think the counter-intuitiveness stems from the fact that the loop is no better than random selection for each individual attempt, but is way better for the riddle's success. Remember, the riddle is only concerned about ALL 100 people succeeding. It doesn't care for the 50 people who succeed on average. Anything that is not 100, is effectively 0. 100 is what the loop algorithm optimizes for. Despite not being allowed to communicate, prisoners can use the design of the riddle itself to collectively follow the same set of loops. Like putting all your eggs into one basket, following the same loops maximizes the chance of all 100 people succeeding. This is illustrated with the probability distribution at 10:24

[@duckymomo7935]

what makes people think they dont get the correct loop is becausee they forget to start on their own number, if they choose a random number then they can get the wrong loop

[@TheGbelcher]

Or you can just buy a new one. The old one always shows up moments after you buy the new one.

[@kevinmacallister4491]

Thanks, this is definitely the trick to why this seems to not make sense.

[@HerculesBallsInc]

Actually not at all. The loop strategy GUARANTEES that the prisoner is on the way to find his number. Because the box that had his number is by definition the last in the loop. The only thing that would stop him is if the loop is greater than 50 numbers. Random choosing strategies NEVER guarantee getting your number. If you phrase the problem this way, everyone thinks the answer is obvious.

[@kevinmacallister4491]

@@HerculesBallsInc only if he chooses the right loop, so it's not a guarantee

[@dothingsthatmatter8567]

This is why I have such a deep appreciation for those who are great at math. I have no idea how this was figured out and I don't need to because we have people like this! Thank God!

[@ngotranhoanhson5987]

can someone explain to me at 8:35 I dont understand that much, the unique loops of 100, and total permutations relate to the possibility of getting 100 numbered loops

[@swapneel3610]

@@ngotranhoanhson5987 It is indeed bit tricky to understand, you can use a simpler analogy to understand it, suppose you roll a dice, the total possible outcomes would be 6, and lets consider we have to find the probablity of getting a even number. Here the sample space would be 6 (All possible outcomes), and the set of all even numbers on the die {2, 4, 6}, is of size 3, hence the probablity 3/6 i.e. 1/2. Similarly in the above case, the sample size would be all the possible ways we can put 100 numbers in 100 boxes and that is 100!. As we are finding the probablity of finding a loop of length 100, we will need a set of all possible unique loops of length 100. As explained in the video, that set will contain 100!/100 loops, so we divide that with the sample size of 100!. Hope that's helpful.

[@bartjansen2294]

Great explanation, actually totally makes sense! The visualisation with the numbers and loops only ending with your own number was so clarifying.

[@assemblywizard8]

for me it was clearer to see it in this way: the reason the loop ends with your own slip is because, excluding the starting box, you only open a box if you already have its corresponding slip, so no other loop is possible (no slip can point to a box you already opened because you already got that slip pointing to that box)

[@LowKickMT]

@@assemblywizard8 well you also will always find your slip if you can open 100 of 100 boxes. you will also not open the same box twice because you already opened it and see that it is open. this alone isnt really the "aha" moment

[@AFriskyGamer]

@@assemblywizard8 That is another great way to word it that would have helped my brain when it was imploding in on itself

[@marutraj795]

At around 11:30, the person asks what if they are on a different loop… a loop is closed, and therefore if you start from a box, the loop has to end on the same box… the corresponding number will lie anywhere in between the loop….. since each number is unique and necessarily part of a box, there has to be at least one loop….

[@alonsoquesada1136]

Love how Destin just says "Teach me", that's humbleness right there

[@Plant_Parenthood]

That's what I love about Destin. He is always so humble and willing to learn from other people. He would be a good dude to hang out with, I feel. Btw, "humility" is the form of the word you were looking for.

[@siinxx7656]

he's a humble guy, just not feeling very smart that day following the answer

[@Billy.Nomates]

​@@Plant_Parenthoodboth are acceptable,both refer to being modest

[@Plant_Parenthood]

@@Billy.Nomates So it is! My bad. Disregard my earlier statement.

[@tylerdavis3]

I guess humbleness technically isn’t wrong but humility would be better in your context.

[@theAkornTree]

I was surprised by the "what if you're on the wrong loop" question, because that part seemed the most intuitive to me. The part I struggled with was how the likelihood of everyone winning was ~30%. The math to reach that number makes sense, but the result still seems really unintuitive to me.

[@noahwelikson1100]

It’s easier if you don’t think about the prisoners at all, just loops. If there is no loop > 50, every prisoner makes it. And there’s a lot more ways to make small loops than big loops, so any given random loop is more likely to be small than large

[@theAkornTree]

@@noahwelikson1100 Yeah, that's what I figured. ~30% feels way too low.

[@daminkon246]

Yeah wtf when he asked that question I went "bruh" out loud

[@jacobboehm983]

I felt the exact same way. The probability works out from graph theory and combinatorics though. You have to sum the number of ways you can make unique loops (cycles) of greater than half the number of boxes and divide by full number of permutations of the boxes

[@Jordan-tr3fn]

I asked about this. You could (maybe - not sure) derive it from normal distribution.. all the loops are random so it follows a random distrib meaning 68% of the time the loops will be in the 1 standard deviation but as the group needs to be right (100% of the prisoners need to find the right number) they will be right only 31% of the time (100% of them) and 68% of the time they will be in the 1 standard dev.. (meaning a part of them will lose so all of them lose)

[@Alena-cq1ze]

I was confused when the guy asked how you know you're in the right loop because my mind never questioned that part. I was more fascinated by the graph, and how it worked that either everyone succeeded, or less than half did. After some thinking, that made sense to me too.

[@griffinbur1118]

But in some ways, it's a sign of your good intuition that you didn't consider that: it's a blunder (in the sense that it just violates the conditions of the problem: if you go to box n_1, it directs you to box n_2, box n_2 directs you to box n_3, and box n_3 directs you back to box n_2, the slip reading n_2 has appeared twice). I think the video doesn't point this out as forcefully as it could (the ensuing explanation is pretty indirect).

[@ZzZ-qd1zo]

The reason that either most people succeed or less than half is because if a loop of over 50 exists, that means at least 50 prisoners will be on that loop, and will all fail.

[@shreya1159]

But I thought that about 3 second before he told

[@melsch8740]

OK but why is it a given that my number is in my loop?

[@AMITKUMARSHARMA92]

​@@griffinbur1118that will cause a Pigeon-Hole issue and that means number is missing , failing everyone , everytime

[@chavab8753]

After watching 3 times and then doing something else entirely, I finally understand the logic. The heart of this problem is in the sentence that begins at 5:30 minutes. Listen carefully to the few seconds before that and follow the logic for WHY all other prisoners in that loop will fail to find their number. That is what I originally missed. In this case, I chose several prisoners in this loop and convinced myself that, indeed, ALL prisoners in that loop fail to find their number, missing it by 1 box. Their number is in the 51st box that prisoner would have opened. I need to try a similar exercise for the nice guard to breaks the loop into two loops by switching just the contents of two boxes, but I think my brain is strained enough for one day. Hope this is helpful for others. It was a revelation to me. Thanks to the poster of this video!

[@TimeBucks]

This is super cool.

[@ruqaiyaansari8642]

It's amazing things

[@azimshamuhammad3302]

DO NOT POST ANY REPLY! DO NOT MENTION ANYTHING ABOUT TIMEBUCKS OR I WILL REJECT YOUR SUBMISSION, JUST THUMBS UP AND THATS IT

[@syedshabih3639]

Nice

[@MidnightSt]

i've known about this puzzle, and the solution, for literally a decade (at least). Until now, I never understood WHY it works. Now I do. Thank you.

[@ConstantChaos1]

It's a trick, the prison is in the u.s. they are all gunna be executed anyway

[@musicexams5258]

@J Boss Well in the case of the prisoners, the warden will notice the box swapping And probably execute them all In the case of the sympathetic prison guard, the warden probably thinks "oh they're just checking the boxes"

[@HawkOfGP]

@J Boss Part of the premise was that the prisoners must leave the room exactly as it was when they entered it.

[@ekelrock9940]

@J Boss I think the issue is that the sympathetic officer needs to intentionally know about the >50 loop and needs to swap 2 boxes that render both halves of the loop to be less than 50 (versus just swapping 2 boxes randomly). I don't know the math to solve it but, logically, if Derek were to swap boxes 78 and 57 he would've made an even longer loop (combining the small loop to the right). Or, if he were to swap boxes 80 and 42 (or any 2 boxes that are close together on the loop) then he would create a loop of 3 but potentially leave a loop greater than 50.

[@VimalKumar-ts7xn]

the question at 10:40 is explained by that a prisoner will always start from the box of his own number assuming that the last slip of the loop will be his numbered slip as explained at 5:17 , so it is a 100% chance that the prisoner numbered slip exists in the loop, but only a 31% chance that the loop is smaller than 51. I think this is the simple explanation to this question.

[@B-Lev]

Almost, it’s a 31% chance EVERY prisoners loop is 50 or less. So for all 100 of them

[@wanhaiqal9442]

Thank you so much i had a hard time digesting that one. I initially thought that the explanation only justifies the existence of the loops, but does not ensure the prisoner to be in the 'correct loop'.

[@Trachta10]

If the loop has all the numbers, Why do you care that the last number is yours or not? Your number can be inside the loop. if you randomly choose any box at the beginning the probabilities of finding your number are the same as if you choose to start with your number box.

[@minhtrinh3646]

I still dont understand why you have to find your number in the loop you pick? You have 31% to have largest loop at 50. You still have to choose the RIGHT LOOP.

[@Trachta10]

@@minhtrinh3646 I am convinced that this is incorrect, because it assumes that the boxes will be ordered in a "loop", so in reality it is pre-ordered, that is, it is not random. That 30% chance is only possible if someone previously ordered the boxes as a "loop". In a real case where the numbers are ordered randomly, this strategy is useless.

[@ashoykh]

Really great explanation! Another way to see that boxes always form a loop when you follow the algorithm is that a box will always have a different number inside of it then the box's index (unless you happened to start with a box that has your number in it). There for since the boxes always have a different number inside it (and numbers aren't repeated), you will always be taken to a new box (until you've made a loop)

[@lukegorman4523]

The loop strategy was very easy for me to understand once you gave the solution, because it is actually the same concept that is used to solve a Rubik's Cube blindfolded. The pieces can be moved around into many different permutations, but they from loops (called cycles) which you can memorize the order of to solve it without even looking at the cube. Very interesting how two unrelated problems can be solved in the same way.

[@mikebarrientos5085]

So in the rubik's case, there's also a 31% chance of solving it blindfolded?

[@lukegorman4523]

​@@mikebarrientos5085 No, because you can look through the whole thing. Essentially, the 31% chance is if there is a cycle greater than half of the total number of pieces, and that is not important in the rubik's case, because there is no constraint about only looking at half the pieces.

[@mikebarrientos5085]

@@lukegorman4523 ohh makes sense

[@rajithanw555]

I still cant understand how you'd solve rubiks cube blindfolded. I mean i can do it in 40 seconds with carefully watching the thing.

[@oscaar_3985]

Wow I’ve always been scared of trying to learn blindfolded but this comment gave motivation, thank you:)

[@MoshpitMaestro]

I love the Destin cameos. His reactions very much mirrored my own, despite him being a LOT more knowledgeable about math than I am.

[@sac58999]

I agree. When Destin said "Teach me." I thought, "...and that's why I like him." Their honest amazement and fascination with learning make these two great teachers for the layman.

The puzzle is even more interesting if it's modified to say that the first prisoner is allowed to check all boxes and make one swap. Then telling that there exists a strategy which guarantees success sounds even more incredible.

[@markmuller7962]

I love it Still we most hope that the prisoner in question swaps the longer loop instead of a smaller one

@@markmuller7962 prisoner only needs to check 50 boxes to guarantee they can always swap in a way that ensures no chain is longer than 50 after the swap. The only reason the first prisoner needs to be allowed to check all boxes is so that they themselves are guaranteed to find their own number. The puzzle can also state that all prisoners can check only 50 boxes, but the first one can do a swap, while the rest must find their number.

[@yosefsl30]

it is not 100% guaranteed, still needs to choose the correct loop...

[@markmuller7962]

@ I know how the riddle and the riddle solution works. Thing is, if the first prisoner swaps a loop of 4 boxes he only creates 2 loops of 2 boxes while the eventual 50+ loop stays intact and it's gonna kill the prisoners. The benevolent guard in the video either knows all the numbers inside the boxes or gets "lucky" by swapping the cards of the longest loop

[@markmuller7962]

@@yosefsl30 Right

[@ZoeAntich]

WOW!!! That just blew my mind!!!! I was watching solving mystery videos because my teacher recommended it for the topic we were learning, and i came across this video and thought, "I MUST watch this", and it left me speechless. I subscribed just because of this.

[@Hannah123605]

Classic Destin strategy - when stumped with a problem, ask with true interest and humility say “teach me”

[@whiitehead]

I actually had to pause the video and just sit back because my mind was so blown. I don't remember a video ever having this kind of effect on me. Very well explained. Thank you

[@fares238]

I think the easy way you can understand the loop is if you think from the perspective of the warden messing up the boxes. Instead of imagining the numbers are just randomly placed inside the boxes, imagine that everything is ordered, all the boxes from 1 to 100 are shut and contain their respective slip number, then a warden enters and changes them. In this case he'll go for the first box, say box 66, opens it and finds slip 66, he'll go for box 2 opens it, takes slip 2 and places slip 66 inside. Now he has the number 2 slip and needs to place it somewhere else, so he goes for box 10, opens it, takes slip 10 and places slip 2. He can keep doing that as many times as he likes but it will always end up with the first box (number 66) empty and needs a slip to be placed inside thus closing the loop. If a prisoner comes in he can follow the same trail the warden left depending on where he starts. If the prisoner is number 66 and starts with box 66, he'll move in reverse of the warden's trail and he'll end up in box 2 with his slip inside.

[@tholespe]

great explanation! That explains it to me how there must be loops in the set of boxes although all the slips of paper were distributed randomly. Thank you!

[@Prakriti_Senpai]

Hnn

[@laxminarayanbhandari855]

@@repentandbelieveinjesuschr9495 Shut up

[@sproins]

@@repentandbelieveinjesuschr9495 reported for spam

[@abinanthanva289]

this is actually a smart way to explain ngl thanks

[@calereliya]

It actually amazes me that anybody could not see why if you start with your own number, you'll eventually find your slip. That was one of the few parts of this that as soon as you said it, I was instantly "well yeah, obviously." Fascinating how different minds see different things.

[@Vecna-v8r]

Thank you😊 Helped me a lot To understand

[@tusy007]

The final clue that helped me understand why is one guaranteed to start on a loop which contains his slip was in a comment by @Graham Blake. Because every slip is unique and every box is opened by finding it's corresponding slip, you can't find a slip that points to an already open box other than yours :). So you are bound to eventually open a box containing your slip pointing to the box you first opened. You just don't know how many boxes will you need to open to find it. It can take you anywhere from 1 to 100 boxes. The mind boggling thing is, by this method you don't lower your individual chances to find your slip. It's still 50%. But you increase your collective chance to succeed by an unbelievable margin.

[@kugelblitz2001]

I didn't get this when he first mentioned it, but the explanation at 11:00 is really good.

[@touristguy87]

...this just sounds wrong, I wonder if that's merely an intuition or logical as well. Mainly because you're assuming from an fact back to a theory which you say is truth because of a fact, you literally say this. Facts don't prove truths. Facts are only facts. They cannot prove truths, they simply *are* truths. It is up to the rational mind to form logical, rational and truthful connections from facts to conclusions. case in point: "Because every slip is unique and every box is opened by finding it's corresponding slip,..." That's not true. The assumption is that each box is opened by having the associated prisoner open it, and with each prisoner taking a loop each box is assumed to be opened at least once. The problem is that the exact sequence of numbers is unknown, and probably is random as long as the prisoners chosen to do the loops are chosen randomly. "... you can't find a slip that points to an already open box other than yours :)." And so that is grossly incorrect. A simple example is to have two prisoners do the search, with the 2nd prisoner having the 1st prisoners' number in their box, which would be box N in the 1st prisoners' search. The 1st prisoner will have found the 2nd prisoners' number in box N-1. The boxes for both prisoners were opened by the 1st prisoner before the 2nd prisoner even entered the room. What I wonder is if this procedure is followed by all 100 prisoners, whether there isn't just a generic path that all must take with their individual path-lengths dependent on their number, but in general, they just trade places in the list of prisoners that search. But their path is predetermined and the probability of overall success can be estimated fairly well by the number of steps the first five or so prisoners have to take to find their number. Essentially what are the salient characteristics of the 2 path-prisoner sets that are guaranteed to fail vs the 1 that is guaranteed to pass? tell me the punchline and I will tell you if the joke is funny

[@daveworkman5213]

@@touristguy87 the boxes are closed back up in the same sequence before each new prisoner enters the room..

[@touristguy87]

@@tyrannusrex31 I never said that there was, so I have no idea where you got the idea that I'm missing any of that.

[@touristguy87]

@@daveworkman5213 thanks for being the 2nd person who's told me the obvious

[@SaltExarch]

Ironically, this also changes what the prisoners want to see when they're looking for their number. If you're picking randomly, you want to find your number as quickly as possible. But after coming up with this strategy, you want to see chains that are decently long, but still under 51. You don't actually want to see short chains. It's good for you, personally if your number is in a short chain, but the more short chains there are, the higher the probability that one of the other, longer chains is gonna be above 51 numbers. Since it doesn't really matter how fast you find your number since all prisoners either fail or succeed together, what you want to see is a long chain, because every other prisoner in your chain is also GUARANTEED to find their number with the same strategy. In fact, taken to its logical extreme, although the individual prisoner would probably be stressed if they're on box 49 and still haven't found their number, if they hit box 50 and find their number, they can celebrate right then and there, because a chain of 50 is a guaranteed win, because none of the other chains can be more than 50 in a set of 100.

[@mikecostanza303]

How is your last sentence not just a 50% chance of picking the right loop of 50 that contains your number? Then multiplied by 100 prisoners..

[@moehead8257]

@@mediagirl eitherway it will be the 50th box no matter where u start, cuz its a loop

[@frodogaming8905]

There's only 1 chain of 100 numbers, its where the prisoners' number is on that chain that is how long their loop is.

[@SaltExarch]

@@mediagirl There's no way to start on the wrong loop, since they're picking the loop that contains their number by default. They're picking the box with their number on it, meaning the loop will always end with their number, since that's where it loops back around. The only factor is then whether or not their loop is less than 51 numbers long, because they can only open 50 boxes. If one loop is 50, then no other loop can be 51 or higher, meaning every prisoner is going to pick their loop, and they will always be able to complete their loop, since none are more than 51.

[@SaltExarch]

You all need to watch the video again. You are ALWAYS, 100% of the time, going to pick your own number's loop if you start on the box with your number on it, because that's what makes it a loop. You literally cannot start on the wrong loop if you pick your own number, since picking your number MEANS starting on your own loop. The only question then is "if I'm only allowed to open 50 boxes, is my number's loop less than or equal to 50 boxes?". If you find a loop of 50, you know it is, and you know everyone else's is too, because you can't have a loop of 50, and then a loop of 51 or more, if there are only 100 boxes.

[@MarkPariente]

An intuitive way to understand why this strategy results in such a higher chance of success is that there is additional information present in the system in the form of "mappings" between the number/label of each box and the number contained inside the box. When every prisoner chooses random boxes this information is not used. The presented strategy uses the mapping information to allow prisoners to coordinate (create a shared fate) which increases their chances of success.

[@56jasa]

Thank you, this was a very helpful explanation!

[@gianpierocea]

I haven't even read yet but first thing when I saw this,I wonder if there is an information theoretic view of seeing this!

[@Vvopat96]

I was thinking it that same way, they have shared information, same roads

[@Angie_P]

thank you for that! This helps to visualize the solution.

[@childofvenus3781]

I finally understood WHY it still seems Impossible/Wrong even AFTER learning the answer to the riddle. And my god it's profound. Because when someone asks (After learning what the solution is) that "What makes you think you'd necessarily be on the correct loop?" and then you respond with the logic of it, The logic of it, [even though it's accurate and mathematically sound/proven] is "CIRCULAR"; so no wonder it seems false to us because circular logic usually IS false, Except when talking about the things that are circular in nature, like a freaking loop. Remember how in your video "Math's fundamental flaw" You explained how paradoxes arise from self reference? Well a loop, by definition Must reference itself. So because our brain will still want to reject it, we say "Yeah i know it's not random and you're opening the box with your number on it but what prevents you from ending up on the wrong loop?" And the response is "Because you're opening the box with your number on it". See the "Problem"? that was part of the question but that is apparently now the ANSWER ?!! Even your explanation in this very video can be summarized in "Because a loop is a loop, it must be a loop". It's kind of like learning God is real and asking him how do you exist?! and God telling you "Because I am God and i am infinitely intelligent and powerful i created myself into existence". The logic could be true but it still seems like WTF...!

[@tim..indeed]

I feel like I'd rather die than trying to explain to my co-prisoners why this works.

[@wengserena7768]

🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣

[@secheltfishmarket6419]

"Why this works"??? Do you not watch the same video I did. It fails 69% of the time.

[@WaynaPicchu05]

@@secheltfishmarket6419 "So you're telling me there's a chance!"

[@itskelvinn]

@nijuo joing comes off as fake and forced for the video tbh. But then again destin always interrupts his guests on his videos and blurts out dumb off topic jokes like he’s Jimmy Fallon or something

[@SquirrelTheorist]

@@itskelvinn rly?

[@matrixphijr]

TED-Ed did one of their really good riddles about this, too. It involved band members trying to find their instruments hidden in boxes. The irony is I had learned this principle in my math classes but forgotten about it since, so I wasn't able to answer the riddle.

[@fos1451]

He show the Ted Ed videos in the intro, frankly many people still don’t understand after watching that, it took me a while to be able to grasp the logic behind the riddle and actually teach others about t

[@had0j]

i think minutephysics also has a video about it, instead with normal people and 1-dollar bills

[@glennpearson9348]

Demonstrating, yet again, just how elegant and, frankly, satisfying mathematics can be. What got me through applied mathematics in college was the idea that there may be 100 different ways to get to a solution, but there's only one specific solution. Now, the solution may involve a set of solutions (a la differential equations), but if you're a fan of mathematics, you'll know what I mean. Enjoyed this one, Derek. Great work, as always.

[@z1ph0n3]

Thank you for the "initial" headache! Great explaination.

[@erwanglasziou4196]

I remember encountering a similar problem when organizing a secret santa : we noticed how the loops emerged and tried minimize the number of small loops so there would be more group-mixing during the reveal.

[@Wrotan]

That's much more friendly than the prisoner execution version!

[@stevedalrymple4939]

I love Destin’s “teach me”. He knew instantly that he couldn’t work it out off the top of his head, and was keen to learn.

[@WarlordM]

Yeah that's how I treat all these silly brain things

[@chrismorgan4375]

Yeah, that's how you get smarter everyday.

[@bronzejourney5784]

Thats %95 scripted.

[@WarlordM]

@@bronzejourney5784 lol

[@bronzejourney5784]

@@WarlordM I mean obviously. Thats not how real human beings interact. Noone has ever said "Teach me" in any scenario throughout the history of humankind.

[@rakha8198]

The easiest way for me to understand this strategy is that when choosing boxes randomly, the prisoners as a group need to be lucky 100 times to win their freedom (each person basically come in with only 50/50 chances) While with the loop strategy, the prisoners as a group only need to be lucky ONCE. That is because if the boxes arrangement doesn't produce a loop longer than 50, then they're all GUARANTEED to win. Therefore they only need to be lucky when the boxes were shuffled, and the probability of the arrangement they need come out of the shuffle is about 31%.

[@ocadioan]

The interesting thing is that if you drop the number of prisoners to 2, the probability goes to 50% if you use this strategy, as opposed to 25% by choosing randomly. If both prisoners agree to pick either their own or the other's box, then both ensure that they don't accidentally pick the same box, meaning that if the correct number is in the box they chose, they both find it, while if the wrong number is in the box, neither of them finds it.

[@Gnaaal]

​@@ocadioan Good observation, tbh I felt this to be more counterintuitive than with 100 prisoners before I read the second sentence :)

[@mperhaps]

Where are you getting this 50/50 business. Remember the chances randomly had 31 zeros to the right of the decimal. That's stretching the definition of the word lucky a bit far.

[@Gnaaal]

@@mperhaps It's because only having 2 prioners is a completely different scenario than having 100. Doing the strategy will remove the 2 possible ways the 2 prioners can chose their 1 box that we know will always fail, i.e. both opening 1 or both opening 2. Out of the two remaining, alternatives, there is a 50% chance since either both pick the correct box (1->1, 2->2) or both pick the wrong one (1->2, 2->1).

[@noisypl]

> Therefore they only need to be lucky when the boxes were shuffled Not exactly. Prisoners can be the masters of own destiny, by adding an arbitrary number to every box, and therefore create a new shuffe. 12:56

[@ffc1a28c7]

10:32 regarding Dustin's comment, to get onto a self-closed loop from outside the loop would require the same number appearing twice (say the loop you're going into is ab...ca, and you go from d to a, then c and d contain a).

[@zarblitz]

You said it was so counter-intuitive that even when you know the answer it doesn't make sense, but your explanation was perfect. I thought it was perfectly intuitive at every step. Your explanation required no temporary leaps or assumptions. Just solid fact after fact.

[@erics2133]

Exactly. I didn't see how this could work until he pointed out that the boxes/slips formed a loop. Since every box had one and exactly one corresponding slip, you can't have a loop that doesn't loop back to the starting point, since you would have to have a box with two ways to reach that box in order to do that. Likewise, the slip can't be on some other loop because otherwise, you'd have dead ends, which in this case would be a box without a slip of paper, which can't happen. This was really easy for me to visualize since I've been programming since every serious programmer rolled their own linked list at least once.

[@splicexjms8117]

Is the loop way described the only strategy? Or is it simply the best strategy? Are there other agreed upon sequences the prisoners could use for a reasonable chance at survival?

[@nathanieljones8043]

This video didn't even need to be more than ten minutes long he capping

[@christopherlperezcruz1507]

I'm not buying it at all. Its an artificial construct. There is no relation between the box numbers and the slips. Also no information is transfered. Sure the prisioners are all argreeing to do a distinct and unique set of random boxes and that seems to be key here but to have that mean the chance of success of each individual prisoner is 99% seems extreme.

[@josephyoung6749]

Prisons in the US don't make sense even when you're aware of them... Nearly one out of every 100 people in the United States is in a prison or jail right now, and 3% of the population has been to jail or prison at one time. So 1 in every 30 people you've met today has been locked up. The US has more people locked up than Russia and China combined, and 1/4 people incarcerated around the world live in the US. This is the largest number of un-free people in the history of humanity. Happy 4th!

[@rikroy2564]

Excellent video, but I think the question at 10:35, although answered with a very effective visual demonstration, is even more intuitively clear to answer. Simply ask the person to give you an example of such a loop where you start with your number but the loop ends without reaching it. They cannot, because it is logically not possible. By definition, the loop ends only when a paper slip redirects you to the original box, and if you started with the box that has your number, then that paper slip is your slip. In a worst-case scenario, the slip chasing takes you on a scavenger hunt across all 100 boxes (although the 50 box limit would have stopped you before then), but there is never any case where you enter a "wrong" loop. By definition, if you start with your box, then you're in the correct loop, because eventually *some* box has to have your slip, and then the slip for that box has to be somewhere else (because it's clearly not in the mentioned box, since that box contains *your* slip) and this goes on across either all 100 boxes or less. Problem solved.

[@itskelvinn]

Yep I would ask destin to come up with a scenario where the loop is closed, but doesn’t include yours. Like actually have him make up the numbers in the boxes and get that scenario He would soon realize you can only close the loop where it started (which is your number). It’s pretty intuitive

[@aramus2158]

That's exactly what I thought. The visual demonstration didn't stick to me quite well so I instead tried to imagine a loop where you don't end where you started. I imagined 5 boxes to make the problem more simple, that way it's easy to see how you would always eventually end up with the initial number.

[@GamezGuru1]

This is also the key to the probability being so high with this strategy - because all 100 prisoners start on a different number, you collectively guaruntee to enter every possible loop, and therefore find every number (unless there is a loop greater than 50).

[@billnoyce5026]

I was imagining a "loop" that starts at my box, and then instead of coming back to my box, it reenters the loop at some later point. For example, 1..2..3..4..2. But that's impossible, because it would require two different boxes to contain the same number ("2" in the example). Thus the loop must eventually come back to my box.

[@TheCasualGentleman]

Haha I got to that guy's reaction and now I'm not thinking these guys are as qualified to talk about statistics if they're racking their brains *that* hard thinking that the most unintuitive part of this problem is that your loop always contains itself. I guess if you aren't realizing that each box only has one slip pointing to it... But that was vital to the premise!

[@jordanwoods3553]

I feel like the really cruel thing to do would be to allow the prisoners to look at 95-97 boxes, with the same thing. If they figure out the stragegy, it's like a >95% chance of success. If they guess randomly, it's

[@jackinkc767]

What is the probability of prisoner compliance and success if the sheets of paper were face up?

[@Astro-X]

thats awesome! A much better example to use with people

[@bensch182]

exactly! try to convince your inmates now! the indivudual 97% seem so much more intuitive than trying to grasp even only that you'll always be on your own loop.. :D for nerds who want to avoid using their calculator: (97/100)^100 = 4.75525% (rounded since there was a 0 after the last digit)

[@pwndbaby]

if they opened them randomly at 95 choices each, that would give them a 0.592% chance. similarly, you could do this with 4 boxes and 10 friends. let them open up 3 boxes each to try to find something in one of the boxes. If the goal is to win if every one finds what was in the box, then they would have about the same odds.

[@Duval-In-The-Wall]

What’s mad is that you can have a prisoner check 99 boxes, and STILL have only a 36% chance if you do it randomly lol

[@abstract_guy1052]

Everyone's gangsta until the first prisoner fails.