The Riddle That Seems Impossible Even If You Know The Answer

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The 100 Prisoners Riddle feels completely impossible even once you know the answer. This video is sponsored by Brilliant. The first 200 people to sign up via brilliant.org/veritasium get 20% off a yearly subscription.
Special thanks to Destin of Smarter Every Day (ve42.co/SED), Toby of Tibees (ve42.co/Tibees), and Jabril of Jabrils (ve42.co/Jabrils) for taking the time to think about this mind bending riddle.
Huge thanks to Luke West for building plots and for his help with the math.
Huge thanks to Dr. Eugene Curtin and Dr. Max Warshauer for their great article on the problem and taking the time to help us understand it: ve42.co/CurtinWarshauer
Thanks to Dr. John Baez for his help with finding alternate ways to do the calculations.
Thanks to Simon Pampena for his input and analysis.
Other 100 Prisoners Riddle videos:
minutephysics: • Solution to The Imposs...
Vsauce2: • The 100 Prisoners Puzzle
Stand-up Maths: • The unbelievable solut...
TED-Ed: • Can you solve the pris...
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References:
Original paper: Gál, A., & Miltersen, P.B. (2003). The Cell Probe Complexity of Succinct Data Structures. BRICS, Department of Computer Science, University of Aarhus. All rights reserved. - ve42.co/GalMiltersen
Winkler, P. (2006). Seven Puzzles You Think You Must Not Have Heard Correctly. - ve42.co/Winkler2006
The 100 Prisoners Problem - ve42.co/100PWiki
Golomb, S. & Gaal, P. (1998). On the Number of Permutations on n Objects with Greatest Cycle Length k. Advances in Applied Mathematics, 20(1), 98-107. - ve42.co/Golomb1998
Lamb, E. (2012). Puzzling Prisoners Presented to Promote North America's Only Museum of Math. Observations, Scientific American. - ve42.co/Lamb2012
Permutations - ve42.co/PermutationsWiki
Probability that a random permutation of n elements has a cycle of length k greater than n/2, Math SE. - ve42.co/BaezProbSE
Counting Cycle Structures in Sn, Math SE. - ve42.co/CountCyclesSE
What is the distribution of cycle lengths in derangements? In particular, expected longest cycle, Math SE. - ve42.co/JorikiSE
The Manim Community Developers. (2021). Manim - Mathematical Animation Framework (Version v0.13.1). - www.manim.community/
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Special thanks to Patreon supporters: RayJ Johnson, Brian Busbee, Jerome Barakos M.D., Amadeo Bee, Julian Lee, Inconcision, TTST, Balkrishna Heroor, Chris LaClair, Avi Yashchin, John H. Austin, Jr., OnlineBookClub.org, Matthew Gonzalez, Eric Sexton, john kiehl, Diffbot, Gnare, Dave Kircher, Burt Humburg, Blake Byers, Dumky, Evgeny Skvortsov, Meekay, Bill Linder, Paul Peijzel, Josh Hibschman, Timothy O’Brien, Mac Malkawi, Michael Schneider, jim buckmaster, Juan Benet, Ruslan Khroma, Robert Blum, Richard Sundvall, Lee Redden, Vincent, Stephen Wilcox, Marinus Kuivenhoven, Michael Krugman, Cy 'kkm' K'Nelson, Sam Lutfi, Ron Neal
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Written by Derek Muller and Emily Zhang
Filmed by Derek Muller and Petr Lebedev
Animation by Ivy Tello and Jesús Rascón
Edited by Trenton Oliver
Additional video/photos supplied by Getty Images
Music from Epidemic Sound and Jonny Hyman
Thumbnail by Ignat Berbeci
Produced by Derek Muller, Petr Lebedev, and Emily Zhang

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29 июн 2022

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Комментарии : 35 тыс.

@pyguy9915 Год назад

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

@joekemp83 Год назад

That calculation applies only for n>50.

@veritasium Год назад

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 Год назад

@@veritasium Agreed

@gc852 Год назад

@@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 Год назад

@@gc852 Only applies to continuous functions.

@wetbadger2174 Год назад

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 Год назад

Underrated comment

@ashcheung4325 Год назад

Yeah

@tolep Год назад

You need a nerd with charisma.

@margaretjones5572 Год назад

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 Год назад

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

@DrDJX Год назад

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 Год назад

Best comment, 10/10

@alveolate Год назад

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 Год назад

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

@R.B.90 Год назад

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 Год назад

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

@reifrei1170 4 месяца назад

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

@markgunther2502 3 месяца назад

White collar criminals no doubt.

@shardulthakur7362 3 месяца назад

Lmao laughed too hard at this

@shardulthakur7362 3 месяца назад

May be all are in for stock market manipulation

@hollamonm 3 месяца назад

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 3 месяца назад

Political dissidents: First time being genocided? :Ü™

@joseph-fernando-piano 3 месяца назад

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 3 месяца назад

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

@bloxer9563 3 месяца назад

Maybe

@RockyRoad213 2 месяца назад

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

@justanordinaryguy658 2 месяца назад

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

@joshjason9460 2 месяца назад

What if the evil warden gave an empty box

@gregsquires6201 Год назад

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

@pokejinwwi Год назад

There’s always that one guy who refuses to listen and wants to be the leader xd

@lavarel Год назад

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

@tvao9010 Год назад

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

@pokejinwwi Год назад

@@lavarel fair enough

@gobbedy Год назад

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

@fixed-point Год назад

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 Год назад

Really?

@AthAthanasius Год назад

@@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.

@kjbhappy123 Год назад

@@Ryuseigan this would imply that everyone in his loop would get their number (on the 50th shot). The remaining 50 numbers can only form a loop of max length 50 so everyone there also finds their number.

@shinichigojir12 Год назад

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 Год назад

@@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.

@cultofmel 2 месяца назад

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 2 месяца назад

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 2 месяца назад

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 2 месяца назад

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

@nicolass.5849 Месяц назад

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

@cultofmel Месяц назад

@@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.

@tfdtfdtfd Месяц назад

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!

@magdasg9571 Год назад

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

@spyro440 Год назад

Well - are you Korean? ;)

@94XJ Год назад

@@spyro440 you win. 🤣🤣🤣🤣

@DarkSideofTheWest Год назад

Lmaooooooo

@Garvit_M Год назад

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

@jackinkc767 Год назад

Thanks for the suggestion. Also, bring canned fool.

@ZamanAristoOrCleon Год назад

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

@Andy-lm2zp Год назад

or the last!

@funnygreenguy Год назад

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

@ramuroy6088 Год назад

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

@funnygreenguy Год назад

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

@Spectre0799 Год назад

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)

@TheDrCN 2 месяца назад

When you first gave the solution, I sketched out on a piece of paper a version of the problem with 4 prisoners, 4 boxes, 2 attempts, and I feel like I understood the entire thing almost instantly with no further explanation required. I think lowering the numbers down to something more manageable makes the problem much more comprehensible. I mean, you could even write out 4! configurations if you wanted and prove that it works for all of them, whereas 100! is so large as to be impossible to visualize. I think it would have been helpful to include this simpler example in the video.

@suspicioussand 2 месяца назад

I think mathematicians made the number big to appear extra smart

@sjenny5891 2 месяца назад

Make it seem more complicated so people over think the answer.

@amieres Месяц назад

In mathematical problems it is always a good a idea to go to the extremes to test/understand the problem and solutions. Go small AND go big.

@sicsempertyrannus226 Месяц назад

Agreed. It's the inverse of the Monte Hall problem. Take a deck of cards and demonstrate Monte Hall to someone and they'll understand why it works.

@shishirrai9878 10 дней назад

By keeping it 100 boxes, you can show how big of a probability you skewed. Like my bro said 1mm to the length of observable universe 😉

@MikhaeylaKopievsky 2 месяца назад

I feel like there needs to be a 'wow out loud' like 'laugh out loud', because when you said "that's like scaling up a millimetre to the diameter of the observable universe", that's exactly what I did.

@jskrabac Месяц назад

Yeah, but WOL is just as many letters as WOW 😆

@DrunKao 11 дней назад

@@jskrabac lol

@bscorvin Год назад

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 Год назад

_"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 Год назад

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 Год назад

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

@Drumaier 11 месяцев назад

A legit concern.

@gerritcasper8730 11 месяцев назад

You would end up with a 15.35 % of winning, not bad

@Bismuth9 Год назад

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 Год назад

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 Год назад

They are called *odd* for a reason

@HYPERWATER Год назад

Raoodnm

@josenobi3022 Год назад

@@jynx619 how did you calculate that probability

@janbacer Год назад

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

@mattsnyder4754 11 дней назад

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.

@geraldcollins5684 23 дня назад

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.

@charliehorse8686 Год назад

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

@davidjames1684 Год назад

No letters, just numbers, ha ha.

@StabbyJoe135 Год назад

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

@andrzejbozek Год назад

@christiankrause1594 Год назад

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 Год назад

@@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 Год назад

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 Год назад

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

@ELYESSS Год назад

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

@flashstar1234 Год назад

Yeah same makes perfect sense to me

@pirojfmifhghek566 Год назад

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

@CertifiedSlamboy Год назад

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

@vpelss 5 месяцев назад

For me the best way to visualize the loops is to start with all the boxes with the same number inside. 100 one box loops (pointing to themselves). Then randomly swap two of the boxes contents. You now have 98 one box loops (pointing to themselves) and one two box loop (the swap). Keep doing this to visualize the small loops getting larger. It is kind of like mixing toffee with the links between the boxes.

@bartoszporzezinski4842 13 дней назад

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.

@inemanja Год назад

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 Год назад

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

@aaron2112 Год назад

@Michael Ritsema sure dude. Whatever.

@Houstonruss Год назад

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

@Brormable Год назад

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

@aspiringdiamond Год назад

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

@neildmoss Год назад

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 Год назад

* Tips Hat*

@tim40gabby25 Год назад

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

@eliaskjrbo8142 Год назад

I live in Aarhus!

@osiris1102 Год назад

@@ApequH tips fedora m'skyum

@neildmoss Год назад

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.

@simonnguyen692 Месяц назад

this reminds me of blindsolving a rubik’s cube. if you label the solved state of the cube in an organized way, you can trace the pieces in a sequence when it becomes scrambled, and completing that sequence will return the pieces to their solved position

@blackbird1132 15 дней назад

same

@kaustubhrao5653 2 месяца назад

Love it!! For some reason, linked lists came to my mind and it dawned on me. I didn’t fully workout that the probability for loops >50 would be 1/3

@justinc2633 Месяц назад

2/3

@ltmcolen Год назад

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

@MusicSounds Год назад

literally this time

@-Jethro- Год назад

I see what you did there!

@donc-m4900 Год назад

But why where they in jail to begin with 😂

@abinbaby4044 Год назад

Or out of the loop!

@alveolate Год назад

@@abinbaby4044 actually, into the loop xD

@GuitarGuise Год назад

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 Год назад

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

@brendawilliams8062 Год назад

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

@sonkeschmidt2027 Год назад

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 Год назад

@@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 Год назад

@@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.

@z1ph0n3 Месяц назад

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

@anonymanonym1364 Месяц назад

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

@michaelgove9349 Год назад

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 Год назад

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 Год назад

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

@balintvarga5146 Год назад

Absolutely stunning example.

@LiveHappy76 Год назад

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 Год назад

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

@wouterpomp5014 Год назад

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

@kevinz8554 Год назад

Life do be like that sometimes

@aarondavis8943 Год назад

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

@everstanding400 Год назад

Imagine knowing this for a fact and no one listens 🤣

@idiatico Год назад

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 Год назад

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

@calereliya 5 дней назад

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.

@plamaoverstrike 3 месяца назад

That's easy to understand, loop strategy sets a fixed probability that will never change while random strategy shuffles the odds all the time

@Robert08010 Год назад

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 Год назад

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

@Solo_man69 Год назад

The wardens watching too much saw and squid game

@teeemm9456 11 месяцев назад

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

@Solo_man69 11 месяцев назад

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

@wearwolf2500 Год назад

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 Год назад

I think this might be a better explanation than the video

@happywhale1786 Год назад

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 Год назад

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 Год назад

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 Год назад

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.

@pradyumnachakraborty3262 6 дней назад

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 5 дней назад

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?

@ffc1a28c7 6 дней назад

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).

@mdtexeira Год назад

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

@micahturpin8042 Год назад

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 Год назад

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 Год назад

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

@gavissing5225 Год назад

@@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 Год назад

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

@jenius00 Год назад

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 Год назад

@markmuller7962 Год назад

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 Год назад

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

@hansschwarz1338 Год назад

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 Год назад

@@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.

@DP-zw8sb 23 дня назад

Please solve this ,added to solution above video mentioned, if prisoners opens the the 2nd number by doubling the present Box number if not possible, open same numered box and every 3rd box by tripling if not same. So every member escapes same loop happening if a loop is 51 or more.

@RGByvemar 4 месяца назад

I think it would be incredibly interesting to see this experiment filmed according to the strategy that gives roughly 31% odds as opposed to a random selection strategy that most scenarios might include

@miloszforman6270 4 месяца назад

Matt Parker has a video on YT where he does this with only 10 participants. From my point of view this has only limited entertaining value, but anyway. I found the computer simulations much more expressive.

@BigPapaMitchell Год назад

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 Год назад

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 Год назад

@@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 Год назад

@@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 Год назад

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 Год назад

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

@brianrussell463 Год назад

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

@shimassi9961 Месяц назад

Towards the end you presented two alternatives with a benevolent/malevolent guard. What would happen in the case of a malevolent prisoner (someone trying to intentionally sabotage this). Would he be best served picking at random, picking at random then following the loop or some other strategy (e.g. perhaps start following a loop but stopping and switching to a new one after so many steps)?

@TheNathan696969 5 месяцев назад

thinking outside the box, inside the box 😂 very easy to understand once you mentioned the loop I understood everything. I can not fathom how people would come to the summation of being on the wrong loop or how more prisoners gives diminishing returns 😅 it was pretty self explanatory. Loved it

@Campfire_Bandit Год назад

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 Год назад

I actually think 5 second of thought is still too short

@ericlevy3317 Год назад

The best teachers never stop being students.

@RJFerret Год назад

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 Год назад

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

@yossam6722 Год назад

A true master is an eternal student.

@WiiAndii Год назад

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 Год назад

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

@Vfulncchl Год назад

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_ Год назад

That's a good way of thinking about it.

@filipposchiabel Год назад

@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 Год назад

@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.

@rueby2kool 11 дней назад

STEINER VOICE: When you don't use this strategy your odds drastic go down. This is a great video! Very interesting and incredibly well editted

@MissShaneice 5 месяцев назад

the board that snaps in place is an amazing thing. genius product! i did watch the video, and i get it, but it's a lot. my brain gave up a long time ago. lol.

@rashadisayev 11 месяцев назад

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 10 месяцев назад

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 10 месяцев назад

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

@rashadisayev 10 месяцев назад

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

@JJforShie1 10 месяцев назад

@@Ren-fo4lg lol me too

@yourtinerary 10 месяцев назад

This comment is 🙌🏻

@mattsnyder4754 Год назад

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 Год назад

@daburnd Год назад

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 Год назад

@@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 Год назад

@@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 Год назад

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

@MinecrafterLawl3456 Месяц назад

On checking 50 boxes, the chance goes up to 100% after allowing for communication afterwards so long as each prisoner keep track of their loop number. There can be only one loop larger than 50 so by addjng the remaining loops number up we get a number X. Taking 100-X-50 we get the length of chain remaining in the loop. We shift everyone down the loop by that number and everyone gets out

@Slinky0205 10 месяцев назад

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

@athletico3548 6 месяцев назад

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 6 месяцев назад

@@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 6 месяцев назад

@@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 6 месяцев назад

@@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 6 месяцев назад

@@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

@chriswasia413 Год назад

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 Год назад

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 Год назад

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

@muhammadmaazwaseem7452 Год назад

Following

@Liberty46 Год назад

@@ace10414 he lied

@imranq9241 Год назад

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

@taitsmith8521 5 месяцев назад

I like that the guy who came up with the problem is lauded, while the peraon who solved it is written off as "a colleague ". What you really mean is the janitor figured out the solution and he's embarrassed .

@hansshekelstein9450 Месяц назад

Lol dude what a reach

@heatherschoppmann9971 Месяц назад

This makes total sense! Thanks we loved it

@user-hw9xe4lm2s Год назад

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 Год назад

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 Год назад

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 Год назад

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 Год назад

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.

@tomgray8156 Год назад

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 Год назад

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

@Legendendear Год назад

@@kevinbean3679 69? Nice!

@AvanaVana Год назад

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 Год назад

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 Год назад

@@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.

@LotusMouse Месяц назад

I’ve actually seen the Ted Ed version of this with the band members, and I’ve been wondering how accurate the strategy is. So finding this video was a pleasant surprise to come across

@NickName-jz7mn 5 месяцев назад

Even when the prisoners pick at random, their numbers are still on a loop, just not coordinating with others to all start at with different number (not the same as any other prisoner). I think that is what seems so "off" with this. I LOVE VERITASSIUM

@thetaomegatheta 5 месяцев назад

'Even when the prisoners pick at random, their numbers are still on a loop' Not guaranteed to be on the right loops, and they do not proceed to follow those loops.

@kenthomas4668 4 месяца назад

@@thetaomegatheta they are always on the right loop, the question is how long is their loop

@thetaomegatheta 4 месяца назад

@@kenthomas4668 'they are always on the right loop' That's trivial to disprove. Consider the situation where all loops are of length 1. Consider now that a prisoner picks a box at random and does not chooses the box with their number. They are now not on a loop with their box, and, therefore, not on a loop with their slip.

@jamesbuchanan1913 Год назад

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 Год назад

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 Год назад

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 Год назад

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

@GameFuMaster Год назад

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

@supersonicgamerguru Год назад

@@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.

@testingreadaboutit Год назад

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

@gaminawulfsdottir3253 Год назад

Hah! I was thinking the same thing!

@anwinkrishna2849 2 месяца назад

this guy is the bestttttttt.... we need more content like this .... i love interesting stuff like this

@markgreen5339 6 месяцев назад

I have a question. While I am far from this figuring out type of logic, I do understand it. Could a similar strategy be employed using the same process? Would it be the same probability if the prisoners stated beforehand that the first to enter would pick box one and then follow the same loop strategy, and the 2nd to enter would select box 2, and so on?

@miloszforman6270 6 месяцев назад

You mean that they should _not_ start with their own number? This cannot work.

@billrexhausen4221 Год назад

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

@manstuckinabox3679 Год назад

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 Год назад

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

@rinnegone377 Год назад

Why are the replies gone?

@irrelevant_noob Год назад

@@rinnegone377 Happens sometimes. I'm guessing either they were removed individually, or the accounts have been poofed. 🤷‍♂️

@MK-fg8hi Год назад

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 😂😂

@Lindeman08 Год назад

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 Год назад

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

@AnthonyGoodley Год назад

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

@lanceslance2930 Год назад

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

@magica3526 Год назад

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 Год назад

For me he has a false title

@yusufuveysdurudeniz2341 5 месяцев назад

What seems wrong was about we are thinking like solving a random coordinat system problem however problem or equations contain more symetry than we are aware of. Therefore, we are dealing with x,y expression of a circle's equation when be get back to our model of coordinat system problem.

@SwiftDustStorm Год назад

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 Год назад

Linked lists :))

@codahighland Год назад

@@rachadelmoutaouaffiq5019 The worst kind of list!

@markingraham4892 Год назад

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

@codahighland Год назад

@@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.

@SwiftDustStorm Год назад

@@markingraham4892 care to elaborate?

@hunterbroadnix9609 Год назад

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 Год назад

Do you get executed if you mess it up?

@JA-nv4zb Год назад

that is so cool

@chapteronefrog Месяц назад

It's basically just opening a box that lights up the next one, the circle just gets bigger as you increase the number of boxes bc you're not changing the percentage of boxes the prisoners can check. If you increased the percentage as the amount got bigger, then it would change the rate if success but bc it's 50% for EVERY amount, the success rate stays relatively the same. The only thing that makes it smaller is the increased number of possible loops

@prathammathura 4 месяца назад

this is definitely one of the best video ever i seen

@TomasJuocepis Год назад

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 Год назад

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

@TomasJuocepis Год назад

@@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 Год назад

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

@markmuller7962 Год назад

@@TomasJuocepis 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 Год назад

@@yosefsl30 Right

@tegxi Год назад

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 Год назад

Why can’t u have no slips pointing to 1

@Mrityunjay7 Год назад

@@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 Год назад

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

@Random-zt4ig Год назад

Your explanation is fantastic. Traveller!

@reidflemingworldstoughestm1394 Год назад

@@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.

@julixpinguimon8023 6 месяцев назад

I love thinking of this like its super sound because, if the chance of the largest loop being

@MatthewNichols0 3 месяца назад

Perhaps it could have been explained better why you are guaranteed to be in "your loop". I'd explain by saying that the algorithm can only terminate when you find a paper that has the number of a box you already opened. Since there is exactly one paper for each box, the only time this is true is for the very first box you opened. Every other box you already opened was due to you finding a paper (the one and only paper) for that box.

@jootpepet Месяц назад

Your explanation isnt better tho you just worded it in a more confusing way lol

@mustang8206 Месяц назад

Your explanation is worse. You have to be on your loop because the loops ends when you get to number of the orignal box, since you started at your number and so the only way it ends is when you get number.

@MatthewNichols0 Месяц назад

@@mustang8206 With your explanation, It could also end at any other box you've opened (so, it could be A->B->C->B). The key is that it cannot end at any of those other opened boxes because you already found the one and only paper for those boxes in the box just before it on the chain.

@mustang8206 Месяц назад

@@MatthewNichols0 No it can't. Not sure if you watched the video but there's only one of each number

@grys9245 Год назад

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 Год назад

Good stuff.

@Harshil_Uppal Год назад

Bruh how do you even code that?

@ramsyfpp6418 Год назад

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

@HEYJO77 Год назад

nice

@Linaiz Год назад

@@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.

@theAkornTree Год назад

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 Год назад

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 Год назад

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

@daminkon246 Год назад

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

@jacobboehm983 Год назад

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 Год назад

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)

@chriflu Месяц назад

Little-known fact: When Peter Bro Miltersen originally published this riddle, his name was just "Peter Miltersen". However, after his statement that he was going to leave solving the puzzle "as an exercise to the reader", so many readers went like "Bro!!!" that he was henceforth given the middle name "Bro".

@miloszforman6270 Месяц назад

Idiotic. Nobody in the academic world would dare to use a trash people expression like "bro!" unless he was willing to be henceforth ignored in this social environment.

@Rehankhan-us8tv 3 месяца назад

Mr beast should do this

@InterestingMindsYT Год назад

and what if you work with time, lets say some people are slower then others, they also say nothing about time!, lets say every number is a = minut, and the second person to move in only can go in if the other out. what if they make a plan, that the amount of minuts somebody is gone is equel to the box that contains there specifik number. and if they dont find it they need to wait exactly 102minuts. ( because its more then 100 boxes = 1 box a minute) by that way the poeple that are waiting only have to count everytime and remember only 50 numbers ( and they have really long time for that) if they know 50 numbers thats not there number or a little bit less) the chances of cracking this code will also improve by multi trillions? correct me if am wrong

@entropie-3622 Год назад

Technically that counts as communication. In practice the warden can easily set things up to prevent this anyways for example by giving each prisoner a fixed time window in which they have to be done and always waiting for the whole window even if a prisoner is done early before letting the next prisoner in.

@Karperteamdelo Год назад

@@entropie-3622 yeah thats true! but you can also say if they randomize the boxes everytime they enter, then this solution wil also not work

@Karperteamdelo Год назад

and lets say it even did, they said nothing about time?

@entropie-3622 Год назад

@@Karperteamdelo The rules do generally say that communication is forbidden, not just verbal communication. Essentially this means any transfer of information between prisoners is forbidden no matter what medium is used to transfer the information, this includes subtle means like using time as well.

@SnailHatan Год назад

Communication is inherently impossible in the scenario, so they couldn’t do that.

@smartereveryday Год назад

Thanks for teaching me.

@jhaz89 Год назад

You're welcome

@daivomjoshi56 Год назад

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

@Ryanisalive Год назад

There you are!

@alberteinstein9626 Год назад

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

@strikerj_ Год назад

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

@zekeblume566 2 месяца назад

Ik I’m late, but I finally got it when I realized that there MUST be loops. By picking the box labeled with your number, you’re just guaranteeing that you sure in your own loop

@cs6496q 2 месяца назад

I have a question. What if you start with box 75, that goes to box 34, that goes to box 92, ect. But in the final box, let's say 20, the slip says 34, which isn't where you started, meaning 75 is in a different loop. Is there a possibility of that happening?

@joso7228 2 месяца назад

@@cs6496q NO because box 75 has slip 34 already.

@AGill-mx8bj 4 месяца назад

Bro, you are just a genius. Most people can’t solve this problem. There is just hard. I am watching this so I can come up with an algorithm so a computer can do it.

@lukegorman4523 Год назад

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 Год назад

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

@lukegorman4523 Год назад

@@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 Год назад

@@lukegorman4523 ohh makes sense

@rajithanw555 Год назад

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 Год назад

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

@Lord_and_Savior_Gay_Jesus Год назад

I like how _Smarter Every Day_ asked *"teach me."* More people need this response in everyday life. Humility and the willingness to learn something you don't know.

@TheBeardedMann Год назад

Well, yeah, that's how he gets Smarter Every Day :)

@vectoralphaAI Год назад

The smartest people are the ones who admit they don't know and are willing to learn.

@JensPilemandOttesen Год назад

All smart people have that attitude. Cause or effect??

@fancen Год назад

why humili

@Vvopat96 Год назад

That's good way to think to become smart, better to think that you're dumb than that you are smart to become smart. People how think they are smart often think that they know better which leads them to not learn from others. I think that's the reason also why many smart people are humble, they become smart because they are not locked into their believes and stay open for new ideas and change their mind often.

@sorensouthard927 5 месяцев назад

Basically it boils down to a function of the loops rather than however many given instances of choice. It's just the chance of a loop greater than half of the total numbers emerging.

@KipIngram Месяц назад

I saw a video yesterday about how "absolute chaos" is impossible. That is, in any complex system, there is an unavoidable amount of "order" that will be present, no matter what. These loops seem like the same sort of thing. The 100! bit makes this seem wildly unpredictable, and yet these loops can't be avoided - all you can do is control their length.

@miloszforman6270 Месяц назад

You mean that these loops form a kind of order?

@kristjanbrezovnik6485 Год назад

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 Год назад

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 Год назад

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

@JackFate76 Год назад

@@DevinDTV But it‘s humble and epic.

@Dylan_Did_A_Thing Год назад

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 Год назад

I felt the same

@Perceptious37 Месяц назад

probability is so weird. There was a mathimatical analysis on the "lets make a deal" game show showing you were almost 17% more likely to open the top winning door after the already opened the door you didnt pick if you swapped your choice to the other door when given a choice. You initially made your pick with a 33% guess, but now its a 50/50 and you have more information means that your initial choice was actually less likely to be the correct choice.

@miloszforman6270 Месяц назад

Lol. There are several YT channels where you can cause fierce debates with your "MHP trolling". I doubt that this one here is the right channel.

@phillard8653 6 месяцев назад

This loop logic is very reminiscent of solving rubik's cubes blindfolded, but in that case this type of method is used more for ease of memorization than any probability. You just have to memorize the order of steps in a loop.

@rtripp25 Год назад

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 Год назад

@@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 Год назад

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

@ibrahim-sj2cr Год назад

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

@ibrahim-sj2cr Год назад

@@nejnovejsi and thanks

@MidnightSt Год назад

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 Год назад

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

@musicexams5258 Год назад

@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 Год назад

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

@ekelrock9940 Год назад

@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.

@TooBoredToDoAnything 3 месяца назад

Why does the loop always close? Each slip can be found only once. If I open my box (42), and there's a slip to another box (13), then I know that no other box points to 13. I can have bad luck and open other 98 boxes without success. But it also means that, at the point of opening the last box, I can NO LONGER expect slips to the 98 of the 99 already-opened boxes, because I found the slips them. The only slip that remains is for the box that I opened initially, without a slip, which is my box 42.

@XtreeM_FaiL 3 месяца назад

Next time just follow the simple rules. Don't improvise!

@ananykashyap7864 Месяц назад

Just take a pen and paper, try with less no. of boxes like 16 or 20. Now write all the numbers randomly and also write down the box no. From 1 to 16 to each box, the limit for each prisoner now is only 8 chances, now try and see. It'll make a loop always. And maybe in 3 or 4 tries you'll see all the prisoners survived.

@dannypowell594 Месяц назад

This is awesome! Thank you

@MoshpitMaestro Год назад

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 Год назад

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.

@alonsoquesada1136 Год назад

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

@Plant_Parenthood 11 месяцев назад

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 9 месяцев назад

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

@Billy.Nomates 9 месяцев назад

@@Plant_Parenthoodboth are acceptable,both refer to being modest

@Plant_Parenthood 9 месяцев назад

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

@tylerdavis3 8 месяцев назад

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

@manuelhaulik7317 24 дня назад

Interesting and paradox thing is that the prisoners should wish to find their number in the last box they open by this method because then they 100% know they will survive since there is no chance to be a bigger loop than 50-number one

@anthonyat2401 4 месяца назад

Fascinating; thank you.

@dothingsthatmatter8567 Год назад

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 Год назад

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 Год назад

@@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.

@Astraeul Год назад

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 Год назад

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 Год назад

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

@fortyyearfitness Год назад

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 Год назад

@@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.