The Riddle That Seems Impossible Even If You Know The Answer

チャンネル登録 14M

再生回数 12M

99% 371 000 1

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...
▀▀▀
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/
▀▀▀
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
▀▀▀
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

に公開

2022/06/29

共有:

ダウンロード:

読み込み中.....

追加:

私のプレイリスト

後で見る

コメント数

pyguy 年前

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

Joe Kemp 年前

That calculation applies only for n>50.

Veritasium 年前

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 _ 年前

@Veritasium Agreed

G C 年前

@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.

Joe Kemp 年前

@G C Only applies to continuous functions.

Greg Squires 年前

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

PokeJin WWI 年前

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

Lavarel Sieghart 年前

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

Tvão 年前

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

PokeJin WWI 年前

@Lavarel Sieghart fair enough

Guillaume Perrault Archambault 年前

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

J GA 3 ヶ月前

Here's a more direct explanation of why there are loops: At each step the prisoner is redirected to another box and there are two possible outcomes: the box either points them back to their own number, in which case there is a loop, or it redirects them to another box. The heart of the matter is that the second scenario cannot happen indefinitely, since there are only 100 boxes. Indeed, the worst case scenario would be that the prisoner needs to visit the remaining 99 boxes before it discovers its own number, thus finding the loop.

Jimmy Liu 3 ヶ月前

Right. If you know some abstract algebra, it's the same thing as saying that no orbit of any permutation can have infinite length

John Yeager 3 ヶ月前

Whatvwas thst again?

Guille 3 ヶ月前

your explanation made me understand it! thanks!!!!

Ceylenium 3 ヶ月前

Ohhhh okay this makes sense of the reasoning behind how the loop even started now- I had to go searching in the comments for something like this cause I just couldn't wrap my head around why a loop would exist, even though what he was saying made some sense. Thank you!

Ultimine 2 ヶ月前

W explanation

TheLostCause 3 ヶ月前

It just dawned on me why this intuitively makes sense to me. When I was younger I discovered that if I miss one part of a matching question(and I didn't use an answer twice) I had to miss at least one more, and the ones I got wrong formed a chain of wrong answers that eventually always loops back to the right answer for the first incorrect question.

Jimmy Liu 3 ヶ月前

WOW that is amazing! It's the exact same set-up as the video's scenario

Gux The Artist 2 ヶ月前

wrong. You could open the whole number of answers, not just 50% of them. Nothing proveen here or in this video

henry langsford 2 ヶ月前

@Gux The Artistdid you not watch this properly or are you just a little simple

128Gigabytes 2 ヶ月前

@Gux The Artistwhat does open the whole problem of answers mean? Open every box? True that could happen, the video even says so itself. It could happen, but its way less likely with this method than by picking at random

Rashad Isayev 5 ヶ月前

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

Lethal Wolf 3 ヶ月前

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 3 ヶ月前

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

Rashad Isayev 3 ヶ月前

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

Tanner Vogan 3 ヶ月前

@Ren lol me too

YourTenary 3 ヶ月前

This comment is 🙌🏻

Jaxson Bateman 3 ヶ月前

That was awesome. I don't think I'd say it seems impossible after knowing the answer, but it does highlight how small changes - in this case, a strategy change - turns everything on its head. In a real life application the biggest issue would be getting the 100 prisoners to get along with it - I'm sure some of them wouldn't understand why it would work, and thus you wouldn't be able to be sure that they'd go along with it.

OverPower 2 ヶ月前

That is a good point about convincing the others that the loop strategy is best. Because for each prisoner that doesn't use that strategy, it halves your chances. If, say, 96 prisoners go along with the plan, that seems great, but that means that four prisoners will be guessing randomly, which means that your chances of success have gone from 1/3, to 1/48.

sankyumiku3939 5 ヶ月前

I think the more basic and simplified answer to the “How do you know your number will be on the loop?” question is that it is a loop. Your number is the starting point and the ending point that’s why it’s a loop. If you reversed the loop, your number is guaranteed to be next in line. The main problem is just that you don’t know how long it’ll take for you to finally come across your number, and it might be more than 50 times.

Trip Fontaine 3 ヶ月前

Another good variation on what you are saying is to tell doubters "If you think someone's number can be outside its own loop, draw a diagram of what is happening there." The diagram cannot look like a loop and will have dead ends that make no sense.

Lance Bryant 3 ヶ月前

This is an excellent explanation.

OptimusPhillip 3 ヶ月前

My first thought when the person asked that question was "Is it really not obvious?" Because it feels obvious to me. Like... that's how loops work?

鬼仌 2 ヶ月前

@OptimusPhillipcommon sense isn't common nowadays

Mvk Vamsi 2 ヶ月前

Exactly, you have started with your own numbered box, so you either find your own number within 50 times or more, but not out of your own loop. Your loop is completed, with you finding your number.

Magda SG 10 ヶ月前

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

Spyro 10 ヶ月前

Well - are you Korean? ;)

94XJ 10 ヶ月前

@Spyro you win. 🤣🤣🤣🤣

PolarEyes 10 ヶ月前

Lmaooooooo

Jack InKC 9 ヶ月前

Thanks for the suggestion. Also, bring canned fool.

Slinky 3 ヶ月前

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

Selchie 2 ヶ月前

It finally started to make sense to me when I realised this one thing. It seems so unintuitive because, since the numbers are placed randomly, and the numbers on the box are placed arbitrarily, there should be no correlation between them. But, the correlation is that they're both ranges from 1-100. You have a hidden array of numbers 1-100, and you're assigning each of them 1 through 100. Basically creating an arbitrary combination of the same numbers on top of it, which obviously creates some real correlation. Thinking of this made it seem much more intuitive, and less random.

p1kles 4 ヶ月前

Yeah, this makes sense. You can only ever get on your loop. It's not that you're choosing a loop at random and then picking right, it's that, because your number has to be in the box "behind" the one you chose on the loop, you will automatically already be on the loop. You literally cannot fail by choosing your own number as the box to start on. This was a cool video.

Gux The Artist 2 ヶ月前

It's called "false logic". It is consistent in itself, but false. The calculations he propose are right, but incomplete. In short what he proposes in this video does not account for reality and has no validity at all, in terms of "best stategy" or playing up the actual probabilities of succeding in the proposed case. It would take a long demonstration to disprove it, suffice is to say his demonstration only succeed in verifying a logic he applies, but not any actual reliable result out of it. "self-referntial logic demonstrating itself, contraddicting reality".

Checkmate 2 ヶ月前

You will only get in your loop because it’s actually just 1 huge loop. How is this not 50/50 though again assuming you will choose the wrong 50 side of your loop.

Funposter 2 ヶ月前

@Checkmate It's 50/50 on an individual level, but the results are no longer related to one other (.5*.5*.5 etc.) As explained in the video, either everyone succeeds, or everyone fails.

Julien Tomezach 4 ヶ月前

I can imagine the excitement of Derek finding a subject so hard that he could not possibly explain it clearly in a

Alena 2 ヶ月前

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.

Jeroen Noël 年前

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

MusicSounds 年前

literally this time

Jethro Stuff 年前

I see what you did there!

Don C-M 年前

But why where they in jail to begin with 😂

Abin Baby 年前

Or out of the loop!

alveolate hermeneutist 年前

@Abin Baby actually, into the loop xD

Cameron Weage 2 ヶ月前

Major props to the motion graphic artist on this episode. I was imagining the steps to make this video in after effects and have a headache lol

Emlu 4 ヶ月前

Actually understood the math. You guys are great at explaining

Wybird666 ヶ月前

Such an elegant solution, and well presented. It really highlights the stark difference in probability when events are not mutually exclusive!

timber72 4 ヶ月前

This doesn't seem impossible at all. You can never, ever be on the wrong loop...even if your loop takes you through all 100 boxes...because you are starting with the box that the correct slip ALWAYS points to...meaning, the correct slip is always at the end of the loop you started, and 31% of the time, all 100 prisoners will find their loop in 50 or less boxes.

MCB400100 3 ヶ月前

You're not telling how it happens. Its a vague discription of loops existing and always being in one. And after u repeat the numbers that are told in the video. Alright, but why is it 31% if everyone does it.

Poise Works 3 ヶ月前

@MCB400100 Because if all the loops are smaller than 51, which has 31% chance of happening, everyone is GARANTEED to find their number. It doesn't matters how many people does it

The Great Fapsby 3 ヶ月前

@MCB400100 The answer is in the video lmao

The Great Fapsby 3 ヶ月前

@MCB400100 it's 31% because there is a 31% that there are no loops of size >50. If there are no loops >50 then all prisoners will find their number on some loop. You can think about it like this. Failure is defined as at least one prisoner fails to find their slip. Using this strategy a prisoner fails to find their slip only if their number is on a loop of greater than 50 length. The probability that any given number is a part of a loop greater than 50 is approx. 69%. The probability of success is equal to 1- probability of failure (Because by definition all possible outcomes must add up to 1 ( Failure +success = 1 ----> Failure + success - failure = 1 -failure -----> failure-failure + success = 1- failure ----> 0+ success = 1 - failure ----> Success= 1-failure) Failure and success refer to P(S), probability of success, and P(F) probability of failure) Therefore P(S) = 1- P(F) = 1-.69 =.31 = 31%. It's less dependent on the individuals doing the trial and just the fact of how many ways a sequence of numbers 1 to n can be divided into loops. N just happens to be the number of prisoners. The other reason why its 31% is because if a loop greater then length 50 exists then all prisoners on that loop will not get there number. Which is kind of why it just simplifies to the probability of the loop existing

4th shot's ecstasy 14 日前

Thanks for all of your amazing and clear explanations of these complicated-for-the-majority contents in your videos. Just subbed. 👍🏻💪🏻👏🏻🙌🏻

WetBadger 年前

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.

Guido 年前

Underrated comment

Perseus Alphietious Darkfire 年前

Yeah

tolep 年前

You need a nerd with charisma.

Margaret Jones 年前

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!!

Marko Mikulicic 年前

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

AsianOtakuGuy 21 日前

To make it more intuitive, let's change the numbers to keys. And forget the "only open 50 boxes" limit for now. All 100 boxes contain a key. All 100 keys are different and will only open the the box that matches the key number. If you find the key that matches your prisoner number, you'll be free. In this scenario, you no longer have a choice. You can't use key #3 to open Box #5 after all. You're forced to start at the box that matches your number. The key you find inside will lead to the next box. You go down the line opening the boxes until you find your number. There can't be duplicates or dead ends, because there are no 2 same keys so you can never open the same box twice. Each box will always lead to a new box until eventually you run into your key. You have to run into your key because you started at your box number. The only question is how many boxes will it take you to find your key? Now the real question becomes, what is the probability that you will find your key within 50 boxes? And the answer is approximately 30.7% of the time.

Sorio99 日前

I feel like the best explanation for why you can’t get in a wrong loop is “Because that would mean two slips of paper had the exact same number on them, and that’s against the rules.”

Visage Liquifier 日前

11:03 - This is a fun question, and an excellent one for people wanting to learn how to code. (.. even though this is essentially the Monty Hall problem with new clothes) Short: Presume you go through the longest possible loop and see every single number, but yours isn't one of them. That's impossible because all the numbers are in the boxes. The only way to end earlier than all numbers (or max tries) is if you find your number sooner. Longer: Every number appears exactly once. Once you open a box and look at the number, you'll never find that number again. If that's the case, the box it leads you to, you'll never inspect again, so opening a box 'consumes' that number. No numbers in the boxes are out of bounds. Every single number in a box is guaranteed to be one of the boxes in the room. If both the above are true, you can never open a box that sends you back to one you already opened (except your number, in which case you're done), so every number is unique and is one of the numbers of a box. There is no way you can't, if you have enough guesses, eventually find your number. Your choices are never duplicates or out of range, so eventually you'll have to see every number. Every number in a box must be a different value from 1-100, which will send you to a different box with a value from 1-100 in it that you haven't seen yet (or your starting point). It's like drawing lotto numbers from a basket. If they are all different and you take out each as you draw it, you'll eventually deplete all of them. You'll see each one, but in a different order every time. Take the numbers 1-5 and scramble them. Label each number from left to right with the values 1-5 in sequence: 5 1 2 3 4 1 2 3 4 5 Use the strategy. You can't help but find the number you are looking for.

CWolf20 2 ヶ月前

It makes total sense but I definitely get how difficult it is to see why. Super clever.

edie 3 ヶ月前

A lot of people say they are confused about how your number has to be on the loop. In fact one person said "what if your number is 1, box 1 takes you to 3 which takes you to 2 which takes you to 3. now you are in a loop without your number". But, one huge part of the problem is that each number slip shows up ONCE. This means that you will ALWAYS have to come back to your number. Let's say your number is 1. box 1 leads you to 2, 2 to 3, 3 to 4, and so on. There is no possible way to loop back to any number other than 1. Which means, you just keep going, up and up, 97,98,99, until the last one, 100 takes you back to 1. The ONLY number you can loop back to is 1. Every other numbered slip has already been used.

Afifyboi 3 ヶ月前

I dont understand what do you do after box 3 says 2

Afifyboi 3 ヶ月前

I understand now

WalkingTaako 8 日前

@AfifyboiJust in case somebody else doesn't get it: Box 3 can't say 2, because 2 is in Box 1, and thus can't be anywhere else at the same time

Mark Green 日前

@WalkingTaako, I am an idiot and it only took me 10 seconds to think of that also. I am just happy you could write it so succinctly.

Dr. DJ X 年前

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.)

Albe Van Hanoy 年前

Best comment, 10/10

alveolate hermeneutist 年前

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 :(

N. Z. Saltz 年前

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

Randy Bhagwan 年前

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 :)

Rick 年前

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

bob czech 4 ヶ月前

Feels like a Benford's law concept for 51 to 100 box loop chances, but I feel like I don't but also do agree with using the loop strat. Even if chances for 100 box loop are 1/100 to 50 box loop at 1/50, it's the same if you go from 1 to 50, so really, it's still a 31.18% chance. However, it feels like it's still a 2^-100 chance because of 100 boxes, and you can only pick 50, but I feel like it isn't impossible to actually have a 1/1 loop and then a 99/99 loop. Would adding 5 still have a loop over 50? Depending on the loop the boxes are in, that is totally possible. If a guard somehow makes a box arrangement that stays as a loop of 100, it is 100% safe to add 10 to the box count because it probably is impossible to do a setup where a loop of 51-100 stays if you add to the perfect number. Also, every box can also be part of a loop of 1, the chance of that happening is: 100/100!.

Demara Darkstrype 2 ヶ月前

So... Something I wish to mention is.... pulling gaming in as a thing: There are people who play video game randomizers as a hobby... And there's a thing called "The Logic" It is the machine learning tool thing that states "You will be able to complete every puzzle that you have been given" And everything you've said leads me to believe "The Logic" can easily be broken with your rules.

emaster01 5 ヶ月前

My favorite part of this video is that, by far, the most counterintuitive and astounding claim in the *entire video* gets only 15 seconds. "Oh you got a malicious guard trying to screw you over? Just randomly choose a number to "renumber" the boxexs and everything just works out. Trust me, I don't need to show you how or why". Great video though.

MonkeyAndChicken 13 日前

It's because the "loops" are based on the relationship between the box numbers and slip numbers, and the loop strategy's success rate inherently depends on that relationship being randomized. It doesn't actually matter which box the prisoners decide to call "Box 1" and "Box 2" and so on... As long as they all agree among themselves on which box is which so they count the same way, their strategy will still work 31% of the time. In fact, even if there are numbers already written on the boxes when they walk into the room, it's actually in the prisoners' best interest to ignore them, and randomly assign their own numbers to each box. The guard can only interfere with the loops if he knows the prisoners' "labeling" scheme in the first place- how else would he know how to create a loop greater than 50 if he doesn't know which physical box each slip inside is pointing to? Just in case this isn't making sense, think of it this way: The warden starts with numbers written on all of the boxes and each slip in its correct box. The warden is nasty and he knows what the prisoners intend to do, so he decides to just move each number forward one box, so Slip 1 is in Box 2, Slip 2 is in Box 3, and so on. This is a loop of size 100, and the prisoners will all definitely fail to find their numbers if they use their strategy. BUT! If the prisoners randomly renumber the boxes, the coherence of the loop of 100 is destroyed just the same as if they had shuffled all of the slips around! Now, they have a 31% chance of success again!

Movies Are Dope 10 日前

I'm so happy. That's the only strategy I guessed in my head & thought "bet that won't be right" I'm excited to find out why that works

Brian Doris ヶ月前

I think when explaining why your loop is guaranteed to have your number in it, you could have said, simply that your loop "CAN NOT" end until you find your number which in turn decides the length of that loop.

Charlie Horse 年前

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

David James 年前

No letters, just numbers, ha ha.

StabbyJoe135 年前

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

Andrzej Bożek 年前

Christian Krause 年前

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.

Charlie Horse 年前

@Christian Krause 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.

Caleb Faithe ヶ月前

This makes sense as a computer simulation with randomly generated numbers. I think if a prison actually were to recreate this scenario, whoever was choosing the numbers at random would end up creating a 100 number loop. Not great for the prisoners

Wesley Khan 2 ヶ月前

So, I tried to solve the problem before playing the video forward and I intuitively thought of the same solution (just guess at work) but I thought if we could move that number to the same box number that would greatly increase the chance of survival and after the end of the loop the box where we started from would contain the number written over the box. Eventually, all boxes would have the same number as written on them.

Bucks 28 日前

Have to leave the room as you found it

Miguel Medina Cobo 3 ヶ月前

10:35 Exactly my question! Now I understand that if you have a loop, you forcibly need your number to come back to the beginning. The only way to avoid a loop is for there to be different numbers inside and outside the boxes, and it is not the case

Kuroro 3 ヶ月前

11:00 The way I would explain that is to go the other way around. The slip with your number is put in an empty box... so that box needs another open box to put its number. Repeat. Once the open box used is yours, then the loop is made. "What if you never use your box?" then you end up with 99 filled boxes and the 100th number only has yours open - the 100 box loop is made. That's essentially what you're explaining, but by making box-number pairings (which isn't the important part)

MichaelodonZ 2 ヶ月前

Its amazing i once see a closed room murder case using this technique, lets say, "Closed room A" With "Key to Closed room B" And to open "Closed room B" You need that key, and in "Closed room B" There is "Key to closed room C" And so on. All room is locked, and theres corpse murdered in each room there. One of my favorite closed room murder case ^^

SmarterEveryDay 年前

Thanks for teaching me.

JHaz 年前

You're welcome

Daivom Joshi 年前

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

A-Ryan 年前

There you are!

Albert Einstein 年前

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

Justinian Siah 年前

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

Hernán Trigo Mogro 4 ヶ月前

Amazing video, there's only one thing I don't get, and I hope someone here in the comments can explain it to me: In the malevolent warden scenario, the prisoners can supposedly rearrange the boxes by "modifying" their numbers (for example by adding 5 to each box number) and get the same ~1/3 probability of success. But if you do that, how can you guarantee that you'll find your actual number in the closed loop you pick? If you're prisoner number 10, for instance, and you add 5 to every box number, instead of opening box 10, you start by box 5 right? That guarantees that the loop you fell into will include 5, but not necessarily 10. Am I missing something here?

Milosz Forman 4 ヶ月前

_"Am I missing something here?"_ Yes. You start with old box 5, which is now box 10 according to the new numbering scheme. Now you have to calculate which _slip_ is pointing to the new box 10. Hint: It's _not_ slip number 5.

Whats Up 4 ヶ月前

it was errored, if the warren simply built one single loop ( box number +1 = number inside) all will fail regardless of the outside number. Vertasim erred in statement saying a arbitrary shift and a systematic shift (+5)

Nigel Hunt 4 ヶ月前

@Whats Up "Vertasim erred"? That's a bold statement. If box 1 has slip 2, box 2 slip 3,....box 100 slip1. (a 100-loop), shifting the box numbers by 1 (box 2 has slip2, box 3 has slip 3...box 100 has slip 100) will result in 100 1-loops (all succeed on first try) so why do you say "all will fail regardless of the outside number"?

Whats Up 4 ヶ月前

@Nigel Hunt I'm not to scared about, error is fairly common. Here is another way to say it, If the prisoners feared sabotage by box number they could actually randomize the box numbers before they start and that would fix the error. The key to understanding the puzzle is to sabotage it by the numbers in the box. The sabotage must contain a 51 or greater loop thus the prisoners lose. It turns out building 51+ loops is much harder than you think which explains the odds.

Nigel Hunt 4 ヶ月前

@Whats Up I'd have thought it quite easy to generate a 51+ loop; for starters there's a 69% chance just by randomly placing slips in boxes, but once that's done you just enumerate the loop sizes by inspection and, if no big loop exists, combine the largest loops two at a time by swapping two box contents until you do. This would be the best way for a malicious prison guard to sabotage the room without making it obvious to everyone that it had been fixed (e.g. slip 2 in box 1, slip 3 in box 2 etc). Undoubtedly highly correlated slips and boxes in a 100-loop are only convertable with specific number offsets that would be less than the 31/69 random split but a systematic shift when a pseudo-random big loop exists (random lesser loops combined by guard) would probably more closely result in the desired random split and in that sense there was no error.

Matthew Queensen 25 日前

The way I explain the solution to people I know is that you're creating a dependent system from what is just chance (independent system). The opening of the box with your number then following the loop creates a systemic way of operating (i.e the system part) and your depending on the other people to follow the same systematic procedure as yourself (i.e dependent ).

Ophat Taerattanachai 27 日前

There is a situation in real life that everyone can find themselves in this kind of loop. That situation is the gift exchange at the party. They write the names in a slip and put them in the same box, and let one person pick out the first slip to get the present from the written name. And then, they continue until the first picker has to give the present to the last picker. However, the gift exchange usually ends up in some smaller loops. So, there will be more than one first picker for everyone to get the present from the other.

Milosz Forman 25 日前

Aka "Secret Santa". I suppose that in most cases _very_ few participants will speculate about the number of loops, their lengths or their probabilistic behaviour. Might be different in the math departments of the universities. But rarely will there be any "Secret Santa" there.

Noxturnix 19 日前

My first idea was to use timing (the time a prisoner spends in a room) to reveal the information about the boxes to other prisoners. I'm not sure if this breaks the rules nor did figure out the algorithm, but I think it can be used as a way to communicate.

Milosz Forman 19 日前

That would be a complicate and error-prone method. It's much easier if the first prisoner writes down all the numbers he finds on a piece of paper together with the associated box numbers and gives this paper to the other prisoners. This way, everyone immediately knows in which boxes they have to look. It's easy for those whose numbers were encounterd by the first prisoner, as they only have to memorize one number. Those whose numbers are in boxes which were not opened by the first prisoner better take this paper with them in order not to forget the correct numbers. They would have to open only those boxes which were not opened by the first prisoner - quite an ingenious strategy. And they must not forget to give the paper to the following prisoner. After all, it's quite easy a riddle, no way does it "seem impossible". Nevertheless, we still only get a chance of survival of 50%. Still much better as these 31.2% which is stated erroneously in the video.

Lushkk 14 日前

@Milosz Forman That would be the case.. if one of the rules wasn't that the prisoners couldn't communicate with each other outside of the strategizing before going in.

Artoria | Kiri 2 ヶ月前

I love maths and number and this was so incredbile and cool to kook at and understand in, really well done.

BSWhiskey 7 ヶ月前

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

Milosz Forman 7 ヶ月前

_"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."

Solo The Man 5 ヶ月前

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

Drumaier J 5 ヶ月前

A legit concern.

Gerrit Casper 4 ヶ月前

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

Maria Chrzanowski 3 ヶ月前

This reminds me of Secret Santa. If you do it the way that whoever is the Secret Santa for a gift gets to open their gift next, ideally you'd want one full loop ending on the person you started with, getting through everybody, but it isn't easy for that to happen.

Ryx 7 日前

Shoutout to my teachers and professors who taught me in my endeavor to getting my degree in mathematics. No way id be able to solve such a problem without them

Plantt 2 ヶ月前

Thank you Veritasium, you made it seem possible 🎉

Bryan Cole 4 ヶ月前

Never would've thought the 50/50 chance method would have such bad odds 😂😂 just shows how bad i am at math 😂

B0ssguy 3 ヶ月前

im proud of myself for thinking of this even tho i didn't how it would work out, it just sounded great when i was thinking of solutions that wouldn't just be random

Fixed Point 年前

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.

ABK 年前

Really?

Ath Athanasius 年前

@ABK 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 年前

@ABK 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.

Marvin_357 年前

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.

Deego 年前

@ABK 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.

Sky Trembath 3 ヶ月前

10:45 That’s the point of the loop strategy though. You have to be on the right loop, because you’re starting with your own number, which will be the last box in the loop. It’s impossible to be on the wrong loop. However, it is possible that the loop is too big for you to make it back to your number in time, and that’s how you lose.

Just a Person Who Comments 2 ヶ月前

This riddle is like a mathematical escape room where the prisoners are just one equation away from breaking free

A Bloke From Stoke ヶ月前

It works because it is stopping choices from being random, meaning there are far less repetitions in choice meaning more boxes are opened.

Todd R 4 ヶ月前

That's much more interesting than I thought it would be. Nice work

Christian Prepper 21 日前

*PLOT TWIST: This video was created as an excuse for Derek to wear his orange prison pajamas.* 😂 (Excellent video! Thanks, Derek)

WhackyCast 年前

Has this been tested in real life? Would be cool to see an actual representation of this.

gaming upgrade 年前

True

Brokkolirepublik Deutschland 年前

Especially the executions

globaltrance86 年前

By "test" you must mean create a live visual representation as some kind of spectacle, because the math has already been tested and verified. Doesn't need to be done "in real life" for us to know it works.

WhackyCast 年前

@globaltrance86 Just because the math has been done, doesn't mean it wouldn't be cool to see it with real people, in a real scenario.

globaltrance86 年前

@WhackyCast Not disputing that, just saying we already understand the probability.

DJ Caësar 9114 3 ヶ月前

Brillantly simple. I feel dumb after watching this and realizing how simple the explanations are, but the guy seems kind so I don't mind.

Duck Yay 4 ヶ月前

i feel like i kinda understand! If you have a loop of fort you automatically KNOW that if everyone follows this, 50 people are guaranteed to win, and so if there is maybe an odd chance the 51st person get all the others you didn’t, you all win, or even a loop of 25, and a loop of 5, and a loop of 20, because everyone in YOUR loop WILL make it, so the bigger the loop up before 51, the better, because that means more people will win. This is really neat!

ASKINATOR 2 ヶ月前

This makes perfect sense in a very confusing complicated kinda way. Like I understand it... but I'm also slightly confused.

pikekeke 2 ヶ月前

So the reverse also holds true. If just finding your number frees you, the others don't have to find theirs, then it's better to open boxes at random than to follow a loop.

Martman321 ヶ月前

Don't think so. Each one still has a 50% shot on their own turn with the loop strategy.

Faith Hellman 4 ヶ月前

I'm not sure what's more interesting, learning this neat strategy, or the fact that everyone doubts it. Dunno, maybe it's just the way my brain's wired, but once I heard it, it just made sense. Like Honestly, might have done that strategy just for myself unintentionally out of a lack of knowing where to start and go. Anyways, all that to say, I find it very intriguing to see how this messes with everyone's head. I personally expected it to be more mind bending. Love it! ☺

Milosz Forman 4 ヶ月前

_"I find it very intriguing to see how this messes with everyone's head."_ Certainly it does not "mess with everyone's head." The title of the video is kind of a clickbait. Derek had pointed that out in one of his videos, that the right title is essential for the number of calls.

Bucks 28 日前

There’s no way the math is intuitive. You intuitively thought 31%?

Faith Hellman 28 日前

@Bucks No, it's more like, without watching the videos explanation of how it worked, I understood that what they described would work, but I couldn't have explained how. Hence why I labeled it as intuitive, since I wasn't quite sure what else to call it. 😅

Nemanja Ignjatović 年前

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.

Michael Ritsema 年前

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

Aaron 年前

@Michael Ritsema sure dude. Whatever.

Houston R 年前

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

Mateusz Konofał 年前

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

Aspiring Diamond 年前

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

di BOI 3 ヶ月前

I always use incomprehensible patterns to make decisions, i can imagine myself doing this

William Flynn 12 日前

I seems to me that the best strategy also ensures that a different starting point i.e. a different loop will always occur. I can’t wrap my head around why this might be significant, but it appears to be.

Milosz Forman 12 日前

A different starting point does not imply a different loop.

Nigel Thorne 2 ヶ月前

Does the changes improve or get worse if each prisoner picks a random offset to add to their numbers? I would have thought that it just meant each prisoner had a different set of loops... so they then get a change each based on if the loop they get happens to have less than 50 steps. You no longer get the all or nothing scenario... so I guess it's worse?

Nigel Hunt 2 ヶ月前

Yes, much worse. They all have to, collectively, be working with the same set of loops so any offset will do as long as they all apply the same number. If each chose a random offset then each 50% chance of success/failure would once more be independent and they'd be doomed.

Melody Wawichi 4 ヶ月前

...I coded a quick simulation that really only counted the number of times all prisoners won vs lost the riddle via this strategy and via random picking. The computer *also* generates around a 30.3-30.7% success rate with the strategy and a 0% success rate with guessing randomly. ...I still can't believe it.

RoboFarmer 3 ヶ月前

You draw boxes funny haha great video, not sure I fully get my head around the logistics of it though.

Matemática Rio com Prof. Rafael Procopio 年前

Incredible video, as always. 👏🏻👏🏻👏🏻

Tomáš Hecht 2 ヶ月前

The solution is OK. The explanation for this very strange result is that different prisoners open the boxes representing the same permutation so their trials are not independent in fact the success at this method is a property of the original permutation namely what kinds of loops are there. I am afraid it is almost impossible to prove that there is no better method But I strongly believe that nothing better exists

Milosz Forman 2 ヶ月前

_"I am afraid it is almost impossible to prove that there is no better method"_ Peculiarly, there is a simple proof which has been published by Curtin/Warshauer (see the introductory text of the video). We define a game with slightly different rules: 1. the lids of the boxes remain open after they have been opened, 2. prisoners are only allowed to open boxes (up to 50 each, as before) unless their number had not shown up yet. It is easy to see that this is a totally random game, no strategy whatever would help to improve the chances. On the other hand, it has the same chances as the standard game when using loop strategy. And any strategy of the standard game can also be used in the new game.

Rts gamer 3 ヶ月前

It works because the prisonners numbers are linked 1 to 100 (dependant) and you are taking advantage of this fact in the loop strategy. You need to hope that all the loops are less than 50.

Godfrey Pigott 3 ヶ月前

Less than 51.

Mikhail Nikitin 2 ヶ月前

I understand the math and the solution to the problem, but I have only one question and it is about the sympathetic prison guard: How do we know that by swapping the contents of two boxes he doesn't accidentally create a longer loop instead of destroying a potential 51+ long loop?

Nigel Hunt 2 ヶ月前

It's assumed the sympathetic prison guard knows/discovers the box contents and so can pick two specific boxes that when the contents are swapped, splits any big loop in half (as demonstrated). When he stressed "just two boxes" he was only saying that it doesn't require more than two (one swap) to make the room safe.

Nicolas K. 5 ヶ月前

If the warden catches on to the pattern as the experiment is being run (or knew beforehand and didn't manipulate the boxes to change the odds), he can cut the experiment short and free each prisoner if the fiftieth prisoner finds their slip - if there were a loop of 51 or higher, someone would have to have lost by now.

Entropie - 5 ヶ月前

In most cases he can do it even earlier. All that is required is that the prisoners so far all found their number all the explored loops combined contain at least 50 boxes. For example if the first prisoner is on a 50 loop (meaning they find their number on the 50th try) then success is guaranteed. Or if the first prisoner is on a 30 loop and the second prisoner is on a 20 loop and so on. Another way to look at it is that each successful prisoner cuts the chance of failure more than in half so success is usually decided by the first couple of prisoners.

Nicolas K. 5 ヶ月前

@Entropie - Damn, that's really cool! Thanks for that improvement

Bismuth 年前

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.

Jynx 年前

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%

John Hunter 年前

They are called *odd* for a reason

HYPERWATER 年前

Raoodnm

Josenobi 年前

@Jynx how did you calculate that probability

Janbacer 年前

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

Julix Pinguimon 11 時間前

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

ReivecS 2 ヶ月前

I won't pretend I knew the answer beforehand, but I am a little surprised at how many people in the comments find this hard to understand. This makes perfect sense. If you have a loop of 50 or less you win so this only boils down to the odds of having that arrangement. The whole idea of "how do you know you are on the right loop" is silly because if the biggest loop is 50 then they are ALL the correct loop.

harsh rana 2 ヶ月前

Alright, suppose I request an additional prisoner, bringing the total count to 101 prisoners. Alternatively, we can consider that one of the 100 prisoners is granted an extra opportunity, allowing them to exchange slips on the 49th turn, effectively closing their loop.. If the number of loops is greater than 50, the chances of winning reach 100 percent. It's important to note that no loop can exceed a count of 50.

Joe Kemp 2 ヶ月前

_"allowing them to exchange slips on the 49th turn, effectively closing their loop"_ And how would these prisoners know they belong to the same loop?

bubacheese1 2 ヶ月前

Me: The experiment specified entering one at a time but did not specify wether multiple prisoners could be in the room at once. Meaning each individual would enter one at a time and wait until all were inside the room, easily figure out everyone’s number. Leave the room one at a time. With no need to communicate after. Probably just a loop hole someone would update the rules to fix idk

Jim Karpule 3 ヶ月前

i used to do this with dvds. my mrs would put a dvd on and then put the box back after the film she left the disk in the player, next time she wanted to watch a film she got the box and put the dvd in the player putting the old one in the new box and so on. to the point none of the dvds had the right disk in (apart from the ones she hadnt watched) i didnt realise this was happening because i played games while she watched movies. when we sat down to watch a film together i realised the film wasnt in the box and not onlyu that but the film that was in the box's origonal box didnt have the film we wanted to watch. so i used this exact method to correct everything without a massive pile of dvds and empty boxes. took like 5 minutes do correct like 100 dvds.

Wouter Pomp 年前

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

Kevin Z 年前

Life do be like that sometimes

Aaron Davis 年前

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

Ever StanDinG 年前

Imagine knowing this for a fact and no one listens 🤣

Roskal Raskal 年前

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

clover 年前

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

Rapid Rabbit 4 ヶ月前

I feel like I am missing a key component here, but the premise says that the numbers are placed randomly. If that's truly the case, then slip 49 can end up in box 49, and when prisoner 36 runs the loop, the loop can terminate, potentially without them finding their number. This causes them to then choose another loop to keep looking, but what means would they use to determine what next box to open? There seems to be an assumption here when we say that everything forms a loop that those loops will terminate with their number being found, just the length of the loop will vary. That seems not to be the case. If just one of them fails, they all die. This seems to be rolling the RNG that a match won't happen.

Milosz Forman 4 ヶ月前

These slips are supposed to be unique: exactly one slip with every number from 1 to 100. This is not exactly said in the video but can be concluded from the picture at 0:41. Therefore there can't be any dead ends but only regular closed loops.

Rapid Rabbit 4 ヶ月前

@Milosz Forman The problem is that you are supposed to go to the box the slip says. If you go to box 49 and it has slip 49 and that’s not your number, you’re done. Unless you randomly choose another box to go to; everyone dies. 49 being in 49 is still completely unique, there is only 1 “49” in the set. This solution requires that each slip in each box does not contain its own number or else there is a loop of one. If that’s you’re number, great, if it doesn’t match your number you have no loop to run; you failed.

Rapid Rabbit 4 ヶ月前

I figured it out, it’s controlled because you always go to the box with your number on it. If on the loop you don’t find your number that’s the 66% chance of failure.

Whats Up 4 ヶ月前

@Rapid Rabbit your loop always has your number (100%). The issue is if your loop is above 50 units in which case you and everyone else in that loop all fail. fyi your first example had two internal #49(s) one to get there and a second in the box. In that example 99 prisoners never touch box 49 only prisoner 49 touches it thus everyone's odds improve.

Andre Baptiste 3 ヶ月前

Like all Veritasium videos, it's starts off good, then I no longer understand them. I honestly think they just make things more complicated than necessary.

Ian Synnott ヶ月前

A beautiful mathematical demonstration... and why the human prefrontal cortex is such a marvel of nature.

Martin P 3 ヶ月前

So this is the case of every time the warden wants to have some fun he randomly distributes the numbers in the boxes. If he is lazy and always goes with the same distribution, we either always have a bigger loop than 50 or not. If the case is not, then the loop strategy will always work. If there is a bigger loop, will the loop strategy increase the prisoners chances or is it the same as if they search at random, i.e. they're doomed?

Nigel Hunt 3 ヶ月前

If a big loop is known to exist the loop strategy is worse than random chance as at least 51 prisoners will fail to find their number and the chance of success is zero compared to (1/2)^100. It wouldn't practically matter though; they're doomed either way. However, some box number reassignments + the loop strategy would restore the 31% chance as was explained when countering the malicious guard.

ElliottR 3 ヶ月前

I think you can increase your chances above 31%. When the first prisoner walks into the room there is a 69% chance that the room has a loop longer than 50. Therefore if he swaps two of the boxes it is more likely that it will break a long chain than create a long chain and therefore he has increased their chances of success. Or am I wrong?

Nigel Hunt 3 ヶ月前

Wrong I'm afraid. Swapping two boxes at random is just creating another random distribution of slips within the boxes still subject to the 31/69% statistical split.

ElliottR 3 ヶ月前

@Nigel Hunt but could it be a case like the monty hall problem. That because the chances of the room having a long loop when you first walk is high swapping it is more likely to help than hinder?

Nigel Hunt 3 ヶ月前

@ElliottR This isn't the Monty Hall problem; in that problem some information was revealed to the contestant when a door was opened. Here no information is divulged. If what you're saying were true then you could simply ask the sympathetic guard to swap the contents of two random boxes before the first prisoner entered the room to save him from the rule break. And in so doing the distribution is changed from a 31/69 split to what? 33/67? Why stop there, just keep swapping random boxes until your chances have increased to 50/50? Nonsense.

Milosz Forman 3 ヶ月前

@ElliottR _"Therefore if he swaps two of the boxes it is more likely that it will break a long chain than create a long chain"_ Provided that we positively know that there is a long loop, swapping two boxes is a reasonable option. Unfortunately, swapping two boxes at random would only give a 18.7% chance to eliminate a long loop under these circumstances. On the other hand, if we do not know if there is a long loop, swapping two boxes might as well create a long loop, so we end up with the same 68.82% failure rate. In all, we have a chance of 18.7%*68.82% = 12.88% to eliminate a long loop. The chance to create a long loop - provided there is none - therefore must be 12.88%/(100%-68.82%) = 12.88%/31.18% = 41.3%.

ElliottR 3 ヶ月前

@Milosz Forman How are you calculating the 18.7%? Surely it will depend how long the loop is to begin with or is this value an average for any loop greater than 50? Also, I am unsure why you are using the 12.88 value in your calculation for creating loops

DooDooBear 年前

It would be awesome to someone like Mr. Beast actually do the experiment with people to show it being done.

Owl of Athena 年前

Omg someone pls send him this video!

[YT] akaRhaegar 年前

I hope mr beast would see this and it would be a pretty good challenge

cubeometry 年前

I thought the same thing 😂 there should be a practicle

Mishajane Chuaseco 年前

I thought so too!

WizzY TheWizzurd 年前

Would be a cool thing to show if it worked, but might be boring with a 69% chance of failure.

stick_figure_animation 3 ヶ月前

I actually came up with the solution in under 15 minutes but I didn't prove why it was a good strategy. I just intuitively thought that a loop with size >= 50 is extremely unlikely and went on with that

Greg K 3 ヶ月前

Could you imagine being the first prisoner, and you didn't find your number... the torture waiting for the other 99 knowing you all gonna die.

A. Teima 2 ヶ月前

From the very beginning I knew the video had something to do with amortization, as there is really no other way one could possibly approach this, but sadly couldn't figure it out, anyway what an amazing video!

Riccardo Mulas 3 ヶ月前

Using the logic, it's enough to be only one loop of more than 50 boxes to guarantee failure, means that with this strategy and knowing for sure that there is a loop longer than 50 we can be 100percent sure that they are going to fail. (If one fails everyone fails)

128Gigabytes 2 ヶ月前

Yes if theres a loop longer than 50 they fail 100% of the time but theres only a loop longer than 50 roughly 2/3 of the time, which is way better odds than everyone picking at random

David C. Duncan Dekker 2 ヶ月前

Choose columns, then alternate them each time, thus creating that "loop" that gets you closer to that 50% rate.

michael gove 9 ヶ月前

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 ?".*

michael gove 9 ヶ月前

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.

Yusuf Ahmed 9 ヶ月前

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

Bálint Varga 8 ヶ月前

Absolutely stunning example.

LiveHappy76 8 ヶ月前

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....

Thomas Rosebrough 8 ヶ月前

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

It's Energy Bob! 4 ヶ月前

I saw this a few months ago and think about it once in a while. It now makes total sense.

Softan J 3 ヶ月前

This makes perfect sense to me. Instead of multiplying probabilities a hundred times over or in other words being lucky a hundred times in a row you only need to be lucky once with the arrangement.

Godfrey Pigott 3 ヶ月前

More than once. But the luck of the first prisoners increases the probabilities of success for the following prisoners.

Nigel Hunt 3 ヶ月前

@Godfrey Pigott Once in the sense that the room is either "safe" (no loop >50) or "deadly" (loop >50) and "you" means everyone collectively?

Steve Desamos 3 ヶ月前

I feel like this could be combined with Monty Hall in some incredibly twisted way

ys f ヶ月前

There's also another one: Jack and Jill have two children, one of whom is a daughter. What are the chances the other is also a daughter? it's not 50%.

Madame Traffic Jam 8 日前

Makes me wonder what a Venn diagram of people who don't understant the Monty Hall problem and people who don't understand this loop strategy would look like lol

Várkonyi Tibor ヶ月前

You must find your number in the loop because these number sequences cannot have a starting point or endpoint or branching, so only closed loops are possible.

Cris 1001 2 ヶ月前

That's cool, it arises out of the fact that, because each number in each box is unique, if you start on your number, you eventually have to find your way back to it, following the "box pointer" strategy. How to see this? Assume you start on your box and are on a line that never leads back to your box. Never, as in, no matter how many steps you take. How could this be that you never find your way back? Only if you are stuck in a loop that doesn't include your box. In other words, your box led to a loop, which then does not point back to your box. But, that means, your box, AND some OTHER box points to that same box in that loop. Maybe try an example. Your number is 69. 69 points to 42. 42 points to 5. 5 points to 42. So you never find your way back. But that cannot be, because both 69 and 5 cannot both point to 42. Because there are only 100 numbers and only 100 boxes, ever number is unique. So you can't have the same number in two boxes. Therefore, everywhere you land is a closed loop. And that loop, if you start on your box, will lead back to your box. It just depends how long it is. That's why the "longest loop" characteristic is key to your chance of winning this.

N Z 年前

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

Jens Pilemand Ottesen 年前

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

ILYES 年前

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

Flashstar 123 年前

Yeah same makes perfect sense to me

Pirojf Mifhghek 年前

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

Overplays AceKing 年前

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

huiyi Zhang 4 ヶ月前

The funny thing about it is that I use that strategy to find my ladtop number when the ladtops get all messed up (number on ladtop and number in each slot for the ladtops)

Lethal Wolf 3 ヶ月前

Scariest scenario is your number is 1. Your box contains 2, next box contains 3 and so on with every box containing its number plus 1. You get to 49 and it says 50. Do you open box 50 or 100? Could be a trap! Plus if you intuited that box 100 will have 1, and are correct, you have to hope the other 99 prisoners will come to same conclusion!

Entropie - 3 ヶ月前

At that point it becomes a problem of game theory. You would need to estimate some a priory chance that the warden set boxes up in a particular pattern to counter the loop strategy. Considering that this pattern would be extremely unlikely to show up randomly even a small chance of the pattern being man made probably justifies adapting the last guess. Though obviously a smart warden would mess the pattern up a little on purpose so that at least someone will fail so you are probably screwed. The proper counter strat is to not rely on the given numbering in the first place and for the prisoners to come up with their own (preferably random) numbering scheme before going into the room.

Milosz Forman 3 ヶ月前

_"You get to 49 and it says 50. Do you open box 50 or 100?"_ Both could be correct, and both could also be incorrect. In such a case when you notice that the slips have been placed not randomly but following a specific pattern, you could try to guess this pattern. So if you noticed a consecutive order after, say, 15 boxes - or some more? - , you might try box 50 next to check if there are two simple loops set up by a benevolent guard. Next check box 100. This way you might kill some "deadly" constellations, but there is also the danger to create such. Difficult question.

Milosz Forman 3 ヶ月前

@Entropie - _"The proper counter strat is to not rely on the given numbering in the first place and for the prisoners to come up with their own (preferably random) numbering scheme before going into the room."_ Still remains the question if such non-random placements could be countered be some strategy with better results than these notorious 31%.

Entropie - 3 ヶ月前

@Milosz Forman If you don't renumber uniformly at random there should always be some residual effect if the initial distribution was not uniformly at random. Further analysis here requires the assumption of a "meta" where the prisoners need to model the behavior of the warden. Of course the warden can also model the behavior of the prisoners. Ultimately whoever is better at accurately modeling the behavior of the other side will probably manage to get a benefit by choosing an appropriate counter strategy.

Quentin D 13 日前

You open 50, like the strategy said. 0:40 "They are RANDOMLY assigned" If you open 100, you have the exact same probability to win or not, but you are killing the entire strategy for your team

Alex Schindler 3 ヶ月前

I understand everything except at @13:32 when it's said that to solve an intentionally long loop you could incrementally renumber the boxes. But that doesn't actually guarantee a solution. I ran a few different simulations that still result in failure where loops are still greater than 50 after re-assigning values, but it at least gives you a chance. If I'm misunderstanding this I'm 100% open to some feedback.

Joe Kemp 3 ヶ月前

He never claimed that a solution is guaranteed, only that they're "back to their 31% chance".

次