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

That calculation applies only for n>50.

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

@Veritasium Agreed

@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.

@G C Only applies to continuous functions.

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.

Underrated comment

Yeah

You need a nerd with charisma.

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!!

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

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

_"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."

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

A legit concern.

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

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

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.

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

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

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

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

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

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

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.

This helped me intuitively believe that this is plausible without doing a ton of math. If all the prisoners decided to pick the same 50 boxes, there would be a 0% chance of success. This simple example proves that you can change the probability of success with a strategy. It just so happens that the optimal strategy is incredibly elegant and yet challenging to work out.

For me I imagined if there were only two prisoners and two boxes. Right off the bat, you can intuitively see that if they agreed to pick different boxes, they would have a 50% chance of success, if they agreed to pick the same box, 0%, and if they picked randomly, 25%.

In this strategy if a box has the same number inside and outside only one prisoner ever touches it. Thus all other prisoners use less boxes Same for paired boxes 2 prisoners only, or 3 box strings(3 prisoners only), 4 ……

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

Well - are you Korean? ;)

@Spyro you win. 🤣🤣🤣🤣

Lmaooooooo

Thanks for the suggestion. Also, bring canned fool.

These math and science videos are surprisingly fun to watch. Just completely different than in class.

I think one way to grasp why this works is to imagine there are several short loops in the set up. If prisoners pick an initial number randomly, but then follow the same strategy, many of them will waste their choices by following a loop which doesn't contain their number. Conversely, by starting with their own number (even though, as explained, it can be an arbitrarily assigned number), they are guaranteeing that they are following a loop which contains their number. Thereby avoiding short loops which don't. Really great puzzle!

Super interesting and logical. What I love is that it shows us to work together rather than individually and that is where we have to link our sciences to philosophy and remember that sciences are merely observations put on paper.

Exactly, this is the problem with modern science. For example, instead of having 10 single biological or medical studies with 10 people each, why not have all 100 people participate in 10 studies simultaneously? The main benefit is not that this would greatly improve sample size, but rather that it would better account for uncontrollable variables in biology, which should be done differently than in physics or chemistry, but unfortunately misguided scientists use the same methods of variable control for biology and psychology as for physics and chemistry.

I seriously admire the mind that thought up this strategy.

The truly sadistic warden would show the prisoners this video. Let them watch it (together) as many times as they want, discuss it as much as needed to ensure they understand what to do, and then label the boxes in binary numbers.

Or they can just set it up like normal as the prisoners still only have a 31% chance of escape. It’s still sadistic. That’s essentially Russian roulette.

The prisoners could just relabel the boxes, though, with a formula of their x and y coordinates?

@Adam Kentthey could just label them one to 100 reading left to right

I could see a variation of this coming up in a sequel to Squid Game. People formed into four groups of 110 players each, each group having the choice of four different box rooms to choose from, and a single player option to stop after opening 49 boxes to report back to your group, but die for choosing that option. First few players know their job is like a soldier, to find out if a room likely has a loop of 50 or over, and if they make it to 49 they have to come back, report such, then die, along with anybody who had already made it through that room, but then the remaining members of that group knowing that they should then try try going through another door and room, say door number 2 to work through the boxes in that room instead, hoping it will be a room without any 50 or more loops. One group, minus a couple of early players who made through and also minus the third one who made it to 49 and choose to report and die, the rest of that entire group then makes it through door number 2 and box room number 2 using the loop strategy. Of course that's our group with most of our favorite Squid Game players. The 2nd group, mainly of just bullies and idiots, all just die by one by one trying to go through one door and randomly choosing boxes. The third group, well there was somebody smart enough to come up with the loop strategy only they had some feisty types in their group that just didn't believe in all that liberal scientific mathematical mumbo jumbo, so it wasn't until after they had fed 25 players into a room before a loop strategy believer finally sacrificed themselves to report back to change rooms. The fourth group, properly choosing to use the loop strategy plus the the sacrifice to report to change rooms strategy, still loses a few more than they should have, because somebody or maybe a couple of players early on got to their 49th box, but didn't want to report back to the group then die, so took their chances and opened a 50th box, died for it, but then leaving the larger group uninformed to change rooms until they had lost a few more players before somebody actually sacrificed themselves to report back that they had made it to 49 boxes and advising the rest to try another room. Anybody making it through an abandoned door and room are, or were, of course all doomed, because it's only when however many people in a group ALL make it though one of the four rooms to choose from, without anybody of them having to open 50 or more boxes in order to find their number. Yep, sounds like a Squid Game episode to me.

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.)

Best comment, 10/10

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 :(

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

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 :)

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

Congratulations for making this video. I can't stop thinking about it. I read the paper by Anna and Peter, but they didn't make it any easier to comprehend the probability distribution, and how the closed systems increase the odds substantially. I'll have to dive deeper into this... Thank you (seriously) for having cause a "glitch" in my brain! 👊🏻😃👍🏻

There is an easier way to think about it. If you use the loop strategy, you make sure all 100 fit in the same family of loops, taking out of the equation the randomness of each prisoner selection and limiting the possible combinations between each prisoner selection of boxes.

4:44 That visual is what really helped me understand this. I'd heard of this riddle before, but I never knew WHY there was such a significant chance of not having any loops longer than 50. It's because a box can only be in one loop. If there's a loop of 5 boxes, the maximum size that any other loop can be is 95, since 5 have been excluded from the pool. (Loop, pooL, coincidence?) If there's another loop of 30 boxes, the new maximum is 65, and all it would take is a loop of 15 or a few smaller loops that add up to that, and boom! Guaranteed success!

Also for those who are wondering where did we got the extra success chance imagine a scenario like this: If you haven't made a strategy before hand you have this "certain failure" scenarios. Think 4 people 4 boxes 2 chances. If they didn't talked beforehand there is a possibility that all people will choose box 1 and box 2 which is a certain failures since 2 boxes cant have the numbers of all 4 people. If you talk beforehand you set a certain path for everyone that eliminates this kind of possibilities. This is what raises the chances of success.

So essentially, this is no longer the probability of each prisoner finding their own number randomly. Instead it is more the probability that there will no loops over 50. Very interesting and well explained! How would the probabilities changes if the boxes were shuffled after each attempt?

Re-shuffling the slips (or boxes) after each man will defeat any strategy, thus bringing down the chance of success to that of random strategy.

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

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

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

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

@Lavarel Sieghart fair enough

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

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.

If you start with the box that has your number on it, the only way to conclude the loop is to find the box with your slip in it that points back to the box you started on. It makes sense. Its just a matter of how many boxes you have to go through to get there

What if I have the number 54, so I open the box 54 that box leads to 63 and so I open 63 and 63 is inside. That's a dead end. And no where to start again.

I found that the underlying concept here is easiest to understand in a simplified example, with for example two prisoners and two boxes, and each is only allowed to open one box. If they both open one box randomly, the chance of them both finding their number is 1/2 x 1/2 = 1/4, or 25%. The other 75% of the time they either both miss, or only one gets their box correct. However, if they both agree to open different boxes, they either both get their number or both miss, with probability 50%, (i.e. depending on whether the first prisoner chooses correctly). By conditioning the second prisoner's outcome on the first, you're essentially consolidating the original probability distribution into a higher variance one with fewer options, where either everyone gets it wrong, or everyone gets it right, with nothing in between. In the context of the original problem, since all 100 have to get their box correct, every other outcome is meaningless, so the loop strategy is the optimal one.

Another way to look at how you know you'll always be on the loop that contains your number is to realize that there is no way to stop a sequence other than at the box you started. 13 -> 42 -> 17 ... Using prisoner 13 for example: box 17 (or any other box) can't have the slip that points to box 42, because box 13 had the slip that pointed to box 42. And so on for box 17 and any subsequent boxes. Meaning you'll never go back to a box you've already encountered the slip for. Also the sequence can't go on forever seeing how there are a finite number of boxes/slips. Therefore, all that remains is to eventually encounter the box with the slip to box 13. For me, this was a clearer way to think about it than the explanation at 11:10.

Thx for this! I was still stuck on how you where guaranteed to be in a loop with your number.. Your explanation cleared that up for me :) By starting on your own number, you will eventually find it in the loop, unless it's too long.

there are loops that don't contain your number. What if I have the number 54, so I open the box 54 that box leads to 63 and so you open 63 and 63 is inside. That's a dead end. And no where to start again.

​@Paul Schaller if you open box 54 and go to box 63 from it, then box 54 contains slip 63. Then box 63 cannot contain slip 63 again.

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

No letters, just numbers, ha ha.

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

xd

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.

@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.

i remember learning about those permutations, factorials, and stuff on out probability and statistics class during 8th grade and it was not fun, but this was actually very interesting

I believe this because my teachers used this tactic to separate groups in college so we wouldn’t divide by friend groups. Problbailty-wise it was only about 1/3 of the time that we ended up separating perfectly 2/3s of the time she would have to reconfigure because someone would end up left out or a loop would be messed up.

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

The reason that it is above 30% all the time is because you can always open half of the boxes. When there is more boxes, one box is a smaller percentage of the entire whole amount of boxes. So a smaller percentage of the whole is able to make the loop above half of the boxes, making them get executed.

Maybe I'm not smart enough to be baffled by the math, but I actually find this really easy to grasp.

I think it's pretty easy to grasp once it's been explained. The clever bit is figuring it out without assistance. This is why I love learning maths puzzles and their solutions. Algorithms are extremely useful things, and the future of our existence relies heavily on them

It’s very easy to grasp once explained to you, but incredibly difficult to be able to come up with this strategy on your own.

Thanks for teaching me.

You're welcome

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

There you are!

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

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

Actually understood the math. You guys are great at explaining

I am terrible at math but this made intuitive sense to me as someone who has had to track down mixups in conference badges/materials where putting the wrong persons stuff into someone else’s envelope sets off a chain of mistakes and this is how we would trace it back

Amazing presentation to an almost impossible riddle! This ilustrates how probability can get weird with coordination of agents.

Has this experiment been tried with real people and real choices?? I would love to take part with 99 other people both doing random choices and using the method explained. It wouldn’t be that hard to setup online and have people log into a “group of 100” and do either a random choice or a planned strategy. It would be really cool. 😃

In the referenced video by Matt Parker he does a real live experiment with a reduced amount of boxes.

This actually made perfect logical sense to me. much more than other prison based riddles

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.

Really?

@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.

@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.

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.

@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.

Awesome video. I feel like this should be added tho…for the sympathetic warden to guarantee success with one swap, he would have to know he was swapping 2 from the >51 loop. If it’s swapping two random numbers, then in doesn’t automatically work. But thanks for this. It’s incredible!

The word "random" is never used. He says "just two boxes" not "just two random boxes". Quite a few have interpreted it this way though which is interesting.

This loop strategy is used as a solving method for 3x3 Rubik's cube blindfolded. The simplest method is called Old Pochmann and as long as you can memorize letters and don't mess up the execution, you can solve a Rubik's cube blindfolded

I have never seen this riddle, but I myself was thinking about number loops starting with the box matching the prisoner's number, before even reaching to that section of the video. I'm not a mathematician or anything, but I thought that because it sounded the most logical, and I've heard of looped numbers before

Now here's the next question... what at the odds if each individual prisoner can communicate back the slips/box numbers as they find them to the rest of the prisoners? Assuming someone with a masters or better in mathematics is in prison for some reason.. like microwaving their ex girlfriend's cat to death.

To better understand what he said about the probabilities being connected, just think of if there are only 2 prisoners that can only open 1 of 2 boxes. While no matter what each individual prisoner only has a 50% chance, it they both pick the same box there will always be one who wins and one who loses, because that one box can only have one of their numbers. So collectively they will always lose. Now if instead they just each decide to pick their own box, each individually still has a 50% chance, but now they will either both win 50% of the time or both lose 50% of the time, because it removes the chance of them overlapping.

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

True

Especially the executions

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.

@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.

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

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! ☺

_"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.

Thanks for the video, so simple but yet so interesting. I also thought on reaching our number first since it is 50% to be there, but didn't think on loops!

I think the easiest way to explain it is that if every prisoner picks their number first, then no prisoner will enter the same loop as another, making it the same chance individually, but higher overall

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

Anything that can be told at all can also be told _in an easy way._ However, it is not always easy to find such an easy way. That's where many teachers strand (provided that they even care). It might take more than 20 minutes in some cases, though.

My big family did name-drawing at Christmas to pick who we'd each buy for, and my older physicist brother woukd map out all the closed loops. So this makes great sense to me!

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.

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

@Michael Ritsema sure dude. Whatever.

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

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

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

To extrapolate this problem further, it’d be interesting to understand at what point prisoners would know they will not die. Like you said at the end, either you fail hard or succeed fully. From the P distribution curve, I’d day the minute 50th prisoner comes out successfully, the rest of them will be guaranteed to find their number, right? But that’s wrong as there could be a 99 number loop and a self contained loop, so one person could still fail. However with your probability distribution, there’s 0% chance that 99 prisoners will succeed and 1 will fail. (I know your graph is right, and it’s a way to visualise probability, just finding it intriguing :))

Yes, 50 prisoners exiting successfully to be sure, but the chances of 49 being successful when a 51-loop is present is miniscule. In the 51-loop case it's approximately a 50-50 chance that your individual number lies in the big loop or not, so the chance of 49 successes is about (1/2)^49. When just 7 prisoners have exited successfully there would only be a less than 1% chance that the room contained a big loop.

The idea that helped me understand this concept is that if someone has a loop of 50 to find their own number, the other loop also have to be 50. Their loops contain all the other little loops other prisoners can possibly have. Correct me if I got it wrong

There might be several other loops shorter than 50 if there is one of length 50. At least, there can't be one longer than 50.

After learning the solution its pretty intuitive for me to understand it. Sure, the INDIVIDUAL probability of finding your paper is still 50 percent... But the net probability is NO LONGER based on multiplying all these different probabilities together, because they are no longer INDEPENDEDNT probabilities. Instead the probability of the GROUP succeeding is the probability that none of the loops are longer than 50. Which is pretty easy to see how thats a pretty decent odds. If the first loop turns out to be, say, 10 in length, then the next person not in that loop has 50 tries out of 90 to find their number. If their loop lasts 30, then the next loop has a 50/60 chance of being valid. And the moment the winning loops total more than 50 in length, you know everyone else wins as well.

An observation that helps me understand the loop result is: for any given set-up, each number must occur in a loop, and in only one loop. In other words, there cannot be a number that is in more than 1 loop.

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.

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

literally this time

I see what you did there!

But why where they in jail to begin with 😂

Or out of the loop!

@Abin Baby actually, into the loop xD

I think I've used this strategy before without realizing it. I don't remember why or how I came up with the idea, but I remember using loops to find a random number.

You are guaranteed to go to the box containing your number in it by starting with the box that has the number on it using that strategy because the loop will never end since it has to end with the box you started with but that just means you will have to find that number in the box right before which contains your number.

If the person doesn't have to show that they have found their number but just to say what box its in, they can choose the box shown in the 50th box they open, this will give success in all loops of 51 as well.

That's an amazing video! Visualizations are very on spot, explaining complex concepts in a simple way. One thing that I missed is maybe a brief note of any this strategy is the best one? Why can't anyone come up with a better one?

This is covered in the paper by Warshauer/Curtin which is mentioned in the introductory text. This paper ("The Locker Puzzle") is avaible on the server of the University of Cambridge. Don't use the Springer/v. Holtzbrinck link, they charge money. There is a short outline of this proof in my answer to @BOB from a day ago.

Sometime, I criticize veritasium's videos as being lazy with the definitions and solutions. Not in this one. Excellent video!!

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

Omg someone pls send him this video!

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

I thought the same thing 😂 there should be a practicle

I thought so too!

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

This reminds me of a game we played in high school, that we called "Gotcha" (it had other names as well). For the younger folks here, this was before kids carried cell phones to school. It was usually about 15 to 20 of us playing. One person would run the game and not play. That person (the "DM") would create a loop and give each player a piece of paper with the name of the next person on it. The object was to find that person and say "Gotcha" at which point they were out of the game. They would give you the paper of the person they were hunting and you then sought out that person. You would win when the paper you were given had your name on it. Papers were handed out near the end of a school day, and the game would start the next day at school. I won once when the person with my name on her paper stayed home sick on the first day of a new game. Nobody got me but I was able work my way through everyone else until I got to her name. The following morning I got her stepping off the school bus. Fun times!

thanks for your content:-) your videos always are so interesting and entertaining, really making me think, thank you :D

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!.

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

Incredible video, as always. 👏🏻👏🏻👏🏻

The reason I found this video was a similar problem: the key differences were that the ‘boxes’ were like bingo numbers and they would be replaced each day and that they stated there was a ‘certainty’ that all prisoners would go free.

Since no time limit was given, i would suggest each prisoner goes in for 100 minutes, starting at 1 and working down a box every minute existing on the minute they find their box. That will signal everyone else to skip that box and keep looking. Doing this over and over would eliminate box by box giving them a much greater chance of winning.

That's a form of communication.

Would be interesting to see the percentage chance if we assumed the loop would be greater than 50, for instance picking any random number and following that until you either get a closed loop, or the right loop, how much different or lower it'd be than the 30.7% chance.

Suppose there is a 51-loop and everyone picks his first box at random, and then follows the loop. So the 49 prisoners who don't have their numbers on this long loop will fail with a probability of 51/100 each, bringing the total probability that they all find their number to a meagre 0.51^49 = 4.7E-15. And the other ones might fail as well. So we won't get even near to these 30.7% chance of the consequent loop strategy.

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

Decent way to think about this problem. This particular strategy makes it a question of “is the room setup in a certain way” Rather than “Are all 100 prisoners gonna correctly guess where their number is?” Because if you increase the number of prisoners by 10 times, but have them duplicates with the other original 100. The 2nd question’s probability would decrease, but the first would remain the same. Theoretically, it might be possible to have a bunch of other strategies that would turn this from Question 2, to Question 1. But loops are probably one of the higher probability strategies. All prisoners selecting the first 50 boxes is a strategy that makes it question 1, but it makes the answer a definitive 0% since it guarantees 50 prisoners find their number, and 50 prisoners don’t. Even though it’s still technically a 50/50 chance for each individual prisoner to find their number.

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

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

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

Yeah same makes perfect sense to me

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

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

To all of my number loving friends: I thought it would be good to try some simplifying assumptions using 10 boxes with each prisoner able to choose five boxes. Just to check and see if it’s reasonable for 100 boxes with 50 box choices. And I thought I would use some common irrational numbers as the basis for how to move the numbers. I moved each number to the right the number associated with an rational number ( since the digits would vary between zero and nine). Ten boxes 1. 2. 3. 4. 5. 6. 7. 8. 9. 0. Pi. 3.141592653589793 Reorganized numbers (move #1 number one box right, move #2 number four boxes right etc. when you reach the end start over. If the box is full take the next empty box to the right: 0. 1. 7. 3. 5. 2. 8. 6. 4. 9 Now starting with box one see what happens: 1 to 09437862. Nine boxes. So they failed. Try again: 1. 2. 3. 4. 5. 6. 7. 8. 9. 0. e. Natural log. 2.718281828459045 3. 0. 2. 5. 7. 4. 6. 1. 9. 8. Ok now starting with box 1 see what happens: 1 to 3208-1. Five boxes 2 to 0813-2. Five boxes 3 to 2081-3. Five boxes 4 to 576-4. Four boxes 5 to 764-5. Four boxes 6 to 457-6. Four boxes 7 to 645-7. Four boxes. 8 to 1320-8. Five boxes 9 to 9. One box. 0 to 9132-0. Five boxes. Success! One more irrational number: 1. 2. 3. 4. 5. 6. 7. 8. 9. 0. Rad 2. 1.414213562373095 9. 7. 2. 8. 1. 4. 3. 5. 6. 0. 1 to 96485–1. Six boxes. Again failure. But one out of three is GREAT! That’s a success rate of 33% based on my three random samples. That’s way better than random. 1/2 to the tenth power is 1/1024. Amazing.

I remember this riddle from a Ted Ed riddle I’ve seen before, and I think I got it right actually! It’s a really cool problem

At first, I was thinking that the first two prisoners could try their luck at getting their number. The plan would be that Prisoner Number One and Prisoner Number Two (relying on them, if they don't find their number, the rest are screwed) would look from the entrance of the room, and would turn the boxes that contained an Odd number inside right side up (again, looking from the entrance of the room), and boxes that contained an Even number upside down. Prisoner One would pick through 50 boxes, turning the boxes accordingly. Once they find their number, they are to finish flipping the boxes, and then leave once they are done. (If this wasn't allowed, I would say it is up to Prisoner #2 to finish). If Prisoner #1 is able to find their number, as well as flip 50 boxes, and Prisoner #2 would have to flip the rest of them as well, then the rest of the prisoners would have a 100% chance of finding their number, if they can remember the box turning rule. Odd= RIght Side up, Even = Upside down. An easy way to remember this is by counting the number of letters in Right Side Up and Upside Down, as the first has an Odd amount of letters, and the latter has an even amount. Again, this all depends on Prisoner #1 and #2. If they are both able to successfully pick their number from their 50 boxes, and be able to switch them accordingly, it guarantees 100% success. Once Prisoner #3 comes in (depending if #1 and #2 screwed things up or not), they should be able to pick the boxes that are right side up and find their number within 50 tries. If they, #1 and #2 are not able to find their number, then they are all dead. This does depends if turning the boxes betrays the rule (which I am sure it does), but this is what I came up with. This was written at minute 2:45 in the video. Gonna see what the vid comes up with now. I have finished the video, and here is my thoughts. Using the Loop strategy would probably be the same outcome as my strategy. Basing on either my first 2 prisoners, or the % of weather the loops are under 51 numbers, either your test or my test should be able to possibly get their own number and escape, or lose trying.

This would be a good strategy for a escape room, with a clue that points to choosing the Box number that correlates the slip number.

i wrote this in bash, i thought there was something wrong with the randomness of shuf because either everyone won, or the first or second prisoner always failed, until i heard your "fail hard or succeed completely" explanation. it works. shuf is fit for purpose.

Indeed each successful prisoner more than cuts the overall chance of failure in half, so most failures will be decided by the first couple of prisoners.

All of this means that, if you are one of the prisoners, and you find that your number is on a loop of exactly 50, you should be celebrating--the other loop can be at most 50, so all prisoners are guaranteed a win. In fact, you should be hoping that your number is on a loop that is as close to 50 as possible while still being less. If your number is on a loop of 49, probability is low that there's a loop of 51, but if your number is in your box, it's much more likely.

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 ?".*

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.

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

Absolutely stunning example.

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....

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

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

Question: if even one of the prisoners finds their number (considering that the number is taken by the prisoner and not left in the box), wouldn't that mean that at least one of the loops is broken? Or is there a rule that you have to keep your number in the box for this strategy to work?

"Finding" does not mean "taking". You can easily see it in the video at 0:56: The slip #1 obviously is still in the box as the lids are closed once again.

Would it increase the odds if someone who finds their number then tells someone with a number in their loop to go next, quickly closing a loop, and then the next members could use the add numbers/rearrange strategy to reset the loops to something more average in length from a smaller population of theoretical lengths?

No, renumbering the remaining boxes after a full loop got removed does not change the probabilities.

I dont know if im a secret genius or something, but this all makes perfect logical sense to me. The chances of a loop containing half or more the total numbers is much lower than the chances of the loops being more numerous and short. Its just cause and effect.

Yeah when you started explaining I was like oh that's brilliant. I understood why it worked somewhat. I knew from the beginning that the strat had to make the probability no longer independent but I couldn't see the how

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.

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%

They are called *odd* for a reason

Raoodnm

@Jynx how did you calculate that probability

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

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!

Great video ! But I’m missing the point at 9:41 : using the strategy, why would the individual probability for s.o. to find the right box be of 50% instead of 30% ? You may have opened half the boxes, but the underlying probability for them to have your number is not the same anymore because of the strategy you’re using, no ? Thanks if you can help !

The probability that someone finds his number is always 50%, independently of the strategy he uses. Could not be otherwise, as he is opening 50 boxes out of 100, and he has absolutely no information about their contents. So his chance to lose the game is 68.82%, while he still finds his number with 50% probability. That means that there is a chance of 68.82% - 50% = 18.82% to find his number even if there is some long loop. Indeed, such a long loop does in most cases not include _all_ boxes: The average size of a long loop - supposed there is one - is 50/0.6882 = 72.65 . So he has a 100% - 72.65% = 27.35% chance to find his number under the condition that there is a long loop ("conditional probability"). For the overall chance for this case we once again get 27.35%*68.82% = 18.82%.

I want an escape room with this as the puzzle. That would be cool

Okay is it weird I got this answer in my head? I was randomly watching a video on electrical currents so these loops made a lot of sense to me.😂

Wait, so do the loops help partly by chunking lots of boxes together and reducing the possibility space more than if everyone just looked at one box?

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

Life do be like that sometimes

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

Imagine knowing this for a fact and no one listens 🤣

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

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

That's actually what we do with pointers in C/C++, OS, MMU... And if you factor in deadlines, the feeling of being a prisoner might become very realistic.

Interesting. Can you use this to predict "lead times" with some % of accuracy?

I think you can even increase the odds slightly if you tell the prisoners that if they don't find their slip to guess the box number of the last slip they found, that way the prisoners can escape even with cycles of 51

"Finding" is not the same as "guessing". These prisoners are required to _find_ their number.

We did such a process while studying probability and the I Ching. Didn’t know there was a riddle about it. Pretty cool. 👌

imagine if the warden rigs it and puts two of the same slip inside of two separate boxes and one number between 1 and 100 does not exist so the prisoner with that number is guaranteed to fail

This is so amazing. I just cannot believe it. I know it is true but it is unbelievable! Thank you so much for sharing this with us.

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

Why are the replies gone?

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

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 😂😂

The "100 prisoner riddle" is a well-known puzzle involving prisoners, hats, and a challenge to escape. Here is a summary of the riddle and a solution strategy: Riddle Summary: There are 100 prisoners standing in a line, each wearing either a red or a blue hat. They can only see the hats of the prisoners in front of them, but not their own or the hats of the prisoners behind them. Starting from the last prisoner in line, a guard asks each prisoner the color of their hat. If a prisoner guesses correctly, they are set free; otherwise, they are executed. The prisoners cannot communicate once the riddle begins. They can hear the responses of prisoners in front of them but not those behind. Solution Strategy: To maximize their chances of survival, the prisoners can agree upon a strategy before the riddle begins. Here's a strategy that guarantees the freedom of at least 99 prisoners: Before the riddle starts, the prisoners agree that the last prisoner in line (the 100th) will act as the "counter." The counter will count the number of prisoners wearing red hats. The counter does not say the color of their own hat but instead announces the parity (even or odd) of the count. For example, if the counter sees an even number of red hats, they will say "even." Each prisoner, starting from the 99th position, will use the information received from the counter to deduce the color of their hat. 1 If a prisoner sees an odd number of red hats in front of them, they know they are wearing a blue hat. They will say "blue" when asked about their hat color. 2 If a prisoner sees an even number of red hats in front of them, they know they are wearing a red hat. To signal this, they will say "red" if they see an odd number of red hats behind them, and "blue" if they see an even number of red hats behind them. 3 Each prisoner follows this strategy, passing the information along the line. 4 By using this strategy, at least one prisoner will always guess correctly, ensuring their freedom. However, it's important to note that one prisoner might still be executed, as this strategy only 5guarantees the freedom of at least 99 prisoners.

It might involve 100 prisoners, but it is not "The 100 prisoner riddle/problem".

I was thinking about this and it really only works if the boxes/papers aren't jumbled at random. Because if it was random you'd have a chance that a box ends up with the correct slip in it which would break your loop strategy

You've described a 1-loop which is perfectly OK as you'd find your slip on the first try. See @4:15