Can you solve the prisoner's riddle?

so if number one prisoner goes in the room he goes to box
1 then whatever number is on that box he needs to go to that box

So you're saying there's a chance....

What you have to do is look in the box of the number you just found and then look in the the box of that number and so on

Clickbait that he doesn't share the solution.

Communication is the answer

Imagine you’re one of the prisoners watching this

The horses name was Friday

If every prisoner chooses 1 3 5 7 9 ...99 they have a 50% of beinf freed

Would this work with any even number group? Just cut the selection in half? Or does it always have to be 100? Sorry, not really a math gal

I think some prisoners in New Orleans solved this riddle, at least for awhile.

Split the boxes each prisoner opens one half and the other the other half.

I understand the chaining mechanism. However, at the end of the day, I am effectively choosing 50 of 100 boxes just the same. I don't see how my odds are different than choosing 50 random boxes.
I can see that it is harmful for everyone to "just pick the first 50 boxes" as this is a guaranteed failure so at least we know that some strategies are definitely worse than others.

Why would they be given a 50% chance when they're only allowed to open 50 of the 100 boxes? I'd say they'd have more like a 25% chance. Their number might be in one of the 50 boxes they weren't allowed access to.

Bro tell us the answer and don’t leave us hanging like that

So what's the answer?

They agree before they go in that they take whatever number they find and give it to the right prisoner that way they find all 50 numbers and they all go free.

I got to 99% anyone else?

Is this like the 3 door gameshow problem, just bigger?? Im good at math for the most part, but like he said, even knowing about probability and what you should do, its still hard to make sense of!!

If they can tell each other when they all finish , then they should team up in teams of 2 , other wise idk 😐

Vsause never did me dirty like this.

PLEASE try to speak with more regular prosody—the constantly interrupted cadence is SUPER distracting

Okay, here's the best strategy for the prisoners, which dramatically increases their chances of survival from a near-zero probability to over 30%:
The Strategy
* Numbering the Boxes: Before anyone enters, they agree to number the boxes from 1 to 100. This is crucial for the strategy. (The problem states the numbers are randomly placed in boxes, but they can agree on a system to identify each box uniquely).
* Each Prisoner's Search:
* When a prisoner enters the room, they first go to the box that corresponds to their own prisoner number. For example, Prisoner #1 goes to Box #1, Prisoner #50 goes to Box #50.
* They open this box. Inside, there will be a slip of paper with a number written on it (this is the number of another prisoner, or possibly their own).
* They then go to the box corresponding to the number they just found. So, if Prisoner #1 opens Box #1 and finds the number "73", they then go to Box #73.
* They continue this process – opening the box indicated by the number they just found – for a maximum of 50 boxes.
Why this strategy works (the mathematical insight):
The numbers inside the boxes form "cycles" or "permutations."
* Imagine the numbers as a mapping: Box X contains number Y. This can be seen as an arrow from X to Y.
* Because each number is unique and there are 100 boxes and 100 numbers, these mappings will always form one or more closed loops (cycles). For example: Box 1 has 5, Box 5 has 20, Box 20 has 1. This is a cycle of length 3 (1 -> 5 -> 20 -> 1).
The Key Insight:
* If a prisoner follows this chain, they are guaranteed to find their own number if and only if their number is part of a cycle that is 50 boxes long or less.
* If their number is part of a cycle that is longer than 50 boxes, they will not find their number within their 50 attempts.
The Probability:
The probability of all prisoners finding their number successfully comes down to the probability that there is no cycle longer than 50 boxes among the 100 numbers in the 100 boxes.
It can be mathematically proven that the probability of a random permutation of 100 elements having a cycle of length greater than 50 is approximately:
P(\text{cycle length} > 50) = \sum_{k=51}^{100} \frac{1}{k}
This sum is approximately ln(100) - ln(50) = ln(2) \approx 0.693.
So, the probability that at least one cycle is longer than 50 is about 69.3%.
Therefore, the probability that all cycles are 50 or less (meaning all prisoners succeed) is approximately 1 - 0.693 = 0.307, or about 30.7%.
This is a massive improvement over the near-zero probability of randomly guessing. While not a guarantee, it's the optimal strategy for this riddle.

Where's the link to the full video?

The prisoners cheat, first two use their blood to mark the boxes eith their number and the last two make sure every number 's been taken so the first ti dont get left out.

Inmates.

lol, for as thorough as your videos usually are, this one really dropped the ball.

This new episode from Viva la dirt aint as good as old ones. Come on Rowan.

I first noticed this riddle while I was watching a TED-Ed video.
Apparently, the goal was to create a loop. I wasn’t very satisfied.

And what prison exactly does this?

I would not have been caught commiting a crime in the first place so i won't be in Prison. Give this ribble to Trump he has a biggly Brain 🧠🪱🧠 & should be in Prison anyways. 😂😂😂

Its pi

Open the boxes numbered after the numbers you just opened, right? Oh shoot I've probably seen this one before

blah

Pretty simple : If you don't find your number in the 50 boxes you open. YOU'RE FREE (from the world)

Is Removing all their shirts an option?

Oh yeah I jas this problem asked inprobabilities classes and I remembered beeing surprised. Later i tried to find it but I didnt remember the setup nor did i find back my notes

The answer is using a pen or marker and having the first 3 prisoners write which number is inside each box on the outside of the box. Rules don't say you can't use a pen or marker.

So you're really just banking on the first 3 prisoners to find their numbers

The probability of picking up the same piece of sand from everywhere on earth is a lot lower than that idgaf💀

The first one scratch every box he open and so on

The trick is to not make 100 individual guesses but to couple the odds of success and failure. Use a strategy that ensures that if it works for one, the odds of it working for the others as well is high.

1 in 3? Lol. They're all cooked anyway. 😂

If they were so smart they wouldn't be prisoners

We are dealing with a very particular configuration: the numbers inside the boxes are not placed arbitrarily. Instead, they form a single random permutation of the integers from 1 to 100. This changes everything. What looks like chaos at first is actually structured randomness. Each box corresponds to a position in the permutation, and each number on a slip points to another position, forming a set of disjoint cycles. These cycles define the fate of the prisoners.
The strategy leverages this structure. Each prisoner starts by opening the box labeled with their own number. If the slip inside shows their number, they’re done. If not, the number they find tells them which box to check next. They continue this process—following the cycle—until they find their number or reach the 50-box limit.

Who tf said its a riddle?
A riddle is like "what has eyes but cannot see? A potato."
THIS is called a logic problem.

I have a riddle for you.
I'm 3rd in a song next to re and fa, and i also rhyme with sea. Telling me who i am can be tricky, but its fun to hear people try. Who are you, and who am I?

....ok I'm sorry i never made a riddle before but i hope its better than this logic problem that's not not a riddle.

Put the lid on the wrong way if you find your own number? Should reduce the odds some what.

Is it about even and odd numbers. I'm not sure about this riddle. It's a good one...

I get it! And I’m a dummy. Maybe that’s why I get it. However the chances of convincing 100 people, let alone inmates, makes it highly doubtful.

The stragety is the loop stragety heres how it goes:the prisoner goes in they find thhe box with there number and like if the box says 3 they go yo box 3 and so on so on

If every prisoner opened only the boxes labeled 1 to 50, they would reach 50% or 1 in 2

I feel like I may be wrong but it sounds similar to the car/goat question on the movie 21. The variable change one. Idk