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: 🤍🤍youtube.com/watch?v=C5-I0bAuEUE Vsauce2: 🤍🤍youtube.com/watch?v=kOnEEeHZp94 Stand-up Maths: 🤍🤍youtube.com/watch?v=a1DUUnhk3uE TED-Ed: 🤍🤍youtube.com/watch?v=vIdStMTgNl0 ▀▀▀ 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). – 🤍🤍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 Produced by Derek Muller, Petr Lebedev, and Emily Zhang
What if we went down? I mean rather increasing the number of prisoners from 100 to 1000 or 10000, what would be the probability when it goes from 100 to 10 or any even number below 100??
What if prisoner no 1 go in and find his no in box no 4 after a loop but when prisoner no 4 goes and open the box there is no number in there so how will he find the next box in the loop 🤔 😉
How am I sitting here, reacting to this video like I'm watching a tense sport's match...?
The one thing i dont understand, why is it guaranteed to be on the loop with your number, if u start with the box labeled with your number? When the numbers are spreaded randomly, you can be on a loop that doesnt include your own number(?) at iSNsgj1OCLA&t=4m50s 4:50
I mean I'm pretty sure we did this in basic training with our jerseys with more than 100 trainees. Really isn't that hard. Just find your damn number and remember where it is. How can it possibly be that difficult 💀
TLDR: there is a 30% chance that the longest loop is 50 or under.
This is literally just an expanded version of the riddle where there's 2 prisoners and they one at a time walk into a room where there's 9, 15, 25... squares with colored markers and one of them has the key under. The first prisoner has to pick a spot on the board and flip only ONE of the markers and somehow tell the 2nd prisoner where to find the key under the spots by that 1 move. If the 2nd prisoner finds the key, they dont get executed. Keep in mind they both can strategize before hand but will not be able to see eachother or communicate until after the 2nd prisoner finishes. And the first prisoner does not leave the spots how he found it
Don't the probability change by what order the numbered prisoners walks in right?
also crazy to think that assuming every prisoners follows this rule. if the first guy finds his on his 50th attempt, he knows everyone will succeed. 2 loops. 50 numbers each
So what if the first prisoner (number 1) goes into the room and randomly chooses box 16 containing the number 16? Would that impact the overall probability? Because, however unlikely, if all  numbers were randomly placed into the 100 boxes there must be the possibility that all boxes contain "their" own number [all boxes point to themselves] and therefore in this case there are no loops per se. I think the part that confuses me is if the boxes are left uncovered after each prisoner leaves the room or are the lids replaced (reset)... I understand that they are reset but I still cannot get my head around cases (as mentioned below) of there being two distinct loops of 50 boxes and all prisoners can survive with a chance of 30.7% (or even greater).
If Slip #1, is in Box #1, and prisoner #1 doesn't pick Box #1 first, the whole experiment fails.
What are the odds of a slip going into its own box? and then, what are the chances of that prisoner picking their own number first?
I often wonder why I have a bit of a jail obsession in my fiction writing but I feel like I'm not alone in it. Riddles like this show all of society has a weird jail obsession. Is there a reason the setup for this riddle has to be a story with such dire consequences? Why can't the setup just be a group of students playing a mathematical game? the prisoner's dilemma also has variations that aren't about prisoners. The paradox about the guy who is going to be hanged on a surprise day could just as easily have a setup that isn't so dire. We could say a kid is going to get a surprise birthday party on one of five days and is trying to figure out when it will be. Why do all these logical puzzles that could just as easily tell a more innocent story have to be about jail?
This is an old computing data structure known as a linked list. I used to use this on indexed file systems before inexpensive databases were available for personal computers. But did I think of it when you explained the problem? Oh hell no.
This was an excellent video
It's a fake!
was ist wenn sie sich vorher absprechen dass die nummer 1 zuerst die zweite Box öffnet und dann die Erste box. und wenn in der 1. die 1 ist wartet er die zweite Zahl in Minuten ab ( zum Beispiel: Nummer 20= 20 minuten warten). Daraufhin geht diese Person raus. Gehen wir also davon aus dass die nächste Person gleich danach reingelassen wird, würde die Person
mit der Nummer die abgewartet wurde reingehen und könnte sie die dritte Box öffnen. Die Zeit in der dritten Box abwarten und dann die zweite Box mit der Nummer der Person öffen. Das heißt also alle würden es schaffen wenn in der ersten Box die 1 ist. Die Wahrscheinlichkeit wäre also 1 zu 100.
what if you agree beforehand that the number 1 will open the second box first and then the first box. and if the 1st is 1, it waits for the second number in minutes (for example: number 20= wait 20 minutes). Then that person goes out. So, assuming the next person is let in right after that, the person would
go in with the number that was awaited and she could open the third box. Wait for the time in the third box and then open the second box with the number of the person. So that means everyone would make it if the 1 was in the first box. So the probability would be 1 in 100.
Whilst a totally different problem, I believe, to 3x+1, I did notice the same looping type thing happening there too. And I noticed that the loop lengths got longer the higher the number, but also within that space of numbers many of the loops were far shorter than the longest, and there was definitely some median length that was interesting.