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13,833,714 Views ⢠Jun 30, 2022 ⢠Click to toggle off description
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:    â˘Â Solution to The Impossible Bet | The ... Â
Vsauce2:    â˘Â The 100 Prisoners Puzzle Â
Stand-up Maths:    â˘Â The unbelievable solution to the 100 ... Â
TED-Ed:    â˘Â Can you solve the prisoner boxes ridd... Â
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References:
Original paper: GĂĄl, A., & Miltersen, P.B. (2003). The Cell Probe Complexity of Succinct Data Structures. BRICS, Department of Computer Science, University of Aarhus. All rights reserved. â ve42.co/GalMiltersen
Winkler, P. (2006). Seven Puzzles You Think You Must Not Have Heard Correctly. â ve42.co/Winkler2006
The 100 Prisoners Problem â 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. â ve42.co/Golomb1998
Lamb, E. (2012). Puzzling Prisoners Presented to Promote North America's Only Museum of Math. Observations, Scientific American. â ve42.co/Lamb2012
Permutations â ve42.co/PermutationsWiki
Probability that a random permutation of n elements has a cycle of length k greater than n/2, Math SE. â ve42.co/BaezProbSE
Counting Cycle Structures in Sn, Math SE. â ve42.co/CountCyclesSE
What is the distribution of cycle lengths in derangements? In particular, expected longest cycle, Math SE. â ve42.co/JorikiSE
The Manim Community Developers. (2021). Manim - Mathematical Animation Framework (Version v0.13.1). â www.manim.community/
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Special thanks to Patreon supporters: RayJ Johnson, Brian Busbee, Jerome Barakos M.D., Amadeo Bee, Julian Lee, Inconcision, TTST, Balkrishna Heroor, Chris LaClair, Avi Yashchin, John H. Austin, Jr., OnlineBookClub.org, Matthew Gonzalez, Eric Sexton, john kiehl, Diffbot, Gnare, Dave Kircher, Burt Humburg, Blake Byers, Dumky, Evgeny Skvortsov, Meekay, Bill Linder, Paul Peijzel, Josh Hibschman, Timothy OâBrien, Mac Malkawi, Michael Schneider, jim buckmaster, Juan Benet, Ruslan Khroma, Robert Blum, Richard Sundvall, Lee Redden, Vincent, Stephen Wilcox, Marinus Kuivenhoven, Michael Krugman, Cy 'kkm' K'Nelson, Sam Lutfi, Ron Neal
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Written by Derek Muller and Emily Zhang
Filmed by Derek Muller and Petr Lebedev
Animation by Ivy Tello and JesĂşs RascĂłn
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
Views : 13,833,714
Genre: Education
Date of upload: Jun 30, 2022 ^^
Rating : 4.929 (7,412/407,495 LTDR)
RYD date created : 2024-05-03T10:05:11.188464Z
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YouTube Comments - 35,554 Comments
Top Comments of this video!! :3
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.
18K |
Memorizing this just in case I'm ever trapped in a prison with a sadistic mathematical prison warden
5.6K |
really sad that these prisoners were so good at math and cooperation, yet still ended up in jail đ˘
1.7K |
A really incredible feature of the loop strategy is comparing how well it works even against random guessing with more chances to open boxes. For example, if each prisoner were allowed to open 99 of the 100 boxes, instead of 50, to find their own number, the total probability of success by randomly guessing is only 0.99^100, or 36.6%! (Whereas the loop strategy gives a comparable chance of success while only opening 50 boxes, and succeeds 99% of the time if you can open 99 boxes) If you were allowed to open 98 of 100 boxes, the chance of winning via random guessing drops to 13.3%, and to below 5% for 97 boxes!
506 |
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.)
33K |
I think the chance of convincing 99 other prisoners that this strategy is their best chance of survival is much lower than 31%.
12K |
The main point here is that "failing hard" comes with no harsher penalty than "failing little"....hence, you can redistribute your loss function to take advantage of this. Great video, btw!
45 |
I was really confused at first on how whatever number you start with is guaranteed to be in your loop, but once I started to type out a comment questioning it I totally realized how it works. In order to finish your loop you have to end up back where you start, and since none of the boxes can be empty, you're guaranteed to be in some sort of loop.
186 |
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.
5.2K |
Imagine being the first inmate and not finding your number. âOof, we triedâ
1K |
So, I think thereâs an easier explanation for why youâre guaranteed to eventually circle back to your own box.
For a loop to close, you have to pick a slip of a box youâve already been to (otherwise youâre just continuing on down the loop).
But the fact that youâve already been to a box requires that youâve already found its slip in a previous box.
The one exception to that, is the box you started with. Which, in this case, is your own number.
9 |
i made a smaller version of this riddle by decreasing the number of participants to 4 and i made artificial intelligence to randomly pick numbers for me
with the loop strategy they succeeded 5 times out of 10
with random picking they didn't succeed in 10 trials.
it was really fascinating.
1 |
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
1.9K |
If you think the riddle is hard, imagine trying to convince 99 fellow prisoners to follow the plan to the letter.
1.1K |
It actually amazes me that anybody could not see why if you start with your own number, you'll eventually find your slip. That was one of the few parts of this that as soon as you said it, I was instantly "well yeah, obviously."
Fascinating how different minds see different things.
1 |
Video: mentions the probability of finding their number approaching a limit
Me: shudders at being reminded about derivatives
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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
3.1K |
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.
1.7K |
When you first gave the solution, I sketched out on a piece of paper a version of the problem with 4 prisoners, 4 boxes, 2 attempts, and I feel like I understood the entire thing almost instantly with no further explanation required. I think lowering the numbers down to something more manageable makes the problem much more comprehensible. I mean, you could even write out 4! configurations if you wanted and prove that it works for all of them, whereas 100! is so large as to be impossible to visualize.
I think it would have been helpful to include this simpler example in the video.
95 |
@pyguy9915
1 year ago
Something seems wrong at 9:00 What is the probability of a loop of length 1? (Can't be 1/1) Length 2?
12K |