@DrDJX
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.)
@magdasg9571
Memorizing this just in case I'm ever trapped in a prison with a sadistic mathematical prison warden
@wetbadger2174
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.
@CreaseQuest
“Teach me” is such a beautiful response to when you don’t understand something.
@gregsquires6201
I think the chance of convincing 99 other prisoners that this strategy is their best chance of survival is much lower than 31%.
@fixed-point
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.
@bscorvin
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
@Datwolfdoood
This is the same method I use at work! I manage inventory for a dealership and if there's a folder with another units stock number, I'll keep following the numbers until I find the right one. It usually a series of 2's or 3's before I find the right papers for the folder I have. I keep an average of 200-300 folders of different units at a time
@Bismuth9
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.
@GuitarGuise
The way I like to think about the solution: you're no longer betting that each individual prisoner will find their number with a pattern that they choose (arbitrary or intentional), but you're betting on the probability that a pattern (a loop exceeding a length of 50) does not exist in the set of boxes. And that's a static property of the set you're betting on, in contrast to rolling the dice every time on 50 different prisoners. So, in a way, you've already succeeded or failed by choosing the loop strategy, whether you know it or not. Random chance no longer has anything to do with the prisoners' choices (unless they mess up the execution of the loop strategy), but entirely on how the box set is arranged and the loop strategy that the prisoners decide to employ at the start. Fascinating mathematics! Thanks for sharing this!
@ZamanAristoOrCleon
Imagine being the first inmate and not finding your number. “Oof, we tried”
@bloxy7891
I cannot imagine being one of the hypothetical 1 MILLION prisoners having to check 500,000 boxes. At that point, just give me the death sentence.
@NZ-fo8tp
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
@BigPapaMitchell
12:20 I have a better intuitive explanation: The only way you could start on a chain and not eventually reach itself is if either that chain forms a line with an endpoint, or that chain loops back on itself in the middle. The first one requires a box to have no number in it, which is impossible, and the second requires that two boxes have the same number, which is impossible, meaning that it must be the case it loops back on itself.
@inemanja
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.
@industrialgoose4756
A good way to think of it is that with this strategy you're betting on the seed rather than your picks. Based on this math there's a 31% chance that the slips in the boxes were distributed in a way that's favorable to you(creating no chains greater than 50 long). Just try to picture a bunch of randomly generated chains in a limited space of 100 items, with no repeating numbers, and it quickly becomes an understandable strategy.
@michaelgove9349
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 ?".
@charliehorse8686
If you think the riddle is hard, imagine trying to convince 99 fellow prisoners to follow the plan to the letter.
@brianrussell463
Destin had my favorite response ever, "teach me!" I love that.
@pyguy9915