The Riddle That Seems Impossible Even If You Know The Answer

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

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.

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

really sad that these prisoners were so good at math and cooperation, yet still ended up in jail 😢

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!

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.)

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

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.

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!

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.

Imagine being the first inmate and not finding your number. “Oof, we tried”

Describing why the prisoners open the box with their number is the hardest part of this video. Destin asked a valid question "How do I know my number box is on the loop which contains my slip?" Once you visualize how it works, that question seems super silly to ask despite the brilliance of said question. It's the natural question to ask despite us all understanding and agreeing to the concept of loops and how they behave in this thought experiment. The first box opened is the only slip number that's guaranteed to be in that loop. Simply put, because loops are loops. If every slip correlates to the new box, you will inherently open boxes until you open a box that points you back to the initial box. Thus, you will find your slip. Thus, you complete the loop. Derek even says "you are always starting the loop directly after the slip with your number" or something similar. If you run the test a couple of times on a loop, you will see the pattern and why it all works. Warning: science below Let's suppose we have a loop of 15 numbers. Using a random number generator (RNG), we get the following 15 numbers: 8 68 1 27 70 94 76 71 12 84 21 45 97 58 37 These 15 numbers are both a collective of boxes and slips. Let's call the box numbers as they are currently listed: 8 68 1 27 70 94 76 71 12 84 21 45 97 58 37 The slip numbers, are exactly the box numbers but with a shift of 1 place. 68 1 27 70 94 76 71 12 84 21 45 97 58 37 8 Thus, the "paired loop" that Derek demonstrated with notecards would look like: 8 68 1 27 70 94 76 71 12 84 21 45 97 58 37 (Box number) 68 1 27 70 94 76 71 12 84 21 45 97 58 37 8 (Correlating Slip) Notice the diagonal correlation. The first box opened is the only slip number that's guaranteed to be in that loop but it's always going to be the last slip that is revealed. In this instance, prisoner 8 starts by opening box 8 and finishes when opening box 37 to reveal slip 8. All 15 prisoners who are a part of this loop will open exactly 15 boxes and always find their slip on the 15th box. In theory, the connecting 14 numbers could be anything but we know for certain what the first box number and the last slip number will be (see below) 8 xx xx xx xx xx xx xx xx xx xx xx xx xx xx (Box number) xx xx xx xx xx xx xx xx xx xx xx xx xx xx 8 (Correlating Slip) Prisoner 68 would start on box 68 generating the following variation of the same loop: 68 1 27 70 94 76 71 12 84 21 45 97 58 37 8 (Box number) 1 27 70 94 76 71 12 84 21 45 97 58 37 8 68 (Correlating Slip) Again, the only guaranteed slip to be in this loop is the number 68 slip which we find in the 15th box. 68 xx xx xx xx xx xx xx xx xx xx xx xx xx xx (Box number) xx xx xx xx xx xx xx xx xx xx xx xx xx xx 68 (Correlating Slip) Really hope this helps to clarify the question that many of us asked to ourselves while watching.

I feel like there needs to be a 'wow out loud' like 'laugh out loud', because when you said "that's like scaling up a millimetre to the diameter of the observable universe", that's exactly what I did.

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

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

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.

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.

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.

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.

Mr beast should do this