Dope Tech Episode 2

<<@pyguy9915 says : Something seems wrong at 9:00 What is the probability of a loop of length 1? (Can't be 1/1) Length 2?>> <<@spamkart says : Why did you make the determination of how many numbers are in a loop? Also, why did you make the rule that your number is definitely in the loop? Also, i would presume that the loop is one hundred. Not then you should continue with the explanation of this equation and riddle.>> <<@melikeypickles says : Idk it's just so attractive to me that I watch videos like this😏>> <<@Charisma-v6s says : Or you could use the Harrison Ford option of skipping the sw-rd altogether and using a g-n. There can’t possibly 100 prison staff, so each prisoner grabs a box and tosses it towards the staff. Then they all run out together. That is the *real* freedom: no one has to use math!>> <<@Charisma-v6s says : Ok, so what would happen if a loop could not be created? If the person who found their number got to keep that piece of paper, thus leaving an empty box?>> <<@sureshsubrahmaniam3225 says : First Prisoner finding his number is not 50% but 1%>> <<@kekstier says : Better explanation (hinted by the final sentences of the video): By using this strategy, the outcome is not determined by every single prisoner getting lucky in finding their number (which has an abysmal chance), but by the arrangement of boxes/slips so the don’t contain a long loop, which has a better chance (30%)>> <<@kevincullers4927 says : Dont mind me, just speaking to you at different times in the form of youtube comments. Can you keep a conversation that transcends time?>> <<@FYTA781 says : I can solve this with higher probability>> <<@michaelfrench7000 says : I can increase the odds to 50%. Yes, I can.>> <<@godfreypigott says : The strategy didn't work for Epstein, and Donny P. Edo was the malicious guard.>> <<@XamseYareCali says : You already have 50/% why 30%>> <<@WoodyC-fv9hz says : Without strategy impossible, with strategy I reckon 30-40% probability to get lucky. Prisoners have enough time and often have access to a library with a computer. I would make a program to test strategies like agreeing to making up box opening patterns. 3X plus 1 comes to mind, primes, parity algorithms, skewing etc. Java or C might do the job. Random checking would be suicide. I am personally a great believer of looping patterns. Problem is, you cannot leave any clues behind during the process.>> <<@iva._to says : ted-ed introduced me to this riddle so i knew the solution, but glad to say this video gave me a deeper and better understanding>> <<@FireDragoncubes says : 6:47 Isn’t that wrong, there’s 100! Then 99! And so on, but aren’t there more possibilities, you could go back to the beginning of the loop and start a new one, too. And there are also other combinations of boxes, there could be 50 and 50 100 99 and 1 98 2 97 3 … And so on, is that not just the possible ways there count be a loop of 100?>> <<@KurzAstroOfficial says : There was a magic in 8:12 The Back side of the white board changes after again turn back>> <<@SinCitySage says : This is why students hate math.>> <<@SimonMeng-l7s says : I want this white board of infinity lol>> <<@Tradv1 says : 0:48 :)>> <<@AsherMcMullinOriginal says : Destin staring in confusion for a second and then asking "teach me" is why I love you guys so much>> <<@vinibyte says : still a monstrous chance to die lmao>> <<@ilTHfeaa says : wait when he started explaining it i was like "did i actually get it right?" okay well.. my strat wasnt exactly the same but like i was close!>> <<@AyyJayy737 says : Peter BRO milterSON🥀😭✌️🥀😭✌️🥀😭😭✌️🥀😭✌️>> <<@sqdancefan says : Is there a strategy for guaranteeing no loops longer than 50?>> <<@dylanvaccaro6919 says : Your math my be correct but good luck convincing 49 other prisoners of this being the best strategy lol>> <<@scottbeare2323 says : What am I missing here? Each prisoner can still only check 50 boxes - and that order is still random. No loop can be longer than 50. So, each prisoner cannot be gaurteed he finds his number. Still seems to be that chanes are 100 factorial…. So, how can there be a loop with more than 50 if each can only check 50 boxes?>> <<@jesisters3408 says : This like Ted Ed’s Prisoner Box Riddle so you basically start with the box labelled with your number, look inside and see what number it has and go to that box. Keep going until you find your number>> <<@TomblePlayz says : Woudl I be right in saying that if 51 prisoners succeed, the game is complete and you are guranteed to win because even if everyone else is on their own loop, it would still be lower than 50.>> <<@Nostromo2144 says : I'd be curious to know if the prisoners could communicate after each attempt, how much could they improve their chances of every one finding their own numbers...? i.e. if the first prisoner comes back with his number <50, then is he allowed to keep looking, get the entire loop(s) mapped out, and thus ensure everyone finds their number? I guess that's just cheating, as he would just be telling them which box to go to directly, if he had an eidetic memory. The issue then would just be the chance of him finding his number in the first 50 checks, which is pure luck I guess, eh. BUT, if he has to stop after finding his number, is allowed to communicate his entire loop, then how would that help the next person and the next and so forth? I guess that's down to random chance again, as the shorter the early loops, the less that will tell succeeding prisoners who are not in any of them, until all numbers are covered in the first few attempts, so still a 31% chance of success, or better...?>> <<@ingainloggningsnamn says : Seem impossible... But ok let's say the first one checks everything from top left and going in a row, second one starts with the second box and third one starts with the 3rd box and so on....? That's like the only strategy I could think of that could increase their odds. But... Does it really increase it THAT much?! As I said, seem impossible.>> <<@HaHa-s3j says : What if the number of boxes allowed was n>> <<@bibi_89890 says : 0:52 Why was slip 67 found first man 😭😭😭>> <<@FreakingCreaking-m7t says : Wait but 100!/100 Isn't just 99! ?>> <<@CloneChannel-h4r says : Show this Video to Mrbeast and he knows what to do 😅>> <<@bilbo100989 says : I mean what stops the prisoners from changing the numbers into there correct boxes as they "look". it didn't say someone would be monitoring... they are criminals after all lol>> <<@davisdilshner929 says : Would this work with the game show deal or no deal?>> <<@etherblood0 says : if unsuccessful prisoners guess the box their last number points at, they will also guess correctly for loops of length 51. So we should be calculating the win probability as the chance of all loops being shorter or equal to 51.>> <<@JackOfShadows1 says : In the real world, the probability remains close to zero because a) you have to convince 100 convicts to do this, b) all 100 must understand the strategy, and c) all 100 must make no mistakes!>> <<@dominicg4009 says : 7:00 box 2 has 98 possible numbers because it can't be it self or there would be a repeated number>> <<@ybloc1428 says : I guess thought experiment wise, I didn't understand how choosing your box ensured your box would ensure your number is in the loop. Now thinking about a loop of 2, going to box 100 and pulling a 1, going to 1 would have 100. If not it would be on a path to 100 because the box is pointing to your designated number crazily>> <<@Drift0009 says : 0:48 predicted 67>> <<@gabrielromano9096 says : If the failure of one prisoner does not make the others fail, should they all go for random, to make it 50%, or follow some other strategy?>> <<@paulksacco says : I STILL can't get my pea-brain to understand how randomness can contain order. But it does seem to me that, by have the randomness influenced by the non-random act of forcing the FIRST guess to be unique; you have removed the FIRST permutation (100!) from the math without the "loop theory.">> <<@SBCASWIMTEAM says : Just gamble.😁👍>> <<@yeayah says : Wrong kid. Nothing says your number has to be in the loop>> <<@Zippythewondersquirrel says : Every person goes in and just chooses one number right or wrong. When it’s over they just give the number they picked to the correct prisoner.>> <<@JeremyBWarren says : This is a really interesting video, but it does a poor job of explaining the probabilities of having loops of various lengths. The host makes a subtle but non-obvious leap in working through the case for L=100, and then he makes a huge leap in just asserting that the result is 1/L for all other Ls down to 50. But L=100 is the edge case, and the logic for L<100 is not the same. In fact, it's the subtle leap he makes in the logic for L=100 that, if explained properly, would make the logic for other cases consistent. And I don't think it's too complicated for the video - but it's necessary to connect things together properly. In particular, the video's explanation ignores the distinction between counts of _unique loops_ and counts of _arrangements of slips in boxes._ Thinking in terms of loops is critical, but this needs to be translated back to slips-in-boxes arrangements ("slip arrangements", for short) in order to calculate the probabilities in the video. When the probability of having a loop of length 100 ("P(L=100)") was given as the # of unique loops of L=100 divided by the total # of possible slip arrangements, it threw me for a loop (pun intended). That's not definitionally how probabilities are done. By definition, the numerator should be a count of _slip arrangements_ that meet the criteria of having a loop of L=100, and it's not immediately obvious (to me, at least) why using the # of unique loops of L=100 works. What's missing is the observation that all unique loops of L=100 have exactly 1 possible slip arrangement each. Think about it: if a loop contains the sequence a->b, it means that box a contains the number b. So if a loop has length 100, then _every_ number has a prescribed following number, and thus _every_ box has a prescribed slip in it. Note that this is different than with the representations of loops given just before in the video (when deriving the count of unique loops of L=100), where each unique loop could be _written_ 100 ways. It's this shift from counting loop representations to counting unique loops to counting slip arrangements that is subtle, but should be made explicit: you can calculate P(L=100) as 100!/100 (# of unique loops of L=100) divided by 100! (# of total slip arrangements) because each unique loop of L=100 has 1 and only 1 slip arrangement. But this is not true in general for L<100. In general, each unique loop specifies exactly which slips should be in the boxes _only for those box numbers (and slip numbers) in the loop._ So there are L boxes that are completely locked in, which leaves 100-L boxes unspecified and 100-L slips to go in them. Thus there are (100-L)! slip arrangements for each unique loop of length L. The calculation of unique loops is also a bit different when L<100. It's not simply L!/L, although the logic is similar. Each loop representation is a sequence of L numbers, so you have 100 * 99 * ... * (100-L+1), or 100!/(100-L)!. (Note that this is just a standard permutation.) It's only the edge case where 100-L = 0 and the denominator disappears (bonus points for showing one reason why 0! must be 1). But just as with L=100, each _unique_ loop has L representations this way, so the # of unique loops of length L is 100!/(100-L)!/L. Putting it all together, we get that P(L) = # of slip arrangements containing a loop of length L / # of possible slip arrangements = # of unique loops of length L * # of slip arrangements per unique loop / # of possible slip arrangements = 100!/(100-L)!/L * (100-L)! / 100! Now we can see that the (100-L)! terms cancel out (in addition to the 100! terms cancelling out as with the L=100 edge case shown in the video) and we are left with P(L) = 1/L. QED! But I made my own subtle leap above :) The # of slip arrangements containing a loop of length L is only equal to the # of unique loops multiplied by the # of slip arrangements per unique loop for L>50. Because if L<=50 then some of the possible slip arrangements we're multiplying out are _duplicates._ If you are talking about L=20, then all of those random arrangements of the other 80 slips are going to include some other L=20 loops, and thus those slip arrangements would also be included when considering those other unique loops. But because of the rules of the game, only L>50 matters, so it's not worth figuring out the _more_ general formula for P(L<=50). Good enough for this problem :) I do really like the Veritasium videos, and I think they could have walked through this in a way that (would be much clearer & entertaining than my explanation above (!) and) wouldn't have been too complicated, and would have made the overall explanation better.>> <<@antenor790 says : 10:42 how to say that you didn't understand without saying that you it. A loop needs to repeat to be a loop. You start from your number and every time it is a loop, because some time you need to found your own number (if you search every box)>> <<@DroneQuadcopter says : Given this task...bI would just ask the guards to put me in the gas chamber.... Or firing squad... There is no way my dumb a s s escape this way>> <<@FrancislouiseIgnatiusp.Francia says : 16:01 how aren't you sure it's most likely most of them are going to be more than five hundred 5000, etc, I mean, it's most likely going, I mean, there's a higher chance every time it's going to be the other and also if that happens, that there's means that at least a single chance of failure>>