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?>> <<@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>> <<@nathJiSang-s5k says : Wow. "Either we all win together or we all lose together" This strategy is pure genius>> <<@squidyhead says : My head hurts>> <<@smitamohite4889 says : 13:50 i didn't expect Matt's entry 😅>> <<@bztrvaldo says : in case i ever become the prison's warden, i have to work on a way to make it harder>> <<@randy5655 says : Imagine a situation where the 100 prisoners go in a room to find a slip inside a box that is their prisoner number. Find the number, you go free. Fail and you are executed. You only get 50 tries. If you find your number you leave the last box uncovered and leave the room. Or else they all stay covered. The next prisoner comes in and repeats the process. As time goes by over 50 boxes will be uncovered. At this point the chances of finding a slip with your number in 50 tries is 100% To counter this, as each prisoner enters the room, the number of tries decreases by one. Therefore, there is a high probability that the last prisoner will see several covered boxes and only have one choice,>> <<@jackrossi4400 says : Stupid does what stupid does and stupid wastes your time!>> <<@Ommprakashsarangi4152D says : New mr. Beast video idea>> <<@mohammadazadi4535 says : If each pairs of prisoners try to find both two number that belongs to them in any first 50boxs and next and when out of the sealed eara and stand near eachother they can chang what they found .>> <<@mohammadazadi4535 says : This video does not explain about freedom of 100 prisoners, isn't it?>> <<@saravanakumar9274 says : Bro really went to prison>> <<@Silas5_1 says : 15:55 well if i had 2 prisoners with two boxes, it's 25% as you would only be able to open one box>> <<@dylancleverdon486 says : i fully understand the strategy when operating with neurtral guards but how does it work when arbitrarily adding 5 to the box to counter the malicious guard. wouldnt you just be guaranteed to be on the wrong loop since now your box isn't even on the loop>> <<@ToastyWaffle708 says : So manny comments are copying each other they made a whole category for these type of comments>> <<@mindin2941 says : Isnt any random strategy ultimately a loop?>> <<@orejovanovic8081 says : Something very wrong at 13:20. If they add 5 on the box (as i understand - starting from your no +5, or going to whatever said in box with your number and adding 5 to that) they are not anymore guaranteed they will be in the right loop.>> <<@TheNameOfJesus says : What are the odds of success if the number of prisoners is 1? (100%) How about 2, 3, 4?>> <<@bobbrown8155 says : You showed the calculation with larger numbers of prisoners (1,000, one million, one billion). What about small numbers of prisoners (2, 4, 6, 8, …). The strategy seems to fail at small numbers. Doesn’t it? It definitely fails at 2 prisoners. The probability is 0.5 * 0.5 = 0.25, which is less than 0.3.>> <<@VivIloveClurk says : All the smart people here and then there's me just graduated 6th and says leave the door open😂😂😂😂😂😂 2:42>> <<@85viveksharma says : I have a question - This is regarding the possibility that the prison warden is evil and he may be likely to arrange the slips in a way that there is that 1 loop bigger than 50. So the prisoners choose to add a fixed number to the box numbers to essentially randomize the distribution and ensure their 31% chance of win. But what if the warden was not evil and in fact all the loops were shorter than 50 initially! Prisoners had a 100% chance of win. By randomizing the distribution, have the prisoners disturbed a favorable distribution (100% chance of win) to a 31% chance? So how do prisoners decide if the warden is truly evil? Should they randomize to redistribute slips to restore their sure shot chance of 31% win or risk it without randomizing? Is randomizing always preferred or subject to some additional information on the intent of the warden?>> <<@Batdihdei says : How does one find this out>> <<@rafafra says : Nah just execute them.>> <<@michaelgoeke6446 says : Hey my intuition was right when I paused the video! I thought of seeded pseudo random number generators. And using the person's number as the starting point... I didn't make the mental leap to it being the "last in a loop" but I have recollections of the loop thing.. you know, from 20 years ago when I played with this stuff.>> <<@PalomaOehler says : It would've been helpful if the problem statement contained that the boxes are numbered, too. It's in the pictures, but not in the text. Otherwise, it's hard for the criminals to agree on a numbering scheme.>> <<@Redrigg07 says : What if❓❓❓❓❓ A prison opens 10 boxes and then he finds his number, then also he continues to 40 time , what the opinion about success and he by hearts all 50 numbers and shares to other prisoners. Then how is it⁉️⁉️⁉️⁉️⁉️⁉️>> <<@bive-channel says : Someone notice the board 7:47 and 8:17>> <<@TP-yc7lv says : But what if its less than 100?>> <<@JoeGb6789j says : I’m not a mathematician, but that does not seem right. If I go into a room with 100 boxes and I have to find the number 3 and it’s all random. I do not have a 50% chance of finding it. If that was true, winning the lotto would be way easier or am I totally wrong?>> <<@MacBor64-q8x says : I HAVE alternate SOLUTION! rotate the box lids so the ext 3 ppl know which box is thheirs. (thhis works if thhe prisoers have their order)>> <<@moaz4700 says : It sounds like loop strategy odds come from committing of the whole group to accept relating their destinies with the rest of the group, which is beautiful.>> <<@Sigmakitty34 says : To finish the loop, you need your number>> <<@chrino21 says : This obviously goes off the rails at around 8:45. Those calculations are for for those loops in THOSE PARTICULAR orders. There are NUMEROUS different orders that a 99-box loop can be in. Admittedly, the results are the same, but it increases the odds ASTRONOMICALLY higher than what this video is purporting.>> <<@wyclef02 says : If the “direction” of your loop is always the number contained in each box, what happens if the second box that you open contains the number it is labeled with? What box do you go to next?>> <<@nuclearheadache says : Here’s a question: how many prisoners have to succeed before you know for sure you’re in the clear? 50? 51? More or less? I imagine there has to be a point where you’ve definitely established that there can’t be any loops larger than 50 in the set.>> <<@EmmanuelMvula-x2w says : Why would a prisoner follow up the line or legobrick if there is no guarantee of finding it>> <<@SpreadTheGospel-y8k says : I'm not gonna try and convert everyone here, but can you please at least try walking with Jesus for a month or 2? not as a rulebook, but as a relationship. Open a Bible, talk to Him honestly, believe he exist, try to live the way he teaches for a season(bars), and see what changes. You don’t have to live it forever if you rather be anything else, but just try an actual relationship with Jesus, even if it's once.>> <<@toddcollins777 says : what if even one box has the same number slip inside? Say box 5 has the $5 slip inside? Every prisoner will get to this box, and then go to a random other box, creating 100 different loops destroying the 31% chance.>> <<@Heyo-j2l says : While it's correct all the possibilities with how to line 100 boxes is correct. For the other ones such as let's say a loop of 80. (With each number being different and having 100 numbers not 80) It wouldnt be 80! For total possibilities. It would be 100x....x20. Why not take this into consideration?>> <<@ClockworkAE-m2w says : Given this. With this strategy held beforehand, if there are no loops bigger than 50, then its 100% sucess. Else fail>> <<@strawberrypudding1017 says : 0:48 i can't believe im laughing i just didnt expect it at all😭😭😭>> <<@randomvids2679 says : why is he talking about arrangements and loops when the boxes are randomized? doesn't make sense at all>> <<@faithli2131 says : Thank you. I was beginning to panic when the guards explained the rules to us, but black eyed Billy snuck his phone in and I was able to find your video. I’m a free man now.>> <<@Roy_No1YorHater says : coming back to some of these good videos before they were bought>> <<@EeZy4Me1 says : My brain isnt braining in this>> <<@吳聖云-u6p says : 每個盒子跟每個編號都會被使用，並且只使用這麼一次。你用掉跟自己一樣編號的盒子後，只有你自己的編號可以指回這個被用掉的盒子。反之，如果第一手不選擇跟自己一樣編號的盒子，就可能進入不包含自己編號的迴圈。>> <<@Manixen says : If you take your number after you after finding your number, won’t the chance increase?>>