Comments by "XtreeM FaiL" (@XtreeM_FaiL) on "The Riddle That Seems Impossible Even If You Know The Answer" video.

  1.  @triganden  also it is easy to counter. Everyone spend same amount of time inside th rom.
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  2. You. You're wrong.
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  3. Succes rate will be close to zero.
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  4. It is called half the USA.
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  5. They can send info after they have enter the room.
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  6. After the first one has enter, there will be no change of information.
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  7.  @cianmorgan6149  You know what boxes the other 99 prisoners will open, but so what? How is that going help anyone? Every one open 50 random boxes and then the chance to win will be practically zero.
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  8.  @cianmorgan6149  They can't communicate after the 1st went in.
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  9.  @cianmorgan6149  How do they know that? Maybe one prisoner enter/day and spend whole day there. They can't transfer any information.
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  10.  @cianmorgan6149  Plan that require information change when it is not possible will fail.
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  11.  @cianmorgan6149  <If they knew how long person stayd in the room, that is exchance of information.
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  12.  @cianmorgan6149  Short memory? Read my replys again.
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  13. If you don't open box with your number, then you have a chance to be in a wrong loop and you waste your boxes. If there are 10 equal loops, you have 1/9 change pick a wrong loop. Next 10 boxes 1/8 chance to be wrong. 1/7 1/6 Last loop and now you can open only 10 boxes. What would you do? Random pick gives 1/6 chance and picking your own number gives 1/1 chance. But what if the loops were not completely equal. last 6 loops have boxes of 10, 10, 10, 10, 9, 11 and your loop is 11 boxes long. You lost!
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  14. If one fail they all fail.
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  15. You don't pick random loop. You pick your own loop.
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  16. Of course you can accidentally pick a wrong loop. You can accidentally trip, fall and die, but that wasn't the plan.
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  17. You start with your own number. That way you're quarantee to be on the right loop.
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  18. Game over.
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  19. No. You will not jump anywhere. You find your number in 50 boxes or less or you will not find it at all.
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  20. Box 2 can't point to 3 because it is already used. Every box leads to a new box unless a box has its own number. No duplicates allowed because then someone is without a number.
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  21. Because your 2 people chose randomly.
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  22. You're not going to get any better.
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  23. Why you start with a wrong box?
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  24. The problem is that you think that infinity is a number.
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  25.  @miloszforman6270  Yes, you can use infinity and get the answer (~30,7%), but your attempt to calculate "infinity/2" is where you went off track.
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  26. Lenght of loop can be anything between 1 and 100, it does not matter. Odds that your number is in the loop is 100%.
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  27. You re-number the boxes and the chance is the same 31%
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  28.  @glenntwitchtvcommentator2446  1-10 that is one loop, but if you re-number boxes you get 2 loops. I tried pen&paper and I am not so sure anymore what the % will be. If I have two 5 box loops (sure win) and I re-number the boxes. I get one 10 box loop. Hmmm.... You got me really confused now. :D
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  29. It is said on the rules.
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  30.  @Thebart3nder  If there is only one prisoner, then there is only one box and the prisoner can open half a box. Now the game is broken.
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  31. Number is always in the last box.
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  32.  @cianmorgan6149  20 or 40 is the information, but that is easy to counter. Everyone gets out after 40min.
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  33. Troll maths at its finest. :D
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  34. Number of people following the same loop depends lenght of the loop. Loop of 2 has 2 people. Loop of 50 has 50 people.
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  35.  @bhorgc7860  More loops can still exist, but it does not matter since every loop is less than 51.
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  36. And david7384 is the best name.
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  37. How did you tought to get out? Doors are controlled outside.
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  38. How? If he follows the system, then he will not find his number.
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  39.  @godfreypigott  Can only open 50.
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  40.  @godfreypigott  They can only open 50.
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  41. You have no way to know is there a loop bigger than 50 or not. The game is predeterminated, but you don't know the outcome. If you know the longest loop then you will no the outcome too. If there is a loop longer than 50, then the strategy will lead to fail and you can go random, but you will be dead anyway.
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  42. You start at your own number. If you start any other number, then all bets are off. If there are 25 long loops 4 times. You're prisoner#1, you open the box1. You will find your number at the box25. 1>2>3>4...>24>25>1
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  43. He did explain it.
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  44. Either the game is rigged or the prisoner did not pick his loop.
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  45. Yes.
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  46. Rules are very detailed. They are not allow to communicate after the 1st enters the room.
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  47.  @Caribeancrysis  Commucation in any form isn't allowed, which means that you can't transfer any information.
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  48. Because numbers are used only once. Boxes always lead to a new box. Eventually you will get back to 1st.
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  49. You have X Boxes. You open half of them. X/2=50%
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  50. It's a loop of one box.
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  51. It improve your odds.
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  52. If at least 50 has found their number or no loop longer than 50 exist, rest will find theirs too and that will quarantee 100% success. Now if one goes random, they only have 50% chance to win.
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  53. Get lot of boxes, place numbers inside them randomly, open a box, follow the loop, use those numbers to fill lotto coupon, wait, check if you won.
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  54.  @adolforodolfo6929  I never said that it would give you better odds. OP ask how you could use loop method for lotto.
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  55. Their number is always in the loop. The success rate depends is there loop longer than 50.
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  56.  @miloszforman6270  If the 1st find his number at the box#50 he knows that they win. Btw. What did you try to calculate?
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  57. Maxing your odds has no value?
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  58. Opening boxes won't affect the outcome. Either there is loop 51 or more or not. Following the strategy quarantee win if maximum loop is 50.
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  59.  @lalovergel87  0,5*0,5*0,5...100 times Stick with the plan. 31% is still way better. If you open the same 50 boxes, then you are quarantee to fail. You can't fit 100 in the 50.
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  60. The loop ends.
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  61.  @miloszforman6270  But only if it is thurday.
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  62. No. You give boxes their number. 1st box is 1, 2nd box is 2 and so on. Of course it won't work if the boxes are changing places all the time.
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  63. Then you're inside inside somebody's box.
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  64. You leave the room.
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  65. Only you wasted time. Twice. 1. watching the video. 2. Posting a comment.
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  66. Very high probability. There are always some people who can't follow simple rules without messing things up.
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  67. Follow the strategy!
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  68.  @13lacle  How is loop strategy communication. Zero information is passed to others.
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  69.  @13lacle  Loop strategy isn't communication. Everything is preset and you can't change that. It would be the same if everyone decides to open the box#1. No information is passed.
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  70. Loop strategy works every time 31% of the time. Best I can offer to you.
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  71. Everyone pick the same box. It is 50% for whole group.
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  72. You have to leave the room as it was. Besides if the boxes and numbers are fixed, you can't change anything anyway.
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  73.  @brianegendorf2023  The rules say you must leave the room as it was.
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  74.  @brianegendorf2023  0:50 They can't communicate in any way with the other prisoners. The room is exactly as it was after every prisoner. Only information other have is that person in a room opens max. 50 boxes. No one knows what numbers person inside saw and what boxes.
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  75.  @brianegendorf2023  The game is to solve the riddle and get a good chance to win. If the game is fair and the boxes are random, then that strategy gives that chance. Rules quarantee success 31% of the time and everyone wins. People play casino games even when they know that house always wins. Odds will change based on your knowledge.
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  76. You can't transfer any information to others once you have entered the room.
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  77.  @carlosmg9718  Why would the order matters? Numbers are arbituary.
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  78.  @carlosmg9718  But you're still opening boxes randomly. 1st go in and out. 2nd goes in and have no idea what has happened. You can't transfer information.
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  79. They can't transfer information after the first one enter the room.
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  80.  NSW HSC Maths  He must have misplaced them in random boxes.
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  81. Before you say there is a flaw. Maybe you should try to understand things first.
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  82. 2nd prisoner drop your odds to 25%.
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  83. Even if you open every box after you have found your number, it ain't chance thing.
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  84.  @helgewollenhaupt4490  Only way to find a closed loop without your number is not to follow the strategy and that will decrease your odds a lot. One person can turn 100% chance to 50% by going random.
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  85.  @helgewollenhaupt4490  "2) you find a closed loop that does NOT include your number." Is not possible if you follow the strategy. You start with your own box and you are in your loop. You arequarantee to find your number, unless you hit 50+. If you don't start at your box, there is a chance that you're not in your loop and then you will not find your number.
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  86.  @helgewollenhaupt4490  yes, last box aka closing the loop is when you find it.
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  87. Watch the video.
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  88. Did I get this right? Every prisoner will follow the loop strategy, but re-number the boxes differently. Then they will have even worse odds than opening boxes randomly. ,3*,3*,3... is clearly a lot less than ,5*,5,*,5...
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  89. Based on the comment section, it's very slim. There are already hundreds of people who can't follow simple strategy.
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  90. That is a one reason not to trust prisoners. They can't follow simplest rule.
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  91.  @entropie-3622  Isn't it worse than random? 69% of the time there is loop over 50. If there are 2 loops, 40 & 60. Isn't it that you have higher chance to be in the longer loop? 60 bad boxes vs 40 good.
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  92. Everybody would turn to dust before the experiment even start.
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  93. You can only find your number in your loop. Opening a random box is only a waste.
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  94. Stick with the strategy, if you go random, you'l get only 25% chance.
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  95. Only the last sentence is true. First will be true only if he finds his number in the 50th box. If he find his number in the 49th box, odds of winning is good, but not 100%
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  96. How original. Only 200+ people have suggested that before you.
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  97. One person loop strategy is not norma, but on math it is.
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  98. 50/100. Btw. Calculate your strategy again. If you open all the boxes you get over 100% 1/100+...1/1
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  99.  @adolforodolfo6929  You missed to point.
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  100. Then you have found your number in the first box.
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  101. 1>2>3>1 is the same as 3>1>2>3 1>2>3>4<1 ain't the same as 1>3>2>4. 1 can't lead to 2 and 3. and 3 can't lead to 2 and 4. Numbers are used once.
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  102.  @michaelcallahan8412  Once the loop is created numbers don't matter, but the order will. Last box lead to first box, 2nd last lead to last, 3rd last lead to 2nd last....1st box start the loop. If you mix the order, you create different and shorter loops. It is said on the video too.
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  103.  @michaelcallahan8412  Lenght matter, yes. If you create two games then 1>2>3>4>1 & 2>1>3>4>2 are the same. Numbers are arbitary.
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  104. Any kind of communication is forbidden.
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  105.  @ninjanomnomSK  That won't help. Other prisoners have no idea how long previous prisoner were inside the room.
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  106.  @triganden  All 100 is at least 50, so you're not wrong.
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  107. If you can change the rules and manipulate the game. -How to enter the vault when you don't know the combination? -Go trough the wall.
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  108.  @Kg_707_g  Only if there are 2 boxes.
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  109. You're right. Your example is flawed.
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  110. It is 50%. He find his own number in 50 box or less or not find it at all.
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  111. Only if he find it in the 50th box.
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  112. 31% is the expectation value before the game starts. Once the slips are in the boxes value is either 0% or 100%, but you can't know which one. If your loop is exactly 50, then you know that you won. You can't end up in the wrong loop if you start at your own box.
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  113. No one take anything.
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  114.  @JayZoop  There is a 100% chance to pick the correct loop.
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  115.  @jasonzupke1474  Ok, you're right. You have ~50% to pick the correct loop, rest of us can follow simple rule and pick the correct loop every time.
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  116. Anyway you like. It won't help in anyway.
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  117. They have to leave the room as it was.
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  118. Only the 1st box you open can have its own number. You always find your number end of the loop. Only way to be on a wrong loop is if you didn't open your box. You're number 1 and open box number 2. The loop will always comes back to box number 2, but in this case your number 1 might no be in the loop. That is why you open box with your number on it.
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  119. 100 loops mean that Box#1 has slip#1 in it, Box#2 has slip#2, Box#3 has #3...
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  120.  @srijandas8783  Every loop has their own set of numbers. Every number will start their own loop. 1>2>3>4>1 Start where ever you want, you will always come back.
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  121. Of course you can, but why?
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  122. Then you leave.
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  123. 4 prisoners open boxes randomly gets only 6,25%
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  124. What remain boxes and why? You're not going to find your number again.
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  125.  @redpoppy3569  Then you have already opened 50 boxes.
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  126.  @redpoppy3569  Rest zero boxes?
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  127. 0,1*0,1=0,01
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  128.  @sheepriderkiller1181  10%change*10%chance=1%chance.
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  129.  @sheepriderkiller1181  Not sure what you are try to count. If you have 90% chance to not succeed X times, but you do it twice then the you have done it 100% succest rate.
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  130.  @sheepriderkiller1181  Flip a coin. You have 50% chance to get 2 same sides in a row. 25% to get 3. 12,5% to get 4. If you have 10 sided die, you have 10% chance to get 1. To get two ones in a row you only have 1% chance. Any double is the same 10% Any doubles twice in a row is 1%
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  131. If half win then the other half win too.
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  132.  @fabiocunha8509  so what? 31%>,00000000000000000008%
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  133.  @fabiocunha8509  Random will lose vs loop system right after you have 2 persons.
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  134.  @fabiocunha8509  "what if they know of the strategy and make sure to make a loop of 51 to guarantee one fail if they follow the strategy. " You re-number the boxes and now you have the same 31% again.
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  135. 0%
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  136. If you start at random box, you lower your odds drastically.
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  137. Everyone pass or none shall pass.
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  138.  NSW HSC Maths  I just explained the rules.
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  139. If you find your number in the 1st box. Opening 49 more won't change anything.
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  140.  @NoOne-yv2ei  Watch the video again.
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  141. 2 people have only 25% if they go random. Possible combinations are. 1-1 1-2 2-1 2-2 1-1 & 2-2 are auto loss. 1-2 & 2-1 gives 50% (0%+0%+50%+50%)/4
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  142.  @joekemp83  That is not what he said. "if both Prisoners chooses a random box, half the time they will loose because they randomly chose the same box. " Strategy gives 50% chance, but random only 25%.
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  143.  @joekemp83  Yes. He said both the random and the strategy gives 50%, which isn't true.
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  144.  @dope414  Try revelation.
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  145. Same as usually when people don't follow strategy. What happens if you are aggreed to meet somebody tomorrow noon, but you skip it?
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  146.  @Bomber-Trebor  Yes I did. Did you watch the video?
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  147.  @Bomber-Trebor  Strategy only works if everyone follows it. Soon as someone desides to go rogue it decrease the odds. He could turn 100% chance to win into a 50% chance. In your OP one guy didn't follow the strategy. Either he is too stupid to understand simple system or he does not want to win.
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  148. I am not sure, but the numbers should not matter. You always find your own number at the last box. If you don't open your box, then there is a chance that you will not find your number.
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  149. Not possible.
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  150. Stick with the plan. Either you're in the winning loop or not. You still (should) have the same 31% chance what you had in the beginning. (No idea how that works if it work) If you chose to ignore the last box, you have 2% change to find your number. At the start you had 50% if you go random. At least 47 other prisoners have the same problem.
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  151.  @niklask7635  You have $0. You take $1000 loan. Now you have 0$
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  152.  @joekemp83  Things go weird if only one person plays the game. ;)
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  153.  @thetaomegatheta  Transfering information IS communication. You can leave after 10 min. if you want, but you will go into different room. No one knows how long you were in.
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  154.  @thetaomegatheta  They can't do that.
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  155. Because you start the loop.
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  156.  @entropie-3622  There is a change to get results over 70% at lest once if you run simulations many, many, very many times.
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  157.  @entropie-3622  If you constantly get 70% then it's a clear sign that the code is wrong.
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  158. Numbers can only exist once. 73 and 50 can't point the same 50.
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  159.  @Kg_707_g  How so? 99 boxes can be randomized.
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  160. Creating two different groups will not increase their chance.
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  161. I wonder who that person is who failed?
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  162. If you start from random box then it will be the same as open all boxes randomly.
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  163. The loop ends.
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  164. *31%
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  165. Every person who do not follow the strategy lowers their chance by half.
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  166. It is a loop of one box.
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  167. Think of it like this. At the start all the boxes are unlabeled. You open a box, there is a number, but does not lead you anywhere. You can pick any box out of 99. Eventually you will close the loop, but the others have noway to know which box is which, but if you numbered all boxes now each box will lead to only one box.
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  168.  @marcinpietrzak9358  Wrong. Loop isn't random, the lenght of the loop is. What my number is does not matter either. It is one between 1 and 100. If it is 1, I pick box1. It it is 5, I pick box5.
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  169. If you can't follow simple rules, then you doomed everybody.
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  170. 50% if they use the strategy. 25% if they choose random.
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  171. Only their own number matter.
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  172. How many boxes you need to open to get over 100%?
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  173. How is that going to change anything even if they could communicate? Everybody in the loop will find their number or none of the will find.
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  174.  @IMA321  Of course you're not start at 15 when it's not your number. You obviously didn't understand the system. You better watch the video again.
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  175. Why?
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  176.  @kartik180rajesh1  Yes, but "Just start a new loop with a random number again" What's the point?
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  177. Sure why not. All you need to know is the winning numbers.
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  178. Next time just follow the simple rules. Don't improvise!
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  179. Even if they could get back in line and enter the room second time, they can't find their number because they have already opened 50 boxes on the first run.
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  180. Then you didn't follow the strategy. If you follow the strategy, you will always be on the loop.
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  181.  @FuSiOn_33301  ...and the 1st box you choose is your number and 25th box is your real number. If you randomly pick 1st box, then you have 3/4 change to be on a wrong loop.
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  182. calculate again. Black Jack is the only beatable casino game. Every other game is a sure lose.
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  183. You can, but it won't help at all.
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  184.  @joekemp83  You put all the lotto numbers in boxes and start opening them. It won't help you like I said, but you can use the strategy. :P
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  185.  @joekemp83  Doesn't matter. Ex. Our local lotto we choose 7 numbers out of 40. So open a box and follow the loop7.
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  186. How did you come up that number?
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  187.  @Wildcard71  That means what?
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  188. It is said on the video.
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  189. You don't find your number in random loop (most of the time). After you have found your number, you can open more boxes, but that won't change anything.
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  190. The job is to find your own number, not to open every box. Naturally you don't want to find your own number that quick.
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  191. If your box contain your number that is isolated loop, lenght of one.
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  192.  @sergiosarmiento1881  Is it? If there are 4 25 long loops. That is 25% chance hit the right loop at first random try. 75% chance to fail. Now you have 25 boxes left and 33,33...% chance to hit the right loop. 75%+66,66...%/2=70% fail. Or should I add them? 75%*66%=~50% Why the odds look very much like the original strategy?
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  193. SImples way to cheat is just tell that you find your number.
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  194. That is not the dumbest comment, but thanks for playing. 0/10
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  195. How much is 1/100 + 1/1?
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  196. How do you use expired milk on your everyday life?
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  197. Just because one loop is 51 does not mean there there will be only one another loop. 51+48+1 51+47+2 51+47+1+1 Only thing that matters is what is the longest loop. If it is over 50,then it is game over.
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  198.  @Kass1231999  All of the prisoners must get their number or they all lose. It would be a total failure.
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  199. What do you mean "proper scientist"? A guy in a white coat?
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  200.  @altos83  It does not have to have less than 50. There can be two loops of 50. As long as there is no loop longer than 50, it's a win.
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  201. 31% survival chance.
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  202. The more boxes you open, the better chances you have unless you can't find your number in the 50th box, then you will lose.
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  203. If you're a number, then you're in a box (or on a box) and somebody comes to find you.
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  204. If there is a loop longer than 50, then it is game over. It does not have to be the first person's loop.
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  205. Yes they can.
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  206. If you can open more than half then you get better change.
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  207. 100% chance.
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  208. With only 2 people the chance is 50% and it goes down from there.
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  209.  @veritasium  What baffles me is, how simple and easy to follow the answer is. It works just like the Clock patiance card game.
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  210. If there is a loop of 51 or more then they lose.
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  211.  @ShannonJacobs0  You can open maximum of 50 boxes and if your number is not there you did not find it and you lose. Knowing that it will be in one of the other 50 boxes won't help you at all. If you know (which you don't) that the longest loop is over 50, then you don't pick your box.
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  212. Ten he have made a mistake.
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  213.  @eagledove9  Start will matter. If the loop is 1>2>3>4-1, but you start at 5 you will not find your number. If you hit your loop then it does not matter, but only way to be sure is to start at your number.
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  214.  @eagledove9  Is there a better system? Idk. Video didn't tell it either. Odds are based on your knowledge. You know you have 31% chance if the boxes are random and you stick with the strategy. If you know that the longest loop is 50 or less. You will get 100% chance if you stick with the strategy. If you know that the longest loop is over 50, then using the strategy give 0% chance. In that case you increase your odds if you not choose your number box.
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  215.  @eagledove9  I tried pen&paper too with 4 boxes too and got 66%, but that can't be right. 2 boxes gives 50%. That's a fact, so 4 can't give better. Looks like a shortcut wasn't the right path. :P
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  216. That's not a loop. That is 3 differents loop starts.
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  217.  @MrSummitville  They can, but they still are starts. 1>2>3>1 Has 3 starting point and it is a loop of 3 boxes. Btw. Loop is always closed.
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  218.  @hugopfeffer4175  Box3 isn't 1, it is 20 and box20 is unknown before you open it.
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  219.  @godfreypigott  To make things clear to everyone, just call them muricans.
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  220. Yes.
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  221. Simplify the problem and use 2 boxes.
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  222. Walking backwards doesn't help a bit.
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  223. They can't communicate in any way.
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  224. If there are no loop over 50, you always win.
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  225. It is not correct.
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  226. All that matter is the maximum loop lenght. If it is 50 or less then it is awin.
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  227. 50/100
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  228.  @sergiosarmiento1881  It is a bad think, because it left high chance of loop over 50 boxes.
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  229. It's never greater than 50% and than require max. one player.
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  230. It has been used in computer science.
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  231. 1/100+1/1 is already over 1.
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  232. 69% change that your battery will die.
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  233. With 2 people you have 50% chance. You get 25% if you go random.
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  234. That will be 0% chance.
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  235. Re-think what you wrote.
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  236. Chance to win increase slightly.
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  237.  @taegrr_yt  Use 2 boxes only.
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  238. Random gives 50% to a one person. 25% to a two. 12,5% to a three. ...
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  239. No. If execution starts right after the first fail, then everybody dies.
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  240. Loop1+loop2+loop3...=100
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