I actually use this to find which disc is in the correct box when playing video games

I voted no, because no matter how good you think your strategy is, it only takes one idiot to screw it up. The odds of there being no idiots in a group of 100 people is greater than the return if you win.

This is the same method I use to find my videogames when I've put them all in the wrong cases. Open Mario Sunshine, find Luigi's Mansion. Open Luigi's Mansion, find Double Dash. Open Double Dash, find Sonic Heroes. Open Sonic Heroes, finally find Marion Sunshine!

people always be like: next, number 26? A:here B:no I'm 26 C:oh I thought I was number 26

You're still going to lose every time because at least 1 of those hundred people will find a way to fuck it up.

Good luck convincing 99 other people to listen to you.

0:37 Me explaining to my parents how my marks are not zero

Naw, fuck that. I'll just keep my dollar.

My mind was legitimately blown. Once you said it, it made so much sense. I had a feeling there was gonna be a trick to this :P

If I get to choose the other 99 people, I might take this bet.

The ones who asked 100 people to bet could've just made a 100 box chain.

but what are the chances that nobody will fuck up and ruin the whole thing? 0.0000000000000000008% of course.

I know a friend who follows this channel but doesn't know I do too. I'm pretty sure he'll ask me to solve this problem tomorrow. Perfect opportunity to "prove" that I'm smarter than him. I just hope he doesn't read the comments.

There is a 100% chance of winning: When it's your turn to find the bill, start from the first perfect square number 1, then 4, then 9... Once you got to 100, turn around and go back to 1 again. This time take the route of natural counting integers: 1, 2, 3, 4, 5... Everytime you open a box, take the dollar bill and put in your pocket, by the end you will have 100 dollars. 100 minus the 1 dollar you paid is 99 dollars total profit. But you might not be satisfied since that if you win the game properly you will earn 101 dollars which is 100 dollar profit. Don't worry. On your way out of the room, beat up the first person you see and he/she will most likely have more than 1 dollar. Take it and run home.

Knowing my luck, my chain will be in the higher half .

Good presentation. The nature of applied probability is so counter-intuitive and fascinating. So it's a good bet to take if you know the strategy.

ok, lets calculate odds that someone will mess up this stragety :D

Even if our odds to win aren't good, I would still pay the dollar because it sounds like it would be a fun game lol.

Too bad this game isn't at casinos.

TedEd already explained this to me once, And I have to say your channel is incredible. I LOVE Physics its basically my life, So thanks for explaining a few of these

best odds= just grab the dollar from each box and leg it

I clicked both Yes and No. -Achievement Unlocked- -25G - Adventurer- #pcmasterrace

I've heard this so many times before but just now am I getting. The diagrams that you drew helped significantly.

Egg

That reminds me of how people will solve a Rubik's Cube blindfolded.

Well, the hardest thing would be to explain to 99 strangers how to co-operate and carry out the same plan.

I have a strategy with 49%(approx) success probability So there are 100 prisoners Before going into the room They can strategize in this way 1)They all will go into the room in the order of their labelled numbers. 2)So, the prisoner labelled 1 will go first 3)He will open 50 boxes out of 100, there is a 50% chance he will find his box 4) Now, he will divide room in 2 parts, the first part will have first group of boxes which will be scattered 50 boxes which he didn't open, the other part will be 49 boxes which he opened and he will arrange the boxes in increasing order according to slips he found. Example: 3,5,6,9 .....(total 49 boxes) 5)Now the next prisoner labelled 2 will go into the room. He would see two groups of boxes, one scattered and one arranged. He will first check the group of arranged boxes, since the boxes were in increasing order, he will only check the first box in order A)If he found box with slip 2, he can happily take his box and get out of the room. B)If the first box in row doesn't have slip 2, it will have 3 or 4 or any other (since boxes were in increasing order). So, now he know that his box is not in the first group among the 49 boxes, then he will check the other group of boxes, and open 49 out of 50 boxes and will find his slip (49/50 probability). In the meanwhile if he found, number 3 box, he will put it with the arranged boxes at starting (since they were in increasing order) 5)Now, the third prisoner will go and repeat the same process A)First of all, take the first box among the arranged ones, if it is his box, take it and leave B)If the box is not his, then he will search the boxes among the scattered boxes (which can maximum be 50), he already opened 1 box, now he can open all the 49 boxes and hence, there is again 49/50 chances he will get his box. Also, he may find box number 4 in scattered ones (if box 4 was not in the arranged ones) which he should put in order in arranged boxes. Other points to be noted: 1)Boxes are to be arranged in increasing order in a row, but there are two sides of a row, so the prisoners can plan that the box with smallest number will be one near the door (the box from which the row starts) 2)No prisoner should do any unnecessary movement of boxes which might shuffle them. Please like if you like my answer👍🏻

The group: all listening to this one guy telling everyone how to win Me: I like this box cause it says 69 oh look a dollar

I didn't pick yes because it was more probable, i picked it because it was one dollar. if there's a chance to win something and i have spare change in my pocket, ill give it a shot. The difference between this and casino gambling is where you stop. Buying a 25 cent lottery ticket one day out of the year is a lot different from going to a poker game with $2000 ready to gamble with.

You should NOT take this bet. It began with him saying ME and 99 others will take a dollar note. Then, at the end he said I will give you 101 dollars each, meaning he is out a lot of money. So, he would just purposefully lose.

I've seen this riddle tons of times before, but this is the first video that really explained how it worked well enough.

Can't believe !!! Same video now can electro-boom as Veritesium made it too. Greatest work dude.

What about this way? If there are 100 boxes and you can talk to the other people before entering the room with the boxes then when the first person enters the room he will look through the first 50 boxes from the top row. If he manages to find the second persons number along with his then he stays in the room for more than 15 minutes. The second person will check his watch, if he is made to to wait more than 15 minutes minutes then he will check the first 50 boxes and if not he will peek through the other 50 boxes and this will continue. If the boxes are not moved after each person enters the room this trick has 50% chance as only the first person does not no where his number is.

My idea was, assuming they go in the order of their numbers, that each person has a given time frame to go through the boxes (let's say, 2 seconds for each peek= 1:20 minutes per person). Not only will they look for their own, but the one that comes after them. If they do find it, they will let the other know by taking longer to finish (let's say twice the time). So, when the first person goes in and they're given 2 minutes to go through the boxes, they will go through the first half of boxes. If they find the dollar bill of the second person in the half, they will wait an extra two minutes. If they don't, they will take 2 minutes, and the second person will know to look for his bill in the other half. They also repeat the same process for the thrid person. The first one has a 50% percent chance of succeeding, but if he does, the rest have a 100% percent chance.

A remarkable puzzle. I considered cycles at some point, but didn't think it would be fruitful, so I didn't pursue that avenue. Instead I wrote a script to generate all possible strategies for N=4 (instead of N=100) to prove that it was impossible to win the wager in principle (at least for N=4). And voila, the best strategy out of them was the one described in the video and it was working very well. The remaining part was easy (figuring out why and computing the probability of a permutation not having a cycle of length > 50). I wish I was able to solve it without a computer, though.

I heard this before and didn’t really get it until this video. Well explained 👏

Since it only costs me 1 dollar, and I could end up with 101 dollars, why not take the bet? Even if it only works 31% of the time, I would only lose a dollar, with the potential to win so much more.

this is so useless, yet so interesting.

This is the first video that explained this in a way I get. Thankyou.

What is the probability of 100 random people actually understanding the tactic and executing it? Infinitesimal. i.e Never take the bet. Even if you know the correct strategy.

Well it is only 1 dollar and u get 100 dollars. Odds of 30% are good

What if he rigs it, sneaks an audio recorder in your planning place, and arranges it so one person screws up. Ex: with this strategy, if your number is 1, he puts 100 in box 1, 99 in box 100, 98 in box 99, 97 in box 96, and so on, where you're stopped at 51 ( If you get the pattern, 1 would be in box 2 ). This strategy can be sabotaged easily by an audio recorder. Best chances are: find the room where he keeps the $10,000 (100 dollars times 100 people is $10,000) and steal his money. You get more profit. Plus, if the people get mad at you, just give them $50 each and say they probably would've lost by one person messing the thing up anyways.

I would probably sign up for this just for the fun of it. It sounds awesome.

I remember I used to like all areas of mathematics except probability, but when I saw this video I instantly fell in love with it.

I use this to find my video games since Idgaf where I put my games XD

There is a major inconsistency in the bet. In the previous video it was said the room was reset after each "peek in 50 boxes" trial. And that is for me once again randomly putting bills in boxes (that are not numbered nor placed in any specific order; you said they're identical so no numbers are allowed on them). So - for me - two things needs clarification: 1. The boxes are somehow numbered or placed ordered (good luck with blank boxes placed randomly in the room and picking up the one with "your number") 2. The bills stay in the same box (it's clearer than "boxes stay in same place").

I never thought someone would come up with this sort of bet but I think I did came up with this sort of strategy for something else that may or may not be similar before. It's called following a pattern and yea, it's a good strategy.

That is so awesomely clever, I'm stunned

I know I'll lose, but's only a 1$ and the experiment is worth it.

lol that's what I do when CDs are put In the wrong cases

Basically, he used the concept of linked list

I actually found this one out before watching this! I'm so proud of myself.

That is astounding. Kudos.

yep i get it .. no

my friend did a scavenger hunt like this in their textbook before. It said go to page __ than that page would say go to page __ etc

This is in a ted-Ed video as well. Check out the Prisoner Box Riddle. It’s tweaked with the same idea though.

About a 30% chance of winning. That means that there is a much higher chance of losing. I would have to argue that while the chances are still better than guessing, it is not worth losing the dollar. Many justify taking this risk as it is only a dollar but the dollar is just an arbitrary amount. That amount can be as large as the person presenting the problem wants it to be therefore it is not worth the risk of losing x amount with a 30% chance of winning 101x. Also this is not worth the risk as the group only gets one chance to succeed. If there were more than one chance then it would be worth doing but it is only done once and the odds are not in the favor of the group.

If you find your bill you have to leave it in the box. So, what's stopping everyone to say they found their bill, even if they didn't?

this is also assuming people don't remove their bill when they find theirs.

You know, the odds are better than than he states, given he doesn't say the boxes get rearranged after each person nor that you cannot share the information you got by going through the chain of boxes.

Even if we don't win, I'd do it. Lose $1 but have a change of winning $100? Sure.

This is the experiment by his technique. Run it repeatedly as an R script. After just 106 trials of the experiment the mean reaches ~0.3113 and should only get closer to .3118 as you continue. The mean should be calculated as the mean of a large sample of X where X=0 if the table output shows 100 "true" values and X=0 if the table output shows anything else. boxes

that is so awesome! thanks alot for that one!

I never would have thought of that but that is very smart!

Good luck trying to get 99 random people to unanimously agree with your idea.

That's a rather ingenious solution. Math, how does it work? lol

This was on TED-ED once, but it was a band and they had to get their instruments out of the boxes.

I love how this solution video has 500k more views then the first video that was actually asking the question.

I'm ok with betting $1 with the chance of winning $100...

"Room is then reset before the next person comes in" You probably should have made it more clear what you meant.

Well this was very cool! I'll have to remember when I'm in this scenario myself

How are the probabilities of the chain lengths calculated? Because if you solely calculate the probability that a chain of longer (or shorter) than 50 exists and base your chances of succeeding on that, you do not take into account the influence of your starting position inside said chain. In other words, if you have a chain of 100 and you start at 'spot' 51 then you will still reach your bill despite it being longer than 50 boxes. In this case the total probability will be considerably higher than 31%. But if the 31% is not calculated based on total chain length but based on the probability of the remaining steps until your bill being smaller than 51, then you are correct.

Absolutely brilliant. I certainly love mental math puzzles like that.

Yeah but you asked if I should take the bet, not if it would be likely to win, so because there's nothing to lose yeah I'll take it

yup. Saw this in a riddle book with 10 musicians and instruments... same concept

The setup video did not mention that there would be numbers on the boxes or a discernible order to the boxes in the room. Without that ordering you can't perform this algorithm.

please explain this: Lets say no chain has length greater than 50 - which happens with prob=31% The prisoner still has to choose the right chain, that contains his number. Lets say there are 3 chains of length 45, 15, 40 : The prob of choosing the right chain will vary depending on which chain - therefore each prisoner does not have a mere 31% of succes if the strategy is followed.

$101 if you everyone finds their own bill... one at a time... How long would this take exactly? I'd rather pay $101 to not do this.

Confused as to how this works. So basically you’ve changed each persons individual odds from 50 percent to 69 percent, which is only a slight improvement, because it’s still 69/100 to the power of a 100 - which is still incredibly low. Only takes one person out of a hundred to not find their note, and that’s still bound to happen.

Ted Ed riddles had one like this one except they had musical instruments and 10 boxes instead of 100. From which you could pick only 5.

never put your faith in 99 other people At least one if gonna fuck up

I am sitting here and scratching my head .. How exactly does this work ? WHAT ???? Can someone explain in more detail why this is working ???

I thought about another solution: We can’t communicate with other players by words, but we can communicate them by time. Like, if we’d set “room preparing” process as a fixed number, for example “x”, than we could create a strategy which uses time as a basis communicator (1 box per 10 mins). For example, if we know, that mister 27 spent 40 minutes to find his number, we subtract the x (for example 10 mins), and can tell for sure the box number “3” on the list (any order group chooses) contains bill number 27. For the last half of group chance of winning is 100%. For other half is increasing by 1% each session. I think it is counted by factorials and shit, but you got the point.

This explanation made a lot more sense than the other one. It all has to do with chains. As long as no chain is above 50 you all win, else many will not find their bills.

Did he explain how he worked out that about 31% of the chance that there is no chain > 50?

I have another riddle that you could possibly make a video for There are 100 prisoners in a jail. The warden tells them that the next day they will be brought into a room with no mirrors They will line up one behind another and given hats. Each hat will be either black or white, but it will be completely random who gets which hat. They can't see the hat on your own head. The warden will then go from the back of the line to the front asking them which color hat they have. If a prisoner says the correct color they are free, but if they say the incorrect color they will be killed. The prisoners are given all night to plan and they come up with a plan so that 99 of them are guaranteed to survive. What's the plan?

My main worry isn't that there is a 70% chance of losing, its the fact that someone will ignore the strategy and just start randomly opening boxes when they go in.

Here is a simpler explanation: But first you need to understand that a "chain" consists of imaginary links we create by letting the bill inside the box lead us to the next box. There are ONLY two scenarios when you use this strategy. 1- (31%): No chains longer than 50; but rather multiple smaller chains [all of which loop back in less than 50 steps]; meaning that the moment you enter that door and choose the box with your number on it, you automatically picked the chain that contains your bill, and so did all 100 participants. 2- (69%): There contains a chain longer than 50; in this case you WILL lose because all those bills inside that long chain will never be found [because you're only allowed to search 50 boxes]. Now the whole premise of this is that a 31% of guaranteed win [despite seeming low], is infinity better than (1/2)^100=0.000...008%

Watched this video when it first came out, and now here I am watching it after I have taken abstract algebra. It makes so much sense!

The correct answer is NO, because there is exactly 0% chance that 100 people will all follow a coordinated strategy perfectly.

this explains it much better than the ted ed riddle video

From the perspective of the group, you're participating in an activity where the amount of risk you must put in is just one (presumably US) dollar, so of course it would be worth taking the bet. The one orchestrating the whole activity is putting a lot on this, with a risk of 100^2 + 100 dollars versus a reward of just 100 dollars.

Also... it's like, $1.... For what is sometimes pocket change, I have the chance of $100... Even if the probability is essentially 0, and I knew that going in, I would STILL take it... I mean, not that *I* ever do this, but that is essentially what people do with the lottery...

Holy shit, I would have never thought of that xD

the same matrix is used to decode passwords of whatever kind. just replace the boxes with letters and numbers. also it is used in logistics. where to go first and next and so on and even with conditions to comply.

Thinking about the problem, came up with communicating even more information to the persons waiting for their turn. You can combine my strategy with the one that is presented in the video, but some things must be allowed for my strategy to work and it is VERY time consuming too (which is ok, if it improves the odds). So here is how it works. First of all the 100 people must have a very good feeling for time and second they all have to see each other while waiting. Before they started, they all made sure, that they count the boxes in the same way (e.g. if it's a square, the first one is the nearest on the left and they count them going right, one up, left, one up and so on). Let's say using the technique of the video, the first person finds his dollar in the 34th box with respect to the counting strategy i mentioned. Then the first person makes sure, he stays in the room 34 minutes in total and then leaves the room, for the next person to take the turn. Each person in the waiting room counted as well, now knowing that the 34th box does not contain their dollar. This gives the second person one less box to choose from. The counting and staying in the room is repeated for each turn, revealing one more box to be not the right one each turn. Ideally the last person picks his dollar with his first guess.

how would we know if the gamemaster didnt arrange the boxes so that it is a complete chain of 100 boxes lol

I have a better plan to win at this All you have to do is take the 50 dollar bills you get to check and split :D As long as the other people don't know your name or adress, you are 49 dollars up 100% of the time :D

This is a good solution to a problem which, if stated differently, will help recover lost computer memory effectively once it has leaked because the computer always has a finite (if not fixed) amount of RAM. Think of the people as the apps that let go of memory carelessly. The memory is of course the money because we usually count it as the biggest deal breaker when buying a PC.

This is amazing! Wow! Mind blown!