The Poisoned Wine Problem
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- Опубликовано: 27 авг 2024
- You have 1000 bottles of wine, but one is poisoned! With only ten... test subjects, how do you guarantee which bottle is poisoned?
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This episode of Scam School was made by:
Brian Brushwood
Brandt Hughes
Bryce Castillo
Jeff Schuessler
Shot on location at The Rusty Mule in Austin, TX 
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Why not just make all the advisors drink their own wine
Agreed. Fuck math
It takes two days for the poison to kill you!
But they have to show up to give you the wine 3 days early for your prisoners to test.
The bottles arrive, but very few parcels are hand-delivered by the people sending them
I guess didn't want 999 guests to know someone was trying to poison him.
I wasn't thinking binary. I was thinking a matrix. Divide the prisoners by 2 and the wines by 5, then shift the samples and batch one place in the rotation every 6 hours. You'd have just enough time to find the bottle based on the order in which the 8 prisoners died.
I like this method more.
+
This method is invalid if you consider poison acts differently in different people, taking more or less time to kill. Specially being a slow-acting poison as the scenario depicts. For something that takes about 48 hours to kill, 6 seems like a small window to work with.
A slow acting poison would be something that your body metabolizes into a toxic substance. The speed would be determined by the rate at which the toxin makes it to the liver. Since everyone has about the same rate of flow for filtering blood, it needs a far smaller window than one would expect.
However, this was only a quick first thought and the binary thing is way more efficient and accurate.
I'm glad you mentioned the metabolism factor. I would like to point out that genetics can and do cause significant variation in how fast different people metabolises chemicals-- it's not as simple as the blood flow to the liver (which btw can vary based on other factors).
In more detail, people can be categorised as slow, ultrarapid, and intermediate metabolisers based on how much of the relevant enzyme needed to process the chemical substance is getting produced by the body. To give a perspective on how significant this effect can be, the exact same amount of drug can be an overdose, an underdose, and a proper dose in these three types of people.
This is still one of my favorite puzzles from scam school... I use it on friends when we go drinking, and even assign it as a problem for students in my classes.
Sweet! I really dig it, too.
You can illustrate this really well by simplifying to the case of 2 prisoners with 4 bottles. Label the bottles 00, 01, 10, 11. Prisoner A tastes 01 and 11, prisoner B tastes 10 and 11. There are 4 outcomes (A, B, both, or neither die) each of which corresponds to exactly 1 bottle.
B live B die
A live 00 10
A die 01 11
For 3 prisoners you can do the same with up to 8 bottles, and so on
Why don't they just tell them to drink the one they brought then u know who poised it
Xanderzone 1504 OMG
Xanderzone 1504 that's what the fuck I thought the solution was??
Xanderzone 1504 because it takes two days to poison them. They would be dead long after the party was over
Lascactus Smith ya but no one innocent would die
because there will be people who pass out after half a bottle...and others complaining why they can't have more than one bottle
Ahh I'm here first I have to be funny
1000 bottles of wine on the wall, 1000 bottles of wine, take one down, pass it around, fuck it's the poisoned one
drakoz254 o
White tiger from CGPGrey's STV/AV videos? You better not be trying to reinstate the FPTP voting system.
Nobody ever wants to read someone saying "first".
you are the first person who is funny while claiming to be first.
you get my upvote good sir.
drakoz254 ljookj
Im actually so proud to have figured this one out, once Brian gave the 1024 clue I was so surprised his programmer friend didn't get it as it made the binary link even more obvious! Maybe it's studying systems and computer engineering that helped me get this one so quick not often I solve a scam school puzzle first try
"Did you know I have been drinking for the past seven hours?" - Jason Murphy, 2017
I managed to get the 2^10 part but binary never even crossed my mind. I like that problem
You don't have to know binary. As long as you mix the wines so that wine from each bottle is served to a unique combination of prisoners, the poisonous wine can be identified by which prisoners die. Counting in binary is simply a cool way of ordering the various unique combinations.
Rick MacKenzie but the prisoners die in 2 days time. If you have access to ALL 1000 bottles of wine presumably the party has already started and you have 1000 people expecting wine. So tell me please how is this solution going to give you an answer there and then so you can serve wine at your party. Because their solution you’d have to wait 2 days for the prisoner to die and point out the poisoned bottle.
As soon as he said 1024 I thought "Binary"
I've watched this 60 times with the time travel trick you taught in episode 1273!
Ethan Noumenon what episode was that?
It doesn't matter, since he went back in time which caused Brian not to release that episode.
"simple act of time travel" lmao what's simple about time travel?
1273! is a really damn big number
Steins;Gate. . .
For some reason I laughed at the intro, I just imagine Brian running up to people's windows yelling random things at them. XD
He said "if you get one drop of this poison, 2 days later you immediately drop dead. So all you have to do is decide to the wine and give one bottle to each prisoner every 10 minutes, then wait 2 days for someone to drop dead, and you can say " well that died at 10:37, so the wine they had 2 days ago at 10:37 was poisoned"
make all the advisors drink a sip of their own wine, and then imprison all of them for 3 days until one dies......
so I simply cannot drink the wine in front of me
Truly you have a dizzying intellect
but i know you know that I know that I simply cannot drink the wine in front of me... so I surely cant drink the wine in front of you!
I actually paused several times to try to figure it out. After dividing it by eleven, I realized that it didn't matter how many prisoners it killed. But then I got stuck on linearly overlapping so even at 1024 and I realized it was 2^10 I still couldn't figure out how I was going to overlap it.
I like this one though. It's easy to remember.
The real fun here is to get someone to tell the riddle and then watch the awkward moment when they explain the solution. Tell it to your friends if you want them to bug you the entire night
Was thinking that maybe each bottle could be labelled 1-1000 then each bottle could be given to a different combination of prisoners, then by similar logic the poisoned one can be deduced based on which combination of prisoner dies
1000 bottles of wine on the wall, 1000 bottles of wine, take one down, pass it around, 999 bottles of wine on the wall.
If you do this there would be no wine left because you would have every prisoner drink a bottle 9 times, and there's 1,000 bottles so not only would you run out of wine, they would die from alcohol poisoning
only a single drop per bottle.
@@ghoulishgam3r508 But there is supposed to be enough wine for 1000 guests.
@@nerychristian they only drink a sip, they don't drink a whole bottle.
As a variation you can actually make your way through at least 59,049 bottles and find the 1 poison one among them with 10 prisoners if you had a month instead of 2 days. [Assuming the same 2 day wait time on the death]
Explanation below...
Take all of the bottle and divide them into 3 groups. Group A, B, and C respectively. Take 2 prisoners Give 1 a taste from all of the A bottles, and give the other a taste from all of the B bottles. Then u wait. If the 1st dude dies, it's in group A, of the 2nd Dude dies it's in B and if no one dies it's in C. Wash. Rinse. Repeat. Worst case scenario someone dies in every heat so you can make it through 1 x 3¹⁰ bottles or 59,049 bottles!
Your method can do far better. You're clearly going pure math, since you're ignoring the impracticality of feeding someone 20,000 drops of wine. So, just divide the wine into n+1 equal groups on each pass (11 for 10 prisoners). After 2 days you know which group contains the poisoned bottle. You you have a small chance of not losing a prisoner, and that improves every time you one dies. However, you get at least 10 days which is enough to process a hair under 40 million bottles (39,916,800 / 11 / 10 / 9 ... / 2 = 1). Best case, you can process infinite wine, given that no-one ever gets poisoned.
I think there's another solution given one modification to the problem: You have infinite time. Give each prisoner a drop from 100 bottles. Wait two days and you've narrowed down which 100 contain poison. Give each of the remaining 9 prisoners drops out of 10 bottles. Narrow down to ten (if no one dies, it was the 10 no one drank). Out of the survivors, pick 5 to get 2 drops each to narrow it to two. Pick your least favorite prisoners to each get one drop out of the last two bottles in order to narrow it down to one. Problem solved in 8 days with (at most) 4 prisoner casualties.
That's certainly a practical solution, but you can make it simpler instead: Divide the bottles in 2 equal groups (let's say groups A and B), give sample of all the bottles of the A group to the 1st prisoner. Divide each of the 2 groups into 2 groups (4 total, let's name them A1, A2, B1, B2) and take A1 and B1 and give samples of both to the next prisoner. Divide the 4 groups into 2 again (8 total: A11, A12, A21, A22, B11, B12, B21, B22) and give samples of A11, A21, B11, B21 to the 3rd prisoner. Repeat until the last prisoner: Each time you divide the groups, the number of groups is doubled, and at the end you will have 1024 groups.
Then, when the poisoned prisoners die: If the 1st prisoner die, the poisoned bottle belongs to group A. If the 2nd prisoner lives, the poison belongs to A2 etc etc. and at the end you will have the poisoned bottle.
Both solutions follow similar principles. The solution with binaries is better and more practical, but it's quite hard for someone to think especially if he doesn't have knowledge of binaries. My answer is lengthier but otherwise simpler.
(computer science thing, with explanations for those that are clueless)
In computers, all systems send 1s and 0s. These aren't always carried perfect, especially if, say, between the hard drive and motherboard. Thus, there are these things called "parity bits", a few extra bits that are sent within the data that help check that the data is correct at the receiving end. It relies on basic logical principles, and works on a similar kind of principle, give or take. This is a thing i remember from computer architecture, and not all schools teach the same class the same, so...
Parity bits are also used in data security against device failure, but that's for later
You could also give the first prisoner 500 sips of wine, and the second prisoner 250 sips of wine, and on like that until the last prisoner. Each time a prisoner dies or lives the number of bottles in question is cut in half.
Having not watched the whole episode or read the comments, here's my solution. Lay out all of the wine in a grid, have one prisoner drink a drop of wine from each bottle (you can use 5 of them to speed up the process) along one axis at set intervals until all samples have been had, do the same on the other axis with your remaining prisoners. Make sure you denote at what time they drank what bottle and keep them isolated on which line they're on, so when 2 of them die it will tell you exactly which bottle it is and you only lose 2 drops from each bottle and 2 prisoners.
There is a way to do this, similar to the given solution. The key to the solution is the relationship between the number of prisoners and the number of bottles. 1000, as in the number of bottles, is 10, the number of prisoners, to the power of 3. In 3 tests, over a few hours, you can figure out the bottle like this:
Lay out the bottles in a grid 10 x 100. Along the short side, line up the prisoners and have them walk down their column of 100 bottles, taking a sample from each.
Some short time later, an hour for instance, line each prisoner up on the 100 wide side. The first prisoner will sample from all bottles in the first 10 rows, the second prisoner from the second set of 10 rows and so on.
Lastly, some short time later, an hour for instance, again line up each prisoner along the long side. This time have each prisoner sample from only one row in front of them, move them down 10 rows and have them repeat. The first prisoner will sample from the 1st, 11th, 21st.....rows until finished.
Possible outcomes:
Someone will die 2 days later from the first test. This represents the column of 100 bottles where the poison is located. Every time, this will happen.
But, an hour later, a prisoner dies. This indicates which group of 10 consecutively placed bottles in that column are suspect.
An hour later, another prisoner dies. That prisoner could only die if he was the one to drink from the particular bottle in that group of 10.
In the second possible outcome, Someone will die from the first test. Every time. This tells us the column of 100 bottles that contains the poison.
An hour later, no prisoner dies. This can only happen if the first prisoner also drank the poison in his 10 rows from the second test. We now know which 10 consecutive bottles in a column are suspect. An hour later, a prisoner dies. The bottle from those 10 that corresponds to that prisoner is the poisoned bottle.
In the third possible outcome, Someone...first test...every time. This gives us the suspect column of bottles. An hour later, another prisoner dies. This gives us the 10 consecutively placed bottles in the column that are suspect. An hour later, no prisoner dies. In that group of 10, which ever the 2nd dead prisoner drank from in the third test is the poisoned bottle.
Last possible outcome. Someone...first test...every time. One prisoner dies in the first test. No other prisoners die. We know which column, we know which group of 10, we know which specific bottle.
Example: 3rd prisoner dies. One hour later, 5th prisoner dies. One hour later, 2nd prisoner dies. After the first test, we know the bottle is in the 3rd column. One hour later, we realize that the poisoned bottle is in bottles 41 through 50 in that column. After an hour, we now know that it was the 2nd bottle in that 10, or bottle number 42 in the 3rd column.
another solution I have found is to when the guests are walking through the door ask them to take a sip from the bottle of wine the one who has the poison wine most likely will not take a sip
I knew it was gonna be a math equation that has a combination of prisoners died, but my method would be to just have every advisor drink their own wine
Got a simpler solution, each prisoner gets a drop from different 100 wines, And each of them give 90 of their wine drops, 10 each to remaining prisoners. So now each prisoner drinks 190 drops, 100 of theirs and 9x10 from different guys. And then the prisoner who got 10 gives 1 drop to each guy exept the one he got the 10. So each of them drinks 191 drops
So the solution works, if for example bottle number 55 is poisoned it means the 1st guy dies becouse its in his 100 bottles, then the 6th guy dies becouse he got bottle numbered 50-60, then the 2th guy dies becouse he got bottle number 55
It makes all the combinations unique, if someone does not know binary code this works too.
Another way of solving the riddle is to divide the wine to the prisoners in groups of 100 per prisoner, then let the prisoners have sips of each bottle of wine in there 100 at set intervals of 14 minutes and 24 seconds throughout a single day (since the party is in 3 days you have 1 to spare), record when each prisoner had a sip from what bottle of wine and when one prisoner dies check in the records what bottle that prisoner had a sip from exactly 2 days prior.
I thought you'd give each prisoner 100 bottles, then you got 900 safe and 100 risk
You then take the 100 and give to the remaining prisoners: 8 get 11 bottles and one gets 12, that leaves you with 88 or 89 safe and 12 or 11 at risk
Then you give to the remaining 8 accordingly
But this process would take at least a week
I immediately realized the solution would be some kind of system where based on which combination of prisoners died you would know which exact bottle it was. However I was not able to figure out what the system would be to do so.
DAMN YOU BRUSHWOOD!!!!
I had heard this same riddle except it was 500 crates of wine. And I actually solved it using logic in a way the guy who told it had never heard it before. It took me like 20 minutes but felt super smart. NOW I COULDN'T SOLVE IT AGAIN!
just make each of the 1000 advisors sip their own glass of wine. the one who dies is the one who was a traitor
If the poison is a little more fast acting, then there is another solution. Give 500 drops to one prisoner and 500 to another. One of them dies, and you have now cleared half the wine and have 9 prisoners left. repeat this process several times and you will be left with only 1 prisoner and 2 bottles. Give the last prisoner a drop from only 1 bottle. If he dies, then that bottle is the poison one, and if he lives, the other remaining bottle is the poisonous one.
Alternate Solution is to have 2 prisoners each sample 500 bottles, then whichever of those 2 prisoners dies you continue to process the portion that was allotted to them. You then have those 500 bottles split among another 2 prisoners, 250 each, and repeat, the last prisoner will have a 50/50 of drinking the poisoned wine, and if they live it is the bottle they did not choose, and vice versa.
If you know the exact dying time, you could also give them each 100 drops in 5 minute intervals (or such) and see, when one is dying.
I was like, oh thats easy, I give the first one 512, the next 256, the third 128 and then he said 'it needs 2 days... Partys gonna start in three'
my favourite bar scam is knocking the bartender out, and stealing the whiskey. 100% succes rate
label the bottle for each advisor and have the advisors drink their own wine
I would make each advisor take a sip of their own bottle of wine. You wouldn't care if the liar died anyway:D
Here is a lot easier way to solve it, but it would take 20 days: you separate the bottles, 500 in one pile and 500 in another. give one prisoner 500 drops, one from each bottle in one pile. if he dies, then that pile has the poisoned wine and that group is thus the poison group (or the other group is the poison group if the prisoner lives). You break that poison group into half, 250 bottles in each group, and you give the prisoner 1 drop from every bottle of one of those groups. This continues (up to 10 times) where you have narrowed it down to 1 bottle. Like i said the only flaw with this is that it would take up to 20 days, but you do have enough prisoners to accomplish this. and it's less complicated, in my opinion, than doing binary.
I was thinking along the same lines as this... I was think more general though: give each bottle to somewhere between 1-10 prisoners and make sure no other bottle goes to the same set of prisoners
This is a good puzzle, almost everyone in the comments don't get it.
Dyllan Jensen Almost everyone in the world whose had sex with an actual woman doesn't get it. That's like if I ask you a question about a jelly donut and the answer turns out to be "change the springs on the Floyd rose tremolo".....unless you know about guitars it makes no sense because it's useless to know unless you play guitar. Binary is useless to almost everyone.
It is actually the opposite of a good puzzle because it requires knowledge that is too specific. Good puzzles require little to no specific knowledge. General knowledge and the ability to think outside the box should be sufficient. Exception is If this puzzle was being presented to programmers who are used to thinking of things in terms of binary.
I have a solution that will put rest to all the arguments..
Imagine the bottles in a grid. Each row and each column labelled 1 to 10. So each bottle has a row no and a column no. Now give each prisoner wine from one row only. 6 hours later give each prisoner wine from one column only. 2 days later you'll know the row no and 2 days and 6 hours later you'll know the column no of the wine. Bam!
@@christopherharrison7931 the other key is when you finally give the answer people understand wtf it means.
I'm pausing at 4:04: I've got an idea. What If it's solved via the binary system? Say each bottle is given a number from 1 to 1000. Then each prisoner has a number that is a power of 2. So the prisoners would be numbered, 1, 2, 4, 8, 16, 32, 64, 128, 256, and 512. Prisoner 1 has every odd bottle. Prisoner 2 has bottles 2 and 3, 6 and 7, 10 and 11, and every other number where the '2' bit is a '1'. Prisoner 4 takes bottles 4, 5,6,and 7, 12, 13,14, and 15, and every number where the '4' bit is '1'. Basically assign each prisoner to a bit, and have them drink the wine that correlates with their bit. Then you can just add the numbers of the prisoners who died. So if prisoners 512, 16, 3, and 1 died, then the poisoned bottle number would be 512 + 16 + 3 + 1, or bottle number 532. I know this would work, but I'm just curious to how you solved it. I'ma watch the rest of the video now.
I like approach better. :P
tell the 'survivors' they must drink their own wine in front of the king, and arrest those who don't do it, then have a prisoner drink the wine of those that refused to drink
Haven't watched the solution yet, but here's mine.
Make every prisoner drink from 100 bottles of wine. Then 2 days later, take the bottles from the prisoner that died, and make all the 9 left drink from 10 bottles (leaving 10 untouched). Then two more days later take the bottles from the prisoner that died (or if no-one died, the 10 untouched ones) and make the 8 prisoners drink one each (leaving two bottles untouched) and also make half of the prisoners drink from one of the untouched bottles. If one dies, it was the bottle he originally drank from, if 4 die it was the untouched bottle that they drank, if no-one dies it was the other untouched bottle.
This solution takes one week to complete
They only have 3 days to do it, though.
Binary code, of course! **looks around to check if anyone saw the dumbfounded look on my face when Brian explained it**
To those saying you could just give 50/50 and see who dies and repeat till you're down to one, this does not work due to the time restriction. You could only do it once before the party started. The most you could get out of this would be 666 bottles if you divide the initial group into thirds.
I have another potential solution:
First, you divide the wines up into 10 batches of 100. Each prisoner drinks from their batch of 100.
8 hours later, you divide up each batch of 100 into 10 batches of 10. Each prisoner drinks from their ten assigned batches of 10.
8 hours later, you divide up each batch of 10 into 10 batches of 1. Each prisoner drinks from their one hundred assigned batches of 1.
8 hours later, you take the same batches of 1 but you give them one prisoner over. So prisoner 7 drinks from the batches previously given to prisoner 8, etc.
Then you see when the prisoners die. Suppose the poisoned bottle is number 273. After 48 hours, prisoner 2 will die. 8 hours later, prisoner 7 will die. 8 hours later, prisoner 3 will die. So you know that bottle 273 is poisoned.
"But why do you give them the shifted batches at the end?" I hear you ask. That is to defend against the following scenario:
Suppose that prisoner 2 dies, then prisoner 7 dies, and then *no third prisoner dies* - was the poisoned bottle 272 or 277? You'll learn that from whether 1 or 6 dies at the end. If 1 dies, then it was 272. If 6 dies, then it was 277.
Suppose it's 333: prisoner 3 dies, then nobody, then nobody, so you know it must be 333. Then 4 dies for fun. Similarly, if it's 338, then 3 will die, then nobody, then 8, then 7. If it's 344, you'll have 3 die, then 4, then nobody, then nobody again, and the only way for that to happen is if it's 344. 343 will be 3, then 4, then nobody, then 2.
Of course, all of this relies upon the poison taking 48 hours to take effect. If it's a 72 hour poison, then the first prisoner won't die until the party starts, and you won't have time to figure out the rest.
Just ask all the advisors for the solution to the problem. The advisor who doesn't show up to the party sent the poisoned bottle
just make every advisor to drink a drop of his own wine
There's a practical problem with this solution:
Everyone's going to bring their wine on the day of the party, so waiting two days to find out which is poisoned is too late.
Better to make everyone drink their own wine, or mix it all into one batch, switch it with another batch while nobody's looking and serve it to see who declines to drink.
To stop the others from saying to sip their own wine, say that the bottles were sent in anonymously
This isn't something that a bunch of tipsy folks aren going to come up with in a bar. Maybe Brian needs a separate devious puzzle channel.
As soon as he said thinking like a programmer I got annoyed - been studying games programming at uni for 2 years now, I should have been able to solve this from the start!
Everyone drinks from their own bottle.
Togepod pretty obvious
good luck trying to control a 1000 people party making sure the assassin actually drinks and does not swap his bottle
There is a rule saying that's not allowed...
Yeah. just duct tape everyone's bottle to themselves.
Igor Freitas what kind of KING OF ALL LANDS can’t control 1000 people drinking wine do that shit 1 at a time make it a week long party for the inconvenience LOL
I don't know how potent the poison is but you could possibly mix all the wine together in order to diffuse the poison into each bottle making it to weak to kill anyone
Other way to solve it, with a similar conclusion it would be : Each prisoner drinks from 100 bottles and 10 bottles more x each of the 10 prisoners bottles so another 100 bottles more.(up to this point he has drunk from "200" bottles, 190 if we have in mind that he drinks twice from his own bottles). (up to this point if HE and only He dies we know that the poison was in his own ten first bottles). If then he drinks from 1 more bottle x prisoner ("10" more , 9 technically)... so each bottle has being tested for 3 different prisoners . Depending on the combination of prisoners dying you will get which is the poisoned wine...
Let's say prisoner 1 prisoner 5 and 3 have died... then you know it was in those 100 bottles (of the prisoner 1) lets say between the 50-60number because the prisoner 5 also dies (and he tried those 10 bottles of prisoner 1) and the prisoner number 3 also died... so you know it was the 53 bottle. :)
In the event that only the prisoner 1 dies... you know it was the bottle that just him and nobody else tried.
The problem is the same as this people are having 153,135, 351,315, 513, 531... are the same combination of people dying... so you have to make somebody else try the bottles with all numbers that can have multiples combinations.
It's gonna be a nightmare moving all those bottles around! :)
I thought by 3 days he meant you get 3 tastings so my solution was:
- first every prisoner drinks 100 bottles. So you narrowed it to 100 bottles of wine and 9 prisoners left.
-then every prisoner drinks 11 bottles and one drinks 12 bottles. So now it's narrowed to at the most 12 bottles and 8 prisoners left.
-then on the third day you let one prisoner drinks the first 6, another drinks the last 6 and then out of the last 6 prisoners, one drinks bottle 1 and 7, one drinks 2 and 8, one drinks 3 and 9. etc. You look at which combination dies and which half dies and you know the poisoned bottle.
Simple: you just have each advisor take a sip of their own wine 3 days before the party.
Either 1 advisor will refuse (because he knows his is poison) or he'll drink it to bluff, and die in 2 days, 1 day before the party.
you cant do that because only 10 prisoners are allowed to die
There is a similar puzzle. You have a jar of 100 labeled bills and someobody stole a random amount.
How do you find out every single labeled number they stole by weighting it once?
You only got a
i actually went 'keep divinding by 2 bottles till you have 10 outcomes' wich turns out to be, the tenth is 1, so it just ramps up (seeing we are working with drops here) so if person 1 dies, 512 bottles are gone, if person 2 dies 256 are gone , etc...
I have a solution that I think would waste less wine. First, you give to each prisoner ten batches of ten different wines, labelled 1 through 10 followed by the number of the prisoner given to, who are also labelled 1-10, so the first batch of the 1st prisoner is 1-1, so each batch of 10 wines is labelled b-p (b:batch, p:prisoner) they take a drop of every bottle, then, you switch the batches of 10 wines so now the n-th prisoner gets the n-th batches of every prisoner, for example, now, your prisoner 2 will get the 2nd batch of the prisoner 1-10. Now they repeat the drinking, you do not need to label again, because now the b stands for the label of the second prisoner who drank from that batch. Now, arrange the bottles of each batch from 1-10, each prisoner will have 10 of each number, so in total there will be 100 bottles of each number, now give all the 1 bottles to the 1st prisoner, all the 2 to the 2nd and so on. Follow the label with a 3rd tag, so it would look something like b-p-l (batch, prisoner, last). Assuming the deaths are in order, you will have the code of your bottle given by the dead prisoners, the first death will correspond to the first dead prisoner, the second to the second and the third to the third. If only one prisoner dies, the code will be his number repeated 3 times, if two die, say a and b, there are three options, they will not die simultaneously, so let's say a dies first and b second, so the options are: aab, abb, aba. Now, given that the poison is deadly by timing, if by the time a second one should die there is only one dead, it is aab, if otherwise, it means that the first and second one died, but the third didn't, so it is either abb or aba, in either case, as deadly as the poison is, you should be able to tell who took double and thus know if it is abb or aba.
An equivalent solution (another way to see the same solution) would be: Suppose we have 1024 bottles (for simplification). Then we divide the 1024 bottles into two groups of 512 bottles (the left group and the right group). Each group of 512 bottles we divide into two groups of 256 bottles (the left groups and the right groups). And so on, so that later each group of 2 bottles we divide into two groups of 1 bottle (the left groups and the right groups). That way we will divide the bottles into 512, 256, 128, 64, 32, 16, 8, 4, 2, and 1 (10 divisions in total). The 1st prisoner drinks a drop of each of the 512 bottles of the left group in the 1st division. The 2nd prisoner drinks all bottles (a drop, of course) of the left groups of the 2nd division (the division into groups of 256 bottles) (he will drink 512 bottles in total). The 3rd prisoner drinks all the bottles of the left groups of the 128 division (he will drink 512 bottles in total as well). And so on, so that the 10th prisoner drinks all single bottle of the left groups of the 10th division (the division into groups of 1 bottle). Depending on which prisoners die, you can trace the poisoned bottle going left or right in each division.
This wouldn't work however, because you only have a few days time, and it also takes a few days for the poison to take effect.
So, in the riddle you would narrow it down to a group of 512 bottles, and then it's already the day of the party
All 10 prisoners would try their portion of the wine at the same time. The ones that are going to die will die at the same time 2 days after drinking. Look at the following image for inspiration: i.stack.imgur.com/9XeYs.png. Think about it and think about how you can get to the poisoned bottle after checking who died or not 2 days later. I'm sure it works.
Just make each advisor sip their wine
Music from the Dave Allen show. Lovely. Very funny man.
Reminds me of some classic jokes that I remember 40 years later.
This works mathematically, but would result in all 10 prisoners needing to drink roughly 500 bottles in a single day. If on the other hand you knew exactly how long the poison took to act you could give each prisoner 100 sips of wine evenly spaced out throughout the day and narrow it down to a single bottle based on his time of death.
I've got a better solution. Make 10 batches of wine bottles each with 100 bottles, then start from batch one, take a drop of wine from every bottle of that batch into the cup (because according to the puzzle one drop is enough to kill a person). If we do that for every batch, we have 10 cups of wine in which only one has poison. Give those wines to every prisoner and then you'll know which batch should eliminate based on who died after having the wine.
I have liked this video under protest.
Because the answer is fucking great and I hate you for it.
Pigeon Man Glad my riddle was a fun one
I got to it a bit differently, but ended up with basically the same answer.
How's this for a solution... Imagine the bottles in a grid. Each row and each column labelled 1 to 10. So each bottle has a row no and a column no. Now give each prisoner wine from one row only. 6 hours later give each prisoner wine from one column only. 2 days later you'll know the row no and 2 days and 6 hours later you'll know the column no of the wine. Bam!
Obviously won't work if we don't have that extra time.
That only works for 100 bottles right?
you give 100 to each prisoner. one dies. you take the bottles from the one that died and divide it amongst the remaining prisoners. each one gets 10 with 10 left over. you take the 10 that has the bad wine and divide it again until you're down to one
I figured you could make 10! combinations, so label bottles 1,2,3 etc. Then 1+2,1+3,1+4 etc and go through as many combinations as you need until 1000 bottles have been labeled, and based on the combination of dead prisoners you will know which bottle is poisoned. A nice bonus if you feel generous is to leave one bottle untested, and if no one dies then that untested bottle is poisoned
I came up with a slightly more complicated solution, but with the same mindset.
I think you could extend it even more, if you use the same method in batches of the same size of "zero"s that batch would have + 1 (so 1010101010 > 6, 1111111111 > 10
Then day after you have everyone drink one bottle from the batches they didn't drink from, as they will die 1 day later than the ones who drank the first day (and you have 3 days, poison uses 2)
You could probably expand it even more by doing the binary trick again
Alternate solution:
If the wine takes 2 days to kill, have 10 prisoners sip 10 bottles of wine each. (as seen around part 1:48 in the video). 12 hours later, have the same 10 prisoners take one sip from each of those 10 bottles. One prisoner will die in 2 days, and you will know which batch of 10 is poisoned. 12 hours later, you will know which bottle itself is poisoned.
This solution is much simpler for the simple minded (or at least the non tech savvy) to get it Brian. :)
That was a good one! Using the marker as means of identification while simultaneously being able to test all the possibilities
LOL 0:41 The king is having a sh*td*ck. I wonder what good the wine will do.
I need more Scam School in my life.
like everyone else, i was like "have everybody drink their own wine, so the end result is 10 non-people prisoners you can use your kingly rule to do whatever to and 1 dead treasonous regicide-happy adviser"
I know im 3 years late but an easy way to decode binary is labeling right to left ...64, 32, 16, 8, 4, 2, 1 and you find the combination for what makes the number so 5 is 101 because its 4+1. If a number isnt used you put a 0, if it is you put a 1. 79 would be 64+8+4+2+1 = 1001111.
“Shitty math riddles guaranteed to bore your friends at a bar! Try them!”
You could also give the 9 prisoners left each 10 bottles which would make a total of 989 bottles of wine known good. The 11 bottles left is tested in two groups of 5 and 6 prisoners. Some of the prisoners survive. The big question: Who poisoned their bottle of wine?
I think a tip for viewability is there's no real need to mention binary here. I'd also present the problem with 1024 prisoners in the first place I think because it's pretty hard if you say 1000
The simple way to understand this is, for a given bottle, we have 2 choices for each prisoner. We can either have the prisoner drink it, or not drink it.
Since there are 10 prisoners, there are 2^10 = 1024 possible unique sets of prisoners that could drink from the bottle.
So as long as we have
Alternate solution... or possibly the same solution from a different angle. Think of the 1000 bottles as being in a 10x10x10 cube, so each bottle has 3 dimensional coordinates, from bottle (1,1,1) to bottle (10,10,10). At 6am on the first day, Prisoner 1 tastes from every bottle with a 1 in the first spot, so (1,y,z). Prisoner 2 tastes from (2,y,z), all the way to prisoner 10 tasting from bottles (10,y,z). At noon on the first day, do the same thing, but focusing on the y coordinate. Prisoner 1 tastes from (x,1,z), blah, blah, blah, prisoner 10 tastes from (x, 10, z). At 6pm on the first day, do the same thing focusing on the z coordinate. Prisoner 1 tastes all of (x,y,1), blah, blah, blah, and prisoner 10 tastes all of (x,y,10). Two days later, you can zone in on which bottle is poisoned by who dies at which time.
Actually, dang... if it's bottle (1,2,1) for example, you wouldn't know if it's actually bottle (1,2,1) or bottle (1,2,2). If it was (1,1,1) you'd know, or bottle (1,2,3), but one digit repeating results in enough ambiguity for there to 2 potential poison bottles.
OR: You could give Prisoner 1 half of the bottles, if he dies, the other half is NOT poisoned. Pick the poisoned half and hand half of THAT to prisoner 2, keep doing this until you narrow down to the 1 bottle wich you can throw away.
Give Prisoner 1 a drop from Bottle 1 (0000000001)
Give P2 a drop from B2 (0000000010)
Give P1 & P2 a drop from B3 (0000000011)
And continue so that each bottle is tasted by a different combination of prisoners all on the first day. You only waste a maximum of ten drops from each bottle. You will know which bottle was poisoned by who lives and who dies. It is like assigning a binary number to each bottle.
Why not just ask ask all the advisors to taste the wine and make sure it tastes goods for king brushwood then the one that refuses is the killa
The pellet with the poison is in the vassal with the pestle. The chalice from the palace has the brew that is true.
My answer was similar. Have the first person sample half the bottles. The next person samples half the bottles that are already sampled and the ones that aren't. These two alone narrow it down to 250 people. Just keep doing this until it's down to 1. But his way was much easier to keep track of.
bro you werent listening to the riddle, it takes 2 days for the poison to work, you dont have that many days
This only goes to show how bad mathematicians can be at solving real life problems. Of course in REALITY the answer is simply to demand that everyone sip his own wine. And even within the context of the riddle, which suggests that only ten prisoners may die, the king could still demand it. None (neither a prisoner nor a guest) need die. The point is that the guilty party would REFUSE, thus live to stand trial and no doubt become a prisoner himself.
The binary option is perfect. My idea was to use the method show in the begining to narrow it down too 100. Then after 1/2 an hour give them a drop from from a crossection of each of the other hundreds. Then over the next 5 hours continue to do this. Then on the 3rd day track when and in which order the prisioners died and compare it to the matrix you would find your bottle.
wow the answer was really witty and clever
So if we are having the advisors bring the wine 3 days early why didn't we just administer a drop to each advisor of their own wine? This finds the poison and eliminates the traitor at the same time with no margin of error.
Give 1000 bottles of wine to all people and only 1 will die!
you cant do that because only 10 prisoners are allowed to die
I was one of the viewers screaming at the screen. I wouldn't expect a non-programmer to necessarily get it but if those two guys have a programming background, it's surprising that they didn't get it almost immediately.
All the people saying that, "why don't they all drink their own wine" That wouldn't work because there is only 10 people that the king will allow to die, and the other 990 can't die, and if 1 of the other 990 dies that's against the rules.
Here's what I came up with: each prisoner samples 10 wines one each day. When one dies you know it was the wine he drank 2 days ago.
Couldn't you just make each guest take a sip of their own wine bottle, thus killing your killer?