The Poisoned Wine Problem
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- เผยแพร่เมื่อ 21 มี.ค. 2017
- 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 - บันเทิง
![[INTERACTIVE] Can You Solve This 200 Year-Old Riddle?](/img/n.gif)
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.
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.
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.
Why don't they just tell them to drink the one they brought then u know who poised it
Xanderzone 1504 OMG
that was the solution I came up with.
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
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
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
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
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. . .
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"
"Did you know I have been drinking for the past seven hours?" - Jason Murphy, 2017
For some reason I laughed at the intro, I just imagine Brian running up to people's windows yelling random things at them. XD
That was a good one! Using the marker as means of identification while simultaneously being able to test all the possibilities
This is my favorite puzzle so far!
Finally a new video! The only ones that show up in my apps are from 5 years ago!
I need more Scam School in my life.
make all the advisors drink a sip of their own wine, and then imprison all of them for 3 days until one dies......
Holy crap, as a computer science student, this was amazing, thank you foyofor this. This is the most I freaked out about a puzzle solution on scam school ever
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!
Hey Brian, awesome episode had me fooled. i really like your jacket in the closing scene and am wondering what brand it is, goes really well with the modern rogue shirt. many thanks big fan
This was the best I've seen so far! I assumed this was going to be one of the terrible solutions I've seen in the past, as in some of the match stick problems, but it was a brilliant solution! I thought the solution was something dumb like have everyone take a sip of the wine they brought and see who showed up later; in reality a poisoner who wanted to succeed would have an antidote or a twin, so that would have been a lame answer. The answer actually works, and for that I give you: in this moment, it is the best it can be and will ever be for this moment.
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!
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"
Music from the Dave Allen show. Lovely. Very funny man.
Reminds me of some classic jokes that I remember 40 years later.
THIS WAS SO CLEVER
Gonna be in Austin soon. Hopefully I'll be able to meet Brian.
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
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.
I saw this before somewhere, and never, in my life, would I have figured that out or known it
Genius. . .truly
just make every advisor to drink a drop of his own wine
1000 bottles of wine on the wall, 1000 bottles of wine, take one down, pass it around, 999 bottles of wine on the wall.
Couldn't unlock my phone fast enough to get here!
(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
damn you Brian!! damn youuuuuuu!!
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
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 remember seeing this on a different channel, but I still got stumped by it lol
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.
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.
my favourite bar scam is knocking the bartender out, and stealing the whiskey. 100% succes rate
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.
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.
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
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.
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.
Was that guy a computer programmer? And he didn't immediately know the answer when Brian said that it would work with 1024 bottles?
I'm here from the programmers union, and this guy needs to turn in his card and find a different career! He's making us look bad!
2^10! Learn It, Love It, Live It!
LMAO
LOL 0:41 The king is having a sh*td*ck. I wonder what good the wine will do.
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 would make each advisor take a sip of their own bottle of wine. You wouldn't care if the liar died anyway:D
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.
Just make each advisor sip their wine
label the bottle for each advisor and have the advisors drink their own wine
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
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.
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!
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.
"But it's so simple. All I have to do is divine of what I know of you..."
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.
Holy shit that's awesome!
Brian Burshwood, Would you ever do a show over in Australia?! :)
Sorry, mate. He's too scared of venomous things.
Binary code, of course! **looks around to check if anyone saw the dumbfounded look on my face when Brian explained it**
just make each of the 1000 advisors sip their own glass of wine. the one who dies is the one who was a traitor
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
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
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
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
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'
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.
To stop the others from saying to sip their own wine, say that the bottles were sent in anonymously
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.
when he said 1024 I thought I had it, but then I remembered the 2 day wait so now I don't know
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.
The King of All is the sickest fantasy title ever.
Couldn't you just make each guest take a sip of their own wine bottle, thus killing your killer?
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
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
I came up with a slightly more complicated solution, but with the same mindset.
that is an awesome puzzle
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.
Someone tell me if I'm getting this right since I'm feeling a bit confused. The idea is to just use the combinations of 0's and 1's to define who is drinking and who isn't, but with the idea that it'll take two days for it to take effect so it's just looking for the "on" positions to die, and the specifics of that will spell out which bottle due to the sequence in binary, or am i muddling this up somewhere?
Yep, you got it!
Okay, awesome! Thanks for letting me know!
wow the answer was really witty and clever
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.
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?
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
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 did think of having some sort of overlap, didn't have it nailed down all the way though...
No more custom ads in the middle? I know people complained but they were genuinely clever.
when I was asked this riddle by my brother a few years ago, I had the answer before he finished asking it...
where are you getting those kickass T shirts from?
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 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.
That moment when you watch the most recent video and then find yourself 6 more in just clicking random vids on the right...
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...
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"
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.
it's a sub-net riddle. awsome
nice Philip Defranco reference
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.
As a programmer watching this, when he said 1024, it was solved