Originally posted by fred Joly:
Thank you both for your replies.
Yes the "bit pattern" is really clever and yet so "simple".
Now I was wondering what can we do if we know that some bottle(maybe none) are poisened and not just one.
The bit pattern does not work anymore.
How many prisonners do we need?
PS : as i understand most of you are asleep by now.
Here (France) it's about 9 AM. So when you wake up, maybe I will
have come up with a solution....I let you know
You think you know me .... You will never know me ... You know only what I let you know ... You are just a puppet ... CMG
"I'm not back."  Bill Harding, Twister
Originally posted by fred Joly:
I guess it is more complicated than that.
For exemple with 4 bottles , I need only 3 prisonners.
Now i'm trying to extract a method from that.
Let's say 4 bottles B1, B2, B3, B4 and 3 prisonners P1, P2, P3
P1 drinks from B1, B2
P2 drinks from B2, B3
P3 drinks from B3, B4
with that I can know exactly wich bottle contains poison.
Note that P2 can not die alone.
You think you know me .... You will never know me ... You know only what I let you know ... You are just a puppet ... CMG
Originally posted by fred Joly:
Ah yes Jim an Ajay you're both right!
So, without you, I could have thrown away 2
bottles of decently good wine.
Seriously, there is no better solution than as many prisonners
as bottles ?
Originally posted by Anupam Sinha:
A variation on Maureen's solution.
1. Line up the 10 prisoners.
2. Give a drop/glass of wine to everyone from a seprate bottle. Name those set of bottles as SET 1 and each bottle with the prisoner's name.
3. After 10 mins give another drop/glass of wine to everyone from a seprate bottle. Name those set of bottles SET 2 and each bottle with the prisoner's name.
4.  11. Same as above with the bottle set different.
12. Now whenever a prisoner dies you can identify the set as well the bottle according to when the prsioner died.
Originally posted by Ryan McGuire:
If there are 4 independent bits of information (whether a given bottle is poisoned) then you need to do 4 independent binary experiments.
fred Joly wrote:Thank you both for your replies.
Yes the "bit pattern" is really clever and yet so "simple".
Now I was wondering what can we do if we know that some bottle(maybe none) are poisened and not just one.
The bit pattern does not work anymore.
How many prisonners do we need?
PS : as i understand most of you are asleep by now.
Here (France) it's about 9 AM. So when you wake up, maybe I will
have come up with a solution....I let you know
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