Originally posted by mtthw
I can do it with 10. Treat each prisoner like a binary digit. For the nth bottle, give a tiny amount to the prisoners who make up the binary representation of n.

E.g. 500 = 111110100

So give a tiny bit to prisoners 1, 2, 3, 4, 5 and 7.

Once you know who's died, they will give the binary number of the poisoned bottle. You need 10 because 2^10 = 1024.

Any good?

cool

Nice puzzle. One minor quibble: It might be difficult to serve 1000 rounds of wine to ten prisoners in a mere 4 hours (especially since the binary count must be maintained without a single mistake!).

Originally posted by BigDoggProblem
Nice puzzle. One minor quibble: It might be difficult to serve 1000 rounds of wine to ten prisoners in a mere 4 hours (especially since the binary count must be maintained without a single mistake!).

Originally posted by Mathurine
I agree; wine and binary don't mix, but then these things are all hypothetical anyway.

When, exempli gratia, does a bloke find his having to transport a wolf, a goat and a cabbage across a river, &c., &c....?

🙂

Originally posted by BigDoggProblem
I was referrring more to the time it might take to swig 1000 wine glasses, but you could get around this problem by mixing the drinks in advance. Then, each prisoner is presented with their own tumbler (keg?) with a unique combination, and you find out with one sip each which bottle it is. In fact, you could remove the times altogether, and just stipulate a poison that kills instantly.

Originally posted by PBE6
I think that would reduce the answer to 1 prisoner, because he could then take a teeny tiny sip from each bottle until he got to the poisoned one.

Originally posted by PBE6
I think that would reduce the answer to 1 prisoner, because he could then take a teeny tiny sip from each bottle until he got to the poisoned one.

can you give us something less scientifical?