Solution
Possibly Poisoned
Show the problem again
A cellar holds 1024 bottles of wine. At most one of them is poisoned (possibly none). Before a single round of tasting, you assign each taster a set of bottles to drink from. A taster who drinks from the poisoned bottle dies and a taster who does not survives. After the round you see exactly which tasters died, and from that you must be able to say whether any bottle is poisoned and, if so, which one. What is the minimum number of tasters needed?
Worked solution
The answer is 11. There are 1025 situations to tell apart (1024 choices of poisoned bottle, plus no poisoned bottle). The pattern of survivors and deaths among t tasters has 2^t possibilities, so 2^t ≥ 1025 and t ≥ 11 (2^10 = 1024 is exactly one too few). Eleven suffices: number the bottles 1 to 1024 and have taster j drink from every bottle whose binary expansion has a 1 in position j. The dead tasters spell out the binary number of the poisoned bottle, and nobody dying means no bottle is poisoned. The classic ten-taster solution fails here because the bottle with the all-zeros code looks the same as no poison at all.
Source: Variation on the classic binary poisoned-bottle puzzle; the 'at most one' twist is original to AxiomIQ.