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The poisonous wine (Posted on 2005-02-26) Difficulty: 3 of 5
There are 1000 bottles of wine and you know that there is exactly one of them that is poisonous, and will kill the drinker in 7 days. (To be precise, it will take randomly from 7 days to 7 days 23 hr 59 min 59 sec.) Now you have 10 testers who are willing to risk their lives and test-drink those wines. What is the smallest number of days you need to figure out which bottle contains the poisonous wine? In addition, under this strategy, what is the maximum and minimum number of deaths? The extension to this problem is, how will your strategy change if you only have 9 testers?

See The Solution Submitted by Bon    
Rating: 4.0000 (7 votes)

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Solution Solution | Comment 1 of 21

Just kidding, vertigo has a much better solution.


The first idea that popped into my head was:

Number the bottles 1-100 and the testers 1-10

Have testers 1-10 drink bottles 1-10 respectively at 00:00 of day one.  Then have them drink bottles 11-20 at 00:00 on day two, then 21-30 on day three, etc.  When someone dies, the poisoned wine was the wine that person had seven days ago.

If the poisoned wine is in a bottle in the range 91-100, it will take seven days to find out, so the minimum number of days with this method is 16 days: 10 days of drinking, and possibly 6 more days of waiting (At first I thought 7 days of waiting, but the last day of drinking is the first day of waiting).  However, results could conceivably come as early as the seventh day.

With nine testers, the strategy becomes 1-9 drink 1-9 on day 1, 10-18 on day two, 19-27 on day three, ..., 82-90 on day ten, and 91-99 on day eleven.

If nobody is dead as of the eighth day after drinking is over, the 100th bottle is poisoned.  So with nine testers, the minimum is 17 days: 11 days of drinking, and possibly 6 more days of waiting.  Again, results could come as early as the seventh day.

Edited on February 26, 2005, 7:40 pm
  Posted by Dustin on 2005-02-26 18:19:12

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