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?

(In reply to Real Solution by vertigo)

I didn't read the whole question initially.

As for the maximum number that might die following that strategy, realise that there is no wine bottle 1023, which is 1111111111b, and this is the only instance where all 10 would die, so the answer is that as many as 9 could die.

As for the least, numbering the bottles from 1-1000 means at least one would die, but this can be improved by numbering the bottles from 0-999. Then, if bottle 0 is the poisonous one, no tester would die. As I said, I didn't read the whole question initially.

If there were 9 testers, you would need 8 days instead of seven. As you are short by one tester, you need to split the wine bottles into two groups, those with the tenth bit unset and those with the tenth bit set. Then on day 1 you follow the strategy with bottles 0-511, and on day 2 you do the same with the remaining bottles.

If testers die on day 7, the tenth bit was 0, on day 8 the tenth bit was 1.

Posted by vertigo on 2005-02-26 19:11:05