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Old 05-20-2014, 01:15 AM   #1
I Will not Admit my Addiction to RS
 
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a little word problem

There were ten hunters on an island and each of them owned a dog. It was known that at least one dog was sick. Every hunter could check other people's dogs every night but not his own. Hunters did not communicate with each other. A hunter would kill his own dog once he is sure that his dog is sick. No dogs were killed until the 4th night. How many dogs were sick?
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Old 05-20-2014, 01:54 AM   #2
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If he had to shoot his own dog once he was sure his dog was sick, wouldn't he have to check all 9 other dogs and prove they are all healthy to know that it is his dog that is indeed ill
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Old 05-20-2014, 02:02 AM   #3
I Will not Admit my Addiction to RS
 
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the hunter is allowed to check ALL DOGS BUT HIS OWN. so he will know if the 9 other dogs are sick or not
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Old 05-20-2014, 04:24 PM   #4
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Start by assuming that only 1 dog was sick. The owner of the sick dog would see that 9 other dogs are healthy and comes to the conclusion that his own dog is sick (given the information that there is at least 1 dog sick). He would kill his dog on the first night which leads to a contradiction that the killing happens on the 4th night.

Assume now that there are 2 sick dogs. Without loss of generality between the two owners of the sick dog, the sick dog owner would see that there is 1 other sick dog among the nine other dogs. However, both sick dog owners do not know for a fact whether their own dogs are sick. Hence no killing happens on the first night. Again without loss of generality, after the first night the sick dog owner knows that if his own dog was healthy, the other sick dog owner would have killed his own dog on the first night as in the first scenario. Hence the sick dog owner comes to the conclusion that his own dog must also be sick and both dogs are killed on the second night.

By inductive reasoning, on the 4th night, 4 dogs are killed by their owners.
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