# The Puzzle

A village has $$N$$ people, and each of them has one dog. One day, certain number of dogs, say $$K$$, contract rabies. Every morning, all people take their dogs out for a walk, and they can tell which dogs, except their own, have contracted the disease. But they do not tell each other of what they see, but they all agree to shoot their own dog during the night if they can conclude that their dogs have contracted the disease.

Some extra premises include:

1. They do not shoot other people's dogs.
2. The disease isn't contagious. ($$K =$$ const.)
3. The number of rabid dogs is at least 1.

There's nothing on the first night,

There's nothing on the second night,

There's nothing on the third night,

There are several gunshots on the fourth night.

How many dogs were killed that night?

# The (Supposed) Solution

Case with $$N =2$$, and $$K = 1$$

So let's say there are 2 people in the village. Premise 3 shows that there's at least 1 dog with rabies. Then during the first morning, the owner of the rabid dog will see the other dog without the disease. Therefore, the owner of the rabid dog will be able to conclude that his dog is rabid, and therefore will shoot the dog first night.

Case with $$N = 3$$, and $$K = 2$$

During the first morning, the owner of one rabid dog, say A, will see that one other dog with rabies, say B, and one healthy dog. A will conclude that if his dog is healthy, then by premise 3, B will be able to conclude that his dog has rabies and therefore will shoot the dog that night. However, because B does not shoot the dog that night, A will conclude that his dog has rabies. Therefore both A and B will shoot the dog during the second night.

Case with $$N = 4$$ and $$K = 3$$

The owner of one rabid dog, say A, will see that two other dogs with rabies. If his dog was indeed healthy, then from the previous case, the other two owners would shoot their dogs on the second night. However, the shots don't come. Therefore A can conclude that his dog is rabid, and therefore all rabid dog owners shoot their dogs on the third night.

So on and so forth, the answer to this puzzle would be 4 dogs were killed that night.

1. This strongly resembles the three logicians walking into the bar joke.
2. The total number of villagers actually don't matter.

# The (Supposed) Problem with the Solution

Apparently there's a problem with this solution. When $$K=4$$, one rabid dog owner will see three other rabid dogs. From this observation, it is impossible to expect other three rabid dog owners to behave accordingly to predicate cases of $$K=1,2,3$$ because they ALL see three other dogs with rabies. Therefore, the problem is flawed and the shots will never come due to there being zero conclusive evidence to convince any owner of his dog's rabidity.

# My Take

I actually don't really agree with the proposed problem to the solution because even if A sees that three other dogs to have rabies, if A were to assume that if his dog were healthy he can fully expect three other people to behave accordingly with the cases of fewer $$K$$s. But obviously, their behaviors will be different from those of expectation, so does he not have the basis to conclude that his dog is indeed rabid? I don't think the person who brought up the supposed problem with the solution realizes that the basis of this puzzle is from the presumption that each owner assumes that his own dog is healthy, and the contradictions that arise thenceforth.

• The arguments of your fist case (N=2, K=1) are valid for arbitrary N Apr 15, 2020 at 11:56

Your take is correct, an owner seeing $$r$$ dogs with rabies has to conclude $$K\in\{r,r+1\}$$ and if he had a healthy dog he would be watching events unfold as if $$K=r$$ and $$r$$ dogs would be shot on the $$r$$th night.