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Three men are arrested on charges of murder. They make the following statements.

  • First suspect: I'm not guilty of murder.
  • Second suspect: the third suspect is the murderer.
  • Third suspect: the first suspect is the murderer.

We know that two of them are lying and one of them is telling the truth, and certainly one of the three is the killer. The question now is:

  • Which one is telling the truth?
  • What is the probability of each suspect being the murderer?
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  • $\begingroup$ This is a duplicate, but I can't find it. $\endgroup$ – warspyking Jan 2 '15 at 17:03
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Third one is innocent

Explanation follows

First: "I am not guilty"

Second: "Third is the murderer"

Third: "First is the Murderer"

Lets assume First and Second are lying and Third is telling the truth.

Then we get the following statements.

First one is the murderer.
Third one is innocent.
First one is the murderer.

Now, lets assume Second and Third are lying and First is telling the truth.

Then we get the following statements.

First one is innocent.
Third is innocent.
First one is innocent.

Now, lets assume First and Third are lying and Second is telling the truth.

Then we get the following statements.

First one is Murderer.
Third is Murderer.
First one is innocent.

Which contradicts.

So from the first two valid assumptions it is clear that

Third is definitely innocent.

Now take the two valid assumptions

1)First one is murderer and Third one is innocent. So Second too will be innocent.
2)First and Third are innocent. So Two is the murderer.

So the chances for First and Second to be murderer are

50-50

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  • $\begingroup$ tnx but i think you have to make it more visible $\endgroup$ – shayan Jan 2 '15 at 17:41
  • $\begingroup$ like in this statement:First one is the murderer. Third one is innocent. First one is the murderer. why you said first one twice ? $\endgroup$ – shayan Jan 2 '15 at 17:42
  • $\begingroup$ Those are assumptions by all three suspects in the order. By listing all the three assumptions, we can check id any of it contradicts. $\endgroup$ – AeJey Jan 2 '15 at 18:05

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