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 Suspects and Investigations (Posted on 2008-08-29)
An officer has to solve a case with 20 suspects, 10 from colony A, and 10 from colony B. He can solve the case once at least 19 of them answer truthfully during an investigation.

The officer has two identical boxes labeled P and Q, which each have 20 cards, one for each suspect. Before each investigation, he takes one card from each box. He interrogates these two people during the investigation; the suspect from box P will tell the truth, and the suspect from box Q will tell the truth if and only if the suspect from box P is from colony A. (The officer can tell who's telling the truth.)

After each investigation, the officer will discard cards from truthful suspects (from both boxes) and return cards from lying suspects to the original box.

Find the number of possibilities that he can solve the case in 10 investigations.

 No Solution Yet Submitted by Praneeth Rating: 1.0000 (1 votes)

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 Circular? | Comment 7 of 9 |

There is still too little information to construe the problem.  Since the entire matter is not a single "investigation", someone's guess that the term refers to a single interrogation is as good as any.  Also, we are not told whether the officer has any knowledge that the crime was done by a single individual (i.e. only one "suspect" to be identified, or possibly more than one).  We apparently assume that each of the 20 suspects in fact knows who (one or more) is guilty (else how can they "tell the truth" about that which they do not know?). "Tell the truth" means that they consciously assert a true proposition, within their knowledge

The real absurdity of this problem is that "the suspect from box P will tell the truth." combined with the statements that box P has 20 cards, one for each suspect, AND "the suspect from box P will tell the truth (unequivocally)."  This would seem to make box Q and its stipulation regarding truth equivalences superfluous, as also the "colonies" A and B.

Given the stated description, the officer can solve the case on his FIRST interrogation, of the FIRST person drawn FROM BOX P. Simply ask "Who did the crime?" Since (a) that person is assumed to know, and (b) that person will tell the truth, (and even the probably redundant -c- "the officer can tell who's telling the truth"), the case is solved with Probability = 1.0.  I have no idea what "the number of possibilities" refers to or how that is related to the problem (someone says this is supposed to be a "combinatorial" problem).

I realize this is probably not what the proposer intended, but it seems entirely consistent with the problem as worded.  What precludes my interpretation??

 Posted by ed bottemiller on 2008-09-02 19:13:01

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