On a certain game show, five families were asked to match famous names to faces they were shown, and each correct guess got one point. The families made the following guesses (in order):

The Addams' guessed Otto, Roebling, Steinmetz, Tesla, and Westinghouse.
The Bunkers guessed Tesla, Westinghouse, Otto, Steinmetz, and Roebling.
The Cunninghams guessed Roebling, Tesla, Steinmetz, Westinghouse, and Otto.
The Flintstones guessed Tesla, Roebling, Otto, Steinmetz, and Westinghouse.
The Jetsons guessed Tesla, Westinghouse, Steinmetz, Otto, and Roebling.

No two families got the same number of points. Which family walked home with the grand prize?

I personally would have scored somewhere around the Addams.  Figuring on blind luck to get me some points.  Which I believe is about how I solved this problem.  I had not thought of the fact that one family would have had to get them all right (4 right not possible).  And did not know how to limit beyond the 120 possible combinations.  I took a shot and assumed the most frequent response for a picture would be the correct response. That meant Te. was first, Ro. or We. was second, St. was third, St. was fourth, and Ro. or We. was fifth.  With St. appearing twice (not possible) I changed third picture shown to second most frequent choice of Ot.  That left only two possible combinations, one of which left two families with the same score. Is there any mathematical practicality to my method or was it pure, dumb, blind luck as I figure it to be?

Posted by john on 2005-05-09 16:40:46