Date: Mar 14, 2013 12:29 PM
Author: Joe Niederberger
Subject: Re: Please help me with the following question
R Hansen says:

>Setting the limit to 7 would certainly put this into the "trick question" category, like you see people pull on each other at bars.

Hardly.

R Hansen says:

>Given the constraint of "7" races, which, along with every other pertinent condition, was left out of the original problem,

I wouldn't call it a constraint, but its natural in this context to look for the least number of races needed.

As far as some of the other "pertinent condition", I noticed you failed to mention some of them.

1. No ties.

2. Two racers who race each other more than once end up in the same relative order. I thinks that's the intended condition that makes this a mathematical problem. Notice than in race#7 you have the second and third place finishers of race#6 going against each other again. One could institute a rule that race#7 is the official decider but that's not very mathematical.

Any others?

Cheers

Joe N