Formula+for+number+of+pairwise+comparisons

Formula+for+number+of+pairwise+comparisons - Math 103...

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Math 103, Section 11, January 25, 2010 The sum of consecutive integers formula is 1+2+3+4+5+…+L=L*(L+1) 2 where L is the last integer to be added. In an election with N candidates, we substitute L=N-1 in the sum formula above. This gives the formula for pairwise comparisons for N candidates: 1+2+3+. ..+ N - 1 = (N-1)(N ) pairwise comparisons. 2 Examples. 1.In an election with 54 candidates,  a. How many pairwise comparisons are there?  Answer:  1431 pairwise comparisons Solution: N=54.       L=N-1=53, Use the formula: 1+2+3+. ..+ N - 1 = (N-1)(N) 2 Then 1+2+3+…+53=(53*54)/2=1431 pairwise comparisons b. If one of the candidates is a Condorcet candidate, how many points does this  candidate receive?  Answer is 53 points. The Condorcet candidate wins each head-to-head. That is, the Condorcet  candidate is preferred over each of the other 53 candidates in a head-to-head  comparison.  2.How many matches are played in a round robin tournament with 12 teams?
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Formula+for+number+of+pairwise+comparisons - Math 103...

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