Allison M. Pacelli, Alan D. Taylor's Mathematics and Politics: Strategy, Voting, Power, and Proof PDF

By Allison M. Pacelli, Alan D. Taylor

ISBN-10: 0387776451

ISBN-13: 9780387776453

Arithmetic and Politics calls for no must haves in both topic. The underlying philosophy includes minimizing algebraic computations whereas targeting the conceptual facets of arithmetic within the context of real-world questions in political technology. This new addition has an further co-author, Allison Pacelli, and covers six significant subject matters: social selection, yes-no vote casting structures, political energy, game-theoretic types of foreign clash, equity, and escalation. as well as having new chapters (treating apportionment and clash resolution), the textual content has been widely reorganized and the variety of workouts elevated to over three hundred.

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Additional info for Mathematics and Politics: Strategy, Voting, Power, and Proof (2nd Edition)

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Voters 1–4 a b c Voters 5–7 b c a Voters 8 and 9 c b a With the plurality procedure, alternative a is clearly the social choice since it has four first-place votes to three for b and two for c. On the other hand, we claim that b is a Condorcet winner. That is, b would defeat a by a score of 5 to 4 in one-on-one competition, and b would defeat c by a score of 7 to 2 in one-on-one competition. Thus, the Condorcet winner b is not the social choice, and so the Condorcet winner criterion fails for the plurality procedure.

Otherwise, use the Borda count to determine the social choice. 8. Prove or disprove each of the following: (a) Plurality voting always yields a unique social choice. (b) The Borda count always yields a unique social choice. (c) The Hare system always yields a unique social choice. (d) Sequential pairwise voting with a fixed agenda always yields a unique social choice. (e) A dictatorship always yields a unique social choice. 9. Consider the following sequence of preference lists: b a d c c a b d c d b a a d b c (a) Find the social choice using the Borda count.

The lists then become: Voter 1 Voter 2 Voter 3 Voter 4 a b c a b c b c a b c a Notice that we still have b over a in Voter 4’s list. 6. Negative Results—Proofs 27 the first place votes. Thus, although no one changed his or her mind about whether a is preferred to b or b to a, the alternative b went from being a nonwinner to being a winner. This shows that independence of irrelevant alternatives fails for the Hare procedure. PROPOSITION 11. Sequential pairwise voting with a fixed agenda fails to satisfy independence of irrelevant alternatives.

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Mathematics and Politics: Strategy, Voting, Power, and Proof (2nd Edition) by Allison M. Pacelli, Alan D. Taylor


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