A committee with three voters votes over four alternatives x, y, z, w using alternative vote. We know that the preferences of the members are as follows: 1: wxyz 2: yxzw 3: Y e. 3 has as the least-preferred option y but we don't know the first three options. We also know that: xis the Condorcet winner x is the first alternative to be removed under alternative vote (the tie-breaking rule is the number of second choices) Find the three missing preferences of member 3. Note: enter ONLY three small letters, e.g., zxw, with no commas. spaces, etc.

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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A committee with three voters votes over four alternatives x, y, z, w using alternative vote. We know
that the preferences of the members are as follows:
1: wxyz
2: yxzw
3:.. .y (i., e, 3 has as the least-preferred option y but we don't know the first three options. We
also know that:
xis the Condorcet winner
x is the first alternative to be removed under alternative vote (the tie-breaking rule is the number of
second choices)
Find the three missing preferences of member 3. Note: enter ONLY three small letters, e.g., zxw, with
no commas, spaces, etc.
Transcribed Image Text:A committee with three voters votes over four alternatives x, y, z, w using alternative vote. We know that the preferences of the members are as follows: 1: wxyz 2: yxzw 3:.. .y (i., e, 3 has as the least-preferred option y but we don't know the first three options. We also know that: xis the Condorcet winner x is the first alternative to be removed under alternative vote (the tie-breaking rule is the number of second choices) Find the three missing preferences of member 3. Note: enter ONLY three small letters, e.g., zxw, with no commas, spaces, etc.
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