Question

Suppose that there exist three voters, each of whom is given three alternatives: A, B and C. There exist six possible strict preference orderings for these three alternatives: A>B>C, A>C>B, B>C>A, B>A>C, C>A>B, and C>B>A. The first voter has preferences A>B>C. The second voter has preferences B>C>A. Preferences of the third voter are unknown. How many of the six possible preference orderings, if selected by the third voter, would produce a voting cycle? (In a voting cycle, A defeats B, B defeats C, and C defeats A).

Answer 1

For the third voter, two of the six possible preference orderings would result in a voting cycle: C>A>B and C>B>A, as they would complete a cycle where each option defeats another by majority but is also defeated by a different one.

Step-by-step explanation:

The question pertains to a scenario in a voting cycle, where each of three voters holds different strict preference orderings for three alternatives: A, B, and C. The first voter's preference is A>B>C, and the second voter's preference is B>C>A. To determine how many possible preference orderings for the third voter would lead to a voting cycle, we can look at each remaining preference ordering and see if it completes the cycle where A defeats B, B defeats C, and C defeats A.

The six possible strict preference orderings for a voter are: A>B>C, A>C>B, B>A>C, B>C>A, C>A>B, and C>B>A. Since the first voter prefers A>B>C and the second prefers B>C>A, we need the third voter to prefer C over A but A over B, or prefer B over C but C over A to create a voting cycle. Examining the options, we find that the preference orderings C>A>B and C>B>A for the third voter would complete the voting cycle. Therefore, two of the six possible preference orderings for the third voter would produce a voting cycle.