Solution
Counting Agendas That Elect Alpha
Show the problem again
Five voters rank four proposals (best to worst): - Voter 1: Beta, Gamma, Delta, Alpha - Voter 2: Beta, Alpha, Delta, Gamma - Voter 3: Delta, Alpha, Beta, Gamma - Voter 4: Gamma, Delta, Alpha, Beta - Voter 5: Alpha, Gamma, Beta, Delta The chair lists the four proposals in some order $(x_1,x_2,x_3,x_4)$. The votes are held in sequence: $x_1$ against $x_2$, the winner against $x_3$, then the winner against $x_4$; each is a simple majority vote and the last survivor is adopted. Voters are sophisticated: at each vote a voter supports the option that leads to the final outcome he prefers, assuming the later votes are also played that way. Of the $4! = 24$ possible orders, for how many is Alpha the adopted proposal?
Worked solution
The answer is 12. Only the pairwise majority results matter. They are: Alpha beats Beta (3-2) and Gamma (3-2); Beta beats Gamma (3-2) and Delta (3-2); Gamma beats Delta (3-2); Delta beats Alpha (4-1). So the majority relation contains the cycle Alpha, Beta, Delta, Alpha, and Gamma is beaten by both Alpha and Beta. Sophisticated outcomes of such an agenda can be found by the Shepsle-Weingast rule: scan the agenda from the last item backward, keeping a list that starts with $x_4$; an item joins the list if it beats by majority every item already on it; the outcome is the last item to join (this is the backward-induction result). Here the outcome depends only on the item voted on last. If $x_4$ is Delta, Beta and Gamma beat Delta but Alpha does not; Beta beats Gamma, and whichever of Beta, Gamma gets on the list, Beta ends up on it last, so Beta wins. If $x_4$ is Alpha, Delta joins (it beats Alpha), and nothing beats both Alpha and Delta, so Delta wins. If $x_4$ is Beta, Alpha joins, and nothing beats both Beta and Alpha, so Alpha wins. If $x_4$ is Gamma, both Alpha and Beta can join, Delta cannot, and Alpha (which beats Beta and Gamma) always ends last on the list, so Alpha wins. So Alpha wins exactly when Beta or Gamma is placed last: 2 × 3! = 12 orders (Beta and Delta win 6 each, Gamma never). A full backward induction over all 24 orders confirms this. Answer: 12.
Source: Agenda control in sequential binary voting, after Farquharson (1969) and McKelvey and Niemi (1978). Statement written for AxiomIQ.