A group of people wishes to elect one of their members as their leader. Each member of the group can vote for any member in a smaller set of candidates of size . Candidates can vote for themselves. At least votes are needed to elect a leader. Suppose that each of the people votes for one of the candidates randomly. The result shows the approximate probability of election as a function of .

Snapshot 1: There are 7 voters and 5 votes are required for election. For 2 candidates, . For 3 candidates, decreases to . For 7 candidates (that is, everyone is a candidate), then .

Snapshot 2: Again there are 7 voters, but now only a simple majority (4) is needed for election. This is the so-called imperial election problem [1]. With 2 candidates, one of them always gets at least 4 votes, so that the election is certain. For 3 candidates, , and for 7 candidates, .

Snapshot 3: There are 7 voters as in snapshot 2, but now only 3 votes are required for election. With 2 or 3 candidates, one of them always gets at least 3 votes, so that the election is certain. For 4 candidates, , and even for 7 candidates is as high as 0.438.

Snapshot 4: There are 25 voters, and 17 votes (two-thirds of the votes) are needed for election. This is the so-called pope problem [1] that refers to the papal election in 1513. For 2 and 3 candidates, the probabilities of election are approximately 0.109 and 0.0013, respectively.

The Demonstration is based on problem 19 in [1].

Reference

[1] P. J. Nahin, Digital Dice: Computational Solutions to Practical Probability Problems, Princeton: Princeton University Press, 2008.