Example
Suppose there are 5 voters, of whom 3 vote for candidate A and 2 vote for candidate B (so p = 3 and q = 2). There are ten possibilities for the order of the votes cast:
- AAABB
- AABAB
- ABAAB
- BAAAB
- AABBA
- ABABA
- BAABA
- ABBAA
- BABAA
- BBAAA
For the order AABAB, the tally of the votes as the election progresses is:
| Candidate | A | A | B | A | B |
| A | 1 | 2 | 2 | 3 | 3 |
| B | 0 | 0 | 1 | 1 | 2 |
For each column the tally for A is always larger than the tally for B so the A is always strictly ahead of B. For the order AABBA the tally of the votes as the election progresses is:
| Candidate | A | A | B | B | A |
| A | 1 | 2 | 2 | 2 | 3 |
| B | 0 | 0 | 1 | 2 | 2 |
For this order, B is tied with A after the fourth vote, so A is not always strictly ahead of B. Of the 10 possible orders, A is always ahead of B only for AAABB and AABAB. So the probability that A will always be strictly ahead is
- ,
and this is indeed equal to as the theorem predicts.
Read more about this topic: Bertrand's Ballot Theorem
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