# 3 Candidate Condorcet Simulations

DEMOREP1 at aol.com DEMOREP1 at aol.com
Sat Nov 23 22:06:35 PST 1996

```I have done a spreadsheet of the simple 3 candidate (A, M and Z) head to head
example using random numbers to generate the numbers in the left column who
vote one of the 15 types of votes shown (voting for 1, 2 or 3). The random
number generator produces a number between 0 and 1 which is multiplied by 100
and rounded (i.e. a maximum of1500 voters is possible). The zeroes and * are
fillers.

* * A M M Z Z A
15 A 15 0 0 0 0 15
74 AM 74 0 74 0 0 74
19 AZ 19 0 0 19 0 19
78 AMZ 78 0 78 0 0 78
65 AZM 65 0 0 65 0 65

65 M 0 65 65 0 0 0
35 MA 0 35 35 0 0 35
77 MZ 0 77 77 0 77 0
37 MAZ 0 37 37 0 0 37
14 MZA 0 14 14 0 14 0

16 Z 0 0 0 16 16 0
10 ZA 10 0 0 10 10 0
90 ZM 0 90 0 90 90 0
17 ZAM 17 0 0 17 17 0
9 ZMZ 0 9 0 9 9 0

621 * 278 327 380 226 233 323 [Totals]
* * 0 1 1 0 0 1 [1= winning candidate in pairing]
* * 0.0 54.0 62.7 0.0 0.0 58.1 [Pairing Winner percentage]
When a candidate gets both wins, I repeat the winning percentages.
* * 0.0 0.0 54.0 62.7 0.0 0.0

* * 15.5 %Vote1 49.1 %Vote2 35.4 %Vote3
This lists the percentage of the total voters (621 in example) who voted for
1, 2 or 3 candidates respectively.

Should there be a restricted range of the percentage of voters who vote for
1, 2 or 3 candidates to produce more realistic examples (such as- not more
than 30 percent of the voters would vote for only 1 candidate)?

```