Example
Let's work through one more example using the IRV method. Consider the preference schedule below, in which a company’s advertising team is voting on five different advertising slogans, called A, B, C, D, and E here for simplicity. Use the IRV method to determine the winning slogan.
Answer:
Initial votes

3 
4 
4 
6 
2 
1 
1st choice 
B 
C 
B 
D 
B 
E 
2nd choice 
C 
A 
D 
C 
E 
A 
3rd choice 
A 
D 
C 
A 
A 
D 
4th choice 
D 
B 
A 
E 
C 
B 
5th choice 
E 
E 
E 
B 
D 
C 
Note that if we use the plurality method, slogan B would be the winner with 9 firstchoice votes, compared to 6 for D, 4 for C, and 1 for E.
There are total of 3+4+4+6+2+1 = 20 votes. A majority would be 11 votes. No one yet has a majority, so we proceed to elimination rounds.
Round 1: We make our first elimination. Choice A has the fewest firstplace votes (none), so we remove that choice.

3 
4 
4 
6 
2 
1 
1st choice 
B 
C 
B 
D 
B 
E 
2nd choice 
C 

D 
C 
E 

3rd choice 

D 
C 


D 
4th choice 
D 
B 

E 
C 
B 
5th choice 
E 
E 
E 
B 
D 
C 
We then shift everyone’s choices up to fill the gaps.

3 
4 
4 
6 
2 
1 
1st choice 
B 
C 
B 
D 
B 
E 
2nd choice 
C 
D 
D 
C 
E 
D 
3rd choice 
D 
B 
C 
E 
C 
B 
4th choice 
E 
E 
E 
B 
D 
C 
Since there is still no choice with a majority, we need to eliminate again.
Round 2: We make our second elimination. Choice E now has the fewest firstplace votes, so we remove that choice, shifting everyone’s options to fill the gaps.

3 
4 
4 
6 
2 
1 
1st choice 
B 
C 
B 
D 
B 
D 
2nd choice 
C 
D 
D 
C 
C 
B 
3rd choice 
D 
B 
C 
B 
D 
C 
Now B has 9 firstchoice votes, C has 4 votes, and D has 7 votes. Still no majority, so we eliminate again.
Round 3: We make our third elimination. C has the fewest votes so we remove that choice, again shifting everyone’s options to fill the gaps.

3 
4 
4 
6 
2 
1 
1st choice 
B 
D 
B 
D 
B 
D 
2nd choice 
D 
B 
D 
B 
D 
B 
We see that D now has 11 firstchoice votes. Having gained a majority (11 out of 20 votes or ~55%), slogan D is declared the winner under IRV.
Another important voting method,
Example
A group of mathematicians are getting together for a conference. The members are coming from four cities: Seattle, Tacoma, Puyallup, and Olympia. Their approximate locations on a map are shown below.
The votes for where to hold the conference were:

51 
25 
10 
14 
1st choice 
Seattle 
Tacoma 
Puyallup 
Olympia 
2nd choice 
Tacoma 
Puyallup 
Tacoma 
Tacoma 
3rd choice 
Olympia 
Olympia 
Olympia 
Puyallup 
4th choice 
Puyallup 
Seattle 
Seattle 
Seattle 
Use the Borda count method to determine the winning town for the conference.
Answer:
In each of the 51 ballots ranking Seattle first, Puyallup will be given 1 point, Olympia 2 points, Tacoma 3 points, and Seattle 4 points. Multiplying the points per vote times the number of votes allows us to calculate points awarded

51 
25 
10 
14 
1st choice 4 points 
Seattle
[latex]4\cdot51=204[/latex] 
Tacoma
[latex]4\cdot25=100[/latex] 
Puyallup
[latex]4\cot10=40[/latex] 
Olympia
[latex]4\cdot14=56[/latex] 
2nd choice 3 points 
Tacoma
[latex]3\cdot51=153[/latex] 
Puyallup
[latex]3\cdot25=75[/latex] 
Tacoma
[latex]3\cdot10=30[/latex] 
Tacoma
[latex]3\cdot14=42[/latex] 
3rd choice 2 points 
Olympia
[latex]2\cdot51=102[/latex] 
Olympia
[latex]2\cdot25=50[/latex] 
Olympia
[latex]2\cdot10=20[/latex] 
Puyallup
[latex]2\cdot14=28[/latex] 
4th choice 1 point 
Puyallup
[latex]1\cdot51=51[/latex] 
Seattle
[latex]1\cdot25=25[/latex] 
Seattle
[latex]1\cdot10=10[/latex] 
Seattle
[latex]1\cdot14=14[/latex] 
Adding up the points:
Seattle: [latex]204+25+10+14=253[/latex] points
Tacoma: [latex]153+100+30+42=325[/latex] points
Puyallup: [latex]51+75+40+28=194[/latex] points
Olympia: [latex]102+50+20+56=228[/latex] points
Under the Borda Count method, Tacoma is the winner of this vote.
This problem is also worked out in the following video.
https://youtu.be/vfujywLdW_s?list=PL1F887D3B8BF7C297
The final voting method that we will consider was devised by Nicolas de Condorcet, a French philosopher, mathematician, and political scientist. This method is based on pairwise (headtohead) comparisons between choices.
Example
Consider our vacation club one last time. The members of the club were deciding between Anaheim, Hawaii, and Orlando. Here is the preference schedule we created in the last section. Use the pairwise comparisons method and find the Condorcet Winner if there is one.

2 
3 
2 
4 
1st choice 
A 
A 
O 
H 
2nd choice 
O 
H 
H 
A 
3rd choice 
H 
O 
A 
O 
Answer:
We need to to analyze all three headtohead pairings to determine the Condorcet Winner (if there is one).
Anaheim vs Orlando: 9 to 2 preference for Anaheim over Orlando

2 
3 
2 
4 
1st choice 
A 
A 
O 
H 
2nd choice 
O 
H 
H 
A 
3rd choice 
H 
O 
A 
O 
Hawaii vs Anaheim: 6 to 5 preference for Hawaii over Anaheim

2 
3 
2 
4 
1st choice 
A 
A 
O 
H 
2nd choice 
O 
H 
H 
A 
3rd choice 
H 
O 
A 
O 
Hawaii vs Orlando: 7 to 4 preference for Hawaii over Orlando

2 
3 
2 
4 
1st choice 
A 
A 
O 
H 
2nd choice 
O 
H 
H 
A 
3rd choice 
H 
O 
A 
O 
Since Hawaii wins both of its headtohead comparisons, Hawaii is the Condorcet Winner.