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.
If this was a plurality election, note that B would be the winner with 9 first-choice 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.
Solution
Round 1: We make our first elimination. Choice A has the fewest first-place votes, so we remove that choice
\(\begin{array}{|l|l|l|l|l|l|l|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{E} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & & \mathrm{D} & \mathrm{C} & \mathrm{E} & \\
\hline 3^{\text {rd }} \text { choice } & & \mathrm{D} & \mathrm{C} & & & \mathrm{D} \\
\hline 4^{\text {th }} \text { choice } & \mathrm{D} & \mathrm{B} & & \mathrm{E} & \mathrm{C} & \mathrm{B} \\
\hline 5^{\text {th }} \text { choice } & \mathrm{E} & \mathrm{E} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
We then shift everyone’s choices up to fill the gaps. There is still no choice with a majority, so we eliminate again.
\(\begin{array}{|l|l|l|l|l|l|l|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{E} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{C} & \mathrm{E} & \mathrm{D} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{E} & \mathrm{C} & \mathrm{B} \\
\hline 4^{\text {th }} \text { choice } & \mathrm{E} & \mathrm{E} & \mathrm{E} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
Round 2: We make our second elimination. Choice E has the fewest first-place votes, so we remove that choice, shifting everyone’s options to fill the gaps.
\(\begin{array}{|l|l|l|l|l|l|l|}
\hline & 3 & 4 & 4 & 6 & 2 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{B} & \mathrm{D} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{C} & \mathrm{C} & \mathrm{B} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{C} \\
\hline
\end{array}\)
Notice that the first and fifth columns have the same preferences now, we can condense those down to one column.
\(\begin{array}{|l|l|l|l|l|l|}
\hline & 5 & 4 & 4 & 6 & 1 \\
\hline 1^{\text {st }} \text { choice } & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{D} & \mathrm{D} \\
\hline 2^{\text {nd }} \text { choice } & \mathrm{C} & \mathrm{D} & \mathrm{D} & \mathrm{C} & \mathrm{B} \\
\hline 3^{\text {rd }} \text { choice } & \mathrm{D} & \mathrm{B} & \mathrm{C} & \mathrm{B} & \mathrm{C} \\
\hline
\end{array}\)
Now B has 9 first-choice 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.
\(\begin{array}{|l|l|l|l|l|l|}
\hline & 5 & 4 & 4 & 6 & 1 \\
\hline 1^{\text {st }} \text { choice } & \text { B } & \text { D } & \text { B } & \text { D } & \text { D } \\
\hline 2^{\text {nd }} \text { choice } & \text { D } & \text { B } & \text { D } & \text { B } & \text { B } \\
\hline
\end{array}\)
Condensing this down:
\(\begin{array}{|l|l|l|}
\hline & 9 & 11 \\
\hline 1^{\text {st }} \text { choice } & \text { B } & \text { D } \\
\hline 2^{\text {nd }} \text { choice } & \text { D } & \text { B } \\
\hline
\end{array}\)
D has now gained a majority, and is declared the winner under IRV.