## 'Voting Paradox' printed from http://nrich.maths.org/

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Many of the solutions sent in this month applied values to a candidate's position in a voters' list. For example, if someone voted BCA, B would get $2$ points, C $1$ point and A $0$ points. With one of these models applied to the example all candidates would score a total of 3 and therefore come out equal. However, the paradox does not depend on assigning values to positions rather who does better overall and the paradox is that the answer is "no one" not because they all score the same but that everyone does better than everyone else!

Amongst the solutions we have recieved, two were from Justin of Skyview High School and Ling Xiang from Tao Nan School. I have used these as the basis of the following explanation.

As each voter can choose any one of the six orders ABC, ACB, BAC, BCA, CAB and CBA, there are altogether (6*6*6 = 216) different combinations of votes that could be cast.

We say the results are intransitive if, for example, A beats B and B beats C but A loses to C so that it is impossible to decide on a winner. So what combinations of votes would the voters need cast to make up a set of three that are intransitive?

We have to find the total number of intransitive combinations of votes over the total number of combinations of votes (216) to get the probability of the paradox arising.

• If the first voter votes ABC, the second voter has to vote either BCA or CAB for the results to be intransitive.
• If the second voter then votes BCA, then CAB is the only order for the third voter which makes the results intransitive.
• If the second voter votes CAB, then BCA is the only order for the third voter which makes the results intransitive.

So, if the first voter votes ABC, there would be $2$ possible combinations of votes making the results intransitive. As the first voter can vote 6 possible votes (ABC, ACB, BAC, BCA, CAB, CBA), the total number of intransitive combinations of votes possible is ($6\times2 = 12$).

So, the probability of this paradox of collective choice arising is $$\frac{12}{216}=\frac{1}{18}$$

However, if we are going to be really water-tight we should really prove that the results are intransitive if and only if no two voters agree on their first choice, nor on their second, nor on their third.

The proof might look something like this:

|             |              |                                  |
|-------------|--------------|----------------------------------|
| First voter | Second Voter | Third voter                      |
| ABC         | ACB          | ABC or ACB or BAC, BCA, CAB, CBA |

For all the six possibilities forwhat the third voter might vote, all the results would be transitive. So, the results are transitive if two voters agree on their first choice.

If two or more voters agree on their second choice, then either they also agree on the first, which we have just shown to be transitive or if not (for example ABC and CBA) then the 3 candidates are all exactly equal as a result of these two votes. The outcome will be decided by the order chosen by the third voter and the result is transitive.

If two or more voters agree on the third choice, suppose B is placed third by two voters, then A beats B and C beats B by at least 2 choices to 1. The result is transitive whether A beats C (when A beats C, C beats B and A beats B) or alternatively C beats A (when C beats A, A beats B and C beats B).

This means that the results are intransitive if and only if no two voters agree on their first choice, nor on their second, nor on their third.
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