Voting Paradox

1968 /www/nrich/html/content/98/08/15plus3/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <table border="0"> <tbody> <tr> <td><mdo:image height="100" width="100" src="mccain.jpg" alt="mccain"></mdo:image> </td> <td><mdo:image height="100" width="100" src="obama.jpg" alt="obama"></mdo:image></td> <td><mdo:image height="100" width="100" src="hillary.jpg" alt="clinton"></mdo:image></td> <td>Some relationships are transitive , such as `if A> B and B> C then it follows that A> C', but some relationships are intransitive , for example if A likes B and B likes C it does not necessarily follow that A likes C.</td> </tr> </tbody> </table> In a voting system, if A beats B and B beats C we might reasonably expect A to beat C. In this example there are 3 candidates for election. The voters have to rank them in order of preference. Consider the case where 3 voters cast the following votes: ABC, BCA and CAB. In the sense that one candidate is preferred to another : A beats B by 2 choices to 1. B beats C by 2 choices to 1 but A loses to C, again by 2 choices to 1. Three voters go to vote in this election and have to rank the candidates. First, check you agree that each voter has six possible ways in which they can do this. Assuming the voters are just as likely to rank them in one order as another, what is the probability that they all vote in a way that results in a paradoxical (intransitive) outcome? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. <ul> <li>If the first voter votes ABC, the second voter has to vote either BCA or CAB for the results to be intransitive.</li> <li>If the second voter then votes BCA, then CAB is the only order for the third voter which makes the results intransitive.</li> <li>If the second voter votes CAB, then BCA is the only order for the third voter which makes the results intransitive.</li> </ul> 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: <table style="" border=""> <tbody> <tr> <td> First voter </td> <td> Second Voter </td> <td> Third voter </td> </tr> <tr> <td> ABC </td> <td> ACB </td> <td> ABC or ACB or BAC, BCA, CAB, CBA </td> </tr> </tbody> </table> 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> It is <a href="http://nrich.maths.org/public/viewer.php?obj_id=1968">an exercise</a> in simple probability and combinatorics that provides an intriging and paradoxical situation for investigation. <h3>Possible approach</h3> The class could name 3 candidates to rank in order. Then everyone could write down their order of choice. You could then take 3 at a time and the class could discuss whether those three are transitive or not. After discussing several sets of 3 rankings they should be able to make conjectures about when the set will be transitive and when it will be intransitive. <h3>Key question</h3> How many possible sets of choice can be made in total by the voters? How many of these sets are intransitive? <h3>Possible support</h3> See the problems <a href="http://nrich.maths.org/public/viewer.php?obj_id=623&part=index"> A Dicey Paradox</a> and <a href="http://nrich.maths.org/public/viewer.php?obj_id=603&part=index"> Winning Team</a> and the article <a href="http://nrich.maths.org/public/viewer.php?obj_id=1345&part=index"> Transitivity.</a> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> How many different possible ways can each of the three voters vote? For a paradox we are looking for situations in which (A beats B beats C beats A) or (B beats A beats C beats B) etc. Try counting the number of ways in which such situations are possible. To reduce the number of cases to analyse, you might consider what happens if two voters agree on their first choice. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Can you find the probability of it being impossible to decide a winner in an election where there are 3 candidates A, B and C, and 3 voters, who place the candidates in order of preference? This was another tough nut which took 8 months to crack and then yielded simultaneously to the application of minds at opposite points of the globe. Well done Justin from Skyview High School, Billings, USA and again to Ling Xiang Ning from Tao Nan School, Singapore for the solution which follows. 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. 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. So, we have to find the total number of intransitive combinations of votes. First, we have to 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. I just take one combination of votes which two voters agree on their first choice, which is let us say A. <table border="0"> <tbody> <tr> <td>First voter</td> <td>Second Voter</td> <td>Third voter</td> </tr> <tr> <td>ABC</td> <td>ACB</td> <td>ABC or ACB or BAC, BCA, CAB, CBA</td> </tr> </tbody> </table> For all the six possibilities of what 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. Now, we have to find the total number of intransitive combinations of votes. 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 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*2 = 12). So, the probability of this paradox of collective choice arising is 12/216, that is 1/18. </mdoxml> 2 4 0 0 0 1 1 Voting Paradox Some relationships are transitive, such as `if A>B and B>C then it follows that A>C', but some are not. In a voting system, if A beats B and B beats C should we expect A to beat C? Probability Combining probabilities Probability Theoretical probability Decision Mathematics and Combinatorics Permutations Numbers and the Number System Counting