# The mathematics of whatever you want: some educational content regarding political systems

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This time, I go educational, and I go educational about political systems, and more specifically about electoral regimes. I generally avoid talking politics with my friends, as I want them to keep being my friends. Really, politics have become so divisive a topic, those last years. I remember, like 20 years ago, talking politics was like talking about the way to organize a business, or to design a machine. Now, it has become more like an ideological choice. Personally, I find it deplorable. There are always people who have more power than other people. Democracy allows us to have some control over those people in power, and if we want to exercise effective control, we need to get your own s**t together, emotionally too. If we become so emotional about politics that we stop thinking rationally, there is something wrong with us.

OK, enough ranting and moaning. Let’s get into facts and method. So, I start as I frequently do: I make a structure, and I drop numbers casually into it, just like that. Later on, I will work through the meaning of those numbers. My structure is a simple political system made of a juxtaposition of threes. There are 3 constituencies, equal in terms of incumbent voters: each constituency has 200 000 of them incumbent voters. Three political parties – Party A, Party B, and Party C – rival for votes in those 3 constituencies. Each political party presents three candidates in the electoral race. Party A presents its candidate A.1. in Constituency 1, candidate A.2. runs in Constituency 2, and Candidate A.3 in Constituency 3. Party B goes sort of the opposite way, and makes its candidates run like: B.1. in Constituency 3, B.2. in Constituency 2, and B.3. in Constituency 1. Party C wants to be original and makes like a triangle: its candidate C.1. runs in Constituency 2, C.2. tries their luck in Constituency 3, and C.3. is in the race in Constituency 1.

Just to recapitulate that distribution of candidates as a choice presented to voters, those in Constituency 1 choose between candidates A.1., B.3., and C.3., voters in Constituency 2 split their votes among A.2., B.2., and C.1.; finally, voters in Constituency 3 have a choice between A.3., B.1., and C.2. It all looks a bit complicated, I know, and, in a moment, you will read a table with the electoral scores, as number of votes obtained. I am just signalling the assumption I made when I was making those scores up: as we have 3 candidates in each constituency, voters do not give, under any circumstance, more than 50% of their votes (or more than 100 000 in absolute numbers) to one candidate. Implicitly, I assume that candidates already represent, somehow, their local populations. It can be achieved through some kind of de facto primary elections, e.g. when you need a certain number of officially collected voters’ signatures in order to register a candidate as running in a given constituency. Anyway, you have those imaginary electoral scores in Table 1, below. Save for the assumption about ‘≤ 50%’, those numbers are random.

Table 1 – Example of electoral score in the case studied (numbers are fictional)

 Number of votes obtained Party Candidate Constituency 1 Constituency 2 Constituency 3 Party A Candidate A.1 23 000 total score [votes]               174 101 Candidate A.2 99 274 Candidate A.3 51 827 Party B Candidate B.1 6 389 total score [votes]               111 118 Candidate B.2 40 762 Candidate B.3 63 967 Party C Candidate C.1 13 580 total score [votes]               134 691 Candidate C.2 33 287 Candidate C.3 87 824 Total 174 791 153 616 91 503

On the whole, those random numbers had given quite a nice electoral attendance. In a total population of 600 000 voters, 419 910 had gone to the ballot, which makes 70%. In that general landscape, the three constituencies present different shades. People in the 1 and the 2 are nicely dutiful, they turned up to that ballot at the respective rates of 87,4%, and 76,8%. On the other hand, people in Constituency 3 seem to be somehow disenchanted: their electoral attendance was 45,8%. Bad citizens. Or maybe just bloody pissed.

Now, I apply various electoral regimes to that same distribution of votes. Scenario 1 is a simple one. It is a strictly proportional electoral regime, where votes from all three constituencies are pooled together, to allocate 5 seats among parties. The number of seats going to each party are calculated as: “Total score of the party/ Total number of votes cast”. Inside each party, seats go specific candidates according to their individual scores. The result is a bit messy. Party A gets 2 seats, for its candidates A.2. and A.3., party B passes its B.3. man, and Party C gets C.3. into the Parliament. The first, tiny, little problem is that we had 5 seats to assign, and just 4 got assigned. Why? Simple: the parties acquired fractions of seats. In the strictly proportional count, Party A got 2,073075183 seats, Party B had 1,323116858, and Party’s C score was 1,603807959. I agree that we could conceivably give 0,32 of one seat to a party. People can share, after all. Still, I can barely conceive assigning like 0,000000058 of one seat. Could be tricky for sharing. That is a typical problem with strictly proportional regimes: they look nice and fair at the first sight, but in real life they have the practical utility of an inflatable dartboard.

Scenario 2 is once again a strictly proportional regime, with 6 seats to distribute, only this time,  in each constituency, 2 seats are to be distributed among the candidates with the best scores. The result is a bit of an opposite of Scenario 1: it looks suspiciously neat. Each party gets an equal number of seats, i.e. 2. Candidates A.2., A.3., B2., B.3., C.2., and C.3. are unfolding their political wings. I mean, I have nothing against wings, but it was supposed to be proportional, wasn’t it? Each party got a different electoral score, and each gets the same number of seats. Looks a bit too neat, doesn’t it? Once again, that’s the thing with proportional: growing your proportions does not always translate into actual outcomes.

Good. I go for the 3rd scenario: a strictly plural regime, 3 seats to allocate, in each constituency just one candidate, the one with the best score, gets the seat. This is what the British people call ‘one past the post’, in their political jargon. Down this avenue, Party A pushes it’s A.2. and A.3. people through the gate, and Party C does so with C.3. That looks sort of fair, still there is something… In Constituency 1, 87 000 of votes, with a small change, got the voters one representative in the legislative body. In constituencies 2 and 3, the same representation – 1 person in the probably right place – has been acquired with, respectively, 99 274, and 33 287 votes. Those guys from constituencies 1 and 2 could feel a bit disappointed. If they were voting in constituency 3, they would need much less mobilisation to get their man past the post.

Scenario 4 unfolds as a mixed, plural-proportional regime, with 5 seats to allocate; 3 seats go to the single best candidate in each constituency, as in Scenario 3, and 2 seats go to the party with the greatest overall score across all the 3 constituencies. Inside that party, the 2 seats in question go to candidates with the highest electoral scores. The results leave me a bit perplex: they are identical to those in Scenario 3. The same people got elected, namely A.2., A.3., and C.3., only this time we are left with 2 vacancies. Only 3 seats have been allocated, out of the 5 available. How could it have happened? Well, we had a bit of a loop, here. The party with the highest overall score is Party A, and they should get the 2 seats in the proportional part of the regime. Yet, their two best horses, A.2. and A.3. are already past the post, and the only remaining is A.1. with the worst score inside their party. Can a parliamentary seat, reserved for the best runner in the winning party, be attributed to actually the worst one? Problematic. Makes bad publicity.

Scenarios 5 and 6 are both variations on the d’Hondt system. This is a special approach to mixing plural with proportional, and more specifically, to avoiding those fractional seats as in Scenario 1. Generally, the total number of votes cast for each party is divided by consecutive denominators in the range from 1 up to the number of seats to allocate. We get a grid, out of which we pick up as many greatest values as there are seats to allocate. In Scenario 5, I apply the d’Hondt logic to votes from all the 3 constituencies pooled together, and I allocate 6 seats. Scenario 6 goes with the d’Hondt logic down to the level of each constituency separately, 2 seats to allocate in each constituency. The total number of votes casted for each party is divided by consecutive denominators in the range from 1 up to the number of seats to allocate (2 in this case). The two greatest values in such a grid get the seats. Inside each party, the attribution of seats to candidates is proportional to their individual scores.

Scenario 5 seems to work almost perfectly. Party A gets 3 seats, thus they get all their three candidates past the post, Party C acquires 2 seats for C.2. and C.3., whilst Party B has one seat for candidate B.3. In a sense, this particular mix of plural and proportional seems even more fairly proportional that Scenario 1. The detailed results, which explain the attribution of seats, are given in Table 2, below.

Table 2 – Example of application of the d’Hondt system, Scenario 5

 Number of votes obtained divided by consecutive denominators Denominator of seats Party A Party B Party C 1 174 101 111 118 134 691 2 87 051 55 559 67 346 3 58 034 37 039 44 897 4 43 525 27 780 33 673 5 34 820 22 224 26 938 6 29 017 18 520 22 449

On the other hand, Scenario 6 seems to be losing the proportional component. Table 3, below, shows how exactly it is dysfunctional. As there are 2 seats to assign in each constituency, electoral scores of each party are being divided by, respectively, 1 and 2. In Constituency 1, the two best denominated scores befall to parties C and B, thus to their candidates C.3. and B.3. In Constituency 2, both of the two best denominated scores are attributed to Party A. The trouble is that Party A has just one candidate in this constituency, the A.2. guy, and he (she?) gets the seat. The second seat in this constituency must logically befall to the next best party with any people in the game, and it happens to be Party B and its candidate B.2. Constituency 3, in this particular scenario, gives two best denominated scores to parties A and C, thus to candidates A.3. and C.2. All in all, each party gets 2 seats out of the 6. Uneven scores, even distribution of rewards.

Table 3 – Application of the d’Hondt logic at the level of separate constituencies: Scenario 6.

 Party A Party B Party C Denominator of seats Constituency 1 1 23 000 63 967 87 824 2 11 500 31 984 43 912 Constituency 2 1 99 274 40 762  (?) 13 580 2 49 637 20 381 6 790 Constituency 3 1 51 827 6 389 33 287 2 25 914 3 195 16 644

Any mechanism can be observed under two angles: how it works, and how it doesn’t. It applies to electoral regimes, too. An electoral regime doesn’t work in two respects. First of all, it does not work if it does not lead to electing anyone. Second of all, it does not work if it fails to represent the votes cast in the people actually elected. There is a term, in the science of electoral systems: the wasted votes. They are votes, which do not elect anyone. They have been cast on candidates who lost the elections. Maybe some of you know that unpleasant feeling, when you learn that the person you voted for has not been elected. This is something like frustration, and yet, in my own experience, there is a shade of relief, as well. The person I voted for lost their electoral race, hence they will not do anything stupid, once in charge. If they were in charge, and did something stupid, I could be kind of held accountable. ‘Look, you voted for those idiots. You are indirectly responsible for the bad things they did’, someone could say. If they don’t get elected, I cannot be possibly held accountable for anything they do, ‘cause they are not in a position to do anything.

Wasted votes happen in all elections. Still, an efficient electoral regime should minimize their amount. Let’s compare those six alternative electoral regimes regarding their leakiness, i.e. their tendency to waste people’s voting power. You can see the corresponding analysis in Table 4 below. The method is simple. Numbers in the table correspond to votes from Table 1, cast on candidates who did not get elected in the given constituency, under the given electoral regime. You can see that the range of waste is quite broad, from 4,8% of votes cast, all the way up to 43% with a small change. It is exactly how real electoral regimes work, and this is, in the long run, the soft spot of any representative democracy. In whatever possible way you turn those numbers, you bump on a dilemma: either the race is fair for the candidates, or the ballot is fair for voters. A fair race means that essentially the best wins. There is no point in making an electoral regime, where inefficient contenders have big chances to get elected. On the other hand, those who lose the race represent people who voted for them. If we want all the voters to be accurately represented in the government, no candidate should be eliminated from the electoral contest, only then it would not be a contest.

Table 4

 Number of votes, which do not elect any candidate Constituency 1 Constituency 2 Constituency 3 Total elections Scenario 1 23 000 40 762 33 287 97 049 Scenario 2 0 13 580 6 389 19 969 Scenario 3 23 000 54 342 33 287 110 629 Scenario 4 63 967 54 342 39 676 157 985 Scenario 5 (d’Hondt method, pooled) 0 54 342 6 389 60 731 Scenario 6 (d’Hondt method, separately by constituency) 23 000 54 342 39 676 117 018 Percentage of votes cast, which do not elect any candidate Constituency 1 Constituency 2 Constituency 3 Total elections Scenario 1 13,2% 26,5% 36,4% 23,1% Scenario 2 0,0% 8,8% 7,0% 4,8% Scenario 3 13,2% 35,4% 36,4% 26,3% Scenario 4 36,6% 35,4% 43,4% 37,6% Scenario 5 (d’Hondt method, pooled) 0,0% 35,4% 7,0% 14,5% Scenario 6 (d’Hondt method, separately by constituency) 13,2% 35,4% 43,4% 27,9% Average 12,7% 29,5% 28,9% 22,4%

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