Angélica Benito Sualdea and Adolfo Quirós Gracián, teachers of Mathematics in the Autonomous University of Madrid present us with the second of the three challenges that EL PAÍS and the Royal Spanish Mathematical Society they are going to propose in this period of intense electoral activity.
This time it deals with how the circumscriptions affect the application of the D'Hondt method (also called the method of natural divisors or, in the United States, Jefferson's method), which is what governs most of the electoral processes in Spain. Readers can send their answers until 00.00 on Thursday, April 25 (midnight from Wednesday to Thursday, Spanish peninsular time) to [email protected].
Then, to clarify doubts, we add the statement of the problem in writing. Recall that the D'Hondt method proceeds as follows to distribute E seats. The votes of each party are divided successively between 1, 2, 3, … up to E, and the seats are assigned to the parties corresponding to the E greater quotients. To give an example, suppose that three parties, A, B and C, which have respectively obtained 240, 150 and 90 votes, should be divided 4 seats. We would make a table like this (although in fact it is not necessary to calculate it whole):
If we look at the larger quotients, the first seat would be taken by A (by 240), the second by B (150), the third by A again (120) and the fourth by C (90). The great majority of the lack of proportionality in the Congress of Deputies does not come from the use of the D'Hondt method, but from the fact that the election is made in 52 circumscriptions, and therefore they are in fact 52 different elections, in most of the which are distributed few deputies. For example, in the elections of April 28, 21 of the 52 districts have four or fewer seats, and it is impossible for any of them to win five parties.
The existence of 52 circumscriptions also makes it difficult to make predictions about the seats from surveys with a sample of sufficient size for all of Spain, but perhaps not for each province individually. Our challenge is addressed to this difficulty. Suppose we want to distribute, using the D'Hondt method, six seats between three parties, A, B and C, which have respectively obtained 63, 68 and 69 votes (if they seem small numbers, they can be "thousands of votes" without the varies). The first part of the table would look like this:
Since 69/3 = 23 <31.5 = 63/2, the distribution of the 6 seats would give 2 to each match. Let's imagine now that these 200 voters actually live in 2 provinces, 100 in each, and that in each province 3 deputies are elected, but we do not know how the votes are distributed among the provinces. The challenge is to give the maximum and minimum number of seats that Party A can achieve as the 200 voters are distributed. Similar to the previous challenge, we must give an example in which we obtain the maximum, another in which we obtain the minimum, and rigorously explain why A can not achieve more or fewer seats.
We wait for your answers.
The original title of this article included an error in the surname of Victor D'hondt that has already been corrected. We apologize for the failure, which a reader has warned us about.