Democracy. One vote per one person. The ones who get the most votes are elected. Right? Not quite, not in several representatives democracies all around the world.

The d'Hondt formula was developed in 1878 by the Belgian mathematician Victor d'Hondt (1841 - 1901), and is to this day used in Belgium. Most of the other countries using the d'Hondt formula, or method, are in Europe too. Examples are Finland, Switzerland, Portugal, and Austria.

The formula is simply: *V/(s+1)*. *V* is the sum of votes all the candidates of the party have received, and *s* is the number of seats already awarded to the party in question. Seat after seat V/(s+1) is calculated for the parties, to decide which seat goes to which party.

In practice, it looks like this:

Party A has three candidates, A(a), A(b), and A(c). Party B has two, B(a) and B(b). Three seats are being voted on. The tally of the election is:

A(a) 1000

A(b) 400

A(c) 700

B(a) 0

B(b) 3000

To see which candidates are elected, the total number of votes a party, or a candidate list, has received is added up. Party A's candidates have a total of 2100 votes, while Party B has a total of 3000 votes. The candidates are then ranked. They can
be ranked internally by the party itself ("a closed list"), or it can be done by popularity ("an open list"). In this
example, we'll be going by popularity:

A(a) 1000

A(c) 700

A(b) 400

B(b) 3000

B(a) 0

Then, the total sum of votes is calculated. For Party A it's 2100, for Party B it's 3000. It's time to start comparing the
candidates between parties by giving each candidate a comparison score, which is where the d'Hondt formula steps in. No seats
have been distributed yet, so the comparison score for A(a) is 2100/(0+1) = 2100. For B(b) it's 3000/(0+1) = 3000. Therefore, B(b) gets the first seat.

For A(a) it's again 2100/(0+1) = 2100, for B(a) it's now 3000/(1+1) = 1500, so A(a) gets the second seat.

Another way to calculate this: the total vote of the candidate list is divided by the rank of the candidate within that
list. So if a candidate is second on the ranked list, their comparison score is the total vote of the list divided by 2, if
they are 50th it is total vote divided by 50. After all, when person #1 on the ranked list has received a seat, person #2
will be receiving the 2nd one for that party, person #3 the 3rd one, rinse and repeat.

The ones who finally get elected are the ones with the highest comparison scores. Let us observe how our example election
will turn out. The following is a list with the comparison scores calculated for all candidates.

A(a) 2100

A(c) 1050

A(b) 700

B(b) 3000

B(a) 1500

Thus, B(b), A(a) and B(a) are elected. Looking at the actual amount of votes B(a) received, zero, the foremost problem
with the formula can be seen. A strong, very popular candidate can tow an undesirable candidate along them, and the
undesirable can be anyone the party decides to nominate. Therefore one is voting not only the candidate but the party. In an open list this is not always obvious, and the voter's vote can end up benefitting someone the voter dislikes. This can
make a party concentrate completely on a very charismatic and famous candidate while filling the rest of the list with
whoever it pleases. It also favours large parties which can gather several candidates and a large voting base, which will be
reflected in the comparison scores. Naturally, in a closed list system, the voters are voting only for the lists.

A real life example of this can be found in the Finnish parliamentary elections in 2003. Finland uses open lists. In the city of Helsinki the
leader of the Keskustapuolue, who moved to the city for the express purpose of attracting a huge number of votes, was elected
by a landslide (15700 votes). With the party leader, another candidate of the Keskustapuolue was elected, even though the candidate
himself received a small amount of votes (1600) compared to other candidates in other parties. His comparison score was just
so big - candidates with a personal vote of up to over 5000 could not beat him.

The method does enable small parties to achieve at least a slight representation, since the dividing factor becomes progressively bigger the more candidates are elected from a party. That way a small party can have a paltry total sum, but it only needs to be a fraction of the total sum of the larger parties.

To prevent too small parties from having disproportionate representation, some nations have a set limit, a barrage, that requires a list to have a certain percentage of the total national vote in order to be eligible for any seats.

If we disregard the single candidates, we can still calculate how many seats a list will get, and stick closer to the actual
d'Hondt formula. Once more, Party A receives 2100 votes and Party B receives 3000 votes. No seats have been allocated yet, so for both parties
we calculate *V/(0+1)*.

2100/(0+1) = 2100

3000/(0+1) = 3000.

Party B receives seat one.

2100/(0+1) = 2100

3000/(1+1) = 1500

Party A receives seat two.

2100/(1+1) = 1050

3000/(1+1) = 1500

Party B receives seat three.

A common device in a d'Hondt system is an election coalition. In a coalition parties throw their votes together into a
joint list, in effect, they are the one and same party for the purpose of distributing seats. This can be useful for small
parties with a handful of candidates that can attract a lot of votes across party lines, in order to compete with the big parties.

Another similar system, the Sainte-Laguë method, is based on the Sainte-Laguë formula. It is otherwise identical, but the total amount of votes is always divided by an odd number. So for the second seat, then, it's 3, for the third seat it's 5, et cetera.

Sources: Things I learned in my civic studies class, and http://en.wikipedia.org/wiki/D%27Hondt_method .