FIRST HALF

It's the

**closing ceremony**and some awards are going to be delivered between the winners of the various categories.
In the section of best

**scorers**, first**3**qualified players are:
1.-

**Yuki Ogimi**, Japanese footballer, with 12 goals.
2.-

3.-

**Lotta Schelin**, Swedish player, also with 12 goals.3.-

**Christine Sinclair**, forward and captain of the Canadian team, with 4 goals.
There's a

Second place, also with 12 goals, but less assits than Ogimi, is for Swedish player

**draw to 12 goals**in the first place. Tha Japanese**Ogimi**is the winner because she has made more**goal assists**.Second place, also with 12 goals, but less assits than Ogimi, is for Swedish player

**Schelin**.
And the third scorer is

**Sinclair**, the Canadian player, with 4 goals.
The Championship

**sponsor**, Yukon Cars has decided to**give away 40 cars**.
They'll
give a car to the president of the Organizing Committee, another one to
President of the Referees’ Committee, and
the

**38 remaining cars**will be apportioned among the 3 top**scorers**in proportion to the goals scored by each one.
At the last
minute, President of the Organizing Committee decides to reject his gift
because he has just been appointed CEO of Manitoba Motors, a direct competitor
of Yukon Cars.

Thus, his
car will join the other 38 to be distributed among the 3 best scorers.

These news are greeted with joy by all players. Well, all except for

**Christine Sinclair**, who is, in addition to good footballer, a**fond of Mathematics**.
Why do you think

**Christine**didn't like this late change?
SECOND HALF

Apparently it's no sense she's sad because there's one more car to apportion, right?

First of all, it seems that if the sponsor is giving away

**one more car**, this shouldn't harm anyone, on the contrary,**someone will benefit**because of it.
Moreover, as that the top 2 scorers are tied in goals, and the extra car can't be distributed between both players in a fair way, it seems logical that Christine will be who gets the extra car.

Nevertheless,

**Christine Sinclair**is right to be**worried**.
Why? Hasn't she got enough parking space for so many cars?

No, that's
not the problem. Actually, she's going to offer all the cars she gets among her
teammates, who have helped her to score these goals.

Perhaps, like the president of the Organizing Committee, she has either a family meber or a friend who works in Manitoba Motors, the rival brand?

No, not
really. In addition, she loves cars manufactured by Yukon Cars.

Then I don't understand...

Let's see the

**first gifts allocation**to the three players, when the sponsor was going to apportion**38 cars**.
We
see that Ogimi and Schelin should have received 16.2857 cars (mathematicians
would refer to a 'quota' of 16.2857) and Sinclair should have got 5.4286 cars.
As far as it's not possible to

**divide a car**, we determine that the Japanese and the Swedish will get 16 cars, and Christine will get 5 cars (which corresponds to the 'integer part' of their quotas). So far we've distributed 16+16+5 = 37 cars. And the**remainig car**belongs to Christine, because she's the player whose**fractional part**is**the largest**and closest to next unit.
I didn't
understand very well this last deal...

Well,
Ogimi and Schelin, in addition to their 16 cars, should have got 0.2857
additional cars, while Christine should have received 0.4286 more cars. So
Christine is more entitled to take the remaining car than the other two players.
This way of apportionment is called in mathematical terms as the ‘

**largest remainder method**’ or ‘**Hamilton's method**’.
Now I see
it more clearly.

Let's
see what happens when the sponsor decides

**to give away one more car**.
Now
we see that the first 2 players are given 16 cars, and Christine receives 5
cars. With

Ogimi and Schelin should get 0.7143 more cars, while Sinclair should get only 0.5714 more cars. Therefore, the 2 remaining cars are now assigned to each of the first 2 scorers, because their

**the two remaing cars**, we'll use the same method as before.Ogimi and Schelin should get 0.7143 more cars, while Sinclair should get only 0.5714 more cars. Therefore, the 2 remaining cars are now assigned to each of the first 2 scorers, because their

**fractional parts are larger**.
So now we give 17 cars to Ogimi, another 17 to Schelin, and only 5 cars to Christine, don't we?

That's
right. It turns out that increasing one gift, not only

**Christine**doesn't get one more car, but she**loses one**of hers.
But this question will be very unusual, right?

You
shouldn't think so. Actually, this issue has received special attention from
mathematicians, who call it by the name of

**Alabama Paradox**.
Why this name?

Map from the Nations Online Project |

This mathematical paradox was detected for the first time in the

**United States House of Representatives**. In it, the number of each state's**seats**are redistributed every 10 years, regarding on their**population**growths.
After the 1880 census, a study was made on a possible extension of the number
of seats in the House. In this study it was found that the state of

**Alabama**would have got 8 seats in a chamber of 299 representatives, but only 7 with a House size of 300.
And is
there any method of proportional apportionment in which these paradoxes do not
occur?

Lots
of mathematicians have tried to give a solution as fair as possible to the
issue of after
a vote, but they haven't achieved a perfect method.

**allocation of seats**
Thus,
numerous ways have been created to assign seats based on the votes cast, like
the

**Hamilton's method**we've seen here.
We've got some

**, such al***divisor methods***Jefferson's**method (also known as**d’Hondt**method),**Webster's**method (method of**odd numbers or S****ainte-Laguë's**method),**Huntington-Hill's**method,**Dean's**method,**Adams'**method, or the**Danish**method, among others.
In
these systems we can find several

**:***paradoxes***Alabama**paradox, the**population**paradox, or the**new states**paradox**.**That means that every method generates some bias or favoritism to some population vs. other.
And we also have

*, among which we have the***preferential voting systems****simple or relative majority**, the**second round**, the**Borda**count, the**Condorcet**method, the**single transferable vote,**or the**'approval voting'**.
But they are also imperfect. We can remember our story about the Arrow paradox, for example.

But some methods are

**better**than others, right?
Yes,

**Webster's**method (also called**Sainte-Laguë`s**method**or****method of odd divisors**) seems to be the one which produces**minor injustices**, but it's not the most used. That's because there uses to be a**political will**to prioritize the**good governance**instead of the strictly proportional representativeness of the Parliaments, and to benefit the**major parties**, or such sort of things.
Now that
we've seen all the problems of the distribution, as in our case,

**Christine**did have reason to worry about when she heard they would deliver a car more...
No doubt, she would have previously read any of these

**great links**:*'The Constitution and Paradoxes', 'Apportionment: The Alabama Paradox' o 'Apportionment and rounding schemes'.*
And regarding on these issues of

**fair divisions**,**voting**and**paradoxes**, surely Christine have also read some of our stories:*And now, who should kick the penalty?*,*The problem of the fair division*, or*The golden goalkeeper*.
So, should
Christine ask for an apportionment based on Sainte-Laguë method?

Well,
despite being the method that produces less distortions, in this particular
case, its application would cause an

**additional problem**, as*Tom (C.M.) Thomson*discovered (many thanks for the comment). While it's also true that Christine may benefit from the use of that method.
Why?

Well, this
method is used as follows :

We add up
the goals scored by each player, and we calculate the quotients resulting from
dividing the number of goals by the successive odd numbers. And we'll give the
cars according to the highest resulting quotients.

Let's see
how the 38th and 39th cars would be assigned, according to the following table:

The cars
would be assigned to the players depending on the

**highest quotients**: 12 - 12 - 4 - 4 - 4 - 2.4 - 2.4 - 1.714 - 1.714 - 1.333 - 1.333 - 1.333 - 1.091 - 1.091 - etc...
As you can
see, the first 37 are easily given out (yellow cells). But with cars 38th and
39th we have a serious problem, since the three players are equally entitled to
the cars, as they've got the same next quotient: 0.364.

Obviously,
there should be some kind of rule to break ties. But if there's no fixed rule, the remaining 2 cars should be delivered by lot, so Christine would have a

**2/3**chance of getting the 6th car for her.
Then, no doubt, Christine should ask for this method to be used!

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