Adding a weight to one element, but keeping the sum of elements equals to 1 I want to add a weight to one element but keeping the sum of all elements equals to one.
Consider the example below. I have a parliament with 10 seats, which is comprised by 3 parties (A, B, C) with the following distribution of seats:
A = 5, B = 3, C = 2.
Where, party A has 5 seats, party B has 3 seats, and party C has 2 seats.
In this particular case, because the size of the parliament is 10, if I want the information in percentage I just need to divide the parties' seats by 10, such that:
A = 0.5 (50%), B = 0.3 (30%), C = 0.2 (20%).
In the example above, each seat values the same 0.1 (10%). But now I want to give a weight to one of the seats (say, the prime-minister) because I believe it should be counted twice due to its importance. Hence, the party with the prime-ministership should have one of the seats counted twice (0.2 [20%] instead of 0.1).
Let's consider that party C has the prime-ministership. If I just add 0.2 to one of party C's  seats, the sum would be greater than 1. So, in order to keep the sum equals to 1 (100%) I need to normalize it such that the sum is equal to 1. My solution so far is to calculate the expressions below to each party (adding (1/10)*2 to party C):
Party A: (5/10 - 5*((1/10)/(10-1)))
Party B: (3/10 - 3*((1/10)/(10-1)))
Party C: ((2-1)/10 - (2-1)*((1/10)/(10-1))) + (1/10)*2
0.5555556 + 0.2666667 + 0.1777778 = 1.
For this simple example these calculations work well. But I get stuck when trying to add a greater weight to one of the seats (for example, 0.3 instead of 0.2). If I add (1/10)*3 to Party C, the sum again is greater than 1.
Is there a general formula that I can use to weight one of the seats, keeping the sum equals to 1, such that I can play giving a weight of 2 (0.2), of 3 (0.3), of 4 (0.4), and so on to one element?  Also, in the above example I kept things simple, but I have cases with more than 3 parties and parliaments with much more than 10 seats. Therefore, a generalization would be an ideal solution.

Just go back to the seats. If prime minister's seat is thrice as important, then he effectively has 3 seats instead of one. Thus the total number of seats are $N=5+3+(2-1+3)=12$. One seat has $k=1/12=8.3\%$ of votes. And parties have percentages:
$$
A=5k = 5/12 = 41.7\,\%,\\
B=3k = 3/12 = 25\,\%,\\
C=(2-1+3)k = 4/12 = 33.3\,\%,\\
$$
which add up to 1 (100 %)
