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The trick is that by using the Dhondt method and gerrymandering you have to manipulate way less votes than one would actually expect. In my country of 5 million

The trick is that by using the Dhondt method and gerrymandering you have to manipulate way less votes than one would actually expect. In my country of 5 million people last week 11 votes decided whether someone went into parliament and there were 11 excesss votes at one station which were then deleted and guess whose those were

jakehop 6y ago
While an interesting problem to try and solve (in theory, not in practice!), one has to be very much aware of where those 11 votes had to be injected for the “hack” to work. You basically need to have a somewhat precise picture of the proportionality before you could do it I a non-obvious way (= small margin victory). While polling gives you something close to that, I don’t believe that it’s precise enough.
jzwinck 6y ago
If the difference is only 11 people, it doesn't seem too much of a stretch to say the result doesn't matter, both candidates got the same number of votes within a very small margin of error (which will never be zero). Either you decide the two candidates were both good and there was no clear decision to elect one over the other, or you set up a voting system where such close races result in some totally different outcome, such as both people enter parliament for a shorter time, having a tie breaker of some kind, etc.
- Spivak 6y ago
  This assumes that voting in the US is a form of collective decision making, that’s the effect sure, but because politics are so tribal it’s actually a game of taking territory. Within a party equal votes implies what you suggest but that doesn’t hold in the actual elections.
- newacct583 6y ago
  What if the marginal/toss-up victories like this always end up going to one party? Is that still fair?
  - jzwinck 6y ago
    Good question. If it's a US style political system with only two viable parties, you could award near-ties alternately to each party. After all, if 50.1% of people vote for for party A every time, it is nonsense that party B with 49.9% never spends one day in office.

jakehop 6y ago

While an interesting problem to try and solve (in theory, not in practice!), one has to be very much aware of where those 11 votes had to be injected for the “hack” to work. You basically need to have a somewhat precise picture of the proportionality before you could do it I a non-obvious way (= small margin victory). While polling gives you something close to that, I don’t believe that it’s precise enough.

jzwinck 6y ago

If the difference is only 11 people, it doesn't seem too much of a stretch to say the result doesn't matter, both candidates got the same number of votes within a very small margin of error (which will never be zero). Either you decide the two candidates were both good and there was no clear decision to elect one over the other, or you set up a voting system where such close races result in some totally different outcome, such as both people enter parliament for a shorter time, having a tie breaker of some kind, etc.

Spivak 6y ago
This assumes that voting in the US is a form of collective decision making, that’s the effect sure, but because politics are so tribal it’s actually a game of taking territory. Within a party equal votes implies what you suggest but that doesn’t hold in the actual elections.
newacct583 6y ago
What if the marginal/toss-up victories like this always end up going to one party? Is that still fair?
- jzwinck 6y ago
  Good question. If it's a US style political system with only two viable parties, you could award near-ties alternately to each party. After all, if 50.1% of people vote for for party A every time, it is nonsense that party B with 49.9% never spends one day in office.