OK. One could devise a preference system in which 2nd-choices come into play when the 1st preference has more than enough votes. But what’s the point? How many voters are going to have a clear opinion of several candidates along with the patience to mark a ballot accordingly? Party lists are a good simple solution. A separate ballot section could allow voters to affect the list ordering. (Only those voting for the Party X party list would be allowed to affect X’s ordering!)
Yes, with epsilon somewhere around 10^-659. You have to use a normal approximation to the binomial and then use a further approximation log of the the tail of the normal distribution.
Several countries have electoral systems which allow voters to rank the candidates in order of preference, and the voters seem to have both the opinions and the patience required to complete the ballot papers correctly. Ireland, Australia, and the UK (in Northern Ireland) have this system, and I suspect other countries do as well.
(My personal theory, for what it’s worth, is that part of the reason why the US has such low voter turnout is that an electoral system designed to minimise the degree of choice the voter gets to exercise discourages participation.)
As for the point, the point is that it gives the voter much more control over the outcome of the election which, from a civic and democratic point of view, is surely a good outcome?
Neat. I came up with 10[sup]-654[/sup], which I suspect is close enough when it comes to these calculations. I used this approximation:
ln(P(n)) =~ ln(N!) - ln(n!) - ln((N-n)!) + n*ln(p) + (N-n)ln(q)
Combined with Stirling’s approximation:
ln(n!) =~ n*ln(n) - n
Along with an approximation for adding logarithms:
ln(e[sup]x[/sup] + e[sup]y[/sup]) =~ max(x, y) + ln(1 + e[sup]-|x - y|[/sup])
And finally an approximation for logs close to 1:
ln(1+x) =~ x(1 - 0.5*x)
Did the integration manually and got an answer of e[sup]-1506[/sup].
Obviously, by lot would be about as fair as anything. Once a year, probably.
Then the problem is Dems vote for all Dems and Pubs vote for all Pubs which is worst than we have now.
Recall from #73 that the binomial has variance npq, i.e. the standard deviation as a proportion is √(pq/n) = √(.55*.45*300000) = .000908. A deviation of .05 is a 55-sigma event (.05/.000908 ~= 55)
On-line tables of the gamma function may not go up to 55 sigmas, but the following Taylor-series estimates the desired probability for any large sigma. Here y = 55√.5 = 38.89
0.5*(exp(-(y^2))/(sqrt(3.14159265)y))(1 - 1/(2y^2) + 3/(4y^4) + 15/(8*y^6))
This gives the very same 10^-659 figure that Buck Godot quotes. We’ll need to multiply by ten since the question was on the chance that any of 10 districts were skewed, so call it 10^-658.
Normally we would now do “inclusion-exclusion” to subtract the chance that two districts were skewed, but that chance is negligible.
Expand on this?
The country is not a pure division of D v. R. It seems as though the plurality, or at least a race-making fraction of voters are unaligned, so a districting scheme that either allows voters to choose multiple candidates on a single slate or elects the top five vote-getters would mitigate the stranglehold that the Ds and Rs have on the political system and possibly drive it more toward collaboration and compromise rather than bitter contention.
And with something like a top-five system, there is a clear-cut mechanism for term limits: a representative would not be able to be re-elected more than twice in a row in a bottom-two position.
There has been some concern expressed about electing weirdos and nuts (something our system has oh so effectively avoided): I contend that we actually need to do that, because the more we shut them out, the more angry, troublesome and dangerous they become. We need fewer outsiders as well as fewer insiders.
I absolutely support a voting scheme which makes it easier for third parties to gain traction. Whether there is a scheme that both accomplishes that goal, and also fights gerrymandering (which is the topic of this particular thread) I do not know.
I think they are mostly separate matters. Either could be addressed independently of the other.
If I’m a Pub, I use my 7 vote to vote for the 7 Pub candidates. Ditto for the Dems
I’m not sure what you’re saying. If the system was as follows: “there are 7 seats. Each voter gets 7 votes. Each party nominates 7 candidates. Whichever candidates get the most votes win.”, then that’s clearly terrible, because a region that was 52/48 D/R would end up with 7 democrats.
Or are we in agreement? I’ve lost track of who you’re responding to.
This subthread goes all the way back to Saint Cad’s #94 almost four days ago! And yes, Mr. Cad does apparently advocate that with a 52-48 split, the majority party should get all the seats.
What’s the point of multi-member districts, then?
No, no, no. You only get this result if you adopt multi-seat electorates but, for some perverse reason, retain a simplistic majoritarian voting system.
Yes, if you each voter gets to put an X against the names of seven candidates, and voters display strong party loyalty, then the party wth 51% support will win 100% of the representation . But since the whole point of having multi-seat electorates is to avoid this, it would be simply perverse to adopt such a system.
Instead, as suggested previously, you introduce preferential voting; voters number the candidates in the order of their preference. It was suggested earlier in this thread that voters would lack the understanding or the patience to do this, but the system is in operation in many countries and there’s not evidence that voters find it difficult to use. We have no reason to think that US voters are less capable in this regard than others.
Under such a system, if 51% of the voters are disciplined party loyalists and they give their first seven preferences to the candidates of the party, the party can expect to win three or, more probably, four of the seven seats. The voters collectively will determine, through the preferences they express, which of the party’s candidates get elected and which do not; the party cannot determine this. The party will not be disadvantaged by fielding an extremely popular candidate, who might “sweep up” votes from other candidates of the same party, since the suprlus votes, beyond what is required to elect the popular candidate, will transfer to the candiidates getting the voter’s second and subsequent preferences.
Plus, if the voters are not party loyalists, they can express their preference for candidates without regard to party affiliation, given their first preference to a candidate from party X, their second to a candidate from party Y, their third to a candidate from party X, their fourth to a candidate from party Z, and so forth.
What’s not to like?
Ummm … no. I advocated each person gets one vote. Others pointed out that voters should get as many votes as there are seats apportioned which I pointed out would lead the the problem you mentioned.
Ref #106
Fair enough. But then the method you recommend in #94 leads to the very significant problems others point out:
If Party A expects to win only two seats, but wishes to run more than two candidates it must make sure the other candidates are non-charismatic, that they don’t draw votes away from their two winners.
Moreover, if Party A has a very charismatic candidate — call her Ms. Palin for fun — then it shouldn’t run her! She would get most of the Party A votes, and Party A would win only one seat instead of two.
How do you figure? If your Mrs. Lapin is so popular that she gets 60% of the vote, that simply means that the remaining 6 representatives divide 40% of the vote among them. #2 might get 23% of the vote, #3, 9%, by the time you get to #7, you might be under 1%. That would give the rent-is-too-damn-high-party candidate a realistic prospect of gaining a seat in the house.