Proposal: Adaptive Proposal Fees

demo

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Yes, if 51% of the voters are wrong the result is wrong. But what form of voting can sustain 51% of crazy voters? com'on.
Besides the mean average that can somehow sustain 51% of crazy voters, there is also another form of voting that can sustain 51% of crazy voters.

This is the time-splitting selection process. This selection process claims that it is absurd the 51% of the people to expect their decision to last for the 100% of the time.
Suppose we have 10 voters and they vote 1,1,1,1,1,2,2,4,4,32
This selection process divides time in time-windows (for example a time-window could be the human life expectancy) and it applies the result to whatever percentage the votes are.

In our example, and suppose the life expectancy is 100 years the result will be:
For 50 years the result is 1, for the next 20 years the result is 2 , for the next 20 years the result is 4 and for the last 10 years the result is 32. And then from the beginning, the same in the next time window and as long as the votes remain the same.

The crazy voters that voted 1, they have to live 10 years with 32, and this may result for them to reconsider and change their crazy vote in the next time-window.
 
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I didn't mean the mathematical stability. I meant the network stability. What happens to the mean (or to the median) average, in case some nodes (and their votes) go randomly on-line (so the votes of the nodes count) or off-line (so the votes of the nodes are not taken into account in the average) ? In this case the mean average is more stable (because the probability the off-line votes to be the extreme votes is very low) than the median average (because it blinks within the bounds of polarization).
Sorry, but no. Make some computer models and you'll see you are wrong.

The mean average can sustain 51% of crazy voters, as long as there are a few voters who keep their vote to the extreme (which is the correct in our case).
Demo, I am sorry, but I don't want a governance by extremists that balance each other.

I also need to say that your idea of governance is worrying at this point.
 

demo

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Sorry, but no. Make some computer models and you'll see you are wrong.
Have you try it in a computer model?
Thats interesting. It would be nice if you could give us your results.

P.S. @TroyDASH you rate this comment as troll but thats not trolling! The man is a scientist, so he may have done experiments on it.
 
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Have you try it in a computer model?
Thats interesting. It would be nice if you could give us your results.

P.S. @TroyDASH you rate this comment as troll but thats not trolling! The man is a scientist, so he may have done experiments on it.
No, @TroyDASH is right. It is upon you the weight of the proof, as I already presented the evidence before in the graph. Denying what someone else has said, and then asking from him to write you a computer program is trolling. Beside I have already showed you the graph. Just go backward in time, and you see how a median system breaks down respect to a mean system.

Said that it is really 3 lines of python, so I might write it if I have the time.
 

demo

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No, @TroyDASH is right. It is upon you the weight of the proof, as I already presented the evidence before in the graph. Denying what someone else has said, and then asking from him to write you a computer program is trolling. Beside I have already showed you the graph. Just go backward in time, and you see how a median system breaks down respect to a mean system.

Said that it is really 3 lines of python, so I might write it if I have the time.
Still you dont understand what I mean.
I mean a different case.
Suppose we have this set of votes 1,1,1,2,3,7,15,23,26
This set is stable, people dont change their votes (as they do in your example).
But randomly, some of the votes go off line, and they do not count.
Then for random reasons, they may appear again.
So the set may appear as
1,1,2,3,7,15,23,26
or as 1,1,1,2,3,7,23,26
or as 1,1,1,2,3,7,15,23
e.t.c.

We just shoot some votes completely randomly.

In this case, which is more stable, the mean or the median?
Can you prove (for whatever set of votes) whats happening?

I am not asking for a python program, I am asking for a mathematical proof (which may run in a python program)
 
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demo

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No, @TroyDASH is right. It is upon you the weight of the proof, as I already presented the evidence before in the graph. Denying what someone else has said, and then asking from him to write you a computer program is trolling. Beside I have already showed you the graph. Just go backward in time, and you see how a median system breaks down respect to a mean system.

Said that it is really 3 lines of python, so I might write it if I have the time.
Also I wonder this.We are talking here about voting the numbers. And you are saying about the condorcet method, which is used to compare between several choices. Isnt this a different case?

In the condorcet method the only thing that counts is whether something is bigger or smaller than something else. You rank the things, without bothering what is the distance between the ranks. Voting the numbers is not necessarily a condorcet method.

Lets take your clasic condorcet paradox:

If you have 35% of the people that prefer A to B to C,
30% of the people that prefers B to C to A, and
35% of the people that prefer C to A to B.
Then you have 65% of the people that prefer A to B; You have 65% of the people that prefer B to C, and 65% of the people that prefer C to A. And yet every single voter is completely rational.

Ok...but what if instead of forcing the people just to rank, you let them vote the numbers?

If you have 35% of the people that vote A=10 to B=5 to C=1 (they prefer A to B to C)
30% of the people that vote B=10 to C=2 to A=1 (they prefer B to C to A )
35% of the people that vote C=10 to A=5 to B=1 (they prefer C to A to B)

Then there is no paradox here because:
Α=35*10+30*1+35*5=555
Β=35*5+30*10+35*1= 510
C=35*1+30*2+35*10= 445

And there is a clear winner (A). You claim that 65% prefer C to A. Yes they do, but HOW MUCH they prefer it, you refuse to count it. This is the wrong with the condorcet method, it forces you to rank choices, but it does not let you vote the numbers.
 
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I must apologise to @demo as the result ended up being nowhere as trivial as I was expecting. And I am also grateful for what I discovered.

I just run the experiment. 4000 nodes, with random values, with normal (gaussian) distribution of float, taking away 100 of them one by one. And recalculating the median and the average at each step. And then taking the standard distribution of those medians and comparing them with the standard distribution of those averages. If my prediction was correct the std(median)<std(average). Instead on 500 experiments, only 172 resulted the median more stable than the average.

So demo was right in saying the average was more stable... if people voted floats.

But then I run again the experiment taking instead of floats integers between 0 and 10. And now the median was dead fix, and the average was changing all the time. And on 500 experiments all 500 the median was fix and the average changed.

So I run it again taking integers between 0 and 100, and now 333 / 500 had the median more stable.
And with integers between 0 and 1000 the result was 173/500. Just one more than with pure float.

So it looks like what really makes the average more or less stable than the mean is the range in which the people vote. If people vote chosing among 10 options, then the mean is more stable. If they chose among 1000 or more than the average is more stable.

But people when they vote tend on average all to chose among the same values. At least was my experience in the graph I posted above. So I rest my case that the median would be more stable. Also with the average people tend to polarise because they try to influence the result (strategic voting), and then the distribution would not be normal anymore. And the effect on the average would be much bigger.

If you want to play with the experiment you can find it here.

Something which some of you could try, for example, would be seeing how this is true if we change the distribution type.
 
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GrandMasterDash

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I must apologise to @demo as the result ended up being nowhere as trivial as I was expecting. And I am also grateful for what I discovered.

I just run the experiment. 4000 nodes, with random values, with normal (gaussian) distribution of float, taking away 100 of them one by one. And recalculating the median and the average at each step. And then taking the standard distribution of those medians and comparing them with the standard distribution of those averages. If my prediction was correct the std(median)<std(average). Instead on 500 experiments, only 172 resulted the median more stable than the average.

So demo was right in saying the average was more stable... if people voted floats.

But then I run again the experiment taking instead of floats integers between 0 and 10. And now the median was dead fix, and the average was changing all the time. And on 500 experiments all 500 the median was fix and the average changed.

So I run it again taking integers between 0 and 100, and now 333 / 500 had the median more stable.
And with integers between 0 and 1000 the result was 173/500. Just one more than with pure float.

So it looks like what really makes the average more or less stable than the mean is the range in which the people vote. If people vote chosing among 10 options, then the mean is more stable. If they chose among 1000 or more than the average is more stable.

But people when they vote tend on average all to chose among the same values. At least was my experience in the graph I posted above. So I rest my case that the median would be more stable. Also with the average people tend to polarise because they try to influence the result (strategic voting), and then the distribution would not be normal anymore. And the effect on the average would be much bigger.

If you want to play with the experiment you can find it here.

Something which some of you could try, for example, would be seeing how this is true if we change the distribution type.
My thinking is, most people would vote integers or half numbers, though of course some people would invariably be awkward. :) Anyway, it would be trivial to enforce increments of half dash votes, and I doubt anyone (except demo) would complain about that.
 
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My thinking is, most people would vote integers or half numbers, though of course some people would invariably be awkward. :) Anyway, it would be trivial to enforce increments of half dash votes, and I doubt anyone (except demo) would complain about that.
there is definitely no need to enforce anything. Set up a poll among your friends asking according to them how much the politicians should be paid. You can do it on google doc, and then advertise it on facebook. And you set it up in a way that who votes cannot read the votes of others. And then see what votes do the people vote. The numbers repeat a lot. Eventually there is a middle group of people that all think the same value, and then people around. a 10% of people that vote 4003.28 makes no difference at all to the median. but only assure a more smooth transition in case
 

demo

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I must apologise to @demo as the result ended up being nowhere as trivial as I was expecting. And I am also grateful for what I discovered.

I just run the experiment. 4000 nodes, with random values, with normal (gaussian) distribution of float, taking away 100 of them one by one. And recalculating the median and the average at each step. And then taking the standard distribution of those medians and comparing them with the standard distribution of those averages. If my prediction was correct the std(median)<std(average). Instead on 500 experiments, only 172 resulted the median more stable than the average.

So demo was right in saying the average was more stable... if people voted floats.

But then I run again the experiment taking instead of floats integers between 0 and 10. And now the median was dead fix, and the average was changing all the time. And on 500 experiments all 500 the median was fix and the average changed.

So I run it again taking integers between 0 and 100, and now 333 / 500 had the median more stable.
And with integers between 0 and 1000 the result was 173/500. Just one more than with pure float.

So it looks like what really makes the average more or less stable than the mean is the range in which the people vote. If people vote chosing among 10 options, then the mean is more stable. If they chose among 1000 or more than the average is more stable.

But people when they vote tend on average all to chose among the same values. At least was my experience in the graph I posted above. So I rest my case that the median would be more stable. Also with the average people tend to polarise because they try to influence the result (strategic voting), and then the distribution would not be normal anymore. And the effect on the average would be much bigger.

If you want to play with the experiment you can find it here.

Something which some of you could try, for example, would be seeing how this is true if we change the distribution type.
I'm so happy that a person like you joined the community.:)
I wish everyone here to learn to give scientific answers the way you do.
 
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demo

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there is definitely no need to enforce anything. Set up a poll among your friends asking according to them how much the politicians should be paid. You can do it on google doc, and then advertise it on facebook. And you set it up in a way that who votes cannot read the votes of others. And then see what votes do the people vote. The numbers repeat a lot. Eventually there is a middle group of people that all think the same value, and then people around. a 10% of people that vote 4003.28 makes no difference at all to the median. but only assure a more smooth transition in case
The question "how much the politicians should be paid." is not a correct question.
The correct question is "how much the politicians should be paid, as a percentage of the total budget, and what are the percentages you vote for other government expenditures?" .

The citizents must have a clear view that the more they pay for a government expenditure the less remains for the rest government expenditures. And this cannot be seen when you ask them separate questions for each expenditure. You have to ask them for all the expenditures together, and with the help of sliders similar to the below.



 
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Ryan introduces the median average, as a way to have adaptive budget system.
4 hrs 07 mins


Will the Dash community finally vote the numbers?
Let us see...