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#1 Re: Help Me ! » Calculating probability of a condorcet paradox » 2005-12-08 03:07:04

MathsIsFun wrote:

Sure? (Apart from you not rearranging them, but I know what you mean)

I would have thought that each ballot was independent and not linked to WHO voted, and so any counting order didn't matter.

But if it DOES matter, then my approach still has merit, becuase once you know the basic patterns (where the row order does not matter) then you can use perms/combs to calculate how many there would be if the rows DO matter.

So tell me how to do it? smile Please?

#2 Re: Help Me ! » Calculating probability of a condorcet paradox » 2005-12-07 02:42:37

MathsIsFun wrote:

1.x y z
2.z x y
3.y z x

is the same as

1.x y z
3.z x y
2.y z x

No, it does not matter whether you count the third voters preferences before the second. However, if voter 2 had voter 3's preferences and voter 3 had voter 2's preferences it would be a different outcome.

1. xyz
2. zxy
3. yzx

1.xyz
2.zxy
3.yzx

These two are not the same outcome.

#3 Help Me ! » Calculating probability of a condorcet paradox » 2005-12-06 05:05:26

b0blet
Replies: 6

In this problem there are 3 voters listing their preferences over 3 political parties x,y,z. Using majority rule, the condorcet paradox occurs when, for example, x > y, y > z, z > x . This creates a cycle and no one wins the abstract election.

EG.
1.x y z
2.z x y
3.y z x

In this example x > y, y > z, z > x.

EG2.
1.z y x
2.y x z
3.x z y

In this example y > x, x > z, z > y.

I've managed to write out 12 occurrences when this happens. What is the probability of a condorcet paradox occurring and how does one work it out?

I also want to do this for 4 voters and 4 parties but am even more boggled as to how to work it out.

Eg.
1.w z y x
2.x w z y
3.y x w z
4.z y x w

w > z, z > y, y > x, x > w .

Please help me. Thanks!

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