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#1 2005-12-07 04:05:26
- b0blet
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Calculating probability of a condorcet paradox
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!
#2 2005-12-07 09:58:50
- Ricky
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Re: Calculating probability of a condorcet paradox
I don't think I understand the question. What I think you are looking for is if the matrix formed is reflexive, I can't find a case where it's not. If they all have to have different ordering, i.e.:
1. x y z
2. x y z
3. y z x
Is invalid (the order of 1 & 2 are the same), I believe the probability is 1, although I'm not quite sure what a proof of that would look like.
If the order is completely random (and therefore two people can have the same order), then you need to find out the probability for every order to be different. This is the only time a symmetric matrix can occur.
Last edited by Ricky (2005-12-07 10:13:20)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
#3 2005-12-07 16:38:58
- MathsIsFun
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Re: Calculating probability of a condorcet paradox
This is new to me, but looks interesting.
Condorcet Paradox on Wikipedia
I think the first step is to figure how many combinations we really have. I would suggest that wheher Voter 1, 2 or 3 casts a partiular vote doesn't matter. That will let us focus on real differences. In other words:
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
In other words the order of counting is disregraded (agree/disagree?)
So what are the possibilities?
x y z
z x y
y z x
x z y
y x z
z y x
Hmmm ... is that it? Other variations can have the order of rows changed to become either of those two?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#4 2005-12-08 01:42:37
- b0blet
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Re: Calculating probability of a condorcet paradox
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.
#5 2005-12-08 07:35:46
- MathsIsFun
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Re: Calculating probability of a condorcet paradox
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.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#6 2005-12-09 02:07:04
- b0blet
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Re: Calculating probability of a condorcet paradox
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? 
Please?
#7 2005-12-09 21:51:31
- MathsIsFun
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Re: Calculating probability of a condorcet paradox
I have been thinking about this off and on, so I am just going to write down my thoughts.
My first thought is to figure how many different votes one person could make:
1: x y z
2: x z y
3: y x z
4: y z x
5: z x y
6: z y x
(If I have left any out, just tell me, I am only mortal)
Next, we need to tally the combinations (order does not matter) of those possible votes
111 (it is possible that each voter casts exactly the same vote)
112
113
etc...
I think there should be 20.
Now, we need to think how many of them are paradoxes. Divide that by 20 and we have the probability.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman