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Icarus
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Re: paradoxes
« Reply #25 on: Feb 13th, 2008, 8:09pm » |
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Which still leaves you with the apparent paradox. What is wrong with either the original, or your version, of the reasoning? Clearly something is, as each version says that both people have the advantage (in your case, each thinks the other has the advantage).
There are four ways to deal with paradoxes:
1) Note that the "paradox" is not really a paradox, because there is no actual contradiction. Buridan's Ass is an example of this. The poor critter may be left to starve, but there is no contraction in this. The Banach-Tarski paradox is another example. The only thing it contradicts is our intuition.
2) Show that the "paradox" is a result of misinterpreting the situation. This is the case in the "two children" paradox, or Monty Hall paradox. One calculation of the probability makes a false assumption.
If the paradox is real, you have the remaining two means of resolving it:
3) Redefine the situation so that the paradox is prevented from arising. Such is the way Russell's paradox was handled. Set theory was redefined so that either not all relations determined sets (Zermelo-Frankel), or else so that paradoxical relations could not be formed in the first place (Russell-Whitehead).
4) Accept that the language in question allows paradoxical statements. As such, the full language is not an appropriate place for deductive logic, though subsets of the language may be. This is the resolution of many natural language paradoxes. They just exist, so get over it. Just don't trust anything that comes out of them.
For the Necktie Paradox, (2) is the appropriate means of resolution. The men's calculation of the value of their bet is flawed. To resolve the paradox means finding the mistake they made.
So far, all you've done is restate the flawed argument from the other side.
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mikedagr8
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Re: paradoxes
« Reply #26 on: Feb 13th, 2008, 11:31pm » |
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on Feb 13th, 2008, 8:09pm, Icarus wrote:| So far, all you've done is restate the flawed argument from the other side. |
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Temporary could play a ping-pong like that with himself.
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temporary
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Re: paradoxes
« Reply #27 on: Feb 14th, 2008, 8:53pm » |
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I've tried tic-tac toe against myself, but I just can't win, and I know I didn't let myself win. Anyway, there are two perspectives and two people. One perspective gives them both the advantage, but the other gives them both a disadvantage, this is no coincidence. One has an advantage, one has a disadvantage, but you don't know which is which, so it's fair.
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Icarus
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Re: paradoxes
« Reply #28 on: Feb 15th, 2008, 3:24pm » |
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That the situation is fair, and neither has an advantage is clear, since their situations are symmetric. The "paradox" is that the reasoning shows them both to have an advantage (or disadvantage from the view you stated). This is clearly impossible. Therefore, either there is something wrong with their reasoning, or this is a true paradox.
Given that this isn't a true paradox, the problem becomes: what is the flaw in their reasoning? All you have shown is that they can also by the same flaw, reason in the opposite direction. But you have not spotted the flaw itself.
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Benny
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Re: paradoxes
« Reply #29 on: Mar 7th, 2008, 10:11am » |
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I found this paradox with solution. But in the solution there's a sentence i'm not sure about.
Imagine a pack of cards, each of which has one number on one side and the number directly above on the other. There is one card with 1 on one side and 2 on the other, two cards with 2 on one side and 3 on the other, four cards with 3 and 4, eight cards with 4 and 5, and so on, ad infinitum, so that there are 2^{n+1} cards with n on one side and n+1 on the other.
Let us now use that pack to play a game of chance.
Two people, Ann and Brian, stand facing each other,while the host draws one card from the pack and puts it between them, so that each can see the side facing him or her, but not the other side. The winner is the one who sees the lowest number.
What is Anns probability of winning? If she sees a 1, the other side must be a 2, and she has won. If she sees a number n>1, the hidden number, on the other side of the card, is either n-1 or n+1. In the first case she loses, in the other case she wins. Since there are twice as many cards of the second type as there are cards of the first, she has a probability 2/3 of winning.
Unfortunately, the same argument holds for Brian: he also has a probability 2/3 of winning. Since one must win, but not both, the two probabilities should sum to 1, so that 2/3 + 2/3 =1, a remarkable equality!!!
This is a true paradox. The argument is perfectly correct. Still, there is a problem with this argument.
Solution:
The argument is perfectly correct. The only problem with it is that there is no such pack of cards. One cannot physically construct a pack with infinitely many cards. Even if one could, what this argument shows is that one could not draw at random a card from it, otherwise one would end up with a contradiction. In other words, even in mathematics, there is no such thing as drawing a card at random from an infinite pack.
It is the last sentence of the solution,"In other words, even in mathematics, there is no such thing as drawing a card at random from an infinite pack."
Could we say that we can draw cards at random from an infinite pack, but not with equal probability to each card. If we assign a discrete probability measure on an infinite deck the probability of different cards will be different and the argument breaks.
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Eigenray
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Re: paradoxes
« Reply #30 on: Mar 7th, 2008, 12:59pm » |
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Yes, that's right. We can pick card {n,n+1} with probability pn  [0,1], and there is no problem as long as  pn = 1.
However, in the 'paradox', they are picking the card at random in such a way that {n,n+1} is always twice as likely as {n-1,n}, i.e., pn = 2pn-1. This is absurd; there is no way to have pn = 1.
The probability of seeing the number n is (pn+pn-1)/2. The probability of winning, given that you see the number n, is pn/(pn+pn-1). So the probability that you win is
(pn+pn-1)/2 * pn/(pn+pn-1) = pn/2 = 1/2,
as it should be.
For a non-absurd example, we can take p0=0, and pn = 1/2n, n>0. Then the probability of winning given that you see a 1 is 1, while the probability of winning given that you see any other number is only 1/3. But since you see a 1 with probability 1/4, your overall probability of winning is 1/4*1+3/4*1/3 = 1/2.
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Christine
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Re: paradoxes
« Reply #31 on: Mar 7th, 2008, 2:05pm » |
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I wonder whether or not Cantor's diagonal and Russell's self-reference argment are linked or connected.
Could anyone explain the relationship between the diagonal argument and self-reference?
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Icarus
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Re: paradoxes
« Reply #32 on: Mar 7th, 2008, 4:33pm » |
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Russell's demonstration can be thought of as a sort of super-diagonal argument. Let U be the universe. To every set S in U we can assign a function cS : U --> {0,1} defined by cS(x) = 1 if x S, = 0 if x  S. Note that S = cS-1(1).
The functions cS serve in the place of Cantor's exhaustive sequence. Now we go along the diagonal to construct a function c : U --> {0,1} that cannot be cS for any S.
Define c(x) = 0 if x is not a set or if x x, c(x) = 1 if x is a set, but x x. We see that for any set S, if cS(S) = 1, then c(S) = 0. But if cS(S) = 0, then c(S) = 1. Therefore, for no set S is c = cS.
Up to this point, there is a fairly direct connection between the two arguments. Here, they diverge. Cantor's adiagonal number doesn't lie in the sequence, so the sequence cannot be all real numbers. Well and good.
Russell's adiagonal c cannot be any characteristic function cS, so the characteristic functions cannot be all the functions from U --> {0,1}. Alas, this is not so good, as c clearly IS the characteristic function of the set c-1(1).
This is why Cantor's argument simply shows that  is bigger than  , while Russell's argument is a true paradox that required re-inventing mathematics to not allow the shenanigans I pulled off above.
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Christine
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Re: paradoxes
« Reply #33 on: Mar 8th, 2008, 10:54am » |
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i have a question about Russell's paradox:
The set M is the set of all sets that do not contain themselves as members. Does M contain itself?
Are the theory of types or Zermelo's special axiom of comprehention or VonNeumanns proper classes, required to show that the Russell class, {x:~(x e x)} does not exist?
could we say...
~EyAx(xRy <-> ~(xRx)) is sufficient. ie. ~EyAx(x e y <-> ~(x e x)) is a theorem. The Russell class does not exist.
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Icarus
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Re: paradoxes
« Reply #34 on: Mar 8th, 2008, 3:01pm » |
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Zermelo's axiom of comprehension was not created to solve Russell's paradox. Zermelo's solution to Russell's paradox is simply this: drop the axiomatic schema "for all unary relations R,  y x(R(x)  x y) is an axiom". This schema of naive set theory is responsible for the paradox. Zermelo resolved the paradox by dropping it, but this meant he needed to give some other tools of set creation in it's place. That is why he introduced the axiom of comprehension. The axiom itself has nothing to do with the paradox. Similar remarks also apply to the theory of types and VonNeumann's proper classes. Both were introduced as alternatives to give us back as much as possible of what we lost when the naive schema above was revoked, without re-introducing the paradoxes.
And "paradoxes" is why your approach doesn't work. It is not one paradox we must avoid. Russell's is just the simplest. But there are infinitely many alternate versions. We do not have to simply avoid his paradoxical set, but the entire situation that allows paradoxes to arise.
Another paradox - not directly related to Russell's - is this one (in many different variants): "the least integer not describable in nine English words." Note that the sentence describes the number in 9 words. This paradox was resolved by removing relational variables from the mathematical theory itself into metamathematics.
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Eigenray
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Re: paradoxes
« Reply #35 on: Mar 8th, 2008, 6:23pm » |
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Cantor's diagonal argument goes like this: Let S be a set, and suppose f : S   (S) is any function (so for all x S, f(x)  S). Then A={x : x f(x)} is not in the image of f. For, if f(x) = A, then x f(x) iff x f(x), a contradiction.
For example, if we have a set of people S, and f(x) is the set of people that x shaves, then A is the set of people who don't shave themselves. So there can be no x such that x shaves all those who don't shave themselves. But there's no paradox here. We just have Cantor's theorem that for any set S, there is no surjection from S to the power set of S.
But there's a problem if we let U be the universe. For any x U, x = { y : y x} U, so the identity maps U to (U). But now Cantor's argument tells us that A = { x : x x } is not in the image of the identity function, i.e., it doesn't exist! But this is only a paradox if you assume that A does exist (i.e., if you assume A is a set).
If the universe were a set, then we would have by the axiom of comprehension that { x U : x x} is also a set, and this would be a contradiction. So the way out (in ZFC) is just to accept the fact that the universe is not a set.
As for the Berry paradox, it is interesting to try to formalize it. Fix a universal Turing machine U, and write U(x)=y if given program x, U halts with output y. Let K(y) be the length of the shortest string x such that U(x)=y. Then the Berry paradox becomes:
Let xn be a program which computes the smallest positive integer y such that K(y) > n, so U(xn)=y, but for all strings x of length |x| n, U(x)  y.
The problem now is that if xn exists, then surely |xn| = O(log n), so |xn|<n for sufficiently large n, and therefore U(x) U(xn).
The conclusion is therefore that xn does not exist. If K(y) were computable, then we could simply write a program to compute K(1), K(2), ..., until we find y with K(y) > n. So therefore K(y) is not computable. And if the halting problem were computable, then K(y) would be computable, since we could simply run through all programs in order until we find x with U(x)=y. Therefore the halting problem is not computable.
Of course, the easier explanation of the Berry paradox is that if F is a function from strings to integers, then F("the smallest integer which is not F(s) for some string s of length < n") need not actually be the smallest integer which is not F(s) for some string s of length < n.
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Benny
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Re: paradoxes
« Reply #36 on: Mar 9th, 2008, 12:07pm » |
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How do you get the paradox from ""for all unary relations R, Ey, for all x R(x) <=> Ey) is an axiom". ?
It seems to me that "Ey, for all x R(x) <=> Ey)" is not well formed.
Did you mean; .. for some y, for all x (R(x) <=> (x is a member of y))?
It is easy to prove that the naive comprehention axiom is invalid...
From ie. ~EyAx(x e y <-> ~(x e x)) is a theorem, we can assert EF(~EyAx(x e y <-> Fx)),
That is ~AF(EyAx(x e y <-> Fx)) is a theorem.
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Icarus
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Re: paradoxes
« Reply #37 on: Mar 9th, 2008, 2:18pm » |
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on Mar 9th, 2008, 12:07pm, BenVitale wrote:How do you get the paradox from ""for all unary relations R, Ey, for all x R(x) <=> Ey) is an axiom". ?
It seems to me that "Ey, for all x R(x) <=> Ey)" is not well formed.
Did you mean; .. for some y, for all x (R(x) <=> (x is a member of y))? |
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Yes, that is what I meant - and if you go back and look at my post, you will see that that is also what I said. You've misread it.
Quote:It is easy to prove that the naive comprehention axiom is invalid...
From ie. ~EyAx(x e y <-> ~(x e x)) is a theorem, we can assert EF(~EyAx(x e y <-> Fx)),
That is ~AF(EyAx(x e y <-> Fx)) is a theorem. |
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Yes, it is easy to show it NOW, more than a century after mathematicians first formalized things enough to come across the paradox. Remember, you were trained in the formalism, and you have benefitted from a century of discussion of it. To the mathematicians of that day, a rigorous approach to set theory was a new concept, and the discovery of these paradoxes was something entirely unexpected. It took careful study for them to figure out where the problem was, and what were the best ways of fixing it.
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Benny
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Re: paradoxes
« Reply #38 on: Mar 9th, 2008, 4:21pm » |
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No, I don't think I've misread it, I may be wrong, but, perhaps your claim is not correct.
Could you please enlighten me?
"for all unary relations R, Ey, for all x R(x) <=> Ey)", what is that? what does it mean?
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Icarus
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Re: paradoxes
« Reply #39 on: Mar 9th, 2008, 7:03pm » |
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I don't know what it means because YOU and YOU ONLY have made that statement.
My statement does not drop the x and turn the element sign into some meaningless big E.
This is why I asked you to re-read it. I had hoped you would actually do so, and carefully this time.
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Re: paradoxes
« Reply #40 on: Mar 10th, 2008, 11:59am » |
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on Mar 9th, 2008, 4:21pm, BenVitale wrote:No, I don't think I've misread it[...]
"for all unary relations R, Ey, for all x R(x) <=> Ey)" |
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on Mar 8th, 2008, 3:01pm, Icarus wrote:| "for all unary relations R, yx(R(x) x y) is an axiom" |
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x y
is not equivalent to
Ey
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Benny
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Re: paradoxes
« Reply #41 on: Mar 11th, 2008, 12:31pm » |
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Sorry guys. I've made an error. Mea Culpa!
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temporary
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Re: paradoxes
« Reply #42 on: Mar 11th, 2008, 10:39pm » |
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Are you all still discussing the tie paradox, or solving random math problems?
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Benny
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Re: paradoxes
« Reply #43 on: Apr 21st, 2008, 3:48pm » |
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Take two round coins of equal size. Hold one still so that it does not move and then roll the other coin around it. Make sure the rims touch at all times. So my question is, how many times will the moving coin have rotated after it has completed one revolution of the stationary coin?
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ThudnBlunder
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Re: paradoxes
« Reply #44 on: Apr 21st, 2008, 4:23pm » |
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on Apr 21st, 2008, 3:48pm, BenVitale wrote:So my question is, how many times will the moving coin have rotated after it has completed one revolution of the stationary coin?
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Twice.
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Benny
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Re: paradoxes
« Reply #45 on: Apr 21st, 2008, 5:03pm » |
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But, we could argue that relative to a fixed point on the stationary coin, the moving coin only rotates once.
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ThudnBlunder
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Re: paradoxes
« Reply #46 on: Apr 21st, 2008, 5:28pm » |
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on Apr 21st, 2008, 5:03pm, BenVitale wrote:| But, we could argue that relative to a fixed point on the stationary coin, the moving coin only rotates once. |
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I was aware of the ambiguity in your question but presumed you were looking for the more interesting answer.
The same phenomenon explains the difference between solar time and sidereal time.
That is, why the constellations appear roughly four minutes earlier per day.
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| « Last Edit: Apr 21st, 2008, 6:02pm by ThudnBlunder » |
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temporary
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Re: paradoxes
« Reply #47 on: Apr 22nd, 2008, 5:20pm » |
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Not knowing what it is will give you the false result of gaining x or losing x/2. The problem is you change the sum of the envelopes from 3x to 3/2 x when you do that. If you consider that x can be in either one, you either gain x or lose x. If you know how much your envelope has, you consider that switching is best. the fallacy is that you change how much the sum is between the 2 outcomes, which influences a variable that shouldn't change.
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Benny
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Re: paradoxes
« Reply #48 on: May 5th, 2008, 11:27pm » |
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Voting paradox :
There are 3 candidates: Sen. John McCaine, Sen. Obama and Sen. Hillary Clinton.
Let Sen. John McCaine be A, Sen. Obama be B, and Sen. Hillary Clinton be C.
Normally, in a voting system, if A beats B and B beats C we might reasonably expect A to beat C.
But, in the following example, it doesn't work this way :
Consider the case where 3 voters cast the following votes: ABC, BCA and CAB:
A beats B by 2 choices to 1.
B beats C by 2 choices to 1
but A loses to C, again by 2 choices to 1.
Find the probability of collective choice arising by considering all the possible permutations of the votes that could be cast by 3 voters ranking 3 candidates by order of preference in this way.
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Re: paradoxes
« Reply #49 on: May 6th, 2008, 7:19am » |
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on May 5th, 2008, 11:27pm, BenVitale wrote:It's no more paradoxical than rock-paper-scissors, to be honest.
Quote:| Find the probability of collective choice arising by considering all the possible permutations of the votes that could be cast by 3 voters ranking 3 candidates by order of preference in this way. |
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Well, there's only 36 combination to consider, so you can easily write it out.
But I'm not sure what the criterion is for a collective choice. Suppose we have ABC ABC BCA, then we get
A beats B 2 to 1
A beats C 2 to 1
B beats C 3 to 0
We have 4 times A is preferred to someone, 4 times B is preferred to someone, only once C is preferred. A is preferred to everyone by 2 out of 3 voters. Should we go with A, or is it tied between A and B?
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