What is the probability this is correct. Another view on self-reference than SRH

 

Here's a self-reference statement to study.

Now SRH says that the existence of self-reference opens up a fixed-point, and a possible self-dependency and inconsistency.

To study let us imagine that a bag contains coins with a text written on them. There are N coins. Text 't' occurs with frequency F(t).

So picking out a coin we can say that the probability that this coin carries the text 't' is:

p(t) = F(t)/N

No self-reference no paradox.

Now we write probabilities on the coins. We have a function r(t) which returns the probability written by the text.

We pull out a coin and it has "10%" written on it. Very pedantic but r("10%") reads it to give us the probability 1/10.

Now if there are N=100 coins and 10 of then have "10%" written on them, then when we pull out a coin with "10%" on it then r("10%") gives a probability of 10% which is also the probability of pulling out a coin with "10%" written on it. We can call this the coin being correct.

Now if no other coin has the frequency that it occurs written on it then we can say that the probability of pulling out a correct coin is 10%.

No paradoxes yet.

So returning to the top we take 4 coins and write the frequencies on them: {0,25,25,50}. Then we ask what is the frequency of pulling a each text:

0     probability 25%
25   probability 50%
50   probability 25%

So there is no way to pull out a coin with the text on it that matches the probability of that text.

Now it is tempting then to say that the probability of pulling out a coin with the correct text on it is 0%. But now there is the contradiction because that is already taken by the text on the coin "0%" and the probability of that is 25% so it can't be 0%.

Indeed the coin "0%" is already slightly problematic because the existence of the text gives it a non-zero chance of being selected an so "0%" already contains a contradiction. It's not quite right to say it is always false, it is actually a contradiction.

So the problem is that we have 2 ways to refer to the coin "0%". One way (1) is to select that coin from the bag as stated in the game. The other way (2) is to select it as the answer to the question "the probability of pulling out a coin with the correct text on it."

So the answer is none. But by using method (1) the probability of selecting "0%" is 25%. But by answering the broader question of "which coin carries the probability that it is chosen" we select the "0%" coin.

This is actually a bit of a cheat. The context of the 25% is selecting randomly. The context of the 0% is selecting not randomly but according to a meta-criteria. They are different methods!

I wonder whether paradox can be avoided here by maintaining the context? Is this always true with paradoxes?

Rather a different solution to SRH which raises questions about self-referential methods of "selection".

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What I wanted to do was use this simple paradox to explore the full set of "fixed points" and whether each one can be used to generate an inconsistency. #TODO

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So a fixed point for a function f() is an element from the input set which maps to the same element in the output set

p = f(p) where p ∈ Sinput, Soutput

so f(x) in this case takes a quote of a probability and returns the quote of the probability of picking that from the bag. we assume that coins only have probabilities on them.

So in a bag of 100 coins then any coin with a number on it that matches the number of such coins in the bag is a fixed point.

In this smaller bag of 4 coins then: {"0%","50%","50%","70%"} means that 50% of the time you pick a coin with "50%" on.

"50%" = f("50%")

Note that everything is a "mention" of a probability in quotes no actual numbers.

Now the actual scenario involves a meta function g() which returns the probability of a coin being "correct." That is a coin being a fixed point of f().

So g("50%") = "50%" in that because "50%" is a fixed point then every coin with "50%" on when picked is "correct."

However "0%" is not a fixed point. The probability of picking "0%" is actually 25%. So the probability of it being correct is 0%

g("0%") = "0%"

So 0% is a fixed point of g() but not f().

Actually I can't see a problem here. To be completed...

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Is there a paradox actually?

There are 2 questions in the above.
(1) What is the probability of picking a certain probability in the answers. Call this f(p)
(2) To be "correct" the answer you chose must have a frequency equal to itself. That is f(p) = p. Call that g(p)

So we can define f(p) as:
f(0) = 25
f(25) = 50
f(50) = 25
Sum of f(x) for all x = 100

Great defined f with no problem.

Now g(p)
So for all p: f(p) != p. Then g(p) = 0 for all p.

Now that is a perfectly safe definition too.

The problem arises because the original question combines these so that the answer given by g(p) becomes a probability within the domain of f(x).

But isn't this apples and oranges. f(x) applies to the frequency of tokens in the question. g(x) applies to the frequency of a different domain namely the range of f(x). "0%" is a token being used in different domains.

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Let's try this on the classic "this statement is false"

If we really smash this up let us have two sets with names 'T' and 'F'. T ∪ F is everything. T ∩ F is empty. This is supposed to parallel True and False. Naively taking on Tarski here ;-)

So we could say:

What is the statement? "this statement is false"
This is just a sequence of characters.

Call that sequence of characters 'A'

Now semantically we can have a function f(x) which defines x is a member of F.

So the meaning of the statement A is f(A) which says that A is a member of F.

The problem is that we start with A being a member of T. But okay lets put A in the set F.

Great that makes sense. "this statement is false" simply says it is a member of the set F.

But we have missed something. If being a member of set F is supposed to parallel the behaviour of "false" then really it must mean "cannot belong to set T". You get put in set F only when you fail to fit into set T, otherwise you belong in T. I've always noticed this implicit bias to Truth in logic. All statements assume Truth to begin with. "There is an elephant on the Moon" we normally approach by looking on the Moon and if we don't find an elephant we assign it to false. Even if we assume it is false to start with we take the "There is not an elephant on the Moon" to be true. We always start with Truth. Maybe I misunderstand something here.

So actually we try and fit "this statement is false" into the T set. But it says that it belongs to the F set. Now it gets into the F set not because it says it belongs in F !! It gets into the F set because it does not belong in the T set. It says it belongs in the F set and so doesn't belong in T.

This very subtle difference is the difference between something being false because it is just false, and something being false because it is a logical contradiction.

"I am made from six words" is a truthful sentence. "I am made from five words" is a false statement.

"I am made of five and six words" is always false without counting because it is a contradiction. If it is made of 5 then it can't also be made of 6 of the same thing.

"I am a member of F" is the same in the rules above as "I am not a member of T" 

"I am not a member of T" cannot belong in T not because of its assigning membership but because it is a contradiction. Contradictions go in set F.

But we have 2 ways to get into set F now.

f(x) from above which says that x is a member of F.

But also anything that is a contradiction e.g. x & -x. says c(x) which also says x is a member of F if it is a contradiction.

Run out of time. But suppose we separate these two function domains as we did for the probabilities.

A = "I am false" maps to f(A) meaning put me in set F.

We could borrow the - sign to mean put me in set T. So -f(x) means put x in set T.

B = "This statement has both five and six words" belongs in F under c(B).

Now we can mix them.

"I am in both set T and F" belongs to F under the operation c(B) while it also assigns itself to T and F with f(x). This is fine.

C = "I am not in set F" is a set assignment and so is -f(C) which puts it in set T.

OKAY need another look. TBC