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chrisnhowell

Infinite monkeys and Shakespeare

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Posted (edited)
12 minutes ago, wtf said:

They might get lucky with probability zero. 

Do you agree that there are only countably many reals whose decimal representation ends in pi? Because there can be only a finite sequence of digits preceding the point where pi starts. And there are only countably many finite strings of digits. You agree?

To the first statement, yes. They could get lucky with probability 0. But why could they not get lucky with probability 1? As a combined effort, meaning that it is enough that a single one of the monkeys achieves the goal? You seem to focus on the probability that a single monkey will make it. Which is zero. But since we have enough workforce assembled, just one of them might get the job done. Like in a movie with Bruce Willis, say. 

Yes, there are countably many reals with a representation that agrees with \( \pi\) from some point on. I am actually only asking now about the reals for which their representations agree with \( \pi \) entirely. Which means, the monkey completes the job successfully if and only if it types \(\pi \) and nothing else, that's it. That does not make any real difference for the question though. Instead of countably many successful outcomes, it reduces to only one successful outcome, which is a small simplification.

 

 

Edited by taeto

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Posted (edited)
2 minutes ago, taeto said:

To the first statement, yes. They could get lucky with probability 0. But why could they not get lucky with probability 1?

 

Because there are only countably many such real numbers ending in the digits of pi. You can't get around that. The measure of the set of reals whose decimal ends in pi is zero.

You seem to focus on the probability that a single monkey will make it.

No, I have calculated the probability that an uncountable collection of monkeys will make it. The probability is zero. There are uncountably many real numbers, of which only countably many end in the digits of pi.

Perhaps you are imagining that countably many strings generated by one monkey times uncountably many monkeys gives you a larger set of possibilities. But the product of a countable set and an uncountable set equinumerous to the reals is has the cardinality of the reals. 

 Which means, the monkey completes the job successfully if and only if it types π and nothing else, that's it. 

Well there is only ONE such string, namely pi itself. 

Edited by wtf

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3 hours ago, taeto said:

Don't lie please...

If "Spacetime" is modelled at least topologically as R4 as it usually is (disregarding BH's and the time before BB), then some people might argue exactly that.

I notice you maintain that most people think that and don't provide a single reference. I couldn't find anyone other than you claiming the above.

23 minutes ago, taeto said:

Excellent idea. If you claim that any region of R4 is not uncountaby infinite, it is very extraordinary and needs references. I will not hold my breath.

Nice vague statement, to which the only answer is yes or no, depending on what you mean.

 

44 minutes ago, taeto said:

However, the outcome of the experiment does mean the same as a function with domain "one flip" to the set of outcomes {H,T}. Which is what I indicated.

No comment. I'm done with wasting my time.

As you've called me a liar and a troll, why not report the relevant posts?

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2 minutes ago, wtf said:

Because there are only countably many such real numbers ending in the digits of pi. You can't get around that. The measure of the set of reals whose decimal ends in pi is zero.

You realize that I assume an uncountable number of monkeys? 

5 minutes ago, Carrock said:

As you've called me a liar and a troll, why not report the relevant posts?

Done.

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Posted (edited)
7 minutes ago, taeto said:

You realize that I assume an uncountable number of monkeys? 

Yes. If you've read my posts you know that. How could you ask? How many times have I agreed?

I think I might see what you're saying though. You pick an uncountable set of reals. What is the probability that pi is in the set? Is that a better representation of your question? 

I believe the answer is that it's not well defined. I'll have to think about that more. To analyze it, first we have to restrict our attention to the unit interval, so that we have a probability space whose total measure is 1. I've been ignoring that point but in this context it matters.

So we can say reasonably that, say, if you pick a random real in the unit interval, the probability is 1/2 that it's between 0 and 1/2. The probability is 1/3 that the chosen number is between 1/3 and 2/3, say. So we can assign probabilities to SOME subsets of the reals, but not all. What if you pick an unmeasurable set of reals in the unit interval? Then there's no probability at all.

So you'll have to be more specific about this uncountable set of reals. After all: Some uncountable sets of reals include pi; and some don't. Some uncountable sets of reals have a measure; and some don't. So your question is not well-posed. But I do think I understand your question now.

https://en.wikipedia.org/wiki/Non-measurable_set

Edited by wtf

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Posted (edited)
5 minutes ago, wtf said:

Yes. If you've read my posts you know that. How could you ask? How many times have I agreed?

Just thought to ask to be sure. You seem to revert to the question of the expectancy of success for just a single monkey. Whereas I feel that this probability has to be added up over the entire population of monkeys.

Edited by taeto

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1 minute ago, taeto said:

Just thought to ask to be sure. You seem to revert to the question of the expectancy of success for just a single monkey. Whereas I feel that this probability has to be added up over the entire population of monkeys.

I wish you would carefully read what I'm writing. I'm perfect well accounting for uncountably many monkeys in each of my last several posts. Which part is unclear?

You are asking if a randomly selected uncountable subset of the reals contains pi. I clearly answered that the question's not well-posed, since not all uncountable subsets of the reals have a measure assigned to them.

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5 minutes ago, wtf said:

You pick an uncountable set of reals. What is the probability that pi is in the set? Is that a better representation of your question? 

That sounds like an excellent representation! 

And it makes it seem like the answer is indeed that pi will belong to the set with probability zero. 

8 minutes ago, wtf said:

I believe the answer is that it's not well defined. I'll have to think about that more. To analyze it, first we have to restrict our attention to the unit interval, so that we have a probability space whose total measure is 1. I've been ignoring that point but in this context it matters.

Good point. 

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Posted (edited)
9 minutes ago, taeto said:

That sounds like an excellent representation! 

And it makes it seem like the answer is indeed that pi will belong to the set with probability zero. 

No, I said the probability may not be defined at all. You are replying to me way faster than it would take for you to give thoughtful consideration to my posts.

Again: 

* In order to have a probability space, which is required to have a total measure of 1, we need to restrict our attention to the unit interval of reals. 

* Therefore we have to replace pi with a proxy in the unit interval. Pi minus 3 = .141... is adequate for the task.

* So now we ask: If we pick a random subset of the unit interval, what is the probability it contains pi - 3 ?

* The problem is that there's no total probability distribution on the powerset of the reals. Some sets are not measurable at all.

* So this problem is insufficiently specified. 

* Probability theory doesn't apply to the powerset of the reals. You can't ask questions about what percentage of the uncountable subsets of the reals have such or so property. I know of no such theory. What is the probability that a random set of reals is open? Closed? Compact? Connected? These questions have no answers. They can't be well-posed at all.

 

 

Edited by wtf

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10 minutes ago, wtf said:

So we can say reasonably that, say, if you pick a random real in the unit interval, the probability is 1/2 that it's between 0 and 1/2. The probability is 1/3 that the chosen number is between 1/3 and 2/3, say. So we can assign probabilities to SOME subsets of the reals, but not all. What if you pick an unmeasurable set of reals in the unit interval? Then there's no probability at all.

That is a very astute remark, thanks!

It seems silly to assume that if monkeys are typing away each one randomly, then together they will somehow produce a measurable subset of the real numbers between 0 and 1, do you agree? 

It forces the assumption that all subsets are measurable. As in Solovay's model of the real numbers en.wikipedia.org/wiki/Solovay_model.

The Choice Axiom is false in this model, so one has to be careful.

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Posted (edited)
3 minutes ago, taeto said:

That is a very astute remark, thanks!

It seems silly to assume that if monkeys are typing away each one randomly, then together they will somehow produce a measurable subset of the real numbers between 0 and 1, do you agree? 

It forces the assumption that all subsets are measurable. As in Solovay's model of the real numbers en.wikipedia.org/wiki/Solovay_model.

The Choice Axiom is false in this model, so one has to be careful.

One has to be careful of Wikipedia. Solovay's model assumes an inaccessible cardinal. The problem with Wiki is that it doesn't provide context. Solovay's model is "inside baseball" in advanced set theory. It's not grist for the web forum mill.

In any event, questions about the measure in the powerset of the reals of subsets with various properties are meaningless. There's no measure to work with that I know of. 

Edited by wtf

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Just now, wtf said:

One has to be careful of Wikipedia. 

Solovay's model is well-known. I can provide other references if you like, but most are also on the bottom of the wiki page.

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Posted (edited)
6 minutes ago, taeto said:

Solovay's model is well-known. I can provide other references if you like, but most are also on the bottom of the wiki page.

I'm perfectly well-aware of Solovay's model. You are replying too fast to be giving any thought to any of this. You want to discuss inaccessible cardinals? Why are you throwing in out-of-context red herrings?

Edited by wtf

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Posted (edited)
23 minutes ago, wtf said:

I'm perfectly well-aware of Solovay's model. You are replying too fast to be given any thought to any of this.

I am just a monkey typing away :cool:

Maybe there are other forum members reading this, who happen to be unfamiliar with Solovay. I do not know why you think that by pointing to this relevant model, it shows that I am not giving thought. 

Where should I pause to think? I am already aware of all the facts that you presented so far. You are able to combine them better than I can, but once I get the drift, I feel up to date and ready to comment back.

Now you make me feel that you already have the answer, and you are just holding back on it to see how I will improve. But would you mind to skip that, please?

Edited by taeto

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Posted (edited)
6 hours ago, taeto said:

I am just a monkey typing away :cool:

 

A joke that occurred to me in the several hours I spent away from the computer this afternoon. Great minds think alike.

Here's an interesting take on the question. What is the probability that a random subset of the reals contains pi? Well, for each set that contains pi, its complement doesn't. The pi-containing and the pi-not-containing sets are in one-to-one correspondence, or bijection. So the probability must be 1/2. How about that!!

However ... that argument fails. Consider the question of what is the probability that a random positive integer is even. The evens and the odds are in bijection, so the probability must be 1/2. But the multiples of 3 are in bijection with the non-multiples of 3, and the multiples of 4 are in bijection with the non-multiples of 4. In fact there are very sparse sets of positive integers such as the primes that are in bijection with the non-primes. So the naive idea of 1/2 doesn't work.

By the way what IS the probability that a random positive integer is even? Well, it's NOT 1/2. Why is that? Well, it turns out that there is no uniform probability distribution on the natural numbers. And why is that?? The rules of probability theory as formulated by Kolmogorov include the requirement that a probability measure must be countably additive. We want that so that we can say that if we have pairwise-disjoint intervals of length 1/2, 1/4, 1/8, etc., their union has length 1. 

But now what should be the probability of picking a random positive integer? If it's zero, countable additivity says that the chance of picking any integer at all is zero. That's absurd. But if the probability is positive, then the total probability of all the integers is infinite. Either way we are stuck. We are force to conclude that there is no uniform probability measure on the positive integers; and in fact there is none on any countably infinite set. That's a hard fact of life. [Uniform just means each number has the same chance of being picked].

https://en.wikipedia.org/wiki/Probability_axioms

Now there IS a way out of this. We can relax the requirement of countable additivity and require mere finite additivity. Then there IS a finitely additive probability measure on SOME special subsets of the natural numbers. For example if we pick some finite number n, about half of the numbers below n are even; and as n goes to infinity, the proportion of even numbers goes to the limit 1/2. We call this the asymptotic density of the even numbers. Similar reasoning shows that the asymptotic density of the multiples of 3 is 1/3, and so forth. This seems promising!

BUT! Asymptotic density only applies to very specialized sets. For example the primes get very thin as you go further out; and the asymptotic density of the primes is zero. And again in terms of probability, asymptotic density is only finitely additive and is therefore not a full-fledged probability measure.

https://en.wikipedia.org/wiki/Natural_density

With these intuitions in hand, let's get back to the pi-containing sets. What we need now is some kind of measure or weakly additive or vaguely useful notion of a measure that would apply to the powerset of real numbers. I don't know anything about any such measure. I'm sure mathematicians have probably studied such a notion, but I can't get any insight into the question of how many sets of reals contain pi. It seems like a question that would involve some advanced math I'm not familiar with, if a solution exists at all. My guess is that the question doesn't have an answer but I've gotten curious about this so I'll look around.

ps -- I did a little Googling. Questions about measures on the powerset of the reals quickly lead to questions about measurable cardinals. Those are the next large cardinals after the inaccessible cardinals required by Solovay's model. These are all cardinals whose existence is independent of standard set theory. I found one quote in one Mathoverlow page that's understandable and on-point. Mathoverflow is a site for professional mathematicians. This is from  Stefan Geschke's comment underneath Andrés E. Caicedo's green-checked answer:

"... you cannot construct a sigma-additive measure on P(R) without the help of some strong additional axioms."

https://mathoverflow.net/questions/103583/finite-measure-on-the-power-set

What this means is that the question of what proportion of subsets of the reals contain pi is a question of higher set theory at the professional level; and not answerable within standard set theory. As far as I could follow this page.

pps -- Another interesting MO thread relating to this discussion is: "Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable?" Of course by "interesting" I mean, "Interesting to see that professional mathematicians talk about this stuff, even if I can only understand a word here and there."

https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable

 

 

Edited by wtf

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Posted (edited)

    That is a very interesting discussion, thanks!

     I will look at the links. I suspect that a "random subset" can be explained so that if \(t\) is any real number between 0 and 1, and if \(n\) is any positive natural number, then from random and independent binary values \(b(t,n)\in \{0,1\} \) we get a random real \(x(t)\) between 0 and 1, which has \(b(t,n)\) as its \(n\)'th binary digit. The set \( X = \{ x(t) : t \in [0,1] \} \) should be the random set. If this does not work, then it is a little difficult to find a relationship with the typing monkeys question. 

     I find it hard to imagine that the random subset \(X\) is Lebesgue measurable with positive probability in ZFC, since the Choice Axiom produces a lot of non-measurable subsets, likely enough to drown out the contribution from the measurable ones. But in ZF it is at least consistent that \(X\) is always measurable, due to the Solovay model. And it is reasonable to expect that further axioms are needed to make \(X\) provably measurable, and the question makes sense whether \(\pi-3\) is in \(X\) with some probability.

     

Edited by taeto

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