taeto

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.

You realize that I assume an uncountable number of monkeys? Done.

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.

Which to me still seems to argue, however correctly, that it is impossible for a single monkey to achieve the goal, and also for any countably infinite number of monkeys as well. But it does not seem to argue convincingly, why not at least one of uncountably many monkeys might get lucky?

It does not quite fly, I think. For simplicity, I suggest to just focus on the event that the successful monkey manages to type the entire sequence of the digits of \(\pi\) starting from 3.14... and ending, hm, not... For a single monkey, yes, unlikely. The question is what will happen if you have a pretty large crew of monkeys working away. Enough that it seems that in unison they can beat the odds. Really, there is only continuum many possible infinite sequences. Surely if you have a crew of \( 2^c \) monkeys working at it, where \(c\) is the cardinality of the continuum, then a host of monkeys will hit the right answer, right? But is there some reason that you would need that many?

Excellent idea. If you claim that any region of \( \mathbb{R}^4\) is not uncountably infinite, it is very extraordinary and needs references. I will not hold my breath.

Since I am countertrolling you, I can employ the concept of a function being "created" as a rhetorical device, in a display of sarcasm. The concept of a function being "created" is ridiculous enough. So you are absolutely right on all counts. Also the experiment to flip a fair coin absolutely does not mean a function. 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. Why not keep the obsequy so strict and not light-hearted?

YES: finally we are getting somewhere! Still, we could also ask an equivalent of the same question by using nothing but formal expressions in ZFC. Some would necessarily use measure theoretic notions, like, you know, lots and lots of greek letters. I prefer to stick with the monkey business for as much as possible. If a monkey is represented by a real number in \( (0,1) \) and a typed sequence is representing a real number in \( (0,1),\) then the outcome of the entire experiment means a function \( f : (0,1) \to (0,1). \) Let us be generous and assume that as we sit and stare at these words on the screen, the function \(f\) will form in something like a minute. That is, if we want at all to assign a time duration to its creation. No, let us be even more generous and give it five minutes; you should be able to prepare a kettle of water for tea while it works. Don't lie please... If "Spacetime" is modelled at least topologically as \( \mathbb{R}^4 \) as it usually is (disregarding BH's and the time before BB), then some people might argue exactly that.

The embiggenment of "correct" is most melliloquent.

The probability is precisely the same as the probability that the ratio between circumference and diameter of a disc happens, seemingly by coincidence, to be precisely equal to\(\pi.\)

Not at all. Remember that I assume that we already reside over uncountably many monkeys in that version of the problem. I can certainly map the reals injectively to the monkeys. I am pretty sure that wtf means to point to the entire set of possible infinite sequences that can be produced by one monkey. And to realize its magnitude, that is, cardinality. As a simplified example, suppose the typewriter has only two keys, one for 0 and one for 1. The possible sequences of these binary digits that are candidates for being typed are in 1-1 correspondence with the real numbers between 0 and 1, if we interpret each sequence as the sequence of binary digits that follow the decimal point in the binary representation of a real number. If you prefer decimal representation, you may assume that the monkey has the numeric keys from 0 to 9 available instead. Either way, the possible sequences correspond to real numbers from 0 to 1.

You are being a programmer now, not a mathematician . No, the letter \(n\) is not ``defined'' or ``initialized'' to be anything in particular. Yes, I admit that in a completely different question raised by StringJunky, I used the letter \(n\) to mean that natural number which is the length of the finite sequence of letters that each of a number of monkeys would have to type in the hopes of hitting the works of Shakespeare. Incidentally, also this \(n\) is supposed to equal your \(S\). If I said, "if we assume that the number of bolts used in the Eiffel Tower is \(n,\) then what is the probability that \(n\) is a prime?" it does not mean that whenever I mention a natural number called \(n\) at any subsequent time in a different discussion, it has to mean that \(n\) has anything to do with the Eiffel Tower. I was asked by Carrock to number the critters. I didn't really think that I needed to, but this way seems practical, i.e. for each real number between 0 and 1, there is a unique monkey with that number as its label. Got any alternative in mind? By "Shakespeare on repeat" I refer to a commonly used ancient technique on CD players (remember those?) of restarting the same CD at the moment when it reaches the end. The idea is that the monkey should type, starting from its very first typed letter, the complete works flawlessly until the end, immediately followed by repeating the exact same sequence of letters once again, and after that once more, etc. The constant repetition means that the money has to type an infinite sequence correctly, not just a finite one. It is in principle the same or very similar to requiring the monkey to type the digits of \(\pi,\) that is, typing 3.141592653... correctly without stopping.

I did try to be careful by typing: "the combined works of Shakespeare amount to precisely n letters". The point is that this sequence length is the only length you want to worry about for a subsequence to be a candidate for a hit. I have not considered to make any observations of the initial sequence of the first \(n\) letters typed by the monkey, which may be what you intended, possibly. And this is exactly what I replied that I would assume, that my \(n\) is equal to your \(S\). It means that for all possible choices of looking for a subsequence of length \(n,\) we can restrict to such \(n\) for which \(n=S.\) So far so good.

They are not meant to be actual living monkeys, for those who still wonder. You can number the monkeys conveniently by the real numbers in the open interval \( (0,1).\) Test the monkeys one by one instead of in a shared location, thus no extra space needed.

You cut away the context yet again, so here is the paragraph you seem to be replying to, judging from the time stamp: "if you have infinitely many monkeys, each hammering away randomly on a (their own individual) keyboard, and each continuing forever without stopping, then it might be interesting to observe whether any of them might fail to reproduce the works of Shakespeare infinitely many times. We already know that a single monkey will make it with 100% probability. The question is whether any of the monkeys might fail to achieve this, now that there are so many of them. And more precisely, what is the probability that at least one monkey fails, or even, what is the expected number of failure monkeys?" 1) There was no n or S given in the question, so I do not know what the quantities refer to. But I suspect that it means that you are supposed to count every occurrence of when a sequence of n letters matches exactly to the works of Shakespeare, assuming that the combined works of Shakespeare amount to precisely n letters. So yes, that is the idea. 2) No, you would not do that. You would take every sequence of the suitable length n and compare it to Shakespeare. 3) Yes. 4) No. What is the bearing of this on what will happen when you engage a crew of infinitely many monkeys? Unless I am missing something, we already know that the works of Shakespeare will appear with probability 1 infinitely many times in the output of the typings of a single monkey.

Sorry, I was too quick. I realize you are here and there and everywhere in the forum. It is not always easy to keep track of single threads. I suggested, in the same vein as StringJunky's question, that if you have infinitely many monkeys, each hammering away randomly on a (their own individual) keyboard, and each continuing forever without stopping, then it might be interesting to observe whether any of them might fail to reproduce the works of Shakespeare infinitely many times. We already know that a single monkey will make it with 100% probability. The question is whether any of the monkeys might fail to achieve this, now that there are so many of them. And more precisely, what is the probability that at least one monkey fails, or even, what is the expected number of failure monkeys? I thought that you reacted by saying "it depends". And that can make sense. But only if you expect that infinitely many monkeys will fail the test. If the number of monkeys to fail is a natural number, then the answer does not depend on how you order the outcomes. I only saw this now, sorry. But you do not address the question as it was posed. Yes, a single monkey will type the works of Shakespeare on repeat with probability zero. Because the single monkey will be restricted to type a single sequence of letters, whereas there is continuum many possible sequences to choose from. But that was not the question. The question was what happens if you engage not a single monkey, but continuum infinitely many. The number of monkeys will be the same as the number of possible sequences. You expect that the expected number of monkeys to achieve the correct sequence will be equal to 1, no?

You are correct, I asked for "the probability". It is true that such a probability does not necessarily exist. But if "it depends", as you say, then it means that you expect infinitely many monkeys to not have completed the assignment. Otherwise there would be a definite probability, namely 0. How did you see that, and so quickly?

When you typed: "So one monkey, typing an infinite amount of characters, would not only type out the complete works of Shakespeare, She would also type out all the (infinite) characters of Pi. Remember the sequence of Pi never repeats." I tried to point out that it would be incorrect to state that she would type out the entire (uninterrupted) sequence of digits of \(\pi.\) There is a discussion on monkeys typing \(\pi\) on math.stackexchange.com/questions/1232394/infinite-monkey-problem-probability-of-an-infinite-sequence-containing-an-infi?rq=1 StringJunky remarked about the case of infinitely many monkeys: "If a monkey can type Shakespeare, given an infinite amount of time, does it make mathematical sense that an infinite number monkeys would type Shakespeare in an infinitesimal amount of time? Or rather, as quickly as one monkey in that infinite number to type the minimum number of keys for the whole works. Are the odds not 1?" And I tried to answer by saying that if we have an infinite amount of monkeys, then we can even impose a stopping criterion, that each monkey has to stop after only typing as many characters as there are in Shakespeare's combined works, and we are still ensured to have infinitely many of the monkeys returning the perfect script. The case when the monkeys are not allowed to stop ought to result in the outcome that every monkey produces infinitely many copies of the works, right? Now that I type this I get unsure. Let it be a question: Assume that each of a countably infinite crew of monkeys is assigned with the task to type random letters without ever stopping. What is the probability of the event that each monkey returns infinitely many copies of the complete manuscript to the works of Shakespeare?

That is how your statement appeared to read, and I am trying to point out that this would be incorrect. However 'possible' the infinite sequence of digits is, it has probability zero to occur uninterrupted. The most you can do is to have it appear with interruptions between the digits in the form of irrelevant subsequences. I assume one monkey typing away without ever stopping.

It means the function defined on \(\mathbb{R}^n\) that maps \(x\) to \(u(x,t).\)

Basically yes. It is 'possible', because for the experiment to make sense, we have to allow some outcomes to be 'possible', and it does not make sense to allow Tails to be possible at the time of any particular flip without also allowing Heads to be possible. That immediately makes the all Heads outcome a possibility. However, if the experimenter reports that the outcome turned out to be all Heads, your proper reaction is to exclaim "impossible!" The reason is that it is 100% certain that the experimenter will be unable to report the exact outcome of the experiment by making a statement that has a finite length.

Who are your "we"? If striving isn't one's favorite hobby, one can imagine that any current something, call it something(0), came from another something, something(1), which in turn came from a something(2), and so on. Then every something has a beginning and comes from another something. I have to go watch the new episode of the Big Bang Theory now. Will the series never end? But careful with this. Assuming she can type alphanumeric characters instead of spelling out the decimal digits (are numerals needed for Shakespeare?) it is true that she will at some point type "3", and some time later "3.1", and later again "3.14" etc. For every \(n\) she will type the initial \(n\) digits of \(\pi\) at some point. But never the infinite sequence of all the digits.

If each of infinitely many monkeys types a random sequence of \(n\) letters, then with probability 1, every possible sequence of length \(n\) will be typed by infinitely many of them.

Depending on the typing speed of an ever more practiced monkey, this might be close to accurate. I seem to remember typing about 120/min when we had typing class in school. Today's texting phenomena might achieve significantly higher or lower speeds. As a side note, the bird's behavior is also, and more precisely, termed "feaking". The same word is used in slang for another common behavior, which I am not at liberty to elaborate on. It is not a worksafe item everywhere. Anyway, I am getting curious about a ballpark figure for how far the bird would actually get with the rock before the first nine words get typed by a monkey working alone. It seems like a very nice type of applied question, as you have to consider the physical properties of the bird's beak and the brittleness and size of the rock. English words seem to be about 5.5 characters long on average. So the monkey would have to punch about \(9\cdot 5.5+8 = 57.5\) characters right, when we include the blanks between words. If we allow for 64 keys on the typewriter, to accommodate lower and upper case letters as well as the necessary punctuation marks, then the monkey has a chance of one in \(64^{57.5} = 2^{345} \sim 7\cdot 10^{103}\) to make a hit on the first try. Given a solid typing speed, the monkey may achieve this once every \(6\cdot 10^{101}\) minutes, that is about once in \(10^{96}\) years. This is informative.But still leaves the question of how long we will have to wait until there is an actual hit. And how the bird is doing meanwhile with carving the rock.

Some forums (the present one not included, necessarily) have been infested with cranks. I have not had much inhibition to trolling cranks and crackpots. Maybe not having been really good at it, but at least it gave the exercise. Too bad that a site like mathsci.net is no longer(?) alive; always a lol to be had there.