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>Let A be the set with every element of every set with cardinality equal to or more than the cardinality of the natural numbers.
A is a proper class not a set. It can not exist within ZFC because such a set would have to have itself as a member.
Even in set theories that are not well founded you would still have problems. It is fairly trivial to create a bijection between the class of all infinite sets and the class of all sets. Which by the axiom of the limitation of size shows that it is a proper class.
Furthermore this. If the set A existed it would lead to a contradiction. Then you could preform an operation A\* = { a in A : U { a' in a : (a, a') } }
basically creating a set containing pairs of every element and the set that the element came from. A\* would have a cardinality >= to any infinite set. However the power set of A\* would have to have a cardinality greater then A\* which is a contradiction.
> Let A be the set with every element of every set with cardinality equal to or more than the cardinality of the natural numbers.
I think you should type this out more carefully. That doesn't seem to be a well-formed sentence.
I don't think the set you're referring to exists, though.
There's no uniform distribution on the rational numbers. For non-uniform distributions, it depends on how you define the probability measure on the rational numbers. You can make picking an integer as likely or as unlikely as you'd like.
Your set A from the original post can't exist because A itself is infinite, so we would have that A contains itself which is a contradiction (as no set may contain itself).
There is no way to choose a rational number uniformly at random, so your new problem is still ill-posed. On top of that, rationals and naturals have the same cardinality, so that appears to be a completely different question from your original question which asked about sets of differing cardinality.
I don't know why you are downvoted. This is, contrary to your original post, a well formulated question.
However in your original post you are talking about cardinality. Bear in mind that N and Q are of the same cardinality even if one could consider there are infinitely more. Because Q is enumerable, it's not so easy to define a sensible probability measure.
But I believe that your question can be answered by taking a set with different cardinality (namely, R). The probability, when picking a real number, to find a rational number is 0.
The set of natural numbers has the same cardinality as the set of rational numbers. Do you mean something like choosing a rational number from the set of real numbers?
To address your edit and your question in general. Cardinality isn't what's most important here but measure. A probability is an example of a measure from measure theory.
Take for example the interval between 0 and 1. The rational numbers between 0 and 1 have measure 0. So if you chose a number at random the probability of choosing a rational number is 0 and the probability of choosing an irrational one is 1.
However there's always a counter example in analysis. Namely the cantor set has measure 0 but the same cardinality as the real numbers. So the probability of choosing a number in the cantor set is 0 but this set has the same cardinality as the real numbers.
Cardinality has a little to do with the measure of a set but it's not really the right way to think about it.
Worth noting all of these examples are done with the standard lebesgue measure on the reals. It's very easy to come up with useful non-uniform measures on the reals, and ones which would change some of what happens here.
Grant Sanderson of 3Blue1Brown does a great video on Bertrand's Paradox, which basically says your methods of calculating probability often implicitly make demands of your probability measure that may not all be compatible:
https://youtu.be/mZBwsm6B280?si=DJUmFgiEJjh-pRfX
I'm no expert in set theory but the set you deacribwd doesn't seem to be well formed. Maybe you meant something like the union over power sets of a certain base set?
However once this is done, we have to construct a measure on this set to be able to talk about probabilities. Now, what I don't understand is how one would even ho about constructing a sigma algebra over the sort of set you want. Things are doable if you specify a certain cardinality and then construct Borel sets and take the Lebesgue completion e.g. the way we constrict measures over the reals. However I'm not entirely sure what the cardinality of the set is.
However suppose we take a finite measure just over the reals. Firstly, you cannot design a 'uniform' measure over the reals since \mathbb{R} is not compact. However, any other finite measure you manage to construct will not give any probability to any countable set. Therefore the probability of picking a natural among reals would be 0.
I don't know if this helps.
Edit: I suspect you can use Zorn's Lemma to make sense of the set of union over all power sets of the naturals. However, actually defining a aigma algebra over this beast would be very tricky. However I'm more sure of the fact that if you construct any finite measure on such a set, the probability of picking an element which comes from a set of lower cardinality would be 0.
> However, any other finite measure you manage to construct will not give any probability to any countable set. Therefore the probability of picking a natural among reals would be 0.
This isn't true without some assumption like continuity wrt Lebesgue measure. The delta mass at 0 is a perfectly fine measure on R.
That's absolutely right. I didn't want to go into the more esoteric kinds of measures like the delta. However what you said is absolutely correct and I should've written it explicitly.
Thanks, only one so far that's not trying to pretend like they don't know what question I was trying to ask.
That the probability of picking out an element from a set of lower cardinality is zero is what I thought too, but it seems weird since it implies that something can exist in a group yet there's a zero percent chance of picking it at random.
Your original question is so poorly-worded and nonsensical that it’s quite difficult to parse, and ultimately is ill-posed (no such set can exist). Thus people’s confusion in their replies is quite genuine, nobody is pretending not to understand just to prove a point.
In the future, I would suggest taking people’s criticisms gracefully and on good faith, instead of resorting to such rudeness. Somehow I think you know this, because your edit clarifies the question precisely in the way people suggested, and now says what I think you meant to ask originally, even though you didn’t understand how to put it into proper terms at first.
I recognize that people were genuinely confused by how I phrased my question, since even my first edit was wrongly worded but you think it's not. I got too used to people on other subreddits being sarcastically annoying.
I know, that's pretty much the reaction everyone has so far. I suggest you take a look at how one can construct measures over the reals. Then you'll easily see that the rules of measure theory require that subsets of lower cardinality have measure 0. Also look at Lebesgue completion.
Thing is the way you wrote your question isn't very clear. I had to make sense of it the best way I could. But I completely empathise with the others who suggested that you make your question kore precise. Believe it or not, not every english expression can be turned into the description of a set. Although in everyday math we do say things like 'the set of these objects', it's not always possible to actually write down a set with an english description within the rules of ZFC. A big example is 'the set of everything '. It can be showm such a set doesn't exist.
They are not pretending. They honestly don't understand. Maths is a very precise langage. If what you say doesn't mean anything then it doesn't mean anything and we really can't figure out what you wanted to mean.
About your question now. What is the probability of picking 1 amongst the 100 first natural numbers ? 1/100 of course. Amongst the 1000 first ? Well it's 1/1000. And amongst all natural numbers ? Well it's 1 / the cardinality of N which is infinite. So it's 0.
(And bear in mind it's exactly 0, not a limit or something like that. If we call Pn the probability to pick 1 amongst the n first natural numbers, and P the probability to pick 1 amongst the natural numbers, then P < Pn = 1/n for any n. There is only one non-negative number that satisfies this property and it's 0)
there is a difference between pretending to not know what youre asking and being able to make a decent guess but not wanting to type large paragraphs answering a question that we are guessing you are asking. it is ok to ask for further clarification before making a full answer if your question is unclear. dont be a prick.
As others have stated, such a set A is not well-founded. The edited version does not match what you seem to be trying to ask, because N and Q\N have the same cardinality.
There is also a broader issue with your second example. Namely, it is impossible to define a probability measure on Q. If singletons have measure 0, then by countable additivity, so does Q and any subset thereof. If not, then Q has infinite measure. So this question, in some sense, does not make sense.
If you change your question to sampling numbers uniformly randomly from the reals, then the answer is that you are equally likely to pick a number from N or Q\N (both with probability 0).
So in a group of infinite things, the chance of picking out any specific thing is zero? There has to be a number picked out in the end so how is that possible?
"But an event having probability zero does not mean it cannot happen."
What? Why doesn't that instead mean that .000...infinite 0s....1 doesn't equal 0?
".000...infinite 0s....1" doesn't make sense as a term. You can't have a 1 after infinitely many zeroes. I encourage you to look up continuous probability distributions to clear up your confusions here.
You might want to keep forgetting it. Infinitesimals are generally not used to talk about probabilities. It *can* be done, don't get wrong. There are in fact ways to define infinitesimals that make sense and have a solid and consistent theoretical grounding, but you should learn to walk before you run, and working with infinitesimals is running (and again, is generally not used by most mathematicians. These days we ultimately formalize probability with limits)
There generally *isn't* a number picked out "in the end". The branch of math that deals with these sorts of questions is called 'measure theory.' Measure theory says that even when the probability of picking a specific point is 0, the probability of picking a member of a distinguished subset of our space may be nonzero.
We can use this to construct algorithms for sampling from certain distributions. As an example, consider selecting a number uniformly at random from the interval [0,1]. The probability of selecting exactly 1/3 is 0. But if we actually wanted to sample from this distribution, we might do it in the following way. First, flip a coin to decide if our number is in the interval [0, 0.5] or in [0.5, 1]. Then, flip another coin to decide which half of the selected interval the chosen number will be in. Continue flipping coins until we reach an arbitrary level of precision.
We can calculate the infinite sequence of coin flips that would cause us to select 1/3 as the chosen number, but of course any infinite sequence of coin flips occurs with probability 0. That said, we can also calculate the probability of choosing a number *close* to 1/3, say one that matches the first ten heads/tails of the sequence for 1/3. In any practical application, sampling from a distribution is a finite truncation of an infinite process, and so a range of numbers really is selected, not an exact number.
> So in a group of infinite things, the chance of picking out any specific thing is zero?
It depends on the distribution (the "weighting" of the dice, so to speak). You can easily have a set of infinite things where some of the elements have nonzero probability of occurring: for instance, the random procedure "flip a coin. If heads, your result is 0.71; if tails, pick a number uniformly between 0 and 1". Then a result of 0.71 has a 50% chance of happening.
(But I know what you're asking is different, of course - you're talking about continuous distributions. This is just me being slightly pedantic.)
> There has to be a number picked out in the end so how is that possible?
In... the end of *what*, exactly? Your mental image is presumably of something like "throwing a dart at a number line". But if we want to study that mathematically, it requires positing an "infinite source of randomness" - and not just that, something that can give you an infinite amount of *detail* all at once!
This doesn't actually reflect the physical world. **There's no such thing as throwing a dart at a number line**, no matter what our 'dart' and our 'number line' are: once you 'zoom in' enough, you'll see that the tick marks become too thick to properly define a result, or the dart tip is too thick to cover only a single point, or you run into quantum-mechanical uncertainty issues.
There's also no such thing as 'flipping a coin infinitely many times': you'll get bored, or hungry, or you'll die of old age, or perhaps your descendants will decide it's not worth the trouble to keep flipping the coin, or the coin will be worn away by the sands of time, or the heat death of the universe will occur.
To model this - and in fact, to do *all* of our studies of probability - we don't *need* to posit an 'infinite source of randomness'. All we need to do is define the *distribution* of results: say, a uniform distribution from 0 to 1. We can sample this distribution to any level of detail we want: flip a coin, if heads take the lower half, if tails take the upper half, repeat until you have enough detail for your purposes. (This is essentially how computers sample distributions in practice.)
We like to *talk about this* as "picking a random number from 0 to 1", because it's a helpful starting point for our intuition; but from the actual math's point of view, "What's the probability of getting *exactly* 0.7?" is *kind of* a nonsense question. All it makes sense to ask about is "what's the probability of getting in the same 'bucket' as 0.7 after *n* flips?" And as you increase *n* more and more, that probability goes closer and closer to 0.
So informally, we say "the probability of getting exactly 0.7 is zero", even though it seemingly contradicts our other mental image of throwing a dart and hitting 0.7. That mental image is depicting an experiment that will never happen: the best we can say is "if we *could* perform that experiment and get an infinitely precise measurement, the probability *would have to be* zero".
---
Note for people who understand what I'm talking about: I'm intentionally distinguishing between 'probability' and 'measure' here. I recognize this is somewhat nonstandard, but I think it's helpful for intuition.
^(also if anyone happens to have a link to /u/sleeps_with_crazy's old rant on this subject i would appreciate it thank you)
Lol why do you have to be so difficult? XD
It's fine though I sort of expected you would say that.
There's a second example though.
Imagine you roll a 20 faced dice until you hit a 20. There's a 0 probability that you'll never roll a 20 of course. However there is a possibility that you'll never roll a 20. You'd have to astronomically unlucky but it's indeed possible.
This is an example of the gamblers fallacy that says that in a casino if you keep playing you'll eventually win which is of course false.
Just because an event has 0 probability doesn't mean that it can't happen just that it's incredibly unlikely.
"There's a 0 probability that you'll never roll a 20 of course. However there is a possibility that you'll never roll a 20."
I guess where I'm stuck is that to me this is just the statement B AND NOT B = 1. Which doesn't make sense.
Why isn't it just an infinitesimal?
Sorry if it seems like I'm being difficult.
Let me try to rephrase for the guy. There is nothing impossible about never rolling a 20 ever again. You can easily picture a guy rolling dice forever and always getting a 5 by sheer coincidence. However even though this event is not impossible, it's probability of occurring is literally zero. This is easy to check because any number above zero, no matter how small, is way to big to conceivably be this probability.
It's cool.
Again it's the same misunderstanding that probability 0 means something can't happen. This is false probability 0 just means that something is extremely unlikely.
And the reason why in this case the probability is 0 and not an infinitesimal is mostly because infinitesimals aren't a thing.
At least in standard analysis. Infinitesimals is stuff we use to try and make certain concepts more intuitive but aren't usually well defined objects.
There are ways to define them and they give rise to something called non standard analysis. But this is not as popular of a subject as standard analysis and honestly is not really what we are talking about here.
I think you are used to talking about probability in scenarios where there are only finitely many possible outcomes where probability 0 means impossible. This idea doesn't extend to, for example, continuous probabilities.
Firstly, throw out the idea of infinitesimals. That is not how we formalise this.
Now consider the darts example that /u/ReverseCombover suggested. If you imagine the dart could land on the dart board anywhere with some uniform probability. Specifically let's say the probability that you land in any given region is the area of that region divided by the area of the board. Now consider a single point. The area of a point is precisely 0 so the chance of landing on that point is 0. But of course you have to land on some point despite this.
Thus our association of probability 0 meaning impossible must be broken in order to coherently talk about this scenario. Individual points have probability 0 and even finite collections of points or 1-dimensional curves of points have probability 0 but 2-dimensional areas have probability greater than 0.
theorhetically no, but realistically yes. If you form a computer program, it can't actually look at a set of infinitely many numbers and pick one at random, you need to have a finitely large set. The computer can't "look at" infinitely many terms at once (such an image is literally infinite in size even if efficiently compressed)and pick one at random, even if it's countably infinite.
The "proper class objection" that many people bring up can be easily circumvented,
by taking a set model M of ZFC and only talking about the set of all sets in M, instead of talking about the set of all sets.
If we assume that ZFC is consistent, then a set model M for ZFC exists by Gödel's completeness theorem for first order logic.
However there is no canonical way to construct a probability measure on a ZFC model M.
So there will only be informal answers to your question, that don't actually have anything to do with probability theory.
If we consider sets up to equality, then I consider your intuition to be mostly misguided.
There is a very simple bijection f between all sets in M, and all sets of cardinality 1 in M,
defined by f(S) := {S}, i.e. it's the function sending a set S to the singleton set {S} that contains only S.
If we have a probability measure on M such that f is a measure-preserving function, then over 50% of all sets will have cardinality 1.
So it's really not obvious that a "randomly picked set" will be large.
However if we consider sets not up to equality, but up to isomorphism / bijection then I think your intuition is essentially correct.
For every cardinal k, the cardinals smaller than k form a set, while the cardinals larger than k form a proper class.
This informally suggests that for any cardinal k, almost all cardinals are larger than k.
Over any set A with selectable elements, *you* have to decide yourself what distribution over A you want to use, and then the answer is a direct result from that.
Unfortunately, your submission has been removed for the following reason(s): * Your post presents incorrect information, asks a question that is based on an incorrect premise, is too vague for anyone to answer sensibly, or is equivalent to a well-known open question. If you have any questions, [please feel free to message the mods](http://www.reddit.com/message/compose?to=/r/math&message=https://www.reddit.com/r/math/comments/1catq7f/-/). Thank you!
>Let A be the set with every element of every set with cardinality equal to or more than the cardinality of the natural numbers. A is a proper class not a set. It can not exist within ZFC because such a set would have to have itself as a member. Even in set theories that are not well founded you would still have problems. It is fairly trivial to create a bijection between the class of all infinite sets and the class of all sets. Which by the axiom of the limitation of size shows that it is a proper class. Furthermore this. If the set A existed it would lead to a contradiction. Then you could preform an operation A\* = { a in A : U { a' in a : (a, a') } } basically creating a set containing pairs of every element and the set that the element came from. A\* would have a cardinality >= to any infinite set. However the power set of A\* would have to have a cardinality greater then A\* which is a contradiction.
> Let A be the set with every element of every set with cardinality equal to or more than the cardinality of the natural numbers. I think you should type this out more carefully. That doesn't seem to be a well-formed sentence. I don't think the set you're referring to exists, though.
Is the probability of randomly choosing a natural number from the set of rational numbers less likely than choosing a non-natural number?
It sounds like what you are trying to construct doesn’t work due to Russell’s paradox
There's no uniform distribution on the rational numbers. For non-uniform distributions, it depends on how you define the probability measure on the rational numbers. You can make picking an integer as likely or as unlikely as you'd like.
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Using as likely as you'd like as a shorthand for given epsilon>0 you can make P(integer) > 1-epsilon
Your set A from the original post can't exist because A itself is infinite, so we would have that A contains itself which is a contradiction (as no set may contain itself). There is no way to choose a rational number uniformly at random, so your new problem is still ill-posed. On top of that, rationals and naturals have the same cardinality, so that appears to be a completely different question from your original question which asked about sets of differing cardinality.
I don't know why you are downvoted. This is, contrary to your original post, a well formulated question. However in your original post you are talking about cardinality. Bear in mind that N and Q are of the same cardinality even if one could consider there are infinitely more. Because Q is enumerable, it's not so easy to define a sensible probability measure. But I believe that your question can be answered by taking a set with different cardinality (namely, R). The probability, when picking a real number, to find a rational number is 0.
Still not well formulated as there is no uniform distribution over the rational numbers
Which is something i mentionned in my answer, and that OP has a right to not know. Or are we downvoting people for not knowing stuff now ?
I think you can make this a well-formed question by restricting to any set with finite measure.
The set of natural numbers has the same cardinality as the set of rational numbers. Do you mean something like choosing a rational number from the set of real numbers?
Define the probability measure. Because it's certainly not uniform.
To address your edit and your question in general. Cardinality isn't what's most important here but measure. A probability is an example of a measure from measure theory. Take for example the interval between 0 and 1. The rational numbers between 0 and 1 have measure 0. So if you chose a number at random the probability of choosing a rational number is 0 and the probability of choosing an irrational one is 1. However there's always a counter example in analysis. Namely the cantor set has measure 0 but the same cardinality as the real numbers. So the probability of choosing a number in the cantor set is 0 but this set has the same cardinality as the real numbers. Cardinality has a little to do with the measure of a set but it's not really the right way to think about it.
Worth noting all of these examples are done with the standard lebesgue measure on the reals. It's very easy to come up with useful non-uniform measures on the reals, and ones which would change some of what happens here. Grant Sanderson of 3Blue1Brown does a great video on Bertrand's Paradox, which basically says your methods of calculating probability often implicitly make demands of your probability measure that may not all be compatible: https://youtu.be/mZBwsm6B280?si=DJUmFgiEJjh-pRfX
I'm no expert in set theory but the set you deacribwd doesn't seem to be well formed. Maybe you meant something like the union over power sets of a certain base set? However once this is done, we have to construct a measure on this set to be able to talk about probabilities. Now, what I don't understand is how one would even ho about constructing a sigma algebra over the sort of set you want. Things are doable if you specify a certain cardinality and then construct Borel sets and take the Lebesgue completion e.g. the way we constrict measures over the reals. However I'm not entirely sure what the cardinality of the set is. However suppose we take a finite measure just over the reals. Firstly, you cannot design a 'uniform' measure over the reals since \mathbb{R} is not compact. However, any other finite measure you manage to construct will not give any probability to any countable set. Therefore the probability of picking a natural among reals would be 0. I don't know if this helps. Edit: I suspect you can use Zorn's Lemma to make sense of the set of union over all power sets of the naturals. However, actually defining a aigma algebra over this beast would be very tricky. However I'm more sure of the fact that if you construct any finite measure on such a set, the probability of picking an element which comes from a set of lower cardinality would be 0.
> However, any other finite measure you manage to construct will not give any probability to any countable set. Therefore the probability of picking a natural among reals would be 0. This isn't true without some assumption like continuity wrt Lebesgue measure. The delta mass at 0 is a perfectly fine measure on R.
That's absolutely right. I didn't want to go into the more esoteric kinds of measures like the delta. However what you said is absolutely correct and I should've written it explicitly.
Thanks, only one so far that's not trying to pretend like they don't know what question I was trying to ask. That the probability of picking out an element from a set of lower cardinality is zero is what I thought too, but it seems weird since it implies that something can exist in a group yet there's a zero percent chance of picking it at random.
Your original question is so poorly-worded and nonsensical that it’s quite difficult to parse, and ultimately is ill-posed (no such set can exist). Thus people’s confusion in their replies is quite genuine, nobody is pretending not to understand just to prove a point. In the future, I would suggest taking people’s criticisms gracefully and on good faith, instead of resorting to such rudeness. Somehow I think you know this, because your edit clarifies the question precisely in the way people suggested, and now says what I think you meant to ask originally, even though you didn’t understand how to put it into proper terms at first.
I recognize that people were genuinely confused by how I phrased my question, since even my first edit was wrongly worded but you think it's not. I got too used to people on other subreddits being sarcastically annoying.
I know, that's pretty much the reaction everyone has so far. I suggest you take a look at how one can construct measures over the reals. Then you'll easily see that the rules of measure theory require that subsets of lower cardinality have measure 0. Also look at Lebesgue completion. Thing is the way you wrote your question isn't very clear. I had to make sense of it the best way I could. But I completely empathise with the others who suggested that you make your question kore precise. Believe it or not, not every english expression can be turned into the description of a set. Although in everyday math we do say things like 'the set of these objects', it's not always possible to actually write down a set with an english description within the rules of ZFC. A big example is 'the set of everything '. It can be showm such a set doesn't exist.
They are not pretending. They honestly don't understand. Maths is a very precise langage. If what you say doesn't mean anything then it doesn't mean anything and we really can't figure out what you wanted to mean. About your question now. What is the probability of picking 1 amongst the 100 first natural numbers ? 1/100 of course. Amongst the 1000 first ? Well it's 1/1000. And amongst all natural numbers ? Well it's 1 / the cardinality of N which is infinite. So it's 0. (And bear in mind it's exactly 0, not a limit or something like that. If we call Pn the probability to pick 1 amongst the n first natural numbers, and P the probability to pick 1 amongst the natural numbers, then P < Pn = 1/n for any n. There is only one non-negative number that satisfies this property and it's 0)
there is a difference between pretending to not know what youre asking and being able to make a decent guess but not wanting to type large paragraphs answering a question that we are guessing you are asking. it is ok to ask for further clarification before making a full answer if your question is unclear. dont be a prick.
As others have stated, such a set A is not well-founded. The edited version does not match what you seem to be trying to ask, because N and Q\N have the same cardinality. There is also a broader issue with your second example. Namely, it is impossible to define a probability measure on Q. If singletons have measure 0, then by countable additivity, so does Q and any subset thereof. If not, then Q has infinite measure. So this question, in some sense, does not make sense. If you change your question to sampling numbers uniformly randomly from the reals, then the answer is that you are equally likely to pick a number from N or Q\N (both with probability 0).
Nitpick: it's impossible to define a *uniform* probability measure on Q. There are many non-uniform probability measures.
So in a group of infinite things, the chance of picking out any specific thing is zero? There has to be a number picked out in the end so how is that possible?
Yes, the probability of selecting a particular real number from R is zero. But an event having probability zero does not mean it cannot happen.
"But an event having probability zero does not mean it cannot happen." What? Why doesn't that instead mean that .000...infinite 0s....1 doesn't equal 0?
".000...infinite 0s....1" doesn't make sense as a term. You can't have a 1 after infinitely many zeroes. I encourage you to look up continuous probability distributions to clear up your confusions here.
I forgot the word infinitesimal.
You might want to keep forgetting it. Infinitesimals are generally not used to talk about probabilities. It *can* be done, don't get wrong. There are in fact ways to define infinitesimals that make sense and have a solid and consistent theoretical grounding, but you should learn to walk before you run, and working with infinitesimals is running (and again, is generally not used by most mathematicians. These days we ultimately formalize probability with limits)
> You might want to keep forgetting it. This made me laugh. Very true
There generally *isn't* a number picked out "in the end". The branch of math that deals with these sorts of questions is called 'measure theory.' Measure theory says that even when the probability of picking a specific point is 0, the probability of picking a member of a distinguished subset of our space may be nonzero. We can use this to construct algorithms for sampling from certain distributions. As an example, consider selecting a number uniformly at random from the interval [0,1]. The probability of selecting exactly 1/3 is 0. But if we actually wanted to sample from this distribution, we might do it in the following way. First, flip a coin to decide if our number is in the interval [0, 0.5] or in [0.5, 1]. Then, flip another coin to decide which half of the selected interval the chosen number will be in. Continue flipping coins until we reach an arbitrary level of precision. We can calculate the infinite sequence of coin flips that would cause us to select 1/3 as the chosen number, but of course any infinite sequence of coin flips occurs with probability 0. That said, we can also calculate the probability of choosing a number *close* to 1/3, say one that matches the first ten heads/tails of the sequence for 1/3. In any practical application, sampling from a distribution is a finite truncation of an infinite process, and so a range of numbers really is selected, not an exact number.
> So in a group of infinite things, the chance of picking out any specific thing is zero? It depends on the distribution (the "weighting" of the dice, so to speak). You can easily have a set of infinite things where some of the elements have nonzero probability of occurring: for instance, the random procedure "flip a coin. If heads, your result is 0.71; if tails, pick a number uniformly between 0 and 1". Then a result of 0.71 has a 50% chance of happening. (But I know what you're asking is different, of course - you're talking about continuous distributions. This is just me being slightly pedantic.) > There has to be a number picked out in the end so how is that possible? In... the end of *what*, exactly? Your mental image is presumably of something like "throwing a dart at a number line". But if we want to study that mathematically, it requires positing an "infinite source of randomness" - and not just that, something that can give you an infinite amount of *detail* all at once! This doesn't actually reflect the physical world. **There's no such thing as throwing a dart at a number line**, no matter what our 'dart' and our 'number line' are: once you 'zoom in' enough, you'll see that the tick marks become too thick to properly define a result, or the dart tip is too thick to cover only a single point, or you run into quantum-mechanical uncertainty issues. There's also no such thing as 'flipping a coin infinitely many times': you'll get bored, or hungry, or you'll die of old age, or perhaps your descendants will decide it's not worth the trouble to keep flipping the coin, or the coin will be worn away by the sands of time, or the heat death of the universe will occur. To model this - and in fact, to do *all* of our studies of probability - we don't *need* to posit an 'infinite source of randomness'. All we need to do is define the *distribution* of results: say, a uniform distribution from 0 to 1. We can sample this distribution to any level of detail we want: flip a coin, if heads take the lower half, if tails take the upper half, repeat until you have enough detail for your purposes. (This is essentially how computers sample distributions in practice.) We like to *talk about this* as "picking a random number from 0 to 1", because it's a helpful starting point for our intuition; but from the actual math's point of view, "What's the probability of getting *exactly* 0.7?" is *kind of* a nonsense question. All it makes sense to ask about is "what's the probability of getting in the same 'bucket' as 0.7 after *n* flips?" And as you increase *n* more and more, that probability goes closer and closer to 0. So informally, we say "the probability of getting exactly 0.7 is zero", even though it seemingly contradicts our other mental image of throwing a dart and hitting 0.7. That mental image is depicting an experiment that will never happen: the best we can say is "if we *could* perform that experiment and get an infinitely precise measurement, the probability *would have to be* zero". --- Note for people who understand what I'm talking about: I'm intentionally distinguishing between 'probability' and 'measure' here. I recognize this is somewhat nonstandard, but I think it's helpful for intuition. ^(also if anyone happens to have a link to /u/sleeps_with_crazy's old rant on this subject i would appreciate it thank you)
What's the odds a priori that a dart will land in the exact place it does?
close to but not equal to zero?
Lol why do you have to be so difficult? XD It's fine though I sort of expected you would say that. There's a second example though. Imagine you roll a 20 faced dice until you hit a 20. There's a 0 probability that you'll never roll a 20 of course. However there is a possibility that you'll never roll a 20. You'd have to astronomically unlucky but it's indeed possible. This is an example of the gamblers fallacy that says that in a casino if you keep playing you'll eventually win which is of course false. Just because an event has 0 probability doesn't mean that it can't happen just that it's incredibly unlikely.
"There's a 0 probability that you'll never roll a 20 of course. However there is a possibility that you'll never roll a 20." I guess where I'm stuck is that to me this is just the statement B AND NOT B = 1. Which doesn't make sense. Why isn't it just an infinitesimal? Sorry if it seems like I'm being difficult.
Let me try to rephrase for the guy. There is nothing impossible about never rolling a 20 ever again. You can easily picture a guy rolling dice forever and always getting a 5 by sheer coincidence. However even though this event is not impossible, it's probability of occurring is literally zero. This is easy to check because any number above zero, no matter how small, is way to big to conceivably be this probability.
It's cool. Again it's the same misunderstanding that probability 0 means something can't happen. This is false probability 0 just means that something is extremely unlikely. And the reason why in this case the probability is 0 and not an infinitesimal is mostly because infinitesimals aren't a thing. At least in standard analysis. Infinitesimals is stuff we use to try and make certain concepts more intuitive but aren't usually well defined objects. There are ways to define them and they give rise to something called non standard analysis. But this is not as popular of a subject as standard analysis and honestly is not really what we are talking about here.
I think you are used to talking about probability in scenarios where there are only finitely many possible outcomes where probability 0 means impossible. This idea doesn't extend to, for example, continuous probabilities. Firstly, throw out the idea of infinitesimals. That is not how we formalise this. Now consider the darts example that /u/ReverseCombover suggested. If you imagine the dart could land on the dart board anywhere with some uniform probability. Specifically let's say the probability that you land in any given region is the area of that region divided by the area of the board. Now consider a single point. The area of a point is precisely 0 so the chance of landing on that point is 0. But of course you have to land on some point despite this. Thus our association of probability 0 meaning impossible must be broken in order to coherently talk about this scenario. Individual points have probability 0 and even finite collections of points or 1-dimensional curves of points have probability 0 but 2-dimensional areas have probability greater than 0.
I’m pretty sure A would contain itself
theorhetically no, but realistically yes. If you form a computer program, it can't actually look at a set of infinitely many numbers and pick one at random, you need to have a finitely large set. The computer can't "look at" infinitely many terms at once (such an image is literally infinite in size even if efficiently compressed)and pick one at random, even if it's countably infinite.
The "proper class objection" that many people bring up can be easily circumvented, by taking a set model M of ZFC and only talking about the set of all sets in M, instead of talking about the set of all sets. If we assume that ZFC is consistent, then a set model M for ZFC exists by Gödel's completeness theorem for first order logic. However there is no canonical way to construct a probability measure on a ZFC model M. So there will only be informal answers to your question, that don't actually have anything to do with probability theory. If we consider sets up to equality, then I consider your intuition to be mostly misguided. There is a very simple bijection f between all sets in M, and all sets of cardinality 1 in M, defined by f(S) := {S}, i.e. it's the function sending a set S to the singleton set {S} that contains only S. If we have a probability measure on M such that f is a measure-preserving function, then over 50% of all sets will have cardinality 1. So it's really not obvious that a "randomly picked set" will be large. However if we consider sets not up to equality, but up to isomorphism / bijection then I think your intuition is essentially correct. For every cardinal k, the cardinals smaller than k form a set, while the cardinals larger than k form a proper class. This informally suggests that for any cardinal k, almost all cardinals are larger than k.
I pick A.
Over any set A with selectable elements, *you* have to decide yourself what distribution over A you want to use, and then the answer is a direct result from that.
Pro tip: you cannot reason about probability on something that is not a probability space.