# Probability question (easy to ask, hard to answer) **URL:** \ **Category:** Factual Questions\ **Created:** [August 28, 2002, 11:04pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007 "2002-08-28T23:04:58Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 3:32pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/21 "2002-08-29T15:32:02Z")
> [@](#): > > \*Originally posted by CuriousCanuck \* > \*\*I agree with you that for all intents and purposes it may as well be 0, but to argue that because the numbers get so small on the march to the infinite that the probabilities become exactly 0 is not correct. Infinity is a mathematical tool used for ease of calculation (and sometimes to hinder it as well) but it is a tool only. When infinites are used in equations there is always a small but normally allowable margin of error. Which is where the probability lies. Within that margin of error. \*\* The probability of picking a rational number from [0, 1] under a uniform distribution is exactly zero. That’s because in a uniform distribution, the probability of picking a number from a given subset of [0, 1] is just the Lebesgue measure of that set. The Lebesgue measure, like any other measure, has two characteristics:[ol][li]The Lebesgue measure of a set is non-negative, but it may be infinite (the infinity here is a formal symbol, not a transfinite number).The Lebesgue measure of a countable union of disjoint sets is the sum of the measures of the sets.[/ol]For an interval, the Lebesgue measure is just the length of the interval (whether it’s open or closed). Since the interval consisting of a single point has length zero, the Lebesgue measure of a set consisting of a single point is zero. [/li] Take a countable set, such as the rationals. This can be expressed as the union of its points. Obviously, {a} = {b} only if a = b, so this is the countable union of disjoint sets. By 2) above, the Lebesgue measure of **Q** is zero–it’s the sum from 1 to infinity of zero, and that adds up to be nothing (in the formal limit sense, just like any other infinite sum). Since the probability of picking from **Q** is just the Lebesgue measure of **Q** , and **Q** has Lebesgue measure zero, the probability of picking a rational number is zero. As I said before, this was discussed in the other thread referenced in the OP. Please read that thread. ---
**Author:** ![CuriousCanuck](https://avatars.discourse-cdn.com/v4/letter/c/d2c977/32.png) [@CuriousCanuck](https://boards.straightdope.com/u/CuriousCanuck)\ **Post date:** [August 29, 2002, 3:39pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/22 "2002-08-29T15:39:43Z")
I did read it… I just didn’t buy it. But thinking about that it way does make sense. I’m not a mathematician by any means as I am sure you are all aware by my arguements. I can see now that I am probably wrong. But it doesn’t feel right that it should be zero. All my insincts tell me that it is wrong. Maybe I should go take a math degree so that I would know better. 🙂 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 3:46pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/23 "2002-08-29T15:46:13Z")
> [@](#): > > \*Originally posted by CuriousCanuck \* > **I can see now that I am probably wrong. But it doesn’t feel right that it should be zero. All my insincts tell me that it is wrong. Maybe I should go take a math degree so that I would know better.** Welcome to the world of higher mathematics, where your intuition is your worst enemy (at first–then you develop a pretty good mathematical intuition, but that’s different). ---
**Author:** ![erislover](https://avatars.discourse-cdn.com/v4/letter/e/71e660/32.png) [@erislover](https://boards.straightdope.com/u/erislover)\ **Post date:** [August 29, 2002, 3:55pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/24 "2002-08-29T15:55:43Z")
Maybe I’m not seeing this correctly, but it seems to me that the question contains its own answer in a way, we are only being thrown for a loop here by the fact that we’ve demanded a rational number. Why don’t we just rephrase the question with a set of numbers that are the same size as the rationals? It seems to me that it would be the same as the following question: I pick a random integer from the set of all integers. What is the probability that it is less than 10 (or 5 or _n_)? The size of the rationals is the same as the size of the integers. I appreciate that I have not stumbled on anything great here (at all, and quite possibly might have messed something up), but I don’t see where I would have gone wrong. Perhaps in explaining it to me we can all help ourselves 🙂 ---
**Author:** ![Nametag](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/nametag/32/406_2.png) [@Nametag](https://boards.straightdope.com/u/Nametag)\ **Post date:** [August 29, 2002, 4:02pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/25 "2002-08-29T16:02:03Z")
**Ultrafilter** , I’m not a mathematician, but I’m going to try this quibble anyway. On the interval [0,1], isn’t the set of rational numbers finite? If so, isn’t the probability of selecting a given rational number finite and, um, “knowable”? So isn’t your question easier to answer than you think? I imagine that you’re after an answer for an infinite subset, but I really don’t know. Being wrong will be more educational than being right. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 4:16pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/26 "2002-08-29T16:16:25Z")
> [@](#): > > \*Originally posted by Nametag \* > **Ultrafilter, I’m not a mathematician, but I’m going to try this quibble anyway. On the interval [0,1], isn’t the set of rational numbers finite? If so, isn’t the probability of selecting a given rational number finite and, um, “knowable”? So isn’t your question easier to answer than you think?** A finite subset would still have measure zero, and so you’d run into the same problem. But the rationals in [0, 1] include the reciprocals of every positive integer, and there’s an infinite number of those. So it’s not finite. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 4:23pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/27 "2002-08-29T16:23:29Z")
> [@](#): > > _Originally posted by erislover_ > **It seems to me that it would be the same as the following question: I pick a random integer from the set of all integers. What is the probability that it is less than 10 (or 5 or n)?** See **Cabbage** ’s first post. ---
**Author:** ![aahala](https://avatars.discourse-cdn.com/v4/letter/a/a88e4f/32.png) [@aahala](https://boards.straightdope.com/u/aahala)\ **Post date:** [August 29, 2002, 5:09pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/28 "2002-08-29T17:09:56Z")
The probability of choosing any particular number within the range(rational or irrational) is zero, but the sum of the probablity of these two infinite sets is 1. Some number will always be chosen. I take it by saying the distribution is uniform, you are saying there are just as many rational numbers in the lower half as the upper half of the range. Therefore the probability is .5 the chosen number will be in the lower or higher range. ---
**Author:** ![jawdirk](https://avatars.discourse-cdn.com/v4/letter/j/df705f/32.png) [@jawdirk](https://boards.straightdope.com/u/jawdirk)\ **Post date:** [August 29, 2002, 6:11pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/29 "2002-08-29T18:11:08Z")
Why is it that mathematicians feel comfortable allowing a probability 0 event to be possible? Why isn’t this regarded as a simple contradiction? My question is, when you mathematically prove that an event has probability 0 of occuring, but your assumption is that it is possible, why do you not then say that you have reached a point of contradiction, and that your assumptions are therefor self-contradictory? I assume mathematicians would agree to the following: If A is a subset of B, and the probability that choosing a random B will result in the choice of an A is 0, then for any number of random Bs chosen, the number of As chosen will be 0. Why would you regard it as “possible” to choose an A in this circumstance? It just seems to be a simple contradiction to me. How about this question: Suppose we choose a random subset of real [0,1], or equivalently, choose a random member of the power set of real [0,1]. What is the probability that the subset contains a rational number? The answer is 0 right? ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [August 29, 2002, 6:44pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/30 "2002-08-29T18:44:11Z")
I don’t know, I’m starting to come around on this idea, so let me try to answer you, **jawdirk** : > [@](#): > > Why is it that mathematicians feel comfortable allowing a probability 0 event to be possible? Okay, suppose you have a random number generator, with two outcomes, t and f. Say you can repeat the trial as much as you want, and T is the number of times t came up so far, and F is the number of times f came up so far. Suppose you assign P(t) = 1 and P(f) = 0. Then you would expect the following limits at infinity: Lim(T / (T + F)) = 1 Lim(F / (T + F)) = 0 Now, I know you can’t complete infinite tasks, but say you generate random events, and you get this: t t f t t t t t t t t t t t t t t t t t … with t’s going out to infinity. Well don’t the limits hold? Your probability assignments would be correct. And it’s silly to say that something which has happened is impossible, and f did happen. Even if f occurred on every prime: t f f t f t f t t t f t f t t t f t f t t t f … You would, of course, suspect your random event generator of dishonesty, but if you were assured it was legit, then what choice would you have but to keep P(f) = 0 and P(t) = 1? Now, unless I’ve misunderstood the thesis here, then there are impossible events. And these events have probability 0. But, just because something has probability 0 doesn’t make it an impossible event. So, just constructing an impossible event with P = 0 doesn’t cut it - you have to show that _all_ P = 0 events are impossible. ---
**Author:** ![CuriousCanuck](https://avatars.discourse-cdn.com/v4/letter/c/d2c977/32.png) [@CuriousCanuck](https://boards.straightdope.com/u/CuriousCanuck)\ **Post date:** [August 29, 2002, 6:49pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/31 "2002-08-29T18:49:26Z")
See I do understand all of that. But to me saying something can happen but has a probability of 0 just doesnt make sense at least in the real world, which as I am aware is not the case with math or mathematicians 😉 ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [August 29, 2002, 6:59pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/32 "2002-08-29T18:59:02Z")
**jawdirk** : it’s counterintuitive, yes. So are many other properties of infinite sets. But the choice is to either (a) accept that “possible” and “probability \> 0” are _not_ necessarily equivalent when there are an infinite number of outcomes or (b) throw out probability theory entirely. (a) is more reasonable, IMO, than (b). To put it another way, what else should we do when we have an infinite number of outcomes, all equally likely? We can’t say each outcome has a positive probability p, for reasons **jabba** has already stated: no matter how small p is, as long as p is positive we can pick some number n so that np\>1, find n different outcomes, and then say that the probability that one of those n outcomes occurs is greater than 1, which is absurd. So all that’s left is to say that each and every outcome has probability zero, _even though one of them has to occur by definition_. It may look like a contradiction, but it isn’t. It’s just another case of a common-sense result about finite sets failing to be true for infinite sets. ---
**Author:** ![CuriousCanuck](https://avatars.discourse-cdn.com/v4/letter/c/d2c977/32.png) [@CuriousCanuck](https://boards.straightdope.com/u/CuriousCanuck)\ **Post date:** [August 29, 2002, 7:05pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/33 "2002-08-29T19:05:07Z")
OK… dumb question… what is the difference between saying that or that each has an infinitesimally small p such that to add all of these possible n’s = 1 That to me is more accurate then saying p=0. I still know I am wrong, just saying is all. ---
**Author:** ![Jabba](https://avatars.discourse-cdn.com/v4/letter/j/b5e925/32.png) [@Jabba](https://boards.straightdope.com/u/Jabba)\ **Post date:** [August 29, 2002, 7:05pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/34 "2002-08-29T19:05:56Z")
> [@](#): > > _originally posted by jawdirk_ > My question is, when you mathematically prove that an event has probability 0 of occuring, but your assumption is that it is possible, why do you not then say that you have reached a point of contradiction, and that your assumptions are therefor self-contradictory? In a probability/statistics context, an _experiment_ is any action whose outcome cannot be determined in advance. It need not have any connection with a scientific experiment as this idea is commonly understood, but could be simply the action of tossing a coin and noting whether the outcome is a head or a tail or watching a sporting contest and noting the outcome. Of fundamental importance is the notion of a _random variable_. This means roughly performing an experiment and associating a real number with each possible outcome. For example, we could roll a die and let the random variable X equal the number rolled. A _discrete random variable_ is one for which the set of possible outcomes is either finite or countably infinite. For example the random variable X defined above is discrete. As another example, let Y be the number of posts made to the SDMB on any given day. The Y is a discrete random variable: the experiment consists of surveying the SDMB on any given day and counting the messages posted; the possible values for Y are the non-negative integers 0,1,2,3,… The behaviour of discrete random variables generally speaking conforms to our intuitions about probability. In particular, if the probability that X takes the value x is zero, when we write P(X=x)=0, then X cannot take the value x. The situation we are dealing with in this thread is what is known as a _continuous random variable_, roughly a random variable which can take any value in a given range. Specifically we have the_uniform distribution_ on [0,1]. The idea we are tring to capture is that it is equally likely to take any value in the range [0,1]. However if P(X=x) is a non-zero number for any x in the range [0,1] then P(X=y) should be the same non-zero value for _every_ y in this range ( by uniformity). However this quickly leads to an absurdity because then the total probability would be infinite, being the sum ( in some sense; we cannot really add up uncountably many numbers, but this technicality would take us too far afield) of the individual probabilities. We are led inevitably to the conclusion that P(X=x) is 0 for every x in the range [0,1]. On the other hand, if we perform an experiment corresponding to the random variable we will certainly get some outcome. So events with probability 0 are not necessarily impossible ( though the converse does hold). This may well contradict our assuptions about how probabilities behave, but it does not involve us in any mathematical contradictions. The way we get round it is to speak instead of the probability that we get a number in a certain range. Thus P(1/4 [symbol]£[/symbol] X [symbol]£[/symbol] 3/4) = 1/2, and so on. ---
**Author:** ![CuriousCanuck](https://avatars.discourse-cdn.com/v4/letter/c/d2c977/32.png) [@CuriousCanuck](https://boards.straightdope.com/u/CuriousCanuck)\ **Post date:** [August 29, 2002, 7:09pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/35 "2002-08-29T19:09:25Z")
_click_ Thanks Jabba… I dont know why my mind didnt get that. Something just clicked there. Took long enough, eh? :smack: ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [August 29, 2002, 7:12pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/36 "2002-08-29T19:12:51Z")
> [@](#): > > \*Originally posted by Jabba \* > **The behaviour of discrete random variables generally speaking conforms to our intuitions about probability. In particular, if the probability that X takes the value x is zero, when we write P(X=x)=0, then X cannot take the value x.** Oh, wait. Um, er… wait. Is that right? I’m confused now. ☹ How do you figure that, **Jabba**? ---
**Author:** ![Tretiak](https://avatars.discourse-cdn.com/v4/letter/t/a88e57/32.png) [@Tretiak](https://boards.straightdope.com/u/Tretiak)\ **Post date:** [August 29, 2002, 7:46pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/37 "2002-08-29T19:46:22Z")
So, have we answered the OP yet? ---
**Author:** ![jawdirk](https://avatars.discourse-cdn.com/v4/letter/j/df705f/32.png) [@jawdirk](https://boards.straightdope.com/u/jawdirk)\ **Post date:** [August 29, 2002, 9:01pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/38 "2002-08-29T21:01:31Z")
Jabba, that makes sense, but it leaves me with the question of whether probability is the right word to use when speaking of continuous random variables. Suppose I ask the question: what is more probable, rolling a 7 on a six-sided die or generating a rational value from a random selection of a real number on the range [0,1]? Are you going to say that they are equally probable or are you going to say that the question is meaningless because I am talking about two different sorts of probability? If the former, then it seems just wrong to say that something that is possible is equally probable as something impossible. If the latter, then the terminology is unnecessarily confusing. It would seem far less confusing to just say that the probability of generating a rational number over real [0,1] is undefined. I’m not questioning that the Lebesque measure of the rationals is 0, I’m questioning that the Lebesque measure corresponds to probability, which as this example demonstrates, really only fits our intuitions in the context of discrete random variables or ranges within continuous random variables. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 9:05pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/39 "2002-08-29T21:05:22Z")
> [@](#): > > \*Originally posted by Achernar \* > \*\*Oh, wait. Um, er… wait. Is that right? I’m confused now. ☹ How do you figure that, **Jabba**? \*\* Discrete random variables are well-behaved, as you use sums to deal with them, not integrals. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 29, 2002, 9:07pm UTC](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007/40 "2002-08-29T21:07:23Z")
> [@](#): > > \*Originally posted by jawdirk \* > **Jabba, that makes sense, but it leaves me with the question of whether probability is the right word to use when speaking of continuous random variables. Suppose I ask the question: what is more probable, rolling a 7 on a six-sided die or generating a rational value from a random selection of a real number on the range [0,1]? Are you going to say that they are equally probable or are you going to say that the question is meaningless because I am talking about two different sorts of probability?** They have equal probability, but one mi If the former, then it seems just wrong to say that something that is possible is equally probable as something impossible. If the latter, then the terminology is unnecessarily confusing. It would seem far less confusing to just say that the probability of generating a rational number over real [0,1] is undefined. I’m not questioning that the Lebesque measure of the rationals is 0, I’m questioning that the Lebesque measure corresponds to probability, which as this example demonstrates, really only fits our intuitions in the context of discrete random variables or ranges within continuous random variables. \*\* [/QUOTE] [Previous page](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007.md?page=1) [Next page](https://boards.straightdope.com/t/probability-question-easy-to-ask-hard-to-answer/126007.md?page=3)