# Is the probability zero?

aquaregia
Something that I have always wondered: say you know that a robot will push a button during a 2 minute period after a timer has been started, and you know that the robot picks a time to press it at complete random.

Is the probability that the button will be pressed exactly 1 minute after the timer is started zero? I would say this because probability is defined as (# of specified events/# of possible events). Assuming time and movement are continuous, you would have infinite possible events, and 1 specified event (the timer is exactly 1.000...), and 1/infinity=0. But the robot has to press it at some time so say it presses it at exactly the sqrt2 minutes. Before it happened, the probability that the robot would press the button at sqrt2 minutes was 0 but then the event happened. How is that possible?

Homework Helper
This is a question about continuous random variables (and measures). Assuming that the period of 2 minutes is infinitely divisble, then the probability that something happens at precisely x seconds (of the 120) is zero under the uniform measure. That's because we only can talk about an event happening within some subset of the period - the measure of a point is zero.

But don't forgot that there is a difference between reality and a mathematical model of reality.

aquaregia
But there is some exact single instant when the button is finally pressed (at least in a non-quantum mathematical universe), right? Or say we were talking about a perfect mathematical sphere getting dropped onto a perfect plane. There would be some exact instant when the sphere first touched the plane and that time could be any real number. Or a better example might be considering a random length perfect stick which could be any real number between 1 and 2. How could the probability that its length be 1.5 be zero, when it actually COULD be 1.5?

shamrock5585
because nothing can be exactly 1.5 based on our measuring system... is it 1.50000000000000000000001 or is it 1.50000000000000000000000000000000000000000000000000001

it can never be perfect unless there is 1.5 and an infinite number of zeros and infinity is not a real number. This is why humans can never create a "perfect circle" because pi is a never ending decimal

maverick_starstrider
yes but if time is a real number then between any two instants in time (no matter how close together they are) there are an infinite number of other possible times (i.e. between 1.000000000 and 1.000000001 there is also1.0000000005, 1.0000000001, 1.00000000055, etc.) therefore the probability of 1 exact instance in an infinite number of possiblities is $$\frac{1}{\infty}$$ which is zero.

Homework Helper
But there is some exact single instant when the button is finally pressed (at least in a non-quantum mathematical universe), right?

there is, in reality, but that has nothing to do with the fact that you're attempting to model things with probaility theory, and if you choose a continuous r.v. on the interval [0,2] with the uniform distribution, then the probability that it happens at 1, or any other singleton set is zero.

The rest of your post confirms that you don't understand that in continuous r.v.s the probability zero does not mean that it cannot happen. Sorry about the number of negatives in that statement - there is a very significant difference between continuous r.v.s and discrete r.v.s.

shamrock5585
maverick i see what your saying but its not 1 divided by infinity... what i was explaining is that there is no EXACT instance... to have an EXACT instance you would need an infinite amount of zeros after that number... anything that occurs in time-space has a duration! a true impulse is 1/0 which does not exist in reality!

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Homework Helper
It is a misunderstanding that events of probability 0 are impossible.

maverick_starstrider
maverick i see what your saying but its not 1 divided by infinity... what i was explaining is that there is no EXACT instance... to have an EXACT instance you would need an infinite amount of zeros after that number... anything that occurs in time-space has a duration! a true impulse is 1/0 which does not exist in reality!

It need not. Definitions are funny like that. In this case I can say my 'event' is defined as the point in space-time where the variable t=1 (exactly 1) and there you go. It's the time which is the complement of the set (-infinity,1),(1,infinity).

Homework Helper
Thanks CRGreathouse, for stating my point with a minimal number of negatives. Wish I'd've stated it like that.

shamrock5585
It need not. Definitions are funny like that. In this case I can say my 'event' is defined as the point in space-time where the variable t=1 (exactly 1) and there you go. It's the time which is the complement of the set (-infinity,1),(1,infinity).

yes but does anything in reality actually occur at an actual instance or does it have a duration, even if it is in nanoseconds.

maverick_starstrider
I dunno. Ultimately a space-time 'event' is all about abstract notion. Therefore, unless you can say in some irrefutable way the time... say... it takes a wavefunction to collapse then an event can often be nothing but an abstract notion and that's fine.

gel
Mathematically you can suppose that the event occurs at a precise time in the interval [0,2]. The answer to the question is down to http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof" [Broken] of the real numbers .
Given a sequence of events A1, A2, ... each with probability 0, then their union also has probability 0.

However, this is not true for an uncountable collection of events.
For such an uncountable collection each with probability 0, you can't infer that their union has probability 0.

This is down to http://en.wikipedia.org/wiki/Sigma_additivity" [Broken] of probability measures.

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DeaconJohn

Something that I have always wondered: say you know that a robot will push a button during a 2 minute period after a timer has been started, and you know that the robot picks a time to press it at complete random.

Is the probability that the button will be pressed exactly 1 minute after the timer is started zero? I would say this because probability is defined as (# of specified events/# of possible events). Assuming time and movement are continuous, you would have infinite possible events, and 1 specified event (the timer is exactly 1.000...), and 1/infinity=0. But the robot has to press it at some time so say it presses it at exactly the sqrt2 minutes. Before it happened, the probability that the robot would press the button at sqrt2 minutes was 0 but then the event happened. How is that possible?

aquaregia,

Good point. Good question. Well put! For myself, I am primarily an applied mathematician. Have been since an early age. So, your question strikes home to me.

Yes, of course, as you explain, there are very good reasons for declaring the probability is zero that the even will occur between 732 and 733 seconds after the start time (assuming that all one second intervals are equally likely). My guess is that there is probably a mathematical theory lying around somewhere that develops a theory of probability that covers this point. Probably even "non-standard analysis" addresses it, but I'll bet there is something simpler, and more relevant to real life, than non-standard analysis.

I'm not sure, though. There is a scale free approach to Baysian statistics that claims to address such questions. The conclusion there, however, is that not all one second intervals are equally likely. Roughly, they claim that a one second interval twice as far in the future as another interval is half as likely as the closer one. I don't like that aproach, the classic "tram problem" shomws that it has technical flaws. Probably there are other approaches. Probably no known approach is perfect.

Deacon John

Staff Emeritus
The probability distribution function for a continuous random variable is used to measure the probability of an event occurring over some range of values (a range with a non-zero measure, to be precise). The probability is defined as

$$P(X\in(x_0,x_1)) = \int_{x_0}^{x_1} p(x)\, dx$$

If you use this definition to compute the probability of hitting an exact point you will always get zero because $\int_a^a f(x)\, dx = 0[/tex] for any function that is integrable over some neighborhood of a. As Matt and GR mentioned above, this does not mean that the robot cannot punch a button. A zero probability does not necessarily mean an impossible event. It might just mean that you erroneously used the pdf to calculate the probability of an event occurring over a space of measure zero (which is exactly what you are doing). That said, there are ways to use the probability distribution function to predict when the robot will punch the button. The pdf lets you compute statistical measures such as the mean, median, variance, etc. The probability distribution no longer applies once the robot has punched the button. He punched the button at exactly 1.414... minutes. There is no uncertainty. The probability he did punch the button at 1.414... minutes is exactly one, and the probability he punched it at any other time is identically zero. One way to visualize what happens when the robot punches the button at 1.414... minutes is to use the approach many physicists take. Envision the pdf as collapsing to the Dirac delta distribution [itex]\delta(x-\surd 2)$ the instant the robot punches the button at 1.414... minutes. Physicists deal with a similar conundrum in quantum mechanics. The outcome of quantum mechanical problems is probabilistic, dictated by the wave function for the process at hand. Once a measurement is made the only uncertainty left is that inherent in the measurement process. One interpretation of quantum mechanics is that the wave function collapses to that measurement uncertainty.

DeaconJohn
Thanks D H,

I had mis-understood the question. It was more of a "beginner's question" than I thought. No, that's no quite right. It was a question with a simpler explanation than I thought. Unfortunately, it reminded me of a related question (stated in my post) that does not have such an easy answer, as far as I know.

Deacon John

krikker
1/infinity isn't zero. It's infinately small, which for most intents and purposes is zero.

shamrock5585
krikker dont argue that... in a math class you are correct... but in this forum some people will argue stuff like that to the death haha i know from experience.

Staff Emeritus
Whether you are in a math class or a physics class, saying "1/infinity isn't zero. It's infinitely small..." is not correct. In a math class, "1/infinity" is not defined, end of story. In a physics class, 1/infinity is short for $\lim_{x\to\infty} \frac 1 x$, and this is zero.

krikker
Since a probability isn't an actual physical thing I think that it belongs in the math class. Can't prove it though

krikker
Whether you are in a math class or a physics class, saying "1/infinity isn't zero. It's infinitely small..." is not correct. In a math class, "1/infinity" is not defined, end of story. In a physics class, 1/infinity is short for $\lim_{x\to\infty} \frac 1 x$, and this is zero.

I think that my statement is actually the closest you can get to a correct answer. Because talking about an actual physical phenomenon, the math conclusion doesn't suffice. But also, the size of a probability (IMO) doesn't really exist anywhere to be meassured, ruling out the physics view. Only thing we can say for sure is that if it happened, then the probability wasn't zero.

Whether you are in a math class or a physics class, saying "1/infinity isn't zero. It's infinitely small..." is not correct.

True, unless of course it's a class on hyperreal numbers.

In a math class, "1/infinity" is not defined, end of story.

Unless of course it's a class that uses extended real numbers, in which case it's defined to be 0.

Homework Helper
Something that I have always wondered: say you know that a robot will push a button during a 2 minute period after a timer has been started, and you know that the robot picks a time to press it at complete random.

Is the probability that the button will be pressed exactly 1 minute after the timer is started zero? I would say this because probability is defined as (# of specified events/# of possible events).
Although matt grime implied it, no one has specifically stated that this is simply wrong. That definition of probabilty applies only in a space with a finite number of (equally likely) events and is very limited even for "discrete" probability (i.e. with a finite number of possible events). "Continuous" probability distributions require that you have a continuously defined probality distribution. For events defined on an interval, such as this, "length of interval defining this particular event divided by the length of the entire interval" is the simplest such distribution (that was what matt grime said in the very first response to the original post) so all discussion of "1/infinity" is irrelevant. It is true that if you are asking "what is the a probability that the robot will press the button at this particular instant, then you are talking about an interval with length 0 so the probability is 0. (NOT "1/infinity" but "0/length of entire interval".)

The rest of this discussion is also wrong in that it assumes "probability 0" means "can't happen" is true only in discrete probability problems. Equivalently, "probability 1" does NOT mean "must happen". If we had a probability problem in which a real number was to be chosen from all real numbers in the interval [0, 1], with "uniform probability" (all such numbers equally likely to be chosen), then it is easy to calculate that the probability that the number chosen is between, say, 0.25 and 0.30 (in [0.25, 0.30]), is (0.30- 0.25)/(1.00- 0.00)= 0.05.

Similarly, the probability that the number chosen is precisely 0.26, say, (in [0.26,0.26]) is (0.26- 0.26)/(1.00- 0.00)= 0. But since every number is equally likely it certainly is possible that the number chosen is precisely 0.26. "0 Probability" does NOT mean, in this case, that it "can't happen". Similarly, the probability that the number chosee is anything BUT 0.26 is the same as "Probability x is in [0, 0.26) or x is in (0.26, 1]" is (0.26- 0)/(1.0- 0)+ (1- 0.26)/(1.0- 0)= 1. But since 0.26 is possible, it is NOT certain that a number other than 0.26 must be chosen: "Probability 1", in continuous probability problems does NOT mean "certain to happen" and "probabilty 0" in continuous probability problems does NOT mean "cannot happen".

Assuming time and movement are continuous, you would have infinite possible events, and 1 specified event (the timer is exactly 1.000...), and 1/infinity=0. But the robot has to press it at some time so say it presses it at exactly the sqrt2 minutes. Before it happened, the probability that the robot would press the button at sqrt2 minutes was 0 but then the event happened. How is that possible?

Staff Emeritus
Although matt grime implied it, no one has specifically stated that this is simply wrong.
I did, here:
As Matt and GR mentioned above, this does not mean that the robot cannot punch a button. A zero probability does not necessarily mean an impossible event. It might just mean that you erroneously used the pdf to calculate the probability of an event occurring over a space of measure zero (which is exactly what you are doing).

krikker
you are talking about an interval with length 0 so the probability is 0. (NOT "1/infinity" but "0/length of entire interval".)

As I see it you just rephrase the problem to how big a single point is. Also earlier posts used integrals as arguments, but all calculus is based on limit values, and as such face the same fundamental problem of the human brain only fathoming finite values.
In physics you can conclude from observation that the true value is actually equal to the limit. For instance that the arrow reaches it's target even though it has to travel half the remaining distance infinately many times. But that's not math. In that respect I will argue that probabilities don't actually exist, and as such you can't use the physics approach. Rather you need to stick with the two things that you know for sure. Zero means it won't happen, and one means that it will.

Somehow I have the feeling that we are having a discussion where all are bound to be wrong one way or the other. I am, however, open to the possibility that it's only me...

Staff Emeritus
As I see it you just rephrase the problem to how big a single point is.
A point has no size.
Also earlier posts used integrals as arguments, but all calculus is based on limit values, and as such face the same fundamental problem of the human brain only fathoming finite values.
Many of us here at this forum have absolutely no problem fathoming things like limits, derivatives, and god forbid, integrals.

In that respect I will argue that probabilities don't actually exist, and as such you can't use the physics approach. Rather you need to stick with the two things that you know for sure. Zero means it won't happen, and one means that it will.
The physics approach has been very successful. For one thing, without the physics of quantum mechanics I would not be typing this note on my computer. Just because you don't like/understand probability and statistics does not mean it is wrong.

krikker
A point has no size.

But it exists. So what you are saying is that zero doesn't necessarily express nothing? I call that hard to fathom, but if you get it I believe you.

Many of us here at this forum have absolutely no problem fathoming things like limits, derivatives, and god forbid, integrals.

So explain to me purely mathematically: Does the arrow reach its target (based on that specific model of course)?
Calculus is constructed in such a way that we never consider the actual infinites/infinitesimals but only their limits. Good thing is we can fathom it. Bad thing is we can never conclude that the arrow reaches its target, because we don't have the actual number. We only have its limit.

The physics approach has been very successful. For one thing, without the physics of quantum mechanics I would not be typing this note on my computer. Just because you don't like/understand probability and statistics does not mean it is wrong.

I never said I didn't like it. And I never claimed it to be wrong just because I don't understand it. I'm merely trying to make an argument to the best of my ability.

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Homework Helper
Probability is just a mathematical theory, and something we use to model things in 'the real world'. When using continuous pdfs via Lebesgue (can never recall the spelling) measure, the measure of a point of the real line is zero. There really is no way round that as it a simple 'fact'. If you want to argue that this isn't the correct model for whatever your situation is, krikker, then go ahead. It doesn't alter the fact that the (Lebesgue) measure of a point is 0.

Staff Emeritus
So explain to me purely mathematically: Does the arrow reach its target (based on that specific model of course)?
While philosophers might be vexed by Zeno's paradox, mathematicians most definitely are not because any mathematician worth their salt knows that
$$\frac 1 2 + \frac 1 4 + \cdots = \sum_{n=1}^{\infty}\frac 1{2^n} = 1$$

krikker
Probability is just a mathematical theory, and something we use to model things in 'the real world'. When using continuous pdfs via Lebesgue (can never recall the spelling) measure, the measure of a point of the real line is zero. There really is no way round that as it a simple 'fact'. If you want to argue that this isn't the correct model for whatever your situation is, krikker, then go ahead. It doesn't alter the fact that the (Lebesgue) measure of a point is 0.

I haven't really worked with that, but as I understand it, it is used to develop an algebra which includes infinites and infinitesimals. Therefore the measure can be defined as zero, so long as the algebraic operations are based on this definition. I'm not claiming this to be true, I'm in a bit of deep water. Just thought I remembered some of what you said from an old conversation with a mathematician friend of mine.

$$\frac 1 2 + \frac 1 4 + \cdots = \sum_{n=1}^{\infty}\frac 1{2^n}$$

I didn't know this relation to actually be true, since the right side is defined as the limit of the left side. If, however it holds true instead of being just a convention, then I can only express my joy since I use it myself all the time. Thought it was just one of those things engineers did and real mathematicians hated them for. Btw, it makes me even happier because my high school math teacher, who I disliked, made it a point how that relation didn't hold.

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Homework Helper
What do you mean by 'actually true'? (In relation to the sum.) It is certainly true in the real numbers from the definition. Whether you consider that to be an 'actual truth' about 'reality' is almost certainly philosophical.

krikker
My point is that by definition
$$\sum_{n=1}^{\infty}$$
is
$$\lim_{N\to\infty} \sum_{n=1}^{N}$$

I thought that mathematically you couldn't put an equal sign between an endless sum and the limit of that sum. I knew people did it, but I thought it to be a convention rather than 100% correct.

Homework Helper
It is essentially the definition of the real numbers that they are equal (since one is *defined* to be the other). Of course it is non-trivial to show that this makes sense. Fortunately, we can construct the real numbers in many ways - dedeking cuts, equivalence classes of cauchy sequences, or equivalence classes of decimal expansions (or any other base than 10).

Staff Emeritus
$$\sum_{n=1}^{\infty} \frac 1 {2^n} \equiv \lim_{N\to\infty}\sum_{n=1}^N \frac 1 {2^n} = 1$$
Non-mathematicians (and high school math teachers, boo!) are the ones who say $0.999\cdots \ne 1$. The limit of a sequence, if it exists, is a specific number. The limit does not differ from this number by some infinitesimal amount. It is the number.
$$\left(\sum_{n=1}^{\infty}\frac 1 {2^n}\right)-1=0$$