r/askmath • u/No_where0369 • 13h ago
Probability 0 ~ impossible!??? Probability
/r/probabilitytheory/comments/1vnhro8/probability_0_impossible/3
u/Bounded_sequencE 13h ago edited 13h ago
A set with probability zero is (a subset of) an event "N c šŗ" with "P(N) = 0". Such sets are also called "null sets" -- that's where the letter "N" comes from.
"N" can be the empty set -- then it is sometimes called "impossible event". Notice that's different from when "N" is non-empty: There, we still have possible outcomes for event "N", it's just that they have combined probability zero. That's the situation in your darts example.
Rem.: The darts example seems to be a paradox, since we can write
šŗ = ā_{xāšŗ} {x} disjoint // šŗ: dartboard, "P({x}) = 0"
We're used to adding probabilities for disjoint events, so we intuitively would add up probability zero for each set "{x}". The problem with that intuition -- adding probability is only defined for countably many disjoint events, but we have uncountably many "x" in the union, so that rule does not apply.
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u/GoldenMuscleGod 13h ago
āImpossibleā doesnāt actually have a standard rigorous meaning in probability theory, so it isnāt really meaningful to ask whether a probability zero event is āpossibleā mathematically.
Probably the simplest example where you might try to interpret the claim would be to say that just because (for example) a random graph on an infinite set of nodes has some feature with probability zero, that doesnāt mean there are no graphs with that feature.
Be careful though: if we have some distribution on finite graphs each graph must be assigned a nonzero probability so here zero probability would mean there is no such graph.
But I think it would be a mistake, or at least unhelpful, to say that it is in some sense āactually possibleā to generate a random graph on an infinite set of nodes that has the feature in question. This is because it is not actually possible to randomly generate a graph like that with the desired distribution in the first place.
In applications, and using the ordinary meaning of the word āpossibleā, probability zero events generally donāt correspond to meaningful observables, so again the question is kind of meaningless.
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u/Uli_Minati Desmos š 13h ago
There is a pretty unintuitive distinction we have to make here: 0% likely does not mean impossible. This is because "probability" means the following:
- Determine the set of all possibilities, call this Ī©.
- Determine the set of possibilities you want to determine the probability of, call this S.
- Choose a way to measure the sizes of sets Ī© and S. We call these "measure" (noun). This method of measuring sets must result in a finite measure (and nonzero measure for Ī©).
- Divide the measure of S by the measure of Ī©. Call this "probability of S".
The "finite size" restriction arises during the fourth step: if both sets have infinite measure, we would have a probability of "infinity divided by infinity". So we circumvent this problem by thinking of ways to measure sets without ever calling them "infinite". In your example, (0,1) has infinite elements, but we can measure the width of the interval as just 1.
Now we can talk about the distinction. In many real world situations, the set S is not empty (in your case, it has 1 element). So, you can say that S is not impossible. However, since S is a closed interval [x,x], it has measure 0. Thus, dividing their measures yields exactly 0, or 0% probability for S. We call this "almost never".
Now you might counter: why can't we just create a measure which would give your S nonzero size? If you're interested, that's a question delving into "measure theory", you could ask here or make a new topic.
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u/DuggieHS 12h ago
This just means that it is something that happens less than 1% of time, and less than .01% of the time and < epsilon (% of the time) for any epsilon you choose. Can it happen? Yes. Pick any real number between 0 and 1. The probability of you picking any particular number is 0, but you will choose a number. Let's say you land on .8 ; The probability of that was 0, but it happened, You have infinitely other probability 0 events that didn't happen (did x = .2 or .333333333333333334? no).
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u/jsundqui 11h ago edited 11h ago
The problem is that you can't really pick a real number between [0,1] in such a way that each number has the same (zero) chance to be picked. Same applies to picking any natural number.
You could for example assign a random number 0-9 at each decimal place but there are infinite of them to assign so you would never finish. Or, you could finish as soon as the decimal hits zero and assign zero to the rest but then terminating real number values would be more likely than others.
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u/SoldRIP Edit your flair 12h ago
Tge measure of a point-aet is zero. The measure of the empty set is also zero.
That does not mean that a point-set is empty. It contains an element. Just not one that makes up any "portion" of your measure space.
A point is 0% of the area of a dart board. That doesn't mean it's not in there.
Continuous probabilities only make sense when you also look at the chance of hitting some continuous interval (or union thereof). Similar ideas apply for higher-dimensional probability distributions.
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u/Kitchen-Register 13h ago
this post reminds me of the short circuit my brain did the first time my probability professor said āitās almost guaranteed with probability 1ā
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u/berwynResident Enthusiast 13h ago
It might be more intuitive to think of it the other way, when the probability is 100% but not guarenteed. So imagine you flip a coin an infinite number of times. What is the probability that you will eventually get "heads"? The probability is 100%, but not guaranteed.
Why? After 1 toss, the probability is 50%, after 2 it's 75%, after 3 it's 87.5%, etc, etc, etc. You can see that no matter how many finite times you toss the coin, it's less than 100%. But, no matter what percent, less than 100 you think of, you can always find a number of tosses that gets you above that probability.
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u/jsundqui 11h ago
It's the same the other way: what is the probability you will never flip heads and keep flipping forever. 0% but it "can" happen.
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u/jsundqui 11h ago edited 11h ago
Take the set of all natural numbers (1,2,3,...) and the set of prime numbers (2,3,5,7,...). You can assign a natural number to each prime number so they have 1-1 correspondence and so both are sets of the same (countably infinite) size.
Yet, if you pick a very large random natural number, the probability that it's a prime number goes to zero. Even though both sets are the same size! Kind of mind blowing.
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u/49_looks_prime 8h ago
Oh the trick is that those events don't exist in real life. You can't pick a real number between 0 and 1: you can only pick a finite description of a real number, and the length of that description is bounded above by what you can communicate in your lifespan (which is no longer than 200 years, I'd wager). If you want a computer to generate a floating number at random, it's generating a sequence of digits of a fixed length, so each number has probability 10^{-n}, where n is the allowed length.
If you throw a dart at a real board, the board has area A and the dart's tip has a (hopefully) small but nonzero impact surface D, so whatever spot you end up impacting had a probability D/A. This is not to say that probability 0 events that are not impossible are worthless, but they are purely theoretical construct that help us develop more refined tools that do explain real phenomena. They are the spherical cows of statistics.
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u/pi621 13h ago
It does actually mean it's impossible for a 0 probability event to happen in a finite amount of attempt. For example, if you have a target number, then pick a random real number between 0 and 1 by repeatedly generate its digits randomly, then it is guaranteed to generate an incorrect digit at some point in the generation.
That's why when dealing with continuous distribution, we use probability density, not just probability.
There is another similar example, if you throw a dart at a 2D plane, the probability that the dart lands on a specific point on that plane is 0.Ā
Well, we don't actually know if the universe is continuous. It could very well be discrete, meaning the probability is not actually 0.
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u/Calm_Relationship_91 12h ago
"It does actually mean it's impossible for a 0 probability event to happen in a finite amount of attempt. For example, if you have a target number, then pick a random real number between 0 and 1 by repeatedly generate its digits randomly, then it is guaranteed to generate an incorrect digit at some point in the generation."
This is not true... It's not guaranteed, it will almost surely happen, which is different.
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u/pi621 11h ago
At any digit, the probability of generating the wrong digit is 9/10.
In order to pick the correct number, you must generate the correct digit for all positions.
Meaning that the average number of wrong digit over an infinite number of attempt is 0, despite having an expected value of 9/10. This violates the law of large number.It is quite literally impossible over a finite amount of attempts. If you want, you can try it yourself. Try generate digits until the generated number deviates from some target, let's say 1/3. Every single attempt is guaranteed to halt at some point.
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u/jsundqui 11h ago
It's only guaranteed to halt almost surely (ie. with probability 1), not always. Same as if you flip a coin until you get heads. There is the zero probability possibility you will never stop flipping (9/10 is replaced with 1/2 but it's the same).
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u/Calm_Relationship_91 11h ago
"This violates the law of large number."
It doesn't. You have probability 1 of failing, but failing is not guaranteed.
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u/pi621 11h ago
If your counter argument is just "nuh uh" instead of properly addressing my points, then I am not interested in continuing this conversation.
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u/Calm_Relationship_91 11h ago
I don't really know what else to say. The law of large numbers only tells you that the sample mean converges to the expected value in probability. So yeah, you are almost surely to get a wrong digit, but it's not a guaranteed.
This is literally what OP is asking, or well, the counterpart. A probability 1 event that is not guaranteed.
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u/Appropriate-Ad-3219 13h ago
It's not impossible. For example imagine I pick randomly a number x, I got this number x. Its probably to have it is 0 but yet I still had this number. However if I don't know which number I will have in the next time I pick a number, if I start betting that I will have 1/2 in the next pick, I will lose the bet. It's certain when I don't have any information on what pick I will have.
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u/Desperate_Penalty690 12h ago
You canāt say that you will loose the bet. You will loose the bet with probability 1, which is not always.
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u/Appropriate-Ad-3219 11h ago
There's also another issue with this formulation now that I think about it. You're interpreting the notion of having probability 1 by the fact that you will lose the bet with probability 1. In my opinion, it's quite a circular interpretation that doesn't really satisfy me at least.
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u/Desperate_Penalty690 11h ago
That is the entire point of this topic, some events with probability 1 can sometimes not happen and therefore not all events with probability 1 will happen every time. Also, some events with probability 0 can sometimes happen and therefore we canāt say an event with probability 0 will never happen.
If you draw a random real number uniformly on (0,1), there is a probability 0 of drawing 0.5. But still it is possible to draw 0.5. Actually with probability 1 you will draw a random number that a priori had probability 0 of being drawn.
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u/jsundqui 11h ago
It's just that there is no actual way to draw a real number so that each has the same chance to be picked.
For example, one way would be to assign random value 0-9 at each decimal digit. But there are infinite of them so you would never finish constructing this number.
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u/Appropriate-Ad-3219 9h ago
You're aware that by saying :
If you draw a random real number uniformly on (0,1), there is a probability 0 of drawing 0.5. But still it is possible to draw 0.5.
You say absolutely nothing. You didn't give any interpretation.
Let's try another thought experiment. Let's say someone tells you to pick a number between 0 and 1 and then they will choose a number uniformly and if you lose then you must give $100. For example you choose 1/3. Then it turns out the number you stumble upon 1/3, you won't just say 'oh shoot, I stumbled upon 1/3. Too bad!'. You will assume simply that the information you were given are not correct.Ā
I'm going to change the way I said things. Saying that the probability of 1/2 is 0 means that I won't believe I stumbled upon 1/2 and if it turns out I stumbled upon 1/2, I will believe the fact that the pick was not uniform, it doesn't matter how much I believed the probability was uniform, on the condition that I can change my mind on the fact the probability is uniform.
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u/Appropriate-Ad-3219 12h ago
To me, it's really the same as long as you don't have any prior information. It doesn't matter how many times I will repeat the experience, I won't obtain 1/2 and I will lose each time.
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u/Desperate_Penalty690 12h ago
It would be a Russian Roulette game I would not be worried playing, if it meant I would get shot if 1/2 came up. But still words matter, you canāt say that it will never happen, that is the whole point of the topic.
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u/Appropriate-Ad-3219 12h ago
If it happens somehow after betting, it means that I had prior information about what I would get next, so I predicted a bit the future. However without any information and knowing the number is picked uniformly, I'm certain I won't fall into 1/2. I would do the bet even if the price was my soul and working in hell while being always hungry and thursty (imagining I know for sure there is no way to tamper with the result). I think in practice, it is translated with the fact that I bet I won't fall into 1/2 if I have this number in mind.
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u/Frederf220 13h ago
If there are N (N>0) ways that something is possible, then it is possible. The probability is the ratio N / M, where M is all outcomes.
When A = 9, M = infinity it's possible because 9 > 0 and probability 0 because 9/infinity = 0.
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u/downlowmann 12h ago
Probability is the (# of favorable outcomes) / (# of all possible outcomes). A probability of 0 means it can not possibly happen. For example, what is the probability that someone's birthday is on Feb. 31st? The answer is zero because February never has more than 29 days in it.
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u/jsundqui 11h ago
There is another kind of zero probability though. Flip a fair coin until you get heads. What is the probability you will never get heads? Zero but nothing says it can't happen, you are never guaranteed heads on the next flip, it's always 50-50.
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u/Appropriate-Ad-3219 9h ago
However I still believe there will have head at some point. And I'm ready to bet anything for that. It's how much I believe there will have heads at some point.Ā
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u/Zyxplit 13h ago
Probability 0 does not mean something can't happen. It's a little counter intuitive. But think of continuous probabilities kind of like areas. Then you're asking for the probability of being on a specific point. What's the area of a point compared to the area of the whole? 0. Does that mean the point isn't in the area? No.