# Selecting two points in [0-1] being the same p>0?

You and I each choose an undisclosed real number from 0 to 1. Then we compare them to see if we chose the same number.

p(a)=0, p(b)=0 -> p(a=b)>0 ?

I seem to recall that the probability of one instance of choosing a particular number should be zero, but if two people are doing so and the probability in question is changed to whether the two choices are the same number, it seems like this probability of the two numbers being equal would be greater than zero, yet the two selections (a and b) individually have zero probability...?

Orodruin
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it seems like this probability of the two numbers being equal would be greater than zero
Why do you think that? It is not true.

Dale
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P(A=B)=0

Why do you think that? It is not true.

Am I asking the same thing as whether a random number selector acting on the closed interval an indefinite number of times would ever select the same number more than once?

Dale
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a random number selector acting on the closed interval
Are you talking about a machine precision random number generator or a theoretical selector acting on all real numbers in the interval.

..."true random" using natural entropy (same as theoretical?)... rather than algorithm...

Dale
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P(A=B)=0

Orodruin
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P(A=B)=0
Not only that, the probability of picking any number twice or more if you let it run forever and picking one number per second is zero.

Am I asking the same thing as whether a random number selector acting on the closed interval an indefinite number of times would ever select the same number more than once?

PeroK
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You and I each choose an undisclosed real number from 0 to 1. Then we compare them to see if we chose the same number.

p(a)=0, p(b)=0 -> p(a=b)>0 ?

I seem to recall that the probability of one instance of choosing a particular number should be zero, but if two people are doing so and the probability in question is changed to whether the two choices are the same number, it seems like this probability of the two numbers being equal would be greater than zero, yet the two selections (a and b) individually have zero probability...?
If you mean: choose a real number between 0 and 1 on a uniform distribution, then that is practically impossible.

Mathematically you can have a random variable uniformly distributed on ##[0,1]##. But, not every mathematical process can be realised by a physical process.

If two people choose a real number, by whatever means they have at their disposal, then the probability they choose the same number is greater than 0.

Dale
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If two people choose a real number, by whatever means they have at their disposal, then the probability they choose the same number is greater than 0.
By which you mean that people don’t choose real numbers in a manner consistent with a uniform distribution. You do not mean that the probability is greater than zero of getting two independent samples the same from a uniform distribution. Correct?

PeroK
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By which you mean that people don’t choose real numbers in a manner consistent with a uniform distribution. You do not mean that the probability is greater than zero of getting two independent samples the same from a uniform distribution. Correct?
People cannot choose even the whole numbers uniformly, let alone the real numbers. It's impossible.

Dale
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Am I asking the same thing as whether a random number selector acting on the closed interval an indefinite number of times would ever select the same number more than once?

That isn't well defined mathematical question. The mathematical theory of probability (which is called measure theory) has no definitions or assumptions that concern whether events that are assigned probabilities actually happen or not. It doesn't have even any assumptions that say you can take random samples.

When people apply probability theory to a real life problem, they do interpret probabilities as a kind of "tendency" for some "possible" event to actually happen. They assume random samples can be taken and propose specific methods for taking them. However, the question of how mathematics should be applied is not a question that can be settled by mathematics itself. Applications of math involve questions of physics, or economics, or whatever discipline treats the problem at hand.

I know of no physical set up that can take a random sample from a uniform distribution. I know of no physical setup that can take infinitely many samples and stop in a finite time in order to announce a result from doing so. My opinion cannot be confirmed or denied by appealing to probability theory because probability theory says nothing about this.

Probability theory is essentially circular. Probability theory talks about probabilities. It doesn't say how to interpret them. Using your notation, probability theory says P(A=B)=0 for two independently distributed uniform random variables A,B. Probability theory doesn't say than an event with probability zero can't actually happen - because it doesn't say anything about possible events actually happening or not. (It wisely avoids the metaphysical complications of defining "possible" and "actual" and formulating axioms about these concepts.)

Probability theory talks about probability spaces and functions (probability measures) defined on sets of outcomes. When people apply probability theory to specific problems they introduce the concept of possible events actually happening or not. They generally interpret an event that is assigned probability zero to be an event that isn't physically possible. Whether this is correct or not is a matter of physics. It cannot be settled by probability theory.

..."true random" using natural entropy (same as theoretical?)... rather than algorithm...

That doesn't describe a specific physical process. If you describe a specific process then its behavior can be discussed - but that discussion belongs in the physics sections.

jim mcnamara
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Probability theory doesn't say than an event with probability zero can't actually happen
This is a good point. If you have a continuous probability density function then any single number in its range has P(X=x)=0. So every result obtained had probability 0.

jim mcnamara

I'm thinking that there is nothing logical or physical strictly preventing a number being randomly selected more than once because prior values of numbers randomly selected have no influence on subsequent values.

Am I wrong to think that true random implies and entails absence of causality (absolute absence and intractability of ontological history)?

Orodruin
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I'm thinking that there is nothing logical or physical strictly preventing a number being randomly selected more than once because prior values of numbers randomly selected have no influence on subsequent values.
This in no way changes the fact that any countable set of points in [0,1] will have probability 0 of being chosen due to having zero measure. Obviously there is nothing physical if you are speaking actual probability theory. It has nothing to do with physics a priori. That a number can be chosen more than once does not mean that it has a non-zero probability of being chosen more than once. As has already been mentioned in this thread, every real number in [0,1] has probability zero of being chosen if you have a truly uniform distribution. So assume you pick your first number to be ##x_0##. When you pick your second number the probability of getting ##x_0## again will be zero because the probability of getting ##x_0## in the second pick is zero. You can repeat this for any ##x_0##.

That a number can be chosen more than once does not mean that it has a non-zero probability of being chosen more than once.

Thanks, everyone, this answers my question and my misunderstanding.

Dale
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I'm thinking that there is nothing logical or physical strictly preventing a number being randomly selected more than once
But that does not change the fact that P(A=B)=0. Did you not read my most recent post? P(A=B)=0 does not mean that anything logical or physical prevents A=B