Is getting a random integer even possible?
$begingroup$
On googling I got
Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.
What if I want to pick a random integer from the number line ?
The number line consists of infinitely many integers.
My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity.
I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.
Now the probability of getting a random number
P(E) = 1 / ∞ = 0
So I won't be able to get a random number, ever.
This implies, never would I ever be able to get a random number.
That's absurd as many of our science topics do depends on things being random.
probability random integers
asked Aug 3 '16 at 14:27
HritikHritik
971110
$endgroup$
add a comment |
$begingroup$
On googling I got
Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.
What if I want to pick a random integer from the number line ?
The number line consists of infinitely many integers.
My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity.
I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.
Now the probability of getting a random number
P(E) = 1 / ∞ = 0
So I won't be able to get a random number, ever.
This implies, never would I ever be able to get a random number.
That's absurd as many of our science topics do depends on things being random.
probability random integers
asked Aug 3 '16 at 14:27
HritikHritik
971110
$endgroup$
5
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
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Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
3
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34
add a comment |
$begingroup$
On googling I got
Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.
What if I want to pick a random integer from the number line ?
The number line consists of infinitely many integers.
My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity.
I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.
Now the probability of getting a random number
P(E) = 1 / ∞ = 0
So I won't be able to get a random number, ever.
This implies, never would I ever be able to get a random number.
That's absurd as many of our science topics do depends on things being random.
probability random integers
asked Aug 3 '16 at 14:27
HritikHritik
971110
$endgroup$
On googling I got
Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.
What if I want to pick a random integer from the number line ?
The number line consists of infinitely many integers.
My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity.
I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.
Now the probability of getting a random number
P(E) = 1 / ∞ = 0
So I won't be able to get a random number, ever.
This implies, never would I ever be able to get a random number.
That's absurd as many of our science topics do depends on things being random.
probability random integers
probability random integers
asked Aug 3 '16 at 14:27
HritikHritik
971110
asked Aug 3 '16 at 14:27
HritikHritik
971110
edited Dec 12 '18 at 20:38
Hritik
asked Aug 3 '16 at 14:27
HritikHritik
971110
asked Aug 3 '16 at 14:27
HritikHritik
971110
asked Aug 3 '16 at 14:27
HritikHritik
971110
971110
5
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
$begingroup$
Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
3
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34
add a comment |
5
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
$begingroup$
Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
3
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34
5
5
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
$begingroup$
Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
$begingroup$
Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
3
3
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Most typically you choose a random natural from a finite range. If you toss a fair coin, you get a random number that is either $0$ or $1$. If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$. There is no problem having equal probability in these cases. If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$. If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
add a comment |
$begingroup$
Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $forall x:0<f(x)<1$.
Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.
An appropriate function for the natural numbers $Bbb N^+$ is $f(x)=1/2^x$
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Most typically you choose a random natural from a finite range. If you toss a fair coin, you get a random number that is either $0$ or $1$. If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$. There is no problem having equal probability in these cases. If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$. If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
add a comment |
$begingroup$
Most typically you choose a random natural from a finite range. If you toss a fair coin, you get a random number that is either $0$ or $1$. If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$. There is no problem having equal probability in these cases. If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$. If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
add a comment |
$begingroup$
Most typically you choose a random natural from a finite range. If you toss a fair coin, you get a random number that is either $0$ or $1$. If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$. There is no problem having equal probability in these cases. If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$. If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
$endgroup$
Most typically you choose a random natural from a finite range. If you toss a fair coin, you get a random number that is either $0$ or $1$. If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$. There is no problem having equal probability in these cases. If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$. If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
edited Dec 12 '18 at 14:10
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
answered Aug 3 '16 at 14:39
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
add a comment |
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
Actually I'm considering entire integers (and perhaps entire number line would be better) because as randomness exists in nature, it has to be fair with all the numbers present.
$endgroup$
– Hritik
Aug 4 '16 at 15:11
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
$begingroup$
@Hritik But randomness in nature doesn't imply a uniform distribution on a countably infinite set, or on the integers specifically, nor on the whole real line for that matter. Why should it?
$endgroup$
– Ned
Dec 12 '18 at 21:12
add a comment |
$begingroup$
Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $forall x:0<f(x)<1$.
Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.
An appropriate function for the natural numbers $Bbb N^+$ is $f(x)=1/2^x$
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
$endgroup$
add a comment |
$begingroup$
Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $forall x:0<f(x)<1$.
Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.
An appropriate function for the natural numbers $Bbb N^+$ is $f(x)=1/2^x$
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
$endgroup$
add a comment |
$begingroup$
Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $forall x:0<f(x)<1$.
Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.
An appropriate function for the natural numbers $Bbb N^+$ is $f(x)=1/2^x$
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
$endgroup$
Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $forall x:0<f(x)<1$.
Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.
An appropriate function for the natural numbers $Bbb N^+$ is $f(x)=1/2^x$
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
edited Dec 13 '18 at 10:49
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
answered Dec 12 '18 at 10:40
user334732user334732
4,27411240
4,27411240
add a comment |
add a comment |
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5
$begingroup$
There is no such thing as a uniform distribution on the set of integers (for the reasons that you outlined).
$endgroup$
– John Coleman
Aug 3 '16 at 14:29
$begingroup$
Integers are distributed uniformly, aren't they ?
$endgroup$
– Hritik
Aug 3 '16 at 14:30
3
$begingroup$
There is no uniform probability distribution on them (though there are various non-uniform distributions). The fact that the integers are spatially distributed in a uniform way is irrelevant.
$endgroup$
– John Coleman
Aug 3 '16 at 14:32
$begingroup$
Perhaps you are using a different definition of "uniform distribution"? If you are using the standard definition then, as @JohnColeman says, there does not exist a uniform distribution on the integers.
$endgroup$
– Lee Mosher
Aug 3 '16 at 14:34
$begingroup$
Would you like to explain a bit further or quote a reference ?
$endgroup$
– Hritik
Aug 3 '16 at 14:34