Mixing
Probability for an infinite set
$begingroup$
The way probability is defined as the expected value works for finite sets. The probability of getting heads is out of two possible outcomes, heads or tails.
If we asked the probability out of an infinite set, like the set of positive integers, we could take a limit. For example, as this site discusses, if we ask "What is the probability of a random integer being divisible by 5?", we can still answer that question, and the answer if 1/5.
However, if we ask the question of what is the probability that of a random integer is 5, and applied the same process, we would get a limit of zero.
Is it possible to ask what is the probability that a random number is 5 in a way that makes sense mathematically?
probability limits infinity
asked Jan 13 at 15:45
useruser
737413
$endgroup$
|
show 9 more comments
$begingroup$
The way probability is defined as the expected value works for finite sets. The probability of getting heads is out of two possible outcomes, heads or tails.
If we asked the probability out of an infinite set, like the set of positive integers, we could take a limit. For example, as this site discusses, if we ask "What is the probability of a random integer being divisible by 5?", we can still answer that question, and the answer if 1/5.
However, if we ask the question of what is the probability that of a random integer is 5, and applied the same process, we would get a limit of zero.
Is it possible to ask what is the probability that a random number is 5 in a way that makes sense mathematically?
probability limits infinity
asked Jan 13 at 15:45
useruser
737413
$endgroup$
$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
2
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
2
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05
|
show 9 more comments
$begingroup$
The way probability is defined as the expected value works for finite sets. The probability of getting heads is out of two possible outcomes, heads or tails.
If we asked the probability out of an infinite set, like the set of positive integers, we could take a limit. For example, as this site discusses, if we ask "What is the probability of a random integer being divisible by 5?", we can still answer that question, and the answer if 1/5.
However, if we ask the question of what is the probability that of a random integer is 5, and applied the same process, we would get a limit of zero.
Is it possible to ask what is the probability that a random number is 5 in a way that makes sense mathematically?
probability limits infinity
asked Jan 13 at 15:45
useruser
737413
$endgroup$
The way probability is defined as the expected value works for finite sets. The probability of getting heads is out of two possible outcomes, heads or tails.
If we asked the probability out of an infinite set, like the set of positive integers, we could take a limit. For example, as this site discusses, if we ask "What is the probability of a random integer being divisible by 5?", we can still answer that question, and the answer if 1/5.
However, if we ask the question of what is the probability that of a random integer is 5, and applied the same process, we would get a limit of zero.
Is it possible to ask what is the probability that a random number is 5 in a way that makes sense mathematically?
probability limits infinity
probability limits infinity
asked Jan 13 at 15:45
useruser
737413
asked Jan 13 at 15:45
useruser
737413
asked Jan 13 at 15:45
useruser
737413
asked Jan 13 at 15:45
useruser
737413
asked Jan 13 at 15:45
useruser
737413
737413
$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
2
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
2
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05
|
show 9 more comments
$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
2
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
2
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05
$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
2
2
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
2
2
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I think you have to define rather precisely what your random process is for choosing an integer. For example this process allows any non-negative integer to be chosen but gives each one a finite probability:
- Randomly choose a digit from 0 to 9.
- Toss a coin. If you get heads, go to step 1.
- When you finally get tails, string together the chosen digits to make your random integer.
The process is random and obviously able to produce any positive integer or zero, but it's also heavily weighted towards lower numbers. In fact single-digit integers will occur more than half the time (since we might get a string of digits of which all but the last are zero).
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
$endgroup$
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
add a comment |
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1 Answer
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$begingroup$
I think you have to define rather precisely what your random process is for choosing an integer. For example this process allows any non-negative integer to be chosen but gives each one a finite probability:
- Randomly choose a digit from 0 to 9.
- Toss a coin. If you get heads, go to step 1.
- When you finally get tails, string together the chosen digits to make your random integer.
The process is random and obviously able to produce any positive integer or zero, but it's also heavily weighted towards lower numbers. In fact single-digit integers will occur more than half the time (since we might get a string of digits of which all but the last are zero).
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
$endgroup$
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
add a comment |
$begingroup$
I think you have to define rather precisely what your random process is for choosing an integer. For example this process allows any non-negative integer to be chosen but gives each one a finite probability:
- Randomly choose a digit from 0 to 9.
- Toss a coin. If you get heads, go to step 1.
- When you finally get tails, string together the chosen digits to make your random integer.
The process is random and obviously able to produce any positive integer or zero, but it's also heavily weighted towards lower numbers. In fact single-digit integers will occur more than half the time (since we might get a string of digits of which all but the last are zero).
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
$endgroup$
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
add a comment |
$begingroup$
I think you have to define rather precisely what your random process is for choosing an integer. For example this process allows any non-negative integer to be chosen but gives each one a finite probability:
- Randomly choose a digit from 0 to 9.
- Toss a coin. If you get heads, go to step 1.
- When you finally get tails, string together the chosen digits to make your random integer.
The process is random and obviously able to produce any positive integer or zero, but it's also heavily weighted towards lower numbers. In fact single-digit integers will occur more than half the time (since we might get a string of digits of which all but the last are zero).
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
$endgroup$
I think you have to define rather precisely what your random process is for choosing an integer. For example this process allows any non-negative integer to be chosen but gives each one a finite probability:
- Randomly choose a digit from 0 to 9.
- Toss a coin. If you get heads, go to step 1.
- When you finally get tails, string together the chosen digits to make your random integer.
The process is random and obviously able to produce any positive integer or zero, but it's also heavily weighted towards lower numbers. In fact single-digit integers will occur more than half the time (since we might get a string of digits of which all but the last are zero).
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
answered Jan 13 at 16:54
timtfjtimtfj
2,333420
2,333420
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
add a comment |
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
This is a great example, in fact p is greater than zero for any number, but this is okay because it is not the same for every number.
$endgroup$
– user
Jan 14 at 0:44
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
$begingroup$
@user Thanks! Another I thought of was: take a random real number $xin(0,1)$ then round $frac1x$ to the nearest integer below. Then there's a $frac1n$ chance of getting an integer $geq n$. But I thought getting $1$ half the time was a bit excessive!
$endgroup$
– timtfj
Jan 14 at 15:13
add a comment |
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$begingroup$
@ParclyTaxel Not necessarily. It depends what measure you're using.
$endgroup$
– user3482749
Jan 13 at 15:49
$begingroup$
Wouldn't that contradict the fact that, however, it is not impossible to get a 5? What measure could I define to get a reasonable answer?
$endgroup$
– user
Jan 13 at 15:50
$begingroup$
No, because (a) that's not a finite event, and (b) "probability zero" is not the same thing as "impossible" in general.
$endgroup$
– user3482749
Jan 13 at 15:50
2
$begingroup$
@user Formally, yes. To borrow a phrase from Douglas Adams, it means that it's "infinitely improbable". For a specific example: if you choose a number uniformly at random from $[0,1]$, the probability of any particular number coming up is exactly $0$, but one of them must come up, so one of those (uncountably infinitely many) probability-$0$ events must occur.
$endgroup$
– user3482749
Jan 13 at 15:54
2
$begingroup$
Something else: I disagree with what is said on the site with a the link in your question stating in a general sense that the probability of a random integer being divisible by $5$ is $frac15$. Why should the probability be not e.g. $frac12$? Think of picking a random number from listing: $1,5,2,10,3,15,4,20,6,25,cdots$.
$endgroup$
– drhab
Jan 13 at 16:05