If you just graph fish vs. time, then the one-time gift is a constant function, the steady income is linear, and the “growing promise” stream-of-streams is a quadratic. The fact that a steady income will eventually surpass any one-time gift is because any positive-slope linear function will eventually exceed any constant-value function. Likewise, any 2nd-order polynomial with a positive x^2 term will eventually exceed any linear function. You could keep going with higher order polynomials if you want.
A similar analogy would be a race where even a large head start will eventually be surpassed by a slightly faster car.
I guess I know that kind of perspective exists but I don’t want to refer to the “up to a time” thing but the total.
For example taking the even and odd integers, that would seem like taking a stream of fish and putting them into alternate buckets every other day. I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
if I do it in a very naive way in graph vs time then I could conclude that the bucket that I start with has a stricly dominating amount of fish (because the graph stay above or matches the other graph). And it feels odd to say that there are more even numbers than odd ones! Althought I guess it wouldn’t be odd that 0+{1,2,3,4,5...| x is even} is more than {1,2,3,4,5...| x is even}. Or is it because the two-bucket arrangement is not the natural choice of odds vs evens? I guess if I received two fish a day and put them on different buckets things could change?
Oh I guess I notice that if I receive two fish a day that seems more than if I receive 2 fish every other day. That seems to map to 2*ω > 2*(ω/2) = ω.
Every other day, the number is the same. On the days in between, there’s a difference. Even taking that notion of ‘dominating’ into account, the difference is finite, and small: 1. If you had a choice between:
a) getting 1⁄3 a fish on day one, then 1⁄9 a fish on day two and 1/27th a fish on day three...(1/3^n on day n)
and
b) getting 1 fish on day one, then 1⁄2 a fish on day two, and 1⁄4 a fish on day three...(1/2^(n-1) on day n)
Then b is better than a...by more than the difference between 0+{1,2,3,4,5...| x is even} and {1,2,3,4,5...| x is even}. But it also seems like a bigger deal in those cases, a and b, because getting what comes out to a finite amount of fish over an infinite period of time isn’t as good as getting an infinite amount.
And it feels odd to say that there are more even numbers than odd ones!
Flip a coin. If it comes up heads put the first fish in the red bucket, and the second fish in the green bucket.
If it comes up tails do the reverse. (With the integers as a whole, where do you start counting from? The answer to that, determines which measure “seems” greater, on odd time/count steps.)
I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
You can color the positive integers, 1, 2, 3, 4,… using this pattern: red, green, red, green,...
You are obviously partitioning them into two collections (sets, because there’s not repetition).
Are there more odd numbers than even? Well if you count how many odd numbers (which are positive integers) are less than or equal to n, and call this function o, o(n) > e(n) for odd n. But for even n, they’re equal. So in the limit as n increases without bound: f(n) = o(n)/(o(n)+e(n) = 1⁄2.* But if you take steps of 2, starting at an even part, like 0, then the two sets are always of equal size. This seems to formalize the intuition well. Using this stream approach, one can articulate that: a randomly chosen integer is as likely to be even as it is to be odd**, and that this doesn’t change over time (as one iterates further through the integers), but the as you iterate out the integers, the frequency with which you see prime numbers goes down.
**This can be generalized further: a randomly chosen integer, for a given radix r, will have its final digit d, drawn from the a distribution indistinguishable from a uniform distribution over [0, r-1].
(Every instance of the word random, I made above is based on a uniform distribution across the integers. If you have a fair coin, and you flip it, until it comes up heads, then the probability that the number of times you flipped it is an odd number is not 1⁄2. The distribution there is: 1⁄2 for 0, 1⁄4 for 1, etc. The probability a red number (even) is chosen is twice that of a green number (odd) being chosen. Using ‘the sum of probabilities is 1’, it can be deduced that the probability the number is odd is 1⁄3, and the probability it is even is 2⁄3.)
*
o(n) + e(n) = n
for even n, o(n) = e(n). f(n) = k/2k = 1⁄2. (k is n/2, and an integer.)
for odd n, o(n) = e(n) + 1. f(n) = (k+1)/(2k) (n is 2k+1, and k is an integer) The limit as n, and thus k goes to infinity for this case is:
1⁄2 + 1/n = 1⁄2. (The limit from above. The other gets infinitely small.)
Every instance of the word random, I made above is based on a uniform distribution across the integers
Strictly speaking, there is no such thing as uniform distribution across the integers, or any other set that can be matched up 1:1 with integers. There are things that aren’t probability distributions that do have some similar properties, such as “asymptotic density”, but they fail to have the properties we require of probabilities. The “stream” idea seems to be another way to describe asymptotic density.
For example we might like to say that there are as many natural numbers of even length in decimal notation as those of odd length, but the stream idea doesn’t support that. The fraction of even-length numbers oscillates between 11% and 89% infinitely often, and there are other properties that behave even more weirdly.
One paradox that illustrates why there is no such thing as a uniform distribution on natural numbers:
Suppose we both pick natural numbers uniformly at random, independently of one another. You reveal your number. What is the probability that my number is greater than yours? No matter what number you chose, there are infinitely many larger numbers and only finitely many numbers that are not larger. Since my number was chosen uniformly at random, my number is larger with probability 1.
1 number of length 0, 9 numbers of length 1 (and maybe 0), 90 numbers of length 2, 900 numbers of length 3, 9000 numbers of length 4
9*10^(n-1) numbers of length n. For each n the amount of numbers of length just before that is 10 times less and the amount of numbers the next length is 10 times more. If you take a rolling fraction of n odd to all numbers seen it starts to go down when even numbered length is reached and starts to go up when an odd length number is reached.
(Ignoring that most people don’t think of 0 as of being length 0.)
Jump by two orders of magnitude every time and it stays stable:
Starting with nothing:
1 of even length, 9 of odd length.
90 of even length, 900 of odd length.
10% versus 90%.
Start after an even jump:
91 of even length, 9 of odd length.
9,091 of even length, 909 of odd length.
(Starts at 91% even, but drops after a double jump. I don’t know what the limit on this is.)
By comparison, resolving proportion of even numbers versus odd numbers, is much easier, because it’s a simple pattern which oscillates at the same rate, instead of changing.
9*10^(n-1) numbers of length n.
(in base 10)
Well, if a different color is used every time, then the coloring aspect is solved. If you ask about addition though, then things get weird.
Right, that was a math typo. It really oscillates between 9% and 91%.
For example, 909090 of the numbers below a million have even length, i.e. 91%. As you increase the bound toward ten million, this fraction decreases until it hits a minimum of 9%, and then starts increasing again until you reach a hundred million, and so on.
Well, the old solution to what is the limit of: +1, −1, +1, −1, etc. was: (index starts at one, pattern is (-1)^(n+1))
Consider the cases:
a) +1, odd index
b) −1, even index
Average them.
0.
If that was applied directly, it’d be: (9+91)/2% = 50%.
You could argue that it should be broken down differently, because there’s different proportions here though.
You could also declare the answer undefined, and say infinity is about growth, it doesn’t have a value, for x % 2 (or odd or even number as the case may be), and averages are ridiculous. (And once you have a breakdown of cases, and probability what more is there?)
That is one of the puzzle in that 0+0+0+0+0… converges and has a value but +1-1+1-1+1-1… which seems to be like (1-1)+(1-1)+(1-1)+(1-1)… diverges (and the series with and without the paranthesis are not equivalent)
The strram idea gives it a bit more wiggleroom. Getting 1,0,1,0,1.. fish seems equivalent to getting 1⁄2 fish a day but 1,1,1,1,1.. seems twice the fish of 1,0,1,0,1,0,1,0… So which with the other methods are “can’t say anthing” there is maybe hope to capture more cases with this kind of approach.
Too bad its not super formal and I can’t even pinpoint where the painpoints for formalization would be.
That paradox is good in that it cuts to the matter very cleanly.
To my mind it “numbers larger than my number” and “uniform integer” don’t need to be same. There are n smaller numbers and ω-n bigger numbers. (ω-n)/ω is going to be near 1 (infinidesimally so) but not quite up to 1. Maybe crucially (ω-2n)/ω is smaller than (ω-n)/ω ie if I hit a high number my chances are better. I get that standard approach somehow gets into the way of this and I would like to know which axiom I have a bone to pick with.
There is a (from my perspective a problem) that events of 0 probablity can happen and events with probability 1 can fail to happen. The associated verbal language is “almost surely” and “almost never”. Showing me that a thing can almost surely happen doesn’t guarantee it. To my mind this is because some zeroes have rounding to the nearest real and some don’t.
If you just graph fish vs. time, then the one-time gift is a constant function, the steady income is linear, and the “growing promise” stream-of-streams is a quadratic. The fact that a steady income will eventually surpass any one-time gift is because any positive-slope linear function will eventually exceed any constant-value function. Likewise, any 2nd-order polynomial with a positive x^2 term will eventually exceed any linear function. You could keep going with higher order polynomials if you want.
A similar analogy would be a race where even a large head start will eventually be surpassed by a slightly faster car.
I guess I know that kind of perspective exists but I don’t want to refer to the “up to a time” thing but the total.
For example taking the even and odd integers, that would seem like taking a stream of fish and putting them into alternate buckets every other day. I would like to be able to say something like “even and odd integers are of equal amount and they are half of that of all integers”. But I am unsure whether that is accurate and where the foundations would crumble.
if I do it in a very naive way in graph vs time then I could conclude that the bucket that I start with has a stricly dominating amount of fish (because the graph stay above or matches the other graph). And it feels odd to say that there are more even numbers than odd ones! Althought I guess it wouldn’t be odd that 0+{1,2,3,4,5...| x is even} is more than {1,2,3,4,5...| x is even}. Or is it because the two-bucket arrangement is not the natural choice of odds vs evens? I guess if I received two fish a day and put them on different buckets things could change?
Oh I guess I notice that if I receive two fish a day that seems more than if I receive 2 fish every other day. That seems to map to 2*ω > 2*(ω/2) = ω.
Every other day, the number is the same. On the days in between, there’s a difference. Even taking that notion of ‘dominating’ into account, the difference is finite, and small: 1. If you had a choice between:
a) getting 1⁄3 a fish on day one, then 1⁄9 a fish on day two and 1/27th a fish on day three...(1/3^n on day n)
and
b) getting 1 fish on day one, then 1⁄2 a fish on day two, and 1⁄4 a fish on day three...(1/2^(n-1) on day n)
Then b is better than a...by more than the difference between 0+{1,2,3,4,5...| x is even} and {1,2,3,4,5...| x is even}. But it also seems like a bigger deal in those cases, a and b, because getting what comes out to a finite amount of fish over an infinite period of time isn’t as good as getting an infinite amount.
Flip a coin. If it comes up heads put the first fish in the red bucket, and the second fish in the green bucket.
If it comes up tails do the reverse. (With the integers as a whole, where do you start counting from? The answer to that, determines which measure “seems” greater, on odd time/count steps.)
You can color the positive integers, 1, 2, 3, 4,… using this pattern: red, green, red, green,...
You are obviously partitioning them into two collections (sets, because there’s not repetition).
Are there more odd numbers than even? Well if you count how many odd numbers (which are positive integers) are less than or equal to n, and call this function o, o(n) > e(n) for odd n. But for even n, they’re equal. So in the limit as n increases without bound: f(n) = o(n)/(o(n)+e(n) = 1⁄2.* But if you take steps of 2, starting at an even part, like 0, then the two sets are always of equal size. This seems to formalize the intuition well. Using this stream approach, one can articulate that: a randomly chosen integer is as likely to be even as it is to be odd**, and that this doesn’t change over time (as one iterates further through the integers), but the as you iterate out the integers, the frequency with which you see prime numbers goes down.
**This can be generalized further: a randomly chosen integer, for a given radix r, will have its final digit d, drawn from the a distribution indistinguishable from a uniform distribution over [0, r-1].
(Every instance of the word random, I made above is based on a uniform distribution across the integers. If you have a fair coin, and you flip it, until it comes up heads, then the probability that the number of times you flipped it is an odd number is not 1⁄2. The distribution there is: 1⁄2 for 0, 1⁄4 for 1, etc. The probability a red number (even) is chosen is twice that of a green number (odd) being chosen. Using ‘the sum of probabilities is 1’, it can be deduced that the probability the number is odd is 1⁄3, and the probability it is even is 2⁄3.)
*
o(n) + e(n) = n
for even n, o(n) = e(n). f(n) = k/2k = 1⁄2. (k is n/2, and an integer.)
for odd n, o(n) = e(n) + 1. f(n) = (k+1)/(2k) (n is 2k+1, and k is an integer) The limit as n, and thus k goes to infinity for this case is:
1⁄2 + 1/n = 1⁄2. (The limit from above. The other gets infinitely small.)
Strictly speaking, there is no such thing as uniform distribution across the integers, or any other set that can be matched up 1:1 with integers. There are things that aren’t probability distributions that do have some similar properties, such as “asymptotic density”, but they fail to have the properties we require of probabilities. The “stream” idea seems to be another way to describe asymptotic density.
For example we might like to say that there are as many natural numbers of even length in decimal notation as those of odd length, but the stream idea doesn’t support that. The fraction of even-length numbers oscillates between 11% and 89% infinitely often, and there are other properties that behave even more weirdly.
One paradox that illustrates why there is no such thing as a uniform distribution on natural numbers:
Suppose we both pick natural numbers uniformly at random, independently of one another. You reveal your number. What is the probability that my number is greater than yours? No matter what number you chose, there are infinitely many larger numbers and only finitely many numbers that are not larger. Since my number was chosen uniformly at random, my number is larger with probability 1.
What?
1 number of length 0, 9 numbers of length 1
(and maybe 0), 90 numbers of length 2, 900 numbers of length 3, 9000 numbers of length 49*10^(n-1) numbers of length n. For each n the amount of numbers of length just before that is 10 times less and the amount of numbers the next length is 10 times more. If you take a rolling fraction of n odd to all numbers seen it starts to go down when even numbered length is reached and starts to go up when an odd length number is reached.
(Ignoring that most people don’t think of 0 as of being length 0.)
Jump by two orders of magnitude every time and it stays stable:
Starting with nothing:
1 of even length, 9 of odd length.
90 of even length, 900 of odd length.
10% versus 90%.
Start after an even jump:
91 of even length, 9 of odd length.
9,091 of even length, 909 of odd length.
(Starts at 91% even, but drops after a double jump. I don’t know what the limit on this is.)
By comparison, resolving proportion of even numbers versus odd numbers, is much easier, because it’s a simple pattern which oscillates at the same rate, instead of changing.
(in base 10)
Well, if a different color is used every time, then the coloring aspect is solved. If you ask about addition though, then things get weird.
Right, that was a math typo. It really oscillates between 9% and 91%.
For example, 909090 of the numbers below a million have even length, i.e. 91%. As you increase the bound toward ten million, this fraction decreases until it hits a minimum of 9%, and then starts increasing again until you reach a hundred million, and so on.
Well, the old solution to what is the limit of: +1, −1, +1, −1, etc. was: (index starts at one, pattern is (-1)^(n+1))
Consider the cases:
a) +1, odd index
b) −1, even index
Average them.
0.
If that was applied directly, it’d be: (9+91)/2% = 50%.
You could argue that it should be broken down differently, because there’s different proportions here though.
You could also declare the answer undefined, and say infinity is about growth, it doesn’t have a value, for x % 2 (or odd or even number as the case may be), and averages are ridiculous. (And once you have a breakdown of cases, and probability what more is there?)
That is one of the puzzle in that 0+0+0+0+0… converges and has a value but +1-1+1-1+1-1… which seems to be like (1-1)+(1-1)+(1-1)+(1-1)… diverges (and the series with and without the paranthesis are not equivalent)
The strram idea gives it a bit more wiggleroom. Getting 1,0,1,0,1.. fish seems equivalent to getting 1⁄2 fish a day but 1,1,1,1,1.. seems twice the fish of 1,0,1,0,1,0,1,0… So which with the other methods are “can’t say anthing” there is maybe hope to capture more cases with this kind of approach.
Too bad its not super formal and I can’t even pinpoint where the painpoints for formalization would be.
That paradox is good in that it cuts to the matter very cleanly.
To my mind it “numbers larger than my number” and “uniform integer” don’t need to be same. There are n smaller numbers and ω-n bigger numbers. (ω-n)/ω is going to be near 1 (infinidesimally so) but not quite up to 1. Maybe crucially (ω-2n)/ω is smaller than (ω-n)/ω ie if I hit a high number my chances are better. I get that standard approach somehow gets into the way of this and I would like to know which axiom I have a bone to pick with.
There is a (from my perspective a problem) that events of 0 probablity can happen and events with probability 1 can fail to happen. The associated verbal language is “almost surely” and “almost never”. Showing me that a thing can almost surely happen doesn’t guarantee it. To my mind this is because some zeroes have rounding to the nearest real and some don’t.
There are two bones you might pick. One is that probabilities are real numbers. The other is that probabilities are countably additive.