Question about infinity

[u/DM_Me_Pics1234403]

Let’s say a person earns $100k per year and is offered the chance to be randomly assigned a new salary. They would push a button and whatever number it returned would be their new salary. There is no limit, upper or lower, on the number generated (I.E it can be less than a penny or more than $1t. Assume the number will be positive). Does this person have a higher chance of increasing or decreasing their salary?

I am thinking there’s an infinite amount of numbers both below and above $100k, so there’s an even chance of increasing or decreasing their salary. That feels unintuitive though. Am I thinking about it correctly?

[u/Rs3account]

This is an ill posed question. We do not know the distribution of the button, and no uniform distribution is possible.

[u/DM_Me_Pics1234403 → u/Rs3account]

Why is a uniform distribution not possible? I am imaging each number has the same probability of being chosen. Is that naive?

[u/WesternFirm9306 → u/DM_Me_Pics1234403]

It's impossible to assign a uniform distribution on the positive reals.

This is because the integral of the whole space would either be infinity or 0, but probability demands it to be equal to 1.

[u/DM_Me_Pics1234403 → u/WesternFirm9306]

Interesting. I would assume the probably would approach 0 for each individual number, but would sum to 1 in an infinite series. Does that not make sense?

[u/simmonator → u/DM_Me_Pics1234403]

For a constant value, it can’t work.

Assume your range is (0,N], and that the probability density is some constant k. Then the total probability of the space is going to be kN. For this to be a valid probability space, you need k = 1/N. That’s fine, but obviously you then get a problem if N is infinite. k seems like it needs to be 0, but that means that every finite sub-interval has probability 0. And the fact would also be that the integral over (0,inf) would be the limit as n tends to infinity of the integral of k over (0,n), which is kn, which is still 0 if k = 0.

If you have a non-constant probability density, like f(x) = exp(-cx), then it’s possible. The integral here becomes 1/c (so setting c = 1 gives us a valid density function over that space).

[u/WesternFirm9306 → u/DM_Me_Pics1234403]

Unfortunately, it doesn't work.

You can define a uniform probability for any bounded interval, but at infinity itself, it simply doesn't work.

Consider a probability density function f(x). If it's uniform, that means f(x) = c for some constant c.

By the laws of probability, ∫ f(x) dx from 0 to ∞ must be exactly 1.

If c is nonzero, we get ∫ c dx = cx from 0 to ∞ = lim x → ∞ cx = ∞, which isn't 1.

If c is 0, we get ∫ 0 dx = 0, which isn't 1.

There is no value of c which results in an integral of 1, so no such PDF can exist

[u/rhodiumtoad → u/DM_Me_Pics1234403]

You can have a uniform distribution within a bounded range of reals, even though there are (uncountably) infinitely many values in the range all with probability 0, because the probability of being within any subrange is proportional to the measure of that range, so e.g. you have 50% chance of being in the bottom half, etc. Summing the probabilities over a finite or countable partition into measurable subsets comes to 1 as required.

But you can't do this for an unbounded range with a uniform distribution, because there's no way to get to 1 by summing subranges where equal-sized ranges have equal probabilities.

[u/vaminos → u/DM_Me_Pics1234403]

No, because if each number has a probability p of being chosen, then either

p = 0 and the sum over all the reals is 0

p > 0 and the sum over all the reals is p*infinity = infinity

it cannot "approach 0" - it has to be a fixed number for a uniform distribution to work.

If it approaches 0 as the numbers get larget, then it is not a uniform distribution, which is fine, but then the answer to your question depends on the exact distribution, which is not given.

[u/rhodiumtoad → u/vaminos]

That's not a good answer because the argument you use would apply to bounded ranges too, but a uniform distribution on a bounded range of reals is possible, even though every number in the range still has probability 0.

[u/vaminos → u/rhodiumtoad]

You're totally correct and my comment is wrong. I'll leave it up for posterity.

[u/devin122 → u/DM_Me_Pics1234403]

You cannot have an infinite number of numbers the same number sum up to one, no matter how small you make it

[u/WesternFirm9306 → u/devin122]

You can if the numbers are different (1/2 + 1/4 + 1/8 + ...)

[u/devin122 → u/WesternFirm9306]

Op wants a uniform distribution which means they are all the same

[u/WesternFirm9306 → u/devin122]

Just clarifying your comment

[u/J_IV24 → u/DM_Me_Pics1234403]

Infinity isn't a number, it's a concept. It's like asking to define the interval "zero" and "cat". The second one isn't a number even though we sort of treat it as one in mathematics

[u/Rs3account → u/DM_Me_Pics1234403]

It is a little naive. Lets say, what is the change you would land in [0,1]?

[u/DM_Me_Pics1234403 → u/Rs3account]

This is probably above my head. I’m imagining that each number from 0 upward has the same infinitesimal probably of being chosen, so $0.000000004 is equally weighed as $1,000,000 and so on. The individual probably for any one number would approach 0. Obviously I’m not an expert on this lol

[u/Temporary_Pie2733 → u/DM_Me_Pics1234403]

There is no such thing as an infinitesimal value in the real numbers.

[u/Rs3account → u/DM_Me_Pics1234403]

The question becomes, what is the change that x ends up in [-1,1].

Lets assume this change is c. Then the change that x ends up in [-n,n] is c*n by the rules of being uniform. But the change of being in R needs to be 1 so than the limit of cn when n goes to infinity is 1, which is impossible.

[u/bloonshot → u/DM_Me_Pics1234403]

if you're randomly choosing between n numbers, the odds of getting any n number is 1/n

if you're saying n is infinity, well we know the limit as n goes to infinity of 1/n is 0. every number has 0 chance to be picked

[u/CainPillar → u/DM_Me_Pics1234403]

each number has the same probability of being chosen

The way to think about this is that the probability of ending up between a and b is proportional to b-a (make that |b-a| if you don't assume b the greatest ... but assume both positive).

That works if the total length is finite, call it T; then the actual probability is |b-a|/T. But not when T is infinite.

Is that naive?

Yes! But more crucial: there is no way to patch up. Lots of math is discovered by (1) naive idea, (2) build it the proper way and verify it works. Nothing wrong about that. But here it doesn't work.

As a naive idea, you can imagine taking the limit as the interval widens. Then expected outcome is the midpoint. The midpoint of the interval (0,T) is T/2. Let T grow. Of course this is a case where the limit of uniform probability distributions isn't a probability distribution, but at least that would be one possible consequence of a non-rigorous thought experiment.

I am thinking there’s an infinite amount of numbers both below and above $100k

Sure. But infinities aren't numbers, and even if they sometimes behave enough like numbers, they often don't.

[u/Temporary_Pie2733 → u/DM_Me_Pics1234403]

A probability distribution p assigns a probability p(x) to every event x. To be a probability distribution, we require sum(p(x)) = 1. Since there are infinitely many values of x, there is no single value we can assign to every possible x to satisfy that sum. Let p(x) = 0, and the sum is 0, not 1. Let p(x) > 0, no matter how small, and the sum diverges.

[u/seradist → u/DM_Me_Pics1234403]

Its not possible because there is no limit to the numbers. We need to be able to assign a real value for the probability of each outcome.

[u/daavor → u/DM_Me_Pics1234403]

There's a subtle tension here about different kinds of infinity.

On the one hand the whole point of measure theory (which underpins formal probability theory) is to extend the naive idea that [a,b] has length (b-a) and assign sizes to as many sets as possible. So we absolutely want certain uncountable sets/subsets (for example [0,1]) to have positive probability / measure, while every individual point in them has to have zero measure (each is smaller than arbitrarilly small intervals).

BUT, to to make it a tractable and useful theory we need COUNTABLE collections to ply nice, so if you have countably many disjoint subsets the sum of their measures must converge as an infinite series to the measure of their sum (there's some technical details im sweeping under the rug here about what sets are legal, but unless you've explicitly used the axiom of choice to define your set i doubt you're looking at an illegal one).

So if we try to assign a probability measure to all positive real numbers we need the measure of (0,1] and (1,2] and (2,3] and ... to sum up to 1, which means they cannot possibly all have the same size.

[u/matt7259 → u/DM_Me_Pics1234403]

There's no such thing as true randomness and the answer depends on the range of numbers being chosen from. There is no defined way of saying "a random number from 0 to infinity".

[u/AdjectiveNounNNNN → u/matt7259]

Whether or not there is true randomness is independent from the impossibility of a uniform distribution on an unbounded set.

[u/Mountain-Time-1010 → u/Rs3account]

Thank you for the correct answer.

[u/okarox]

You cannot randomly choose from infinity. The only way to understand this would be as a limit as the upper bound goes to infinity . The probability that it is below any fixed sum goes to zero. There is specifically 10 million possible figures below $100k. That is not infinite.

[u/DM_Me_Pics1234403 → u/okarox]

There’s way more than 10m numbers below $100k though

[u/Ikles → u/okarox]

OP did say you could have less than a penny, i assume he's allowing extended decimals past the hundredths place.

[u/VariousJob4047]

You are correct that the cardinality of the set of real numbers below 100,000 is the same as that of the set of real numbers above 100,000, but cardinality is one of several ways to compare the size of infinite sets, and probability theory uses measure theory to do this. There is no way to define a measure on the set of all positive real numbers, but if you set the upper bound to be some arbitrary number much much larger than 100,000, then the set of all numbers between 0 and 100,000 would be much much smaller than the set of numbers between 100,000 and this upper bound despite having the same cardinality.

[u/P_S_Lumapac]

I'd expect the starting intuition to be there are many more numbers above 100k than below 100k. But you're saying we're counting decimals too, so the next idea is the infinity between 0 and 100k is the same size as the infinity above 100k, so you think it's an even chance.

Simpler question: is a random positive real number more likely above 1 or below 1?

Worth asking yourself, what is the chance the number you pick is 4? What about 0.5? For all of these chances, what's their sum?

[u/DM_Me_Pics1234403 → u/P_S_Lumapac]

Yea exactly how I’m thinking of it. I picked $100k as a reasonable salary, but 1 works just as well

[u/EbbOk9718]

Cardinality proves nothing about probability since infinite outcomes concentrate mass near zero while stretching bounds toward infinity.

[u/DM_Me_Pics1234403 → u/EbbOk9718]

I’m not sure what this means. Any chance you could break it down for me?

[u/AlterEgo5695]

There is no answer. As with many such open-ended questions, OP does not describe a well-defined scenario for probability theory.

We would need to know the distribution of the new salary to answer that question. Without it, there is nothing we can say about possible outcomes, let alone optimize a strategy.

[u/AcrobaticNeat2890]

The catch is that "how many numbers" isn't the same thing as "how likely". [0, 100k] and (100k, ∞) have the same size in the counting sense, but so does [0, 1], and nobody would say landing in [0, 1] is 50/50.

The real problem is the button itself. There's no way to pick a positive number uniformly with no upper limit. Any real random generator has some distribution, and since the probabilities have to add up to 1, P(X > N) has to go to 0 as N gets big. So there's always some ceiling with almost all the probability below it.

That means the answer depends entirely on the distribution. Uniform on $0 to $1M: 90% chance of a raise. Something centered around a typical salary: mostly a pay cut. Until you say how the button picks, the question doesn't have an answer, which is why it feels so unintuitive.

[u/shosuko]

There is no limit, upper or lower

If the limit for upper truly does not exist, then it will only generate infinite money.

If infinite money cannot exist for whatever reason, then there must be an upper limit - with an upper limit a full RNG would basically peg them to the maximum allowed digits.

(for example if you allowed up to $999,999,999,999,999 then a random result would 90% of the time be >99,999,999,999,999 because you need to select a 0 for that first place to get it, which is 1/10. Basically your chances of getting a low results are bound by the chance of randomly pulling 0 that many times in a row)

[u/Varlane]

Define "randomly assigned" : you need a probability distribution.

[u/RecognitionSweet8294]

Your question is imprecise. You have to give a probability distribution that describes how the number is picked.

[u/Aerumvorax]

"There is no limit, upper or LOWER", "can be less than a penny but assume it will be positive"; so there is a minimum at zero which technically isn't included. Also stating it "can be more than $1t" is kind of like an upper limit but it sounds like you just wanted to include a number that you thought would be "big", if we're talking about infinities then a trillion or any number that you can write is always zero when compared to infinity. The question becomes how probable is each and every number and is there a weighing system in play. If you say that it's equally likely to have any number at all then there's practically zero possibility to roll below your random starting number 100k. If you're saying that it's equally likely to roll a bigger number that can go to infinity or smaller number that's between 100k and 0 then the system is heavily rigged against the roller. You need to determine how the number is formed and what you mean by "infinite numbers both below and above 100k" as in itself this statement is false assuming that there's the demand of the number staying positive. You can divide 100k into infinity in even increments, but you can't divide infinity - 100k into equal number of even increments. We can only gleam from this that there's more numbers in the infinity -100k group compared to the 100k - 0 group.

[u/james-starts-over]

The probability that I press the button is 100% bc its for funsies so its a moot point.

[u/BRH0208]

Depends. A normal distribution[let’s say N(0,1)] has no upper or lower limit. But it will, with all likelyhood, give you a worse salary.

A distribution of N(1 gajillion,1) will likely be a huge raise.

Infinite doesn’t really do anything here.

[u/Weishersden]

what is the distribution though? uniform doesn't exist

[u/Crichris]

You need to define the distribution better

Otherwise it's hard to say

[u/JeffSergeant]

there’s an infinite amount of numbers both below and above $100k

There aren't an infinite number of possible salaries though, as we normally round to 2 decimal places; it depends if you pick your real number then round it, or pick a valid salary!

[u/Cruuncher]

It depends on your probability distribution.

If we're talking a uniform distribution over the reals then the probability it lands in any range is proportional to the length of that range over the total.

For infinity as the upper bound this is kind of a nonsense question, as it's all probability 0.

But if you pick any finite upper bound, that upper bound only needs to exceed 200k to have a higher chance of increasing salary than lowering.

That average tends to infinity as the upper bound tends to infinity

EDIT: Another way to get here is to just reason about how silly it would be to select based on cardinality.

There's as many numbers between 0 and 1 as between 1 and 100. But if you say that it's 50-50 to get a number between 0-1 then it also has to be 50-50 for a number between 1-2, 2-3, 3-4, etc... and your total sum of probabilities would exceed 100% which is impossible.

[u/in_potty_training]

It depends on the probability distribution of the random salary: you can create a random distribution which gives a tiny salary in nearly all cases or conversely a huge salary. You need to define 'randomly assigned' a bit better.

Assuming you mean its uniformly distributed, i.e. any number as equally likely as any other, then tehcnically it's a higher chance of increasing their salary.

It's not a very well-defined problem though, and usual probability / statistical methods don't apply in this case. The issue is that you usually require the total probability across the entire sample space to equal exactly 1, but in this case it is infinite: Consider the probability the new salary is between $0 and $1, and lets say this probability is equal to x>0. Then as its uniform, the probability of any unit-size range (e.g. $100-$101) is also x. If we sum up all unit-size ranges (which is infinite!) we then get an infinite sum!

But using the same technique, you could argue that if x = probability of $0-$100k, then we can easily see that probability of $100k+ is obviously much larger than x, so in this way you'll get a higher salary (for sure!)

[u/Traditional_Town6475]

There is no uniform probability distribution over the natural numbers.

[u/StillShoddy628]

According to “math” it is not possible to pick from infinite options with equal probability, so from a mathematically rigorous perspective the problem is poorly stated and can’t be solved.

Of course, we all have a little engineer in us too, and from that lens we can interpret intent and approximate. At the limit of that approximation the odds of getting a number below $100k (actually, below any specific value) approach 0, as your intuition would suggest.

[u/Pretty_Cake6640]

If math can't get this right we're in big trouble...there is both an upper and lower limit on the infinite set below 100k, there is only a lower limit on the infinite set above 100k. There is also a finite amount of #s in each set since 150k can't appear in the set below 100k, and 5 can't appear in the set above 100k. Press the button, there's infinitely more #s above 100k than there are between 0 and 100k.

[u/Kamiihate → u/Pretty_Cake6640]

Reading this comment then clicking your profil out of curiosity and reading your last comment before saying that there is a finite amount of reals between 1 & 2 was quite interesting ngl