Chance of Identical Numbers in Infinite Series

[Serene_Chaos]

Ok I am not sure if this is where this thread should go, but anyway.

My question is this:
if we have an infinite series of numbers (or apples, or whatever) and we choose one at random, and then choose another at random, is there a chance that the two will be the same? I mean, is a 1 in infinity chance a chance at all?

 

[VikingF]

If the numbers are chosen from a finite definition area (let's say: S=[0, 10000]), then of course that is possible. Look at it as this: You ask person A to think of a random number between 0 and 10000 and the same to person B. Then you ask them both to write their number on a paper and you look at the results. The possibillity of having the same number is possible, but not very likely... (It's 10000^-2)

 

[Hurkyl]

It depends on how you're picking at random.

Usually, when people say "at random" without giving the distribution, they mean the uniform distribution... but discrete, infinite sets do not have a uniform distribution, so the question is ill-posed.

 

[Serene_Chaos]

well how about my second question, "is a 1 in infinity chance a chance at all?". i supose another way of asking this is "is 10^-∞ closer to 0 or 1?" like, is it a number at all?

 

[Hurkyl]

Well, -∞ isn't a (real) number to begin with.

In the extended real numbers, 10^(-∞) = 0. (But the +∞ and -∞ of the extended real numbers have nothing to do with the concept of the size of a set. The size of an infinite set lies in a different number system entirely, the cardinal numbers)

 

[Intuitive]

Simple VBscript random chance life generator.
I wrote this code so you can see your chances.
You can save it as a .vbs file and double click it to run it.

Evolve = RND*10000
Random_Luck = RND*10000 ' Random uncertainty
Random_Who = RND*10000
Random_What = RND*10000
Random_When = RND*10000
Random_Where = RND*10000
Random_Why = RND*10000
Random_How = RND*10000
If Evolve > 0 Then
If Evolve = 10 Then 
msgbox"Evolution has Spurted"
Else
If Random_Luck = Evolve Then 
msgbox"Luck and evolution itself has changed evolution"
Else
If Random_Who = Evolve Then 
msgbox"Who and evolution has changed evolution"
Else
If Random_What = Evolve Then 
msgbox"What and evolution has changed evolution"
Else
If Random_When = Evolve Then 
msgbox"When and evolution has changed evolution"
Else
If Random_Where = Evolve Then 
msgbox"Where and evolution has changed evolution"
Else
If Random_Why = Evolve Then 
msgbox"Why and evolution has changed evolution"
Else
If Random_How = Evolve Then 
msgbox"How and evolution has changed evolution"
Else
If Random_Luck = Random_Who Then 
msgbox"Luck and Who has changed evolution"
Else
If Random_Luck = Random_What Then 
msgbox"Luck and What has changed evolution"
Else
If Random_Luck = Random_When Then 
msgbox"Luck and When has changed evolution"
Else
If Random_Luck = Random_Where Then 
msgbox"Luck and Where has changed evolution"
Else
If Random_Luck = Random_Why Then 
msgbox"Luck and Why has changed evolution"
Else
If Random_Luck = Random_How Then 
msgbox"Luck and How has changed evolution"
Else
msgbox"No change in species"
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If
End If

Now spin the wheel for four billion years every second, remember, once a hit is made it might replicate on its own.

You can also load the .vbs file on your startup panel so it auto starts itself.

Every reboot will be a white knuckle evolution, Now multiply this by the number of Users.

There are far more probabilities to increase the luck factor.

Frightning isn't it!

 

[RunDMC]

Infinity is a weird number. There is an infinite progression of odd numbers and an infinite number of prime numbers. Does this mean that there is an equal number of odd numbers and prime numbers in the infinite spectrum? I think my brain just exploded.

 

[Intuitive]

Infinity is a weird number. There is an infinite progression of odd numbers and an infinite number of prime numbers. Does this mean that there is an equal number of odd numbers and prime numbers in the infinite spectrum? I think my brain just exploded.

I know the word random seems intimidating, but if (luck) holds out then it can be hit the first try, This is a probability.

It's like playing a slot machine, you have a 1 and 10,000 chance of hitting a Jackpot, But sometimes a person hits it the very first try, This can happen in nature as well.:rofl:

I know this is hard to understand, but, People should not confuse Luck with Randomness.
Sometimes there is a great deal of luck, Even in Nature, Nature has been pulling its own slot machine handle for well over 4 billion years on Earth.

 

[hypermorphism]

Infinity is a weird number. There is an infinite progression of odd numbers and an infinite number of prime numbers. Does this mean that there is an equal number of odd numbers and prime numbers in the infinite spectrum? I think my brain just exploded.

Make a bijection. There is a trivial bijection between odd numbers and the natural numbers, and there is a trivial bijection between the naturals and the primes, so the two sets have the same cardinality.

 

[Joffe]

Yes there is zero chance:

$$\lim_{x\rightarrow\infty} \frac{1}{x} = 0$$

VikingF, you made a very large error in your statement, it is 10000^-1

 

[benorin]

Is there a probability theorist in the house?

What is meant by an infinite series of numbers? Do you mean a countable set of numbers like the integers or the rationals? So long as you ask the (ideal) persons to select at random (ideal) from some finite subset of such numbers (e.g. "Pick an integer between 1 and 11."), then the probability of said persons selecting the same number (or the probability of one of them picking a particular number) is a positive number strictly between 0 and 1.

If, however, you admit an infinite set of numbers (countable or uncountable) from which to choose, such as "Select a real number between 1 and $\pi$," (or "Name an integer, any integer") the probability of the choosen number being any particular number is zero. This is not comforting, since we are forced to conclude that the probability of that number being between 1 and $\pi$ (or being an integer) is 1. In defence of this, I offer that ideal is a term coined by physicsists which variously denotes: massless, frictionless, inelastic, prefectly rigid, ect. and hence it follows that ideal persons understand and follow directions for the sake of experiment. So, how come an infinite sum of zero probability events is 1? Welcome to measure theory.

I now yield my rant that someone better versed in probability theory might take it up.

 

[VikingF]

Yes there is zero chance:

$$\lim_{x\rightarrow\infty} \frac{1}{x} = 0$$

VikingF, you made a very large error in your statement, it is 10000^-1

Not if two independent individuals pick a random number each between 0 and 10000.

In that case, there wil be a probability of picking the same number equal to: 1/10000 * 1/10000 = 1/10000^2 = 10000^-2.

 

[Joffe]

Not if two independent individuals pick a random number each between 0 and 10000.
In that case, there wil be a probability of picking the same number equal to: 1/10000 * 1/10000 = 1/10000^2 = 10000^-2.

No, the first person's guess is irrelevant. No matter what he chooses the second person has a 1/10000 chance of matching it. What you described is the chances that they both pick the number 1 but what you are forgetting is that they could have both gotten 2 or 3 or any other number so you must multiply by 10000.

 

[JasonRox]

Infinity is a weird number. There is an infinite progression of odd numbers and an infinite number of prime numbers. Does this mean that there is an equal number of odd numbers and prime numbers in the infinite spectrum? I think my brain just exploded.

Sounds like you are really interested in this stuff.

An introduction to Set Theory can answer this question for you, just follow hypermorphism. He told you exactly where to look.

 

[VikingF]

No, the first person's guess is irrelevant. No matter what he chooses the second person has a 1/10000 chance of matching it. What you described is the chances that they both pick the number 1 but what you are forgetting is that they could have both gotten 2 or 3 or any other number so you must multiply by 10000.

Of course... You are right. I was a little too quick on that one. :shy:

 

[HallsofIvy]

I mean, is a 1 in infinity chance a chance at all?

"1 in infinity" doesn't make sense by itself. As soon as you have an infinite number of possible outcomes, you must be more careful about your definitions. For example, it is impossible to define a probability distribution over the integers such that each integer is "equally" likely. On the other hand, you can define a probability distribution over "all real numbers between 0 and 1 (inclusive)" by P([a,b])= b-a. Essentially, the probability that a number chosen is in a given set is equal to the (Lebesque) measure of the set.
That means that the probability of choosing a specific number is 0 (the measure of any finite or countable set is 0) but obviously it is possible that any given number can be chosen. So the answer to your question is generally no- an event having probability 0 does not always mean that the event cannot happen.

 

[VikingF]

That means that the probability of choosing a specific number is 0 (the measure of any finite or countable set is 0) but obviously it is possible that any given number can be chosen. So the answer to your question is generally no- an event having probability 0 does not always mean that the event cannot happen.

If the set is finite, the probability of choosing that exact number is NOT 0, because 1/n, where n is a finite number will never lead to the answer 0, but the probability may be very close to 0.

On the other hand, if the set is infinite, then we will get the probability: $$P = \lim_{n\rightarrow\infty} \frac{1}{n} = 0$$, which means that p=0 is the limit the probability is closing into as n is increasing. I don't know if this is what you meant?

 

[Hurkyl]

HoI meant what he said.

With the typical probability distributions one might use on the real line, we really do have P(X = r) = 0 for each real number r.

 

[-Job-]

Suppose the person picks a number between 0 and infinity and then writes the digits one by one.
Let $$X_{i}$$ be the probability that the person picked a number x given that the number starts with the $$i$$ digits the person wrote so far. Notice that $$P\{X_{i} = x\}$$ is the same for all $$X_{i}$$ because the problem is the same. Yet with each digit the person writes we eliminate a whole range of numbers. This means that the set of possible outcomes is getting smaller but the probability of getting a given x stays the same. So this probability would have to be 0.

 

[VikingF]

HoI meant what he said.
With the typical probability distributions one might use on the real line, we really do have P(X = r) = 0 for each real number r.

But let's say we choose one out of 1000 numbers, then the probability of choosing one specific number is 1/1000. If we choose one out of 1,000,000 numbers, then the probability is 1/1,000,000. And as n is increasing, then P will get closer and closer to 0, but how can you say that P will ever reach the actual value 0?

Even if we have 10^1,000,000,000 numbers, the probability of choosing one specific is not any less than 1/10^1,000,000,000.

I can't see how any finite amount of numbers can lead to a probability equal to zero, if any number is possible...?

With infinity, on the other hand, the situation is somewhat different, but infinite is not a "number" that can be placed under the fraction line, without using limits.

 

[Hurkyl]

But let's say we choose one out of 1000 numbers, then the probability of choosing one specific number is 1/1000.

No it's not; I was going to choose a number by flipping a coin 999 times, counting the number of heads, then adding 1. The probability of getting, say, 1000 is 1/2^999. (And, of course, the probability of getting 1001 is 0)

But let's say we choose one out of 1000 numbers, then the probability of choosing one specific number is 1/1000. If we choose one out of 1,000,000 numbers, then the probability is 1/1,000,000. And as n is increasing, then P will get closer and closer to 0, but how can you say that P will ever reach the actual value 0?

I didn't say that. In fact, I (and HoI) have said absolutely nothing that resembles what you are saying here.

 

[VikingF]

No it's not; I was going to choose a number by flipping a coin 999 times, counting the number of heads, then adding 1. The probability of getting, say, 1000 is 1/2^999. (And, of course, the probability of getting 1001 is 0)

I was not talking about flipping coins, but about choosing a random number from a finite interval.


I questioned about this statement:

"That means that the probability of choosing a specific number is 0 (the measure of any finite or countable set is 0) but obviously it is possible that any given number can be chosen. So the answer to your question is generally no- an event having probability 0 does not always mean that the event cannot happen."

I didn't get how a probability 0 can happen at all, and still be a probability 0.

 

[Hurkyl]

I was not talking about flipping coins, but about choosing a random number from a finite interval.

I was talking about choosing a random number from a finite interval too. I just chose to do it in an unusual way.

The thing I'm trying to get you to realize is that you're making a hidden assumption -- when you say "choose a random number from a finite interval", you mean "choose a random number from a finite interval, according to the uniform probability distribution".

Of course, the general agreement is that if you don't say, then you mean the uniform probability distribution... but I'm making a big deal of this because I'm not sure you realize that you're making this hidden assumption.

I didn't get how a probability 0 can happen at all, and still be a probability 0.

Maybe it's just that natural language doesn't handle probabilities very well? Here is what's being said:

It is possible for there to be an event E such that $P(X \in E) > 0$, and yet for each individual outcome $e \in E$, we have that $P(X = e) = 0$.

 

[VikingF]

The thing I'm trying to get you to realize is that you're making a hidden assumption -- when you say "choose a random number from a finite interval", you mean "choose a random number from a finite interval, according to the uniform probability distribution". Of course, the general agreement is that if you don't say, then you mean the uniform probability distribution... but I'm making a big deal of this because I'm not sure you realize that you're making this hidden assumption.

Yes, I understand that. But I thought it was not necessary to add that it was a uniform distribution.

Maybe it's just that natural language doesn't handle probabilities very well? Here is what's being said:
It is possible for there to be an event E such that $P(X \in E) > 0$, and yet for each individual outcome $e \in E$, we have that $P(X = e) = 0$.

Maybe.

I have always heard that a probability P(e)=0 means that the event e is impossible and that a probability P(e)=1 means that the event e is certain. (i.e.: To toss a dice and get 7 has a probability equal to zero, while getting one of the values in the interval [1,6] is equal to one.)

That is way I reacted on the statement.

 

[-Job-]

Let $$f(x)$$ be the probability density function of $$X$$
This means that we can integrate $$f(x)$$ over an interval [a, b] to determine the probability that $$a \leqslant X \leqslant b$$. (notice that integrating $$f(x)$$ over negative and positive infinity, must equal 1)
$$P\{a \leqslant X \leqslant b\} = \int_{a}^{b} f(x) dx$$
However, the probability that $$X = a$$ is given by:
$$P\{X = a\} = P\{a \leqslant X \leqslant a\} = \int_{a}^{a} f(x) dx = 0$$

 

[VikingF]

Let $$f(x)$$ be the probability density function of $$X$$
This means that we can integrate $$f(x)$$ over an interval [a, b] to determine the probability that $$a \leqslant X \leqslant b$$. (notice that integrating $$f(x)$$ over negative and positive infinity, must equal 1)
$$P\{a \leqslant X \leqslant b\} = \int_{a}^{b} f(x) dx$$
However, the probability that $$X = a$$ is given by:
$$P\{X = a\} = P\{a \leqslant X \leqslant a\} = \int_{a}^{a} f(x) dx = 0$$

I see your point, but shouldn't probability theory be redefined in that case? I mean, if P=0 does not mean that an event is impossible?

If we take a look at the Gaussian curve, e.g. for height of a certain number of people, then the probability that a person is say 180 cm can't be 0, just because $$P\{X = 180\} = \int_{180}^{180} f(x) dx = 0$$? Atleast that is not a practical or logical conclusion...

What do you guys say about this?

(NB! There are some years ago since I took Statistics at Engineering school. Got a B back then. )

 

[shmoe]

Have you ever measured a person to be exactly 180cm? You're using the Gaussian as a convenient way to model the real distribution of heights. In reality there are a finite number of people, so the set of heights attained is also finite, and therefore discrete, etc.

 

[VikingF]

Have you ever measured a person to be exactly 180cm? You're using the Gaussian as a convenient way to model the real distribution of heights. In reality there are a finite number of people, so the set of heights attained is also finite, and therefore discrete, etc.

Okay, I understand.
Then a probability of 0 really means that an event is very, very, very unlikely to happen, but then again, not impossible? I.e., every person has to be of some size.

But what can then be said about really impossible events?

I'm asking just to be sure about my probability theory, because I think it is really interesting!

 

[Hurkyl]

I mean, if P=0 does not mean that an event is impossible?

If P = 0, that simply means the probability is zero. What you take that to "really" mean is not a mathematical question.

Incidentally, mathematicians have adopted the term "almost surely" to refer to things that "happen" with probability 1. (More precisely, to events whose probability is 1)

 

[VikingF]

If P = 0, that simply means the probability is zero. What you take that to "really" mean is not a mathematical question.

Maybe not, but it is nice to have a practical understanding of what it really means to have a probability equal to something, at least the probabilities 0 and 1 (and maybe 0.5).

Incidentally, mathematicians have adopted the term "almost surely" to refer to things that "happen" with probability 1. (More precisely, to events whose probability is 1)

I think that's an ugly abuse of the probability 1.
You can't find probabilities that are larger than 1, but an event can SURELY be more certain than just "almost surely", if you see what I mean.

Maybe it's more a topic of the category "Philosophy of Science" that I'm talking about here...

 

[shmoe]

I think that's an ugly abuse of the probability 1.
You can't find probabilities that are larger than 1, but an event can SURELY be more certain than just "almost surely", if you see what I mean.

Things can also "surely" happen. Consider the uniform distribution on the unit interval [0,1]. You will "almost surely" pick an irrational number. However, you will "surely" pick a real number.

You might want to look into measure theory if you want to try to understand the finer points behind probability theory. This is the stuff that they won't bother teaching your engineering stats class, but is really needed if you want to learn some rigorous probability and not just how to apply various distributions to bolts coming off an assembly line.

 

[-Job-]

Discrete & Continuous random variables have different contexts. Discrete random variables coincide more with experience because we don't generally deal with infinite sample spaces in real life. With discrete random variables we assign probability 0 to events that we suppose will not happen, so it's only a discrepancy if we don't keep in mind that the context is different, so things have different meanings. Any time we are dealing with continuous random variables we should be aware that probability 0 doesn't mean the outcome is impossible, only that it is so small that it is irrelevant. Mostly a semantics issue.

 

[wcemichael]

The answer to your questions is, YES. A chance is still a chance, even if it's a showball's chance in hell

 

[SW VandeCarr]

Ok I am not sure if this is where this thread should go, but anyway.

My question is this:
if we have an infinite series of numbers (or apples, or whatever) and we choose one at random, and then choose another at random, is there a chance that the two will be the same? I mean, is a 1 in infinity chance a chance at all?

The answer to your questions is, YES. A chance is still a chance, even if it's a showball's chance in hell

The answer is the same as it was over 5 years ago. It is possible with a probability of zero. (That's not a contradiction).