[unreality]

I have a few problems to scratch your heads over

1) You pick a random number from an infinity of integers, ie: 1,2,3,4,5...etc. What's the chance of picking 11,734?

2) You pick a random number from 0 and infinite integers, ie: 0,1,2,3,4,5...etc. What chance do you have of picking 11,734?

3) You pick a random number between 0 and 1 (including 0 and 1). What chances do you have of picking 0? of picking 1? of picking a whole number in general (0 OR 1)?

4) You have a random number generator that generates a random digit, ie 0-9, with a 1/10 chance per digit. You run the generator to infinity, creating an infinite stream of random digits. What is the probability that there are absolutely no 3's in the digit stream?

have fun

[HoustonHokie]

One of the interesting differences between your first three problems and the last two is that you can actually see the limit approaching 0 in the last two as the number of attempts increases. For example, if you've only got one pull at the random number generator, #4 is 0.9 and #5 is 0.1. With a second pull, #4 is 0.9*0.9=0.81 and #5 is 0.1*0.1=0.01, and so on. While both are, in the end, infinitesimals, #5 approaches 0 much more quickly than #4.

Because the first three are single shot events, the only way to visualize the limit approaching 0 is to imagine the actual number of possible random integers or decimals increasing towards infinity. From that standpoint, you can then see that #2 approaches 0 slightly more quickly than #1 which is the point the OP was trying to make. Similarly, the third part of #3 is twice as likely as the first or second parts, whatever that really means.

An interesting question then forms in my mind: Can we rank the different scenarios as to which is most likely, relative to each other? My guess is that #4 is the most likely event, but what about the others?

[Guest]

In the end, every problem ends up to be (some positive number ) / infinity. A property of infinity is that any fraction or multiple of infinity is still infinity (unless you're using a negative, where you get negative infinity). Therefore, infinity / (some positive number) = infinity. Furthermore, this means that any equation of the form (some positive number) / infinity can be reduced to 1 / infinity. So each of the problems is equally likely.

This does not work if your numerator is infinity or zero.

[Guest]

4) You have a random number generator that generates a random digit, ie 0-9, with a 1/10 chance per digit. You run the generator to infinity, creating an infinite stream of random digits. What is the probability that there are absolutely no 3's in the digit stream?

could you not say that there are an infinite number of answers that could be derived that would have an answer without the digit 3 in them, and if so would that not mean the answer is infinty/infinity or 100%?

[Guest]

could you not say that there are an infinite number of answers that could be derived that would have an answer without the digit 3 in them, and if so would that not mean the answer is infinty/infinity or 100%?

Infinity/infinity is not 1. It is undefined like 0/0. In fact, if you take limits, you can turn infinity/infinity into 0/0.

Example: You have an infinite series of positive integers (1, 2, 3, 4, 5...). You randomly select one number. What is the chance of getting an even number?

You could choose from 2, 4, 6, 8... In other words, there are infinite possibilities. So in this case, infinity/infinity = 1/2.

[HoustonHokie]

could you not say that there are an infinite number of answers that could be derived that would have an answer without the digit 3 in them, and if so would that not mean the answer is infinty/infinity or 100%?

The problem is that the denominator infinity is much larger (infinitely larger) than the numerator infinity. So it still approaches 0. I vaguely remember some techniques for dealing with limits that might eventually reduce the solution to only have infinity in the denominator and some other number in the numerator, but calculus was so long ago...

[Guest]

An interesting question then forms in my mind: Can we rank the different scenarios as to which is most likely, relative to each other? My guess is that #4 is the most likely event, but what about the others?

Just because you are more likely to generate a string of random digits that excludes the number three, than you are of generating a string of equal length of the digits of pi, does not change the fact that you have no hope of doing either infinitely. The two events therefore have identically equal probability (0). It is important to note that saying something has zero-probability does not mean it cannot occur, just that it will not occur with any measurable probability.

Now, consider the following: if you choose a number on the interval [0,1] truly at random, you can never actually conclude which number you have chosen, because you need an infinite amount of information do distinguish, say, 0.5 from 0.5000[etc...]0001. Therefore, it can be argued that it is, in fact, impossible to randomly choose any number from an infinite set.

[Guest]

In the end, every problem ends up to be (some positive number ) / infinity. A property of infinity is that any fraction or multiple of infinity is still infinity (unless you're using a negative, where you get negative infinity). Therefore, infinity / (some positive number) = infinity. Furthermore, this means that any equation of the form (some positive number) / infinity can be reduced to 1 / infinity. So each of the problems is equally likely.

This does not work if your numerator is infinity or zero.

Infinity does not represent any number. Therefore, performing mathematical operations on it is meaningless, even if it is occasionally seen in practice.

[Guest]

only one person got it completely right ;D

1/infinity is not NECESSARILY zero, but is usually defined as infintesimal

congrats Yoruichi!

Nope. Wrong. The answer is zero. Think of it as .000000000000 going on forever then followed by one, but unfortunately, forever never ends (by definition) so we never quite get to that 1.

TO ILLUSTRATE: take my assertion that the answer is 0. It is the same as the assertion that .9999999999.... = 1, which is true, and I will prove it:

We all agree that 1/3 = .3333333..... and 2/3 = .66666666..... (plug it into your calculator if you dont believe me)

Therefore, 1/3 + 2/3 = .3333333..... + .6666666..... = .9999999.... = 1.

More succinctly:

1/3 + 2/3 = (.3333333.....) + (.6666666....)

= 1 = .9999999......

Q.E.D.

I consider this subject closed.

[unreality]

yes, .999999=1, but infitesimal is in fact different than 0. Go back and read my post on page 3 that explains why the probability is not 0

to use your statement: "I consider this subject closed." lmao

Edited July 29, 2008 by unreality

[Yoruichi-san]

Lol...nvm...missed that page...well, Unreality's right, which impresses me for someone his age...;P

Edited July 29, 2008 by Yoruichi-san

[unreality]

w00t hehe. I like math, though, so it doesn't count. Math is pretty much ageless

[Guest]

yes, .999999=1, but infitesimal is in fact different than 0. Go back and read my post on page 3 that explains why the probability is not 0

to use your statement: "I consider this subject closed." lmao

You're right, infinitesimal is greater than zero, since zero is a real number. A probability is also a real number. The probabilities for these events are 0.

It seems counterintuitive that the sum of an infinite number of zeroes can be something other than zero, but that is essentially what a Riemann integral is.

[unreality]

math girl hath spoken ;D I stand by my answer

[Guest]

Guys, the answer is definitely 0. I think you are being overly dismissive of those who are arguing that the answer is 0 just because it doesn’t line up with your preconceived notions. I’ll try to give a more concrete example to help illustrate it. Consider number 4. The probability that a single digit is not 3 is 9/10. The probability that none of 22 consecutive digits are 3s is .9^22 or about .0985 which is less than .1. So the probability that this is not the case is greater than .9. Multiply this number by itself. Now we have a number less than .01 and at least .99. Keep multiplying by this number n times. 1 minus the result will always be at least .999…9 where the number of nines = n. So after infinite such multiplications, 1-result=.999… => -1+result=-.999 => -1+result =-1 => result = 0.

I know that wasn’t as elegant as it could have been, but it demonstrates my point. In response to esjohnson, yes your link shows that 1/infinity is not 0, but that is because it is meaningless to say 1/infinity. And, unreality, you need to look closely at what an infinitesimal is. It is not a real number, so unless you are in a state of unreality, it is not a valid probability. In the set of real numbers, the answer is 0. There are other valid ways of expressing the answer other than 0 (though 1/infinity is not one of them), but 0 is a correct answer. What is throwing a lot of people off is that conceptually, it doesn’t make sense for choosing a number from a set of numbers to have 0% chance of being any one number. The key is that this is a completely impossible situation to begin with so our normal idea of picking one from a limited set doesn’t apply. Hopefully I explained myself well enough that you will now all agree that this subject is closed. If not, make your case for it not being 0.

[bonanova]

it is, in fact, impossible to randomly choose any number from an infinite set.

Interesting statement. Let's pursue it a bit.

The set of reals in the interval [0, 1] is infinite.

I choose the number 0.314.

Prove that it was not chosen at random.

[Guest]

Interesting statement. Let's pursue it a bit.

The set of reals in the interval [0, 1] is infinite.

I choose the number 0.314.

Prove that it was not chosen at random.

.314 is equivalent to .314000… with infinite 0s. The probability of choosing a single 0 starting at a given point is .1. The probability of choosing 2 is .01. One minus this is .99. So as in my other example, we see that the probability of it having infinite 0s at that point is 0. You have to consider the 0s at the end because to do otherwise would be to say that a certain 3 digit decimal is more likely than a specific 20 digit one, meaning it wouldn’t be random. So since the probability of any one value being chosen is 0, we see that the probability of the digits of the randomly chosen number being able to be expressed using the atoms of the universe is 0. So although .314 is as likely as any other value to be chosen, the probability is 0.

[Yoruichi-san]

.314 is equivalent to .314000… with infinite 0s. The probability of choosing a single 0 starting at a given point is .1. The probability of choosing 2 is .01. One minus this is .99. So as in my other example, we see that the probability of it having infinite 0s at that point is 0. You have to consider the 0s at the end because to do otherwise would be to say that a certain 3 digit decimal is more likely than a specific 20 digit one, meaning it wouldn’t be random. So since the probability of any one value being chosen is 0, we see that the probability of the digits of the randomly chosen number being able to be expressed using the atoms of the universe is 0. So although .314 is as likely as any other value to be chosen, the probability is 0.

Actually, since he has already chosen .314 (i.e. the event has already been observed), the probability of him choosing .314 is 1. ;P

[Guest]

Actually, since he has already chosen .314 (i.e. the event has already been observed), the probability of him choosing .314 is 1. ;P

you got me there

[bonanova]

Agree the probability [before the choice was made] of choosing .314 is zero.

Impossibility and zero probability are different concepts.

Anyway, I asked not about its probability, but its randomness.

My question remains for d3k3: prove that .314 was not chosen at random.

[Guest]

hehe ;D as others have said "approaches 0 as a limit" does not mean "equals zero". It is a class of negative-powered cardinals called 'infintesimals', they're part of the aleph ladder (sort of)... ie, the reciprocals of the alephs. Infintesimal. Infinitely small, but still something, just something so small it's infinitely small ;D impossible to grasp

therefore, we can say the probability of picking 11,734 is infinitely small, but NOT ZERO. Think about it. If every number had a probability of 0 of being picked, NONE OF THE NUMBERS WOULD BE PICKED. Since one number is picked, the answer is not 0. It's "infintesimal"

The probability of picking 11,734 is 1/infinity = .00000000.....going on forever........1, which is equal to zero, and here is my mathematical proof. If you agree that .99999..... equals one (by my earlier proof), then:

1 = 1

1 - (.999999....) = 1-1

Therefore .000000000000000......1 = 0.

Q.E.D.

You can't randomly choose a single number from an infinite set. Think about it. A computer can't do it because it wouldn't be able to choose a number that requires more spaces than it can calculate. Ditto for humans.

[Guest]

The chances are 1:infinity

[Guest]

Two things:

First, all this talk about infinitesimals is quite interesting... but irrelevant. Infinitesimals are not real numbers (in the rigorous sense of "real", that is, they are not elements of the set R). A probability is defined as an element of the set [0,1], a subset of the real numbers; so a probability cannot be an infinitesimal. It's just a matter of definition.

Second, SOMETHING WITH PROBABILITY ZERO IS NOT IMPOSSIBLE. To say that something has probability zero means (roughly) that, given infinite time (more rigorously, in infinitely many trials), it will occur only finitely many times; while something with positive nonzero probability will occur infinitely often in infinitely many trials. So, if you show me that something has happened, say, 600 times, that does not prove that it has mathematically nonzero probability. (It seems to SUGGEST nonzero probability, but it doesn't PROVE it.)

[Guest]

Oh, and this is an interesting philosophical point:

You can't randomly choose a single number from an infinite set. Think about it. A computer can't do it because it wouldn't be able to choose a number that requires more spaces than it can calculate. Ditto for humans.

Are you suggesting that the mind is a finite-state device? Evidence for this? And, if this is so, "you can't choose" doesn't mean "it is impossible to choose." To use this argument, you'll need to give evidence that the UNIVERSE is a finite-state device--which may be true, if it's examined at the sub-quantum level, but seems unlikely a priori.

[Guest]

Agree the probability [before the choice was made] of choosing .314 is zero.

Impossibility and zero probability are different concepts.

Anyway, I asked not about its probability, but its randomness.

My question remains for d3k3: prove that .314 was not chosen at random.

You could argue otherwise (although I think it's pretty well agreed upon that the universe is not deterministic), but I think that .314 most certainly was chosen randomly. I can't believe, though, that it was chosen from a distribution uniformly random over the interval [0,1].

If you could give us a few more random numbers from the same distribution, we could do some tests and make a probabilistic conclusion about the distribution's uniformity.

[unreality]

The probability of picking 11,734 is 1/infinity = .00000000.....going on forever........1, which is equal to zero, and here is my mathematical proof. If you agree that .99999..... equals one (by my earlier proof), then:

1 = 1

1 - (.999999....) = 1-1

Therefore .000000000000000......1 = 0.

Q.E.D.

You can't randomly choose a single number from an infinite set. Think about it. A computer can't do it because it wouldn't be able to choose a number that requires more spaces than it can calculate. Ditto for humans.

I agree with all of that, except the assertion that 1/infinity = .00000000[etc]...1

where did you get that from? 1/infinity cannot be expressed by numbers in that way. 1/infinity is not 0. At all. As others have pointed out, though, infitesimal cannot be a probability and is thus 0%, sortof (even though a number is picked, theoretically )... as lytefoot said, something with 0 probability could be picked in an infinite number of picks, which means it's reallly not truly 0 but more of infintesimal, but w/e if people wanna call it 0, same thing for probability-wise

but your assertion that 1/infinity = .00000000[etc]...1 is wrong. That would of course make 1/infinity = 0, which it is not

edit: not ALWAYS. I'm sure sometimes it can be 0, lol. That's how crazy infinity is.

Another way to think of 1/inf is infinity to the negative first power. Aren't all powers of infinity equal to infinity? So infintesimals are the same thing as infinity, in a way

Edited July 30, 2008 by unreality