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 Infinity question Goto page 1, 2  Next
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MTGAP
Daedalian Member

 Posted: Sat Aug 02, 2008 11:35 pm    Post subject: 1 I'm sure there are some math forums that this question would be better for, but I like this site better. Pick a number between 0 and 1. Let's say 0.5. If you randomly selected a number between 0 and 1, the probability of it being 0.5 is zero. Let's say you keep selecting numbers until you get 0.5, at which point you stop. The number of numbers you just selected (which we will call X) is finite. Now repeat that process an infinite number of times, and average all the values of X. What is the result? Is it finite, because you averaged finite numbers, or is it infinite, because on average you have to select an infinite number of numbers before you get 0.5? There are two sub-questions. 1. If you randomly pick infinite numbers between 0 and 1, are you guaranteed to get 0.5 at least once? 2. What is the average of a finite number, say, 1, and infinity? How about the average of an infinite number of finite numbers and infinity? P.S. Sorry for confusion involving numbers of numbers. I hope it's understandable._________________This statement is false.
Chuck
Daedalian Member

 Posted: Sun Aug 03, 2008 12:00 am    Post subject: 2 I don't think that a random number between zero and one can be generated that gives each equal probability.
ralphmerridew
Daedalian Member

 Posted: Sun Aug 03, 2008 2:04 am    Post subject: 3 Check: It is possible to generate a uniform real number between zero and one; it is not possible to generate a random rational number in that range. The probability of getting a 0.5 eventually depends on which infinite cardinal is used. The average of the infinite set must first be defined. If it's indexed in a countable way, then an average can be defined as a limit, but the value of the average will almost certainly depend on the order in which numbers are indexed. (Will vary unless all but finitely many elements of the set are the same.) But the probability of getting .5 in a countable number of attempts is zero.
AZu
Daedalian Member

Posted: Sun Aug 03, 2008 6:44 am    Post subject: 4

Stolen from another site. Hope it is not a chestnut.

 Two Envelope Paradox wrote: You are taking part in a game show. The host introduces you to two envelopes. He explains carefully that you will get to choose one of the envelopes, and keep the money that it contains. He makes sure you understand that each envelope contains a cheque for a different sum of money, and that in fact, one contains twice as much as the other. The only problem is that you don't know which is which. The host offers both envelopes to you, and you may choose which one you want. There is no way of knowing which has the larger sum in, and so you pick an envelope at random (equiprobably). The host asks you to open the envelope. Nervously you reveal the contents to contain a cheque for 40,000 pounds. The host then says you have a chance to change your mind. You may choose the other envelope if you would rather. You are an astute person, and so do a quick sum. There are two envelopes, and either could contain the larger amount. As you chose the envelope entirely at random, there is a probability of 0.5 that the larger check is the one you opened. Hence there is a probability 0.5 that the other is larger. Aha, you say. You need to calculate the expected gain due to swapping. Well the other envelope contains either 20,000 pounds or 80,000 pounds equiprobably. Hence the expected gain is 0.5x20000+0.5x80000-40000, ie the expected amount in the other envelope minus what you already have. The expected gain is therefore 10,000 pounds. So you swap. Does that seem reasonable? Well maybe it does. If so consider this. It doesn't matter what the money is, the outcome is the same if you follow the same line of reasoning. Suppose you opened the envelope and found N pounds in the envelope, then you would calculate your expected gain from swapping to be 0.5(N/2)+0.5(2N)-N = N/4, and as this is greater than zero, you would swap. But if it doesn't matter what N actually is, then you don't actually need to open the envelope at all. Whatever is in the envelope you would choose to swap. But if you don't open the envelope then it is no different from choosing the other envelope in the first place. Having swapped envelopes you can do the same calculation again and again, swapping envelopes back and forward ad-infinitum. And that is absurd. That is the paradox. A simple mathematical puzzle. The question is: What is wrong? Where does the fallacy lie, and what is the problem?
Zag
Tired of his old title

 Posted: Sun Aug 03, 2008 7:54 am    Post subject: 5 Easy answers: MTGAP, it takes you an infinite number of choices (on average) to choose any specific number. Since you are doing an infinite number of trials, you might pick it in a finite number eventually, but it still has to average with all those infinites, and what do you get? AZu, it is a well-known chestnut. The answer is in the distribution of how they choose the amounts. They certainly didn't choose them as "from 1 to infinity, with equal probability." If you knew, for instance, that it was equal distribution from 1 to 100000, then you'd know what to do. Or if it was a non-linear distribution, you'd still be able to figure it out. At the very least, you can make an educated guess at the upper bounds (the gross income of the game show).
ralphmerridew
Daedalian Member

 Posted: Sun Aug 03, 2008 11:06 am    Post subject: 6 Zag: All numbers picked by MTGAP's method are between 0 and 1.
MTGAP
Daedalian Member

Posted: Sun Aug 03, 2008 5:55 pm    Post subject: 7

 Zag wrote: Easy answers: MTGAP, it takes you an infinite number of choices (on average) to choose any specific number. Since you are doing an infinite number of trials, you might pick it in a finite number eventually, but it still has to average with all those infinites, and what do you get?

But once you hit the number you want, you stop. Since you stop, it is by definition finite. So what is the average of all these finite numbers?
_________________
This statement is false.
Lepton*
Guest

 Posted: Sun Aug 03, 2008 6:51 pm    Post subject: 8 Let N+1 be some large even number. Then there are N rationals of the form i/(N+1) in (0,1), where i ranges [1,N]. Therefore the probability of selecting 0.5 (or i such that i = N/2) is 1/N. On average, it will take N times to choose a particular value from a list of N entries. Now, recognize that taking the limit N -> infinity gives an increasingly good approximation of the set of the rationals on (0,1). However, in this limit, the probability of choosing 0.5 in one selection is exactly 0 and the average amount of time required to choose a particular rational is strictly infinite. Imagine throwing a dart at a meter stick. What are the odds that the dart will strike the 50 cm dash exactly, no matter how finely you measure it? (assume no granularity of matter)
Amb
Amb the Hitched.

 Posted: Sun Aug 03, 2008 8:44 pm    Post subject: 9 Imagine you go to 10 decimal places. Each time you pick the next number down in the string, you have a 1 in 10 chance of picking 0. Ie, for one dp, picking 0.5 is a 1 in 10 chance. For two dp, the chances reduce to 1/10 for digit 1 being a zero, and 1/100 for digit 2 being a zero. Multiply the numbers out to infinity and you have an absurd situation where every number through makes the odds progressively smaller to the point of saying it will never happen.
Zag
Tired of his old title

Posted: Sun Aug 03, 2008 10:58 pm    Post subject: 10

MTGAP wrote:
 Zag wrote: Easy answers: MTGAP, it takes you an infinite number of choices (on average) to choose any specific number. Since you are doing an infinite number of trials, you might pick it in a finite number eventually, but it still has to average with all those infinites, and what do you get?

But once you hit the number you want, you stop. Since you stop, it is by definition finite. So what is the average of all these finite numbers?

Why are you assuming you ever hit the number and stop? You are the one who claimed you could do an infinite number of trials. I can claim that you do an infinite number of choosings.

When there is zero chance of an event happening, then, when you posit infinity, it might happen, but only after an infinite number of attempts. But you're the one who posited infinity -- if you're going to play in those waters, you have to stay in those waters (i.e. with the sharks). As soon as you pick a number (say, a googolplex of trials), then I am comfortable saying that it never happens.
Chuck
Daedalian Member

 Posted: Mon Aug 04, 2008 3:55 am    Post subject: 11 To generate a random real number between zero and one with equal probability for each you'd have to generate an infinity of random digits. After each digit you've either failed to generate ½ or you're not done yet.
ralphmerridew
Daedalian Member

 Posted: Mon Aug 04, 2008 12:07 pm    Post subject: 12 Chuck: And? What's so difficult about, theoretically at least, taking the process an infinite number of steps? Or do you agree with Zeno that motion is impossible?
Chuck
Daedalian Member

 Posted: Mon Aug 04, 2008 12:45 pm    Post subject: 13 The time it takes to move half the distance to a goal becomes smaller as you move toward the goal so you can do an infinity of such halvings in finite time. Is that true of generating random digits? Is there no limit to how many you can generating in any arbitrarily small amount of time?
MTGAP
Daedalian Member

Posted: Mon Aug 04, 2008 5:19 pm    Post subject: 14

 Chuck wrote: The time it takes to move half the distance to a goal becomes smaller as you move toward the goal so you can do an infinity of such halvings in finite time. Is that true of generating random digits? Is there no limit to how many you can generating in any arbitrarily small amount of time?

That's irrelevant. You can generate an infinite number of digits instantly. Another way of looking at it is that every possible number ever already exists.
_________________
This statement is false.
ralphmerridew
Daedalian Member

 Posted: Mon Aug 04, 2008 5:26 pm    Post subject: 15 Imagine throwing an ideal dart at a target which is a square with side 1 until it lands on or in the target. Take the x-coordinate of the point for your result.
Chuck
Daedalian Member

Posted: Mon Aug 04, 2008 11:22 pm    Post subject: 16

MTGAP wrote:
 Chuck wrote: The time it takes to move half the distance to a goal becomes smaller as you move toward the goal so you can do an infinity of such halvings in finite time. Is that true of generating random digits? Is there no limit to how many you can generating in any arbitrarily small amount of time?

That's irrelevant. You can generate an infinite number of digits instantly. Another way of looking at it is that every possible number ever already exists.

But how do you randomly choose one of those already existing numbers? It does no good if you don't know which one is yours.
Chuck
Daedalian Member

Posted: Mon Aug 04, 2008 11:26 pm    Post subject: 17

 ralphmerridew wrote: Imagine throwing an ideal dart at a target which is a square with side 1 until it lands on or in the target. Take the x-coordinate of the point for your result.

You'd need to be able to measure the distance along the x coordinate with infinite precision. The dart would be sticking in some point on the board but how do you turn that into your random number?
ralphmerridew
Daedalian Member

 Posted: Mon Aug 04, 2008 11:34 pm    Post subject: 18 When I started with an "ideal dart", I thought it was clear that I was treating this as a theoretical problem.
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 2:21 am    Post subject: 19 In theory the probability of hitting any given point is zero but the dart has to hit somewhere. So in theory something that can't happen is certain to happen.
ralphmerridew
Daedalian Member

 Posted: Tue Aug 05, 2008 2:23 am    Post subject: 20 "Having probability 0" does not mean the same as "can't happen".
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 2:29 am    Post subject: 21 It means it won't happen by chance.
Bicho the Inhaler
Daedalian Member

 Posted: Tue Aug 05, 2008 6:25 am    Post subject: 22 The "it has probability zero, so it can't happen" argument is tempting, but it's totally fallacious, and I'll try to explain why. It's because the concept of probability is useless when applied to an event after it occurs: the event either happened or it didn't happen. To assign a probability to an event, you have to do it before the event occurs. To illustrate this, here's an easier thought experiment: suppose I flip an unbiased coin 200 times. The result: HTTHHTHTTHHHTHTTHTHTTHTHHTTHTTTHHTHTHTHHTTHTHTTHHHHHTHTHHTTHHTTHHHHTTHTHTTTHHHTTHTHTTTHTHTHHHTHHHTHT HTHHHHTTTHHHHHTHTTTTTTTHHHTTTHHHTHHTTHTHTTTTTHTHHHHHTTHTTHHHHTHTTTHHHHHHTTHHHTTTTHTHTHTHHTHHHHTHHHTT Now what was the probability of obtaining that result? 1/2^200, right? That's astronomically small. Nevertheless, it's completely unexceptional, regardless of the fact that the probability of that event was supposedly so incredibly close to zero. The problem is that we can't say it has probability 1/2^200 after it has already happened. There was a time, namely before I started flipping coins, when we could have written down that sequence of heads and tails and said the probability of obtaining that sequence was 1/2^200, but we missed that opportunity, and so we missed our chance to observe an event of probability so small that you'd be more likely to win 5 lotteries in the same day. The same is true with these so-called "events of probability zero." There is no qualitative difference. If I choose a random real number uniformly between 0 and 1 and show it to you, you have lost your right to claim that you've seen a event of probability zero, since it happened before you decided to give it a probability. However, there was a time, namely before I showed you the number, when you could have said "the probability of you showing me this number is zero," and you would have been correct, no matter what number you had in mind. And there's no contradiction here: if you first select a number (by any means) and then select another uniformly random number between 0 and 1, there is no chance at all that the two will match. So it really does have probability zero. Edit: Before anybody says that probabilities can sometimes be assigned after events occur (for example, I can still consider odds on a baseball game I taped from TV if I didn't see it and nobody told me what happened), the "occurrence" I'm talking about is really the knowledge of the result. Probability is a tool for reasoning with limited information, and when you gain complete information, there's no more talk of probability. (In general, when you gain any information about an event, its probability changes, and its old probability is useless.)
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 7:14 am    Post subject: 23 It's easy to make an extremely low but still positive probability event happen. Some sequence of numbers will come up if I toss a die a thousand times. I've never heard of a zero probability event actually occurring.
ralphmerridew
Daedalian Member

 Posted: Tue Aug 05, 2008 9:36 am    Post subject: 24 Chuck, modern probability theory is not strictly frequentist. It's part of measure theory. The perimeter of an shape has area 0, but the boundary still exists.
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 1:30 pm    Post subject: 25 Let's see, then, if you randomly select a point in a figure, the probability of landing on the boundary is zero. But that's not very satisfying because the boundary is there so it seems that it ought to be possible to land on it, but saying there's zero chance of landing on it makes it sound like it can't happen. So instead of saying that the probability is zero we'll call it measure zero instead. That makes it sound like it's not really zero. We'll also just overlook the fact that we can't actually select a random point with uniform probability at all.
ralphmerridew
Daedalian Member

 Posted: Tue Aug 05, 2008 2:38 pm    Post subject: 26 Where am I disputing that the probability is zero? I'm claiming that "probability 0" != "impossible". (The second implies the first, but they are not equivalent.)
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 2:46 pm    Post subject: 27 It's not impossible. You can intentionally choose a point on the boundary. If something has a probability of zero of happening at random then it won't happen.
Jack_Ian
Big Endian

 Posted: Tue Aug 05, 2008 3:20 pm    Post subject: 28 Every point has an equal probability of being chosen, namely 0, but if you randomly choose a point then you have made something with a probability of 0 actually occur.
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 3:25 pm    Post subject: 29 But can you choose a point in such a way that each has equal probability?
GH
Daedalian Member

Posted: Tue Aug 05, 2008 3:52 pm    Post subject: 30

 Chuck wrote: If something has a probability of zero of happening at random then it won't happen.

That's both true and false.

If you had a finite number of possible results, and one result had a probability of 0, then it could never happen. But when you have an infinite number of possible results, the probability of any one happening is calculated as a limit, which leads to assign zero probability for things that can happen.

Here's what I find annoying about that. If I randomly choose any integer, the probability of choosing 11 is 0. The probability of choosing 37.5 is also 0. Clearly, though, they're different kinds of 0. Argh!

Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 3:57 pm    Post subject: 31 My problem is, if you have an infinity of things from which to choose, how can you choose one at random with equal probability for each? I see no way that it can be done.
GH
Daedalian Member

 Posted: Tue Aug 05, 2008 4:05 pm    Post subject: 32 Hm. So your issue is not with the fact that any given element has the same probability is any other element, but that we don't have a realistic notation for the particular element that we've chosen? What if the original problem were rephrased so that we're choosing random integers until we choose 0?
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 4:13 pm    Post subject: 33 How do you choose a particular integer at random from the entire set with equal probability for each? It would seem that it can't be done. What is the probability that it will have more than 1000 digits? A finite number of them have 1000 digits or fewer and an infinity have more so it looks like you'd have a 100% chance of choosing a number with more then 1000 digits. But the same argument applies to any number of digits.
ralphmerridew
Daedalian Member

 Posted: Tue Aug 05, 2008 4:14 pm    Post subject: 34 GH, that wouldn't work because it's impossible to make a fair selection from the integers (or any countably infinite set). And Chuck, why do you keep assuming that "event has probability 0" is equivalent to "event is impossible"?
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 4:18 pm    Post subject: 35 It's not impossible, it just won't be selected at random. Isn't that what zero probability means? You can certainly intentionally select 7 from the set of all integers, but not at random.
Jack_Ian
Big Endian

 Posted: Tue Aug 05, 2008 7:05 pm    Post subject: 36 There are an infinite number of locations where I can place a coin in a room. Even if I remove the coin and replace it, it will not be in exactly the same position. Before I place the coin anywhere, every one of the infinite locations has 0 chance of being selected. This is true even though I am more likely to place the coin on the floor than on the ceiling, since that would be impossible. Once I place the coin in the room, I have made one of these zero probability events occur. Even if I decided to cover a particular spot on the floor, the inaccuracy of my movements would mean that the infinite number of possibilities within 1 mm of the final resting place were equally likely to have been chosen by me and so all had probability zero. Unless, of course, space is granular in nature, in which case there are only a finite number of locations in the room.
GH
Daedalian Member

 Posted: Tue Aug 05, 2008 7:09 pm    Post subject: 37 I don't think that's a fair interpretation of "zero probability." "Zero probability" really just reflects that as the number of equally-likely possible results increases, the probability of any given result decreases. If the number of possible results increases without bound, the probability of any given result approaches zero. Did you look at that Wikipedia page about what the terms "surely" and "almost surely" mean? I found it interesting, particularly because it included examples of flipping a coin and throwing an ideal dart at a unit square!
Chuck
Daedalian Member

 Posted: Tue Aug 05, 2008 7:10 pm    Post subject: 38 But you still haven't generated a real random number because you can never know to an infinity of decimal places exactly where the coin is. We don't even know for sure that there are an infinity of places to put it. Space might be grainy on a low lever.
GH
Daedalian Member

 Posted: Tue Aug 05, 2008 8:08 pm    Post subject: 39 Infinity: not adequate for real-life situations.
ralphmerridew
Daedalian Member

 Posted: Tue Aug 05, 2008 8:29 pm    Post subject: 40 Chuck, from a practical POV, it is impossible to generate a random real number, but I already pointed out that we are talking theoretically.
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