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#1
Old 08-27-2002, 03:22 PM
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if probability = 1; means event must occur ?

i was just trying to teach a friend some probability today... it's been a while since i last touched probability...

got me wondering... if the probability that an event might occur is 1, does that mean that the event must occur ?

take these examples:

1. the probability that when a normal die is thrown, it will roll 1 or 2 or 3 or 4 or 5 or 6. The probability of this event is 1, and i know it must occur.

2. the probability that in 6 throws of a die, i will roll a 6. the probability of rolling a 6, per throw, is 1/6. therefore for 6 throws it's 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6. therefore the probability that this event might occur is 1. however, i know that it is not necessary that i get a 6 in any of the 6 throws, even though the probability of the event is 1. this is where i'm kinda confused about the meaning of the probability of an event being 1.

3. the probability that Opal will get pissed about a list of 2 items.

so can anyone explain to me what the statement:

the probability that an event might occur equals 1

means ? does it mean that the event has to occur ?

or is there a fundamental flaw in my reasoning somewhere ?
#2
Old 08-27-2002, 03:45 PM
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Quote:
Originally posted by xash
the probability of rolling a 6, per throw, is 1/6. therefore for 6 throws it's 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6. therefore the probability that this event might occur is 1.
Your flaw lies here! You can only add probabilites when the events are mutually exclusive. For instance, you can roll a 2 on your first throw, or you can roll a 3, but you can't do both. Thus the probability of rolling a 2 or a 3 on your first throw is 1/6 + 1/6 = 1/3. Now, in this example, you can roll a 6 on your first throw, or you can roll a 6 on your second throw. But you can do both! The events are not mutually exclusive. Thus your formula is in error.

The simplest way to do this problem is consider not rolling 6's. The probability of not rolling a 6 on your first roll is 5/6. Ditto for the other throws. Since these are independent events (eg, what you throw the first time does not affect the probabilities for the second time) you can multiply them. The probability that you do not roll a 6 all six times is:

5/6 5/6 5/6 5/6 5/6 5/6 = 15625/46656

Now, this is the probability that you don't roll any 6's at all. The probability that you roll at least one 6 is the negation of this:

1 - 15625/46656 = 31031/46656 = 66.51%

(Assuming, of course, I did my arthmetic right, an event with probability significantly less than 1.)
#3
Old 08-27-2002, 03:48 PM
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Quote:
the probability that in 6 throws of a die, i will roll a 6.
This probability is not 1. To get the probability of a composite event, you need to multiply, not add. In this particular case, it's easier to calculate the probability that you do not get a 6 in six rolls of the dice. Then the event you're interested in is "no six on the first roll AND no six on the second roll AND no six on the third roll..." etc. This is equal to (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6), or 15625/46656, approximately 33%. The probability of getting at least one six is then approximately 67%.
#4
Old 08-27-2002, 03:48 PM
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You can't add probabilities like that.

You can multiply them, though. But 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 would equal the chance of rolling six sixes in a row. For your question, I would work backwards:

1) Chance of NOT rolling a 6 in one try is 5/6

2) Chance of NOT rolling a 6 in two tries is 5/6 * 5/6 = 25/36 (logically this makes sense, if you count all the possibilities for rolling two dice.)

3) Chance of rolling at least one 6 after two tries = 1 - (5/6 * 5/6) = 11/36

Note that 11/36 is less than 2/6

4) Chance of rolling at least one 6 after three tries = 1 - (5/6 * 5/6 * 5/6) = 125/216

5) Chance of rolling at least one six after n tries = 1 - (5/6)n

6) after 6 tries = 0.6651, just a little less than 2/3

7) After twelve tries = .8878

8) After twenty-four tries = .9874
#5
Old 08-27-2002, 03:54 PM
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#6
Old 08-27-2002, 03:55 PM
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Scooped! phooey

And I made a math error too.

4) Chance of rolling at least one 6 after three tries = 1 - (5/6 * 5/6 * 5/6) = 91/216
#7
Old 08-27-2002, 03:56 PM
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Probability 1 means the event must occur.

2. The probability of rolling at least one 6 in six rolls is obviously not 1. It's 1 - (5/6)^6. (Pure probability)

The average number of 6's rolled in six rolls is 1 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6. (Probabalistic average)

So you are confusing a pure probability with a probabalistic average. Probabalistic averages are not extremely useful -- really they are more confusing than helpful. The most illuminating way to represent the outcome of rolling a die six times and counting the 6's is a graph with 7 points plotted -- the probability of having 0, 1, 2, 3, 4, 5, or 6 6's rolled after six rolls -- all these numbers are between 0 and 1.

When you calculate the probabalistic average you are adding up these seven important values into one number, so you are losing information.

Consider the problem, what is the average number of 6's rolled in seven rolls? Obviously the answer is 7/6. Here it is clear that the answer has nothing to do with a probability, because probabilities are always between 0 and 1.
#8
Old 08-27-2002, 03:56 PM
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Probability usually equals 1 when the event is so defined, as in your first example: when you throw a die, something will come up.

Your second example is screwed up; you can't just add together the probabilities that way (if you throw the die seven times, is p = 1.17?)

The probability that you won't throw a six for any throw is 5/6; the probablity that you won't throw a six in six throws is multiplicative, so that p = (5/6)^6 = 0.335; the probability that you will roll one or more sixes is 1-0.335 = 0.665
#9
Old 08-27-2002, 03:58 PM
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And, in general, even having a probability of one that it will occur does not mean it's guaranteed to occur.

For example, imagine we pick a real number at random from the interval [0,1], such that no number is more likely to be picked than any other number.

(Such a probability function does exist--Lebesgue measure, for example, where the probability of our random number x being in some (Lebesgue measurable) subset A of the reals is equal to the Lebesgue measure of A.)

The probability for any particular number to be picked is zero. However, we know some number b will be picked, so here we have an example of an event with probability zero that actually does happen. Correspondingly, the probability that b will not be picked is one, and that didn't happen.
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#10
Old 08-27-2002, 04:08 PM
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Quote:
Originally posted by Cabbage
And, in general, even having a probability of one that it will occur does not mean it's guaranteed to occur.

For example, imagine we pick a real number at random from the interval [0,1], such that no number is more likely to be picked than any other number. ------8<------------------- The probability for any particular number to be picked is zero. However, we know some number b will be picked, so here we have an example of an event with probability zero that actually does happen ---8<----
I may be playing out of my league here, but doesn't the probability just approach zero? Or is the reciprocal of some type of infinity, or some such thing? I have a hard time accepting that an event with probability of exactly zero can still happen. After all, isn't that the very definition of a probability of zero?
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#11
Old 08-27-2002, 04:14 PM
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Quote:
Originally posted by CookingWithGas
I may be playing out of my league here, but doesn't the probability just approach zero? Or is the reciprocal of some type of infinity, or some such thing? I have a hard time accepting that an event with probability of exactly zero can still happen. After all, isn't that the very definition of a probability of zero?
No, in Cabbage's example, the probabilities really are 1 and 0. If there's a finite sample space, then an event of probability 1 must occur, and therefore an event of probability 0 can't.

Once you start dealing with infinite sample spaces, you're dealing with measure theory, and that's weird, and all kinds of strange stuff can happen. For instance, the odds of picking a rational number from [0, 1] are zero using the Lebesgue measure, and (if you accept the axiom of choice) there is a subset B of [0, 1] such that there is no way of assigning the probability of picking an element of B from [0, 1]. Don't ask; it's really weird.
#12
Old 08-27-2002, 04:27 PM
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i bow to you. all. not only do you guys make it so darn clear, you even respond in like minutes.

though i still have some confusion over Cabbage's post...Cabbage, could you respond to CookingWithGas's post ?

Thank you, Achernar, heresiarch, jawdirk, Nametag, Cabbage, CookingWithGas, even KneadToKnow on preview, thanks ultrafilter

i wish i had studied in the Straight Dope School of Mathematics with you guys as my teachers...

do keep the discussion going... it's really interesting...
#13
Old 08-27-2002, 04:30 PM
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and i forgot to mention our beloved Chronos... it's good to see that you're still roaming the boards...
#14
Old 08-27-2002, 04:31 PM
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Perhaps this has nothing to do with math, but I'd argue the probablity of rolling the number 1 through 6 on a single die roll is NOT 1. It is slightly less.

Imagine an uneven surface, or a wall against which the die may "lean". It's even conceivable that the die may simply come to rest on an edge, though extremely unlikely.
#15
Old 08-27-2002, 04:39 PM
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Quote:
Originally posted by yojimboguy
Perhaps this has nothing to do with math, but I'd argue the probablity of rolling the number 1 through 6 on a single die roll is NOT 1. It is slightly less.

Imagine an uneven surface, or a wall against which the die may "lean". It's even conceivable that the die may simply come to rest on an edge, though extremely unlikely.
Yes, but that isn't math. The mathematical die is really a random variable with a particular distribution over {1, 2, 3, 4, 5, 6}. The physical die is just an imperfect model of that variable.
#16
Old 08-27-2002, 04:42 PM
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That's true, yojimboguy. When doing probability on paper, you assume that the die is "perfect" or "fair". That is, nothing out of the ordinary like what you're describing happens, and that all sides are equally likely to come up. In real life, with something as simple as a die, this is a really good approximation. And anyway, if it did come to rest on an edge, what would you do? You'd throw that roll out and redo it.

Cabbage, is it safe to say that your whole thing about events with zero probability occuring is only an issue when dealing with infinite sample spaces?
#17
Old 08-27-2002, 04:44 PM
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Quote:
Originally posted by Achernar
Cabbage, is it safe to say that your whole thing about events with zero probability occuring is only an issue when dealing with infinite sample spaces?
Yes. I suspect that not all infinite sample spaces have that property.
#18
Old 08-27-2002, 05:29 PM
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Lebesgue measure is a generalization of "segment length". For example, the Lebesgue measure of the interval [2,28] is 28-2 = 26, which is just what you should expect, since that line segment is 26 units long. Since it's a generalization of segment length, there are some pretty bizarre sets that you can also assign a Lebesgue measure to, but the basic idea is still segment length. Also, as ultrafilter, there are even some really bizarre sets that can't be assigned Lebesgue measure (assuming the axiom of choice).

Anyway, while the definition of Lebesgue measure does involve limits, the actual Lebesgue measure of a set cannot "approach zero" or anything like that, the Lebesgue measure of a set is just some real number (or infinity), either zero or not.

It makes a lot of intuitive sense that the Lebesgue measure of a single point is just zero, period, since the "length" of a single point is just zero. That is why the with the probability function I gave above, the probability of picking a single point is just zero; not approaching zero, but just zero. And the probability of not picking a particular point is exactly one.

Also, speaking more generally here, it's important to remember the context of what we're talking about. On one level, probability is just a certain kind of mathematical function satisfying certain properties. There's absolutely nothing in the axioms that say a probability of zero means it can't happen, or a probability of 1 means it's bound to happen.

What those probabilities mean and how to interpret them depends on the actual model you're using the probability function for. Earlier I gave the example of an infinite sample space in which probability one does not mean it necessarily happens. What about a probability function with a finite sample space?

Building on yojimbguy's comment, I see nothing wrong with the following probability function for rolling a single six sided die:

The sample space will be: {1, 2, 3, 4, 5, 6, something else happens}.

I'll choose to define the probabilities of 1,2,...,6 as being 1/6 each, and the probability of "something else happens" as being zero (and extend it additively to assign the probability of other subsets of the sample space). I'll set up my model that way because I'm only interested in the behavior of the die working the way it's intended to. I don't care about the times the die lands on its edge, or, if the die may be particularly fragile, cracks into two pieces when it lands. Either of these events can certainly happen, however; just because I assigned them probability zero doesn't prohibit them from happening, it's just the fact that I'm not interested in them.

So here I have a perfectly valid probability model; it may be a little unconventional, but it certainly satisfies all the probability axioms. It's even an "accurate" probability model, in that it describes the probabilities of a fair die (working the way it's intended to). However, the way I've chosen to assign the probabilities demonstrates that probability one does not mean "guaranteed to happen," e.g., the probability that a 1, 2, 3, 4, 5, or 6 is rolled is one, but that doesn't exclude the possibility that it may land on its edge, or break when it hits the table.

In other words, there doesn't seem to be any mathematically objective part of probability theory defining what probability one or probability zero means; it seems to be all about how you set up your model and how you interpret it.
#19
Old 08-27-2002, 09:20 PM
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no, something with a probability of one need not occur.

if you flip a coin an infinite amount of times. the odds get closer and closer to one that a tails will come up. since you did in an infinite amount of times. the probiblity IS one (a number infinitely close to one IS one... thats the basis of calculus... so its gotta be true)

however. flip it the first time... heads come up. flip it the second time... heads come up.... flip it the billionth time still heads. no law of physics ever says it HAS to be tails. every time you flip it, its got no memory of what happened last. so although its magicly unlikely.... you could flip it infinitely and have it come up heads every time, even with tails haveing a proboblity of one of comeing up.
#20
Old 08-27-2002, 09:46 PM
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owl, your example is that of the probability approaching 1 as a limit; that's different than being exactly 1.

I do add the minor note, that you have a probability of 1 of getting {1, 2, 3, 4, 5, 6} when you roll a mathematical die. However, that means that IF you roll the die, THEN you will get one of the those outcomes. It does NOT mean that you have to roll the die. Semantics, perhaps, but an important distinction.
#21
Old 08-27-2002, 09:52 PM
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I know this wasn't the point you were making, but....

Quote:
Originally posted by C K Dexter Haven
you have a probability of 1 of getting {1, 2, 3, 4, 5, 6} when you roll a mathematical die. However, that means that IF you roll the die, THEN you will get one of the those outcomes.
I believe that Cabbage was saying that this isn't the case. Just because there's a probability 1 that you'll get one of those outcomes, doesn't mean that you'll certainly get one of those outcomes. In this case, it happens to be true, but it doesn't follow from the fact that P = 1.
#22
Old 08-28-2002, 12:55 AM
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Quote:
Originally posted by Cabbage
And, in general, even having a probability of one that it will occur does not mean it's guaranteed to occur.
What other interpretation is there?

Quote:
Originally posted by Cabbage
For example, imagine we pick a real number at random from the interval [0,1], such that no number is more likely to be picked than any other number.

(Such a probability function does exist--Lebesgue measure, for example, where the probability of our random number x being in some (Lebesgue measurable) subset A of the reals is equal to the Lebesgue measure of A.)

The probability for any particular number to be picked is zero. However, we know some number b will be picked, so here we have an example of an event with probability zero that actually does happen. Correspondingly, the probability that b will not be picked is one, and that didn't happen.
IIRC. the result of a number divided by infinity is not zero, it is undefined.

The limit of n/x as x approaches infinity is zero, but not n/x itself.
#23
Old 08-28-2002, 01:10 AM
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Urban Ranger: P(b) is a well-defined quantity, so it has to have a value. By the axioms of probability, it must be non-negative. If we can show that P(b) is less than any positive number (and I think we can) then P(b) = 0. No dividing of infinities required.
#24
Old 08-28-2002, 01:21 AM
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P(b) has a value, with is 1/x where x is the number of real numbers in [0,1]. A value does not have to be numerical, there is nothing that prevents P(b) being a function of some other parameters.
#25
Old 08-28-2002, 03:21 AM
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For a discrete random variable X, if P(X=x)=0 then X cannot take the value x. For a continuous random variable X, P(X=x)=0 for every x.
#26
Old 08-28-2002, 09:01 AM
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Quote:
P(b) has a value, with is 1/x where x is the number of real numbers in [0,1]. A value does not have to be numerical, there is nothing that prevents P(b) being a function of some other parameters.
I see what you're thinking, and that idea works for modeling a number picked at random from finitely many numbers. However, one of the axioms of a probability function is that the range of probabilities must be contained in the real interval [0,1]. Any sort of 1/infinity is not a real number, and generally not defined at all. So, while you may be getting an intuitive feel for what's going on by thinking of it that way, that's really not the way the probabilities are defined in any mathematical sense, and probability is known for having some very counter-intuitive results, so relying on your intuition is liable to lead to mistakes.

Lebesgue measure is defined in terms of limits, but the Lebesgue measure of any (Lebesgue measurable) subset of [0,1] is a real number in the interval [0,1]. The Lebesgue measure of a single point is not "approaching" zero in any sense, it is exactly zero. And Lebesgue measure satisfies all of the probability axioms, so it's a perfectly valid probability model, and one which demonstrates that probability one is not the same as "guaranteed to happen" (GTH).

Having said what I said before about probability functions with a finite sample space, and how you can construct models with such probability functions so that probability one does not mean GTH, I have a strong feeling that you can make refinements to any such model to rectify this situation--simply take the "largest" subset with probability zero, and get rid of it. In other words, for example, take my original sample space {1,2,...,6,something else happens}, and simply reduce it to {1,2,...,6}. Doing this doesn't significantly change your probability model, it simply discards the events that are not interesting. In fact, I suspect this may also be doable when your sample space is countably infinite, though I haven't actually tried proving either of these ideas. Keep in mind, however, that this doesn't demonstrate that, even in the finite case, probability one = GTH; while the mathematical side of probability is rigorous, I am still free to model real world examples in such a way that probability one != GTH.

However, when your sample space is uncountable, there's no way around it--there will be events with probability one that are not GTH (unless your probability function is simply too trivial).
#27
Old 08-28-2002, 09:23 AM
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Quote:
Having said what I said before about probability functions with a finite sample space, and how you can construct models with such probability functions so that probability one does not mean GTH, I have a strong feeling that you can make refinements to any such model to rectify this situation--simply take the "largest" subset with probability zero, and get rid of it. In other words, for example, take my original sample space {1,2,...,6,something else happens}, and simply reduce it to {1,2,...,6}. Doing this doesn't significantly change your probability model, it simply discards the events that are not interesting. In fact, I suspect this may also be doable when your sample space is countably infinite, though I haven't actually tried proving either of these ideas.
Actually, scratch that, forget it. I think my mind got off on a tangent somewhere, and I don't think the above is really relevant to what we're discussing, after all. Reducing the sample space doesn't change the fact that something else can still happen to the die--it's not guaranteed to come up 1-6, even though it has probability one of doing so.

Still, there is a fundamental difference between the finite case and the uncountable case. Maybe what I was trying to say could be rephrased as this: In some cases, such as the above case, probability zero can possibly be interpreted, not as "can't happen", but rather, "not interesting". The "something else happens" is not impossible, rather, it's just that our attention is focused solely on what happens when the die is rolled the proper way with a proper outcome.

This is distinct from the uncountable case. With the picking a real number/Lebesgue measure example; the probability of any particular number being picked is zero, however, taken as a whole set, there's probability one that one of them will be picked. That's a significant and fundamental property of the probability function in this case, the probabilities of zero for any particular number can't be easily dismissed as in the preceding example, since together they end up forming the whole sample space. Anyway, I hope that makes sense.
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Old 08-28-2002, 12:51 PM
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Cabbage,

I agree with what you've written, but at the same time I think it is misleading.

Suppose you choose a random real number on the interval [0,1]. As you said, the chance that it is rational is 0. But also, the chance that it is representable as a mathematical expression is 0. In other words, mathematical expressions are countable, and the real interval [0,1] is not countable, so the probability of a random point on real [0,1] being representable is 0.

So if you were to actually generate a string of these numbers, you would not be able to represent any of the particular values you generated. The best you could do is say they were greater than a certain representable number, and less than another representable number.

So what I am arguing is that the chance that you randomly generate a rational number is 0, and furthermore, just as you would expect, it couldn't happen. Furthermore, the chance that you will not generate any particular rational number is also 1, and furthermore, that will always happen. What will always happen is that you generate a number which is not precisely representable by any finite string of mathematical symbols.
#29
Old 08-28-2002, 01:42 PM
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jawdirk, I'm afraid I don't see how that makes what I said misleading. The point I was trying to make is that any particular number has zero probability of coming up. That means any number, whether it's rational, expressable by some finite slgorithm, or some other number that we can't even generate with any kind of algorithm. The point being that, whatever number is picked (whether it's a "nice" number or not), the probability of it being picked was zero, and yet it was still picked. It's not misleading to say that probability zero does not mean "impossible", and it's not misleading to that that probability one does not mean "GTH", since that's exactly what happens here, and that's the only point I'm trying to make.
#30
Old 08-28-2002, 01:57 PM
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I think that what jawdirk was saying is that it's not meaningful to say you've "picked" a number if that number is not representable. If you've got your random number generator, and I ask you what number it returned, what do you say?
#31
Old 08-28-2002, 02:03 PM
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Quote:
Originally posted by Chronos
I think that what jawdirk was saying is that it's not meaningful to say you've "picked" a number if that number is not representable. If you've got your random number generator, and I ask you what number it returned, what do you say?
I would say that I've picked a number. Yay axiom of choice!
#32
Old 08-28-2002, 02:12 PM
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Quote:
Originally posted by Chronos
I think that what jawdirk was saying is that it's not meaningful to say you've "picked" a number if that number is not representable. If you've got your random number generator, and I ask you what number it returned, what do you say?
I'd say I picked a number. Yay axiom of choice!

On a more serious note, every real number is representable. What's at stake is whether we've "picked" a number if the number we picked is not finitely representable. I have no problem with allowing that case. After all, the set of finitely representable numbers in [0, 1] is countable, so it has measure zero--therefore, it is almost certain (i.e., probability 1) that we will pick a number which is not finitely representable.

Here's a question: what is the probability that we've picked a rational number less than or equal to 1/2, given that we've picked a rational number?
#33
Old 08-28-2002, 02:21 PM
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Yes Chronos, that's the point.

Cabbage, it's not just that it's not a nice number, it's that you can't even say what number you've picked. You can't differentiate these numbers from other numbers that also have (equal) probability 0 of being picked. Between any two representable numbers there will always be an infinite number of unrepresentable numbers, and you will have no way to say which of these was randomly chosen except by narrowing the bounds again.

I could be wrong about this -- I'm just working on a CS Masters not math, but I believe this is territory where the axiom of choice becomes important. I would guess that it is unprovable that it's ok to assume that a particular number has been chosen. In other words, all the numbers do have probability 0 of being chosen, and "choosing a number at random" doesn't violate this, because you can't say which number has been chosen. Unless you embrace the axiom of choice, you can't even assume that a particular number can be chosen.

Please educate me if I'm misinterpreting the axiom of choice.

For those who don't know anything about the 'axiom of choice', it is an axiom, in that a coherent mathematics can either hold it true or false, as far as anyone knows. When the axiom of choice is held true certain unintuitive results arise, and when it is held false other unintuitive results arise, so there is no concensus on which mathematics is 'preferable.' In consequence, mathematicians study mathematics with and without the axiom of choice. In general, denying the axiom of choice limits what can be proven.
#34
Old 08-28-2002, 03:15 PM
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Well, we're obviously talking about a thought experiment in the first place anytime we talk of picking something from any infinite set. The fact is, is that Lebesgue measure does form what can be called a probability function on the interval [0,1]. Is there any way to realize an actual mechanism that will do the picking for us? Of course not, but we can still speak of what would happen if we had such a mechanism, and I still claim that it's perfectly fine to speak of picking a particular real number at random from [0,1], using the axiom of choice (which, really, is commonly accepted). As far as I'm concerned, all I need to say is that some particular number has been picked; I don't have to specify what that number is in order to demonstrate my point that probability zero is not the same as impossible--a particular number was picked, and it had probability zero of being picked.

For that matter, you claim that picking a rational number "couldn't happen". How so? I agree that it's extraordinarily unlikely; it has probability zero of occuring, so it's natural to think of it as being extraordinarily unlikely, but my point all along has been that that doesn't imply that it's impossible.

Quote:
Here's a question: what is the probability that we've picked a rational number less than or equal to 1/2, given that we've picked a rational number?
Good question. It's tempting to say that, intuitively, the probability should be 1/2, but I'm not really satisfied with that. The problem is that, given that a rational number is picked, we've really restricted our sample space to a countable set. Unfortunately, it's impossible to construct a uniform probability function on a countable sample space (which is essentially what the question is asking for), so I'd have to say that it's pretty much unanswerable (under the standard probability axioms, anyway).
#35
Old 08-28-2002, 05:06 PM
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Ugh, I hope the failed post I'm rewriting actually failed:

In response to: what is the probability that we've picked a rational number less than or equal to 1/2, given that we've picked a rational number?

As Cabbage said, it's certainly not 1/2. There is a 1-to-1 correspondence between any two countably infinite sets, so the interval [0,1/4] has just as many rational numbers as the interval [0,1/2]. So any argument used to show that the probability was 1/2 could be used to show that the probability was 1/3, or any other value between 0 and 1.

Cabbage wrote:
Quote:
For that matter, you claim that picking a rational number "couldn't happen". How so? I agree that it's extraordinarily unlikely; it has probability zero of occuring, so it's natural to think of it as being extraordinarily unlikely, but my point all along has been that that doesn't imply that it's impossible.
See that's the crux of it. In this case my intuition seems to go against the axiom of choice, because as far as I am concerned, probability 0 means impossible. As you said, the axiom of choice is commonly accepted, but also, it is commonly rejected. Mathematicians point out that their proof requires the axiom of choice in order to differentiate what can and can't be proven without it. It has been proven that it is an axiom -- it can't be proven to be either true or false. As I said, some of its results are generally regarded as intuitive, and some of its results are generally regarded as absurd.

I have heard that with the axiom of choice it can be proven that there exists a way to cut a solid sphere into a finite number of pieces so that they can be reassembled into two solid spheres with volume equal to the original sphere! Importantly, it is not possible to describe the shapes mathematically -- but it is necessary that they exist if the axiom of choice is true.

So I'm glad we agree that the axiom of choice is the question here.

Here is a question for you:

Suppose a mechanism randomA outputs a random real number on [0,1]. Then let us define randomB as:

x=0;
while (x is rational) repeat {x=output of randomA}
output x;

The question is, does randomB produce a different output from randomA? Note that with probability 1 the while loop iterates exactly once!
#36
Old 08-28-2002, 05:07 PM
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Quote:
Originally posted by Cabbage
Good question. It's tempting to say that, intuitively, the probability should be 1/2, but I'm not really satisfied with that. The problem is that, given that a rational number is picked, we've really restricted our sample space to a countable set. Unfortunately, it's impossible to construct a uniform probability function on a countable sample space (which is essentially what the question is asking for), so I'd have to say that it's pretty much unanswerable (under the standard probability axioms, anyway).
This probably merits a new thread--there sure oughta be a way to do it, and if not, it's gonna engender some interesting discussion. I'll open it tonight.
#37
Old 08-28-2002, 05:40 PM
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Quote:
Originally posted by jawdirk
Ugh, I hope the failed post I'm rewriting actually failed:

In response to: what is the probability that we've picked a rational number less than or equal to 1/2, given that we've picked a rational number?

As Cabbage said, it's certainly not 1/2. There is a 1-to-1 correspondence between any two countably infinite sets, so the interval [0,1/4] has just as many rational numbers as the interval [0,1/2]. So any argument used to show that the probability was 1/2 could be used to show that the probability was 1/3, or any other value between 0 and 1.
No, because the probability depends on the measure of the set, not the size. The two are very different.

Quote:
Here is a question for you:

Suppose a mechanism randomA outputs a random real number on [0,1]. Then let us define randomB as:

x=0;
while (x is rational) repeat {x=output of randomA}
output x;

The question is, does randomB produce a different output from randomA? Note that with probability 1 the while loop iterates exactly once!
With probability 1, the output of randomB is the output of randomA. That doesn't mean it's guaranteed to happen.
#38
Old 08-28-2002, 06:19 PM
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Promised spinoff
#39
Old 08-28-2002, 07:25 PM
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Cabbage, would you say your interpretation "P(x) = 0 does not imply x cannot happen" is somewhat unorthodox? I like it and all, because it's so weird, but my probability book defines a sample space as something like, "The set of all possible outcomes". And I think this definition is not uncommon.
#40
Old 08-28-2002, 07:28 PM
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Quote:
Originally posted by Achernar
Cabbage, would you say your interpretation "P(x) = 0 does not imply x cannot happen" is somewhat unorthodox? I like it and all, because it's so weird, but my probability book defines a sample space as something like, "The set of all possible outcomes". And I think this definition is not uncommon.
This is orthodox. If probability zero is not impossible, then your sample space can contain non-empty subsets of probability zero.
#41
Old 08-28-2002, 11:43 PM
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About the axiom of choice, what I was trying to say is that I'd be willing to bet, if you polled mathematicians on whether it is true or not, a majority of them would say that it is true; it has become a pretty standard axiom in set theory.

Mathematics can get a lot uglier without the axiom of choice:

Given two sets A and B, you cannot compare the cardinalities A and B. In fact, the cardinality of A (or B) is not even guaranteed to exist.

Given an infinite set, it may not have a countably infinite subset.

You can't talk about arbitrary Cartesian products of sets.

Some vector spaces won't have bases.

About the randomA vs. randomB, I agree with ultrafilter's earlier answer.
#42
Old 08-29-2002, 10:40 PM
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Quote:
Originally posted by Cabbage

The probability for any particular number to be picked is zero. However, we know some number b will be picked, so here we have an example of an event with probability zero that actually does happen. Correspondingly, the probability that b will not be picked is one, and that didn't happen.
I'm going to jump in late (and probably over my head) but this is what's bothering me. You claim that the Lebesgue measure is a legitimate probibility distribution, and that P(something)=1 is not necessarily GTH. I don't understand how you claim that something gets picked. That is, if P(b)=0 for all b in [0,1], I could claim that no number ever gets picked at all. That would seem to violate the probability axiom P(S)=1, where S=[0,1], but there you have it.

That is, I think you're claiming that P(S)=1 means that something gets picked, but the probibility that that something gets picked is zero.

Seems like you could save a step by saying that P(b)=0 means nothing ever gets picked, therefore P(S)=1 doesn't mean that probibility =1 means GTH.

Now, getting to my problem with this - either you have to say that P()=1 means GTH, or P()=0 means impossible. You can't have it both ways, can you?
#43
Old 08-30-2002, 09:06 AM
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All right, let me try a slightly different, more concrete approach.

Say your cable goes out, so you call the cable company for a repair man. They tell you they'll send somebody out between noon and 1PM. You ask, "Could you be any more specific?", to which they reply, "I'm sorry, I can guarantee that he'll be there between noon and 1PM, but other than that, I really have no idea what time to expect him."

How can we model the probability of him arriving at a certain time? We know it'll be between noon and 1PM, but, other than that, no time is preferred over any other. So we can model this analogously to my previous example--picking a real number at random from [0,1] (corresponding to the number of hours after noon), only now we don't have to worry about issues of how we actually pick the number--the cable guy simply arrives when he arrives. As before, the probability of him arriving at any particular time is zero; still, he will arrive at some particular time.

Now I know that some may object to this model, possibly by saying "Time isn't continuous," or something to that effect. However, we use, for example, continuous functions in calculus all the time to model the real world; how can you really object to this model?

kellymccauley:
Quote:
That is, if P(b)=0 for all b in [0,1], I could claim that no number ever gets picked at all. That would seem to violate the probability axiom P(S)=1, where S=[0,1], but there you have it.
A probability function is required to be countably additive, but here the sample space is uncountable. It doesn't violate the probability axioms to have a probability function that's zero at every point, yet one on the whole sample space. (In other words, P(a single point)=0, while P(whole sample space)=1).
Quote:
That is, I think you're claiming that P(S)=1 means that something gets picked, but the probibility that that something gets picked is zero.
I'm claiming something gets picked, but not because P(S)=1. I'm claiming that something gets picked because that's given in the problem, just like the cable guy is guaranteed to show up between 12 and 1. The probability any particular thing is picked, but, taken as a whole, we know something must get picked.
Quote:
Seems like you could save a step by saying that P(b)=0 means nothing ever gets picked, therefore P(S)=1 doesn't mean that probibility =1 means GTH.
I don't follow this.
Quote:
Now, getting to my problem with this - either you have to say that P()=1 means GTH, or P()=0 means impossible. You can't have it both ways, can you?
Sure I can, why not? Think of it this way: Things with probability zero come in two categories: "impossible" and "damn near impossible". Things with probability one also come in two categories: "certain" and "damn near certain". I see no contradiction.
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