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Old 29th July 2018, 11:34 PM   #1
Fudbucker
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Infinite Sets and Probability

Suppose I awake in an unknown hotel room. The only clue as to my whereabouts is the fact that I know I'm in one of two sets of infinite hotel rooms: the Ritz set or the Motel 6 set. In the Motel 6 set, for every opulent room, there are a billion dingy ones. In the Ritz set, it's the opposite. The room I wake up in is opulent. Does this allow me to conclude I'm probably in the Ritz set, or does the fact there are just as many dingy rooms as opulent rooms (since there are an infinite amount of both types of rooms in both sets) make a probability calculus impossible?

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Old 30th July 2018, 12:21 AM   #2
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I think this proves at you are immortal.
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Old 30th July 2018, 12:45 AM   #3
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Seriously though, I'm pretty sure we can conclude you are probably in the Ritz. Because although there are an infinite number of both types of room, their ratio is still a billion to one.
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Old 30th July 2018, 01:20 AM   #4
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This is a pretty simple case of Bayes' Theorem. I haven't bothered to do the maths but it's pretty clear that, provided that you're a priori equally likely to have ended up in either hotel, then there's about a billion to one probability you're in the Ritz.

If, on the other hand, you went out last night with ten billion drinking buddies, at some point in the evening youphoned the Ritz and they said they'd reserve one room for the first one of you to get there and all the others would have to go to the Motel 6, and you then all got blind drunk and you can't remember anything more than a vague memory of going to a hotel afterwards, then I'd say there's about a ten-to-one chance you're in the Motel 6.

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Old 30th July 2018, 01:34 AM   #5
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The probability that you are in the Ritz is either 1 or 0, you just don't know which.
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Old 30th July 2018, 05:45 AM   #6
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In neither. It's well-known the only hotel in the world with an infinite amount of rooms is Hilbert's Hotel.

ETA: and what MetalPig and Dave Rogers said..
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Old 30th July 2018, 06:37 AM   #7
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Well no, if you are sitting in a hotel room and it happens to be a room at Motel 6 then the probability that you are in the Ritz is 0.

But if you are sitting in a hotel room and it happens to be a room at the Ritz, then the probability that you are in the Ritz is 1.
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Old 30th July 2018, 06:44 AM   #8
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I guess my question is the same as Fud's: Does the infinite set affect the probability calculation at all?
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Old 30th July 2018, 06:56 AM   #9
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Originally Posted by Robin View Post
Well no, if you are sitting in a hotel room and it happens to be a room at Motel 6 then the probability that you are in the Ritz is 0.

But if you are sitting in a hotel room and it happens to be a room at the Ritz, then the probability that you are in the Ritz is 1.
Since in either of these instances the outcome is completely known, probability theory is therefore inapplicable. Probability theory is applicable when we have incomplete information; for example, if you're sat in a hotel room but you don't know which hotel you're in. What's actually being discussed in this example is the relative probabilities of two future events, one and only one of which will occur when you find out which hotel you're in; one is that you find out you're in the Ritz, and the other is that you find out you're in Motel 6.

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Tony Szamboti: That is right
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Old 30th July 2018, 06:58 AM   #10
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Originally Posted by theprestige View Post
I guess my question is the same as Fud's: Does the infinite set affect the probability calculation at all?
Suppose you're about to look out of the window, and you know that it'll be dark roughly half of the time and light roughly the other half. Does the fact that there's an infinite number of possible instants that you could look out of the window affect the calculation that the probability of it being light is about 0.5? Any probability calculation over a continuous interval is in effect using an infinite set.

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Tony Szamboti: That is right
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Old 30th July 2018, 07:00 AM   #11
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Originally Posted by Dave Rogers View Post
Since in either of these instances the outcome is completely known, probability theory is therefore inapplicable. Probability theory is applicable when we have incomplete information; for example, if you're sat in a hotel room but you don't know which hotel you're in. What's actually being discussed in this example is the relative probabilities of two future events, one and only one of which will occur when you find out which hotel you're in; one is that you find out you're in the Ritz, and the other is that you find out you're in Motel 6.

Dave
It doesn't matter if it is known or not. The probability of being somewhere else than you are is 0, even if you don't know where you are.

And the OP states: "Suppose I awake in an unknown hotel room" so it is not talking about two future events it is talking about present location.

So if you awake in an unknown hotel room and it happens to be in Motel 6 then the probability that you are in the Ritz is 0.

If you awake in an unknown hotel room and it happens to be in the Ritz, then the probability that you are in the Ritz is 1.
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Old 30th July 2018, 07:13 AM   #12
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I mean assuming the room is not going to randomly change hotels between you waking up and finding out the name of the hotel you are in.
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Old 30th July 2018, 07:25 AM   #13
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Originally Posted by Robin View Post
It doesn't matter if it is known or not. The probability of being somewhere else than you are is 0, even if you don't know where you are.

And the OP states: "Suppose I awake in an unknown hotel room" so it is not talking about two future events it is talking about present location.

So if you awake in an unknown hotel room and it happens to be in Motel 6 then the probability that you are in the Ritz is 0.

If you awake in an unknown hotel room and it happens to be in the Ritz, then the probability that you are in the Ritz is 1.
The answer may be predetermined, but it is unknown, and the probabilities still exist. There is a billion to one chance that you'll *learn* you're in one hotel vs the other.
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Old 30th July 2018, 07:34 AM   #14
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I don't know if I am right mathematically, but if all I had to go on is that one piece of information (opulent room) and I had to stake my life on one choice or the other then I would pick the Ritz.
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Old 30th July 2018, 07:36 AM   #15
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If the unknown room you happen to wake up in is in Motel 6 then there is a 0 probability that you will learn that you are in the Ritz (if you learn which hotel you are in at all).

If the unknown room you happen to wake up in is in the Ritz then there is a probability of 1 that you will learn that you are in the Ritz (again, if you learn which hotel you are in at all).
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Old 30th July 2018, 07:36 AM   #16
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Originally Posted by Robin View Post
It doesn't matter if it is known or not. The probability of being somewhere else than you are is 0, even if you don't know where you are.
Suppose someone you know to be honest flips a fair coin and covers it with their hand, then says they'll agree to pay you $2000 if it's heads provided you agree to pay them $1 if it's tails. Do you reason that the odds are overwhelmingly in your favour and take the bet, or do you say that the probability of it being one of the two is already 1 and of the other is already 0 you can't tell which it is, so you don't know what to do?

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Old 30th July 2018, 07:38 AM   #17
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By the way, how are you doing your calculations?

There are the same number of dingy rooms in both hotels.

There are the same number of opulent rooms in both hotels.

There is the same ratio of opulent to dingy rooms in both hotels.

So why aren't you getting 50/50 Ritz or Motel 6?
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Old 30th July 2018, 07:38 AM   #18
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Originally Posted by Dave Rogers View Post
Suppose you're about to look out of the window, and you know that it'll be dark roughly half of the time and light roughly the other half.
Since that one is already 50/50 and his question is about whether or not an having countably infinite instances of two options causes everything to become 50/50 it probably won't help him.

Originally Posted by Robin View Post
It doesn't matter if it is known or not. The probability of being somewhere else than you are is 0, even if you don't know where you are.
Congratulations on being technically correct and also not making any attempt to be constructive in any way.

Originally Posted by theprestige View Post
I guess my question is the same as Fud's: Does the infinite set affect the probability calculation at all?
I think the main problem is it makes it really hard to word it in a valid way. Maybe impossible. We can't really talk about picking a room at random because we can't calculate probability with infinity. Narrowing it down to two hotels might in theory make it possible, but I'm not really certain.

Probability and infinity don't play well together.
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Old 30th July 2018, 07:39 AM   #19
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Some fundamental misunderstanding of probability theory going on here I think.
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Old 30th July 2018, 07:41 AM   #20
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Originally Posted by Dave Rogers View Post
Suppose someone you know to be honest flips a fair coin and covers it with their hand, then says they'll agree to pay you $2000 if it's heads provided you agree to pay them $1 if it's tails. Do you reason that the odds are overwhelmingly in your favour and take the bet, or do you say that the probability of it being one of the two is already 1 and of the other is already 0 you can't tell which it is, so you don't know what to do?

Dave
What does this have to do with the original question?
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Old 30th July 2018, 07:42 AM   #21
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Originally Posted by Robin View Post
There are the same number of dingy rooms in both hotels.

There are the same number of opulent rooms in both hotels.

There is the same ratio of opulent to dingy rooms in both hotels.

So why aren't you getting 50/50 Ritz or Motel 6?
Because you've stated the problem incorrectly. The OP clearly states that the ratio of dingy to opulent rooms is the opposite in the two hotels - one billion to one in the Motel 6, one to one billion in the Ritz.

Dave
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Old 30th July 2018, 07:48 AM   #22
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Originally Posted by Robin View Post
What does this have to do with the original question?
It has to do with your rejection of a posteriori probability. If you don't want to discuss that, then please feel free to ignore it, but it might then be polite if you stopped replying to a question about a well-known aspect of probability theory by saying you don't believe in the underlying theory. Technically, it's off-topic to the thread.

Dave
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Tony Szamboti: That is right
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Old 30th July 2018, 07:49 AM   #23
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Originally Posted by SOdhner View Post
I think the main problem is it makes it really hard to word it in a valid way. Maybe impossible. We can't really talk about picking a room at random because we can't calculate probability with infinity.
Yes, we can. There is an infinite number of positive integers. What is the probability that a randomly chosen positive integer will be divisible by ten?

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Tony Szamboti: That is right
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Old 30th July 2018, 07:50 AM   #24
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Originally Posted by SOdhner
Congratulations on being technically correct and also not making any attempt to be constructive in any way.
It really depends on what point was supposed to emerge from this.

As I said there are the same number of dingy rooms in both hotels, same number of opulent rooms in both hotels, same ratio of dingy to opulent rooms in both hotels - so I am interested in how is the calculation is being done.
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Old 30th July 2018, 07:53 AM   #25
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You're in the Ritz, having vastly over-estimated the number of oppulent rooms in Motel 6.
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Old 30th July 2018, 07:57 AM   #26
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Originally Posted by Dave Rogers View Post
Yes, we can. There is an infinite number of positive integers. What is the probability that a randomly chosen positive integer will be divisible by ten?

Dave
Well there are the same number divisible by ten as there are not divisible by ten, so I guess 0.5.

Again, how is the calculation being made?
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Old 30th July 2018, 07:57 AM   #27
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Originally Posted by Robin View Post
By the way, how are you doing your calculations?

There are the same number of dingy rooms in both hotels.

There are the same number of opulent rooms in both hotels.
Yes, to be precise: those numbers are Aleph-naught.

Originally Posted by Robin View Post
There is the same ratio of opulent to dingy rooms in both hotels.
No, they're reverse.

Originally Posted by Robin View Post
So why aren't you getting 50/50 Ritz or Motel 6?
Because calculations with infinites don't work the same way as with finite numbers.

Maybe this helps.

Both the Ritz and Motel-6 have their rooms numbered 1, 2, 3, etc., ad infinitum, using all the natural numbers. In the Ritz, all rooms of which the number ends with 9 zeros are dingy, the other rooms opulent. In Motel-6, all rooms of which the number ends with 9 zeros are opulent, the rest dingy. (*) The room numbers are only displayed on the outside; there's nothing inside the room that tells you its number.

(1) As soon as you wake up, someone phones you and tells your room number is lower/equal to 10^12. What is the probability now that you are in the Ritz?

(2) As soon as you wake up, someone phones you and tells your room number is lower/equal to 10^15. What is the probability now that you are in the Ritz?

(3) Repeat, with 10^18.

(4) Repeat, with 10^21.

Et cetera.

(*) In the interest of giving a simple description, I've slightly changed the ratios from 1 billion to 1 to 999,999,999 to 1, but that shouldn't distract from the point.
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Old 30th July 2018, 07:58 AM   #28
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Originally Posted by Robin View Post
It really depends on what point was supposed to emerge from this.

As I said there are the same number of dingy rooms in both hotels, same number of opulent rooms in both hotels, same ratio of dingy to opulent rooms in both hotels - so I am interested in how is the calculation is being done.
No, there aren't the same number. Infinity is not a number.

The sets being infinite doesn't matter. We're given the information that there is a fixed ratio of good to bad rooms in both hotels. It is therefore a simple probability of being in one type of room or the other, and given a random room of either type there is a simple probability of being in one hotel or the other.
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Old 30th July 2018, 07:59 AM   #29
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Originally Posted by Dave Rogers View Post
Because you've stated the problem incorrectly. The OP clearly states that the ratio of dingy to opulent rooms is the opposite in the two hotels - one billion to one in the Motel 6, one to one billion in the Ritz.

Dave
No, there are infinitely many rooms.

Motel 6 has infinitely many dingy rooms and infinitely many opulent rooms.

The Ritz has infinitely many dingy rooms and infinitely many opulent rooms.
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Old 30th July 2018, 08:00 AM   #30
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Originally Posted by Dave Rogers View Post
Yes, we can. There is an infinite number of positive integers. What is the probability that a randomly chosen positive integer will be divisible by ten?
That raises a question in my mind. Can we truly choose randomly from an infinite set? Any positive integer we could actually pick would have to be from the lower part of that ordered set.

In other words, no matter which number we pick there would always be more numbers in the set higher that the selected number than there are ones lower than the selected number. Seems biased to me.

Maybe?

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Old 30th July 2018, 08:02 AM   #31
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Robin, by your logic, picking a random number out of all possible integers has a 50/50 chance of it being divisible by 9342193475892734. Because there are an infinite number of integers divisible by that number, and an infinite number not divisible...

Does that seem right to you?
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Old 30th July 2018, 08:03 AM   #32
Robin
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Originally Posted by phunk View Post
No, there aren't the same number. Infinity is not a number.
Put it this way then.

There are as many opulent rooms in Motel 6 as there are dingy rooms.

There are as many opulent rooms in the Ritz as there are dingy rooms.
Quote:
The sets being infinite doesn't matter. We're given the information that there is a fixed ratio of good to bad rooms in both hotels. It is therefore a simple probability of being in one type of room or the other, and given a random room of either type there is a simple probability of being in one hotel or the other.
Are you saying there are more dingy rooms than opulent rooms in Motel 6?

Or that there are more opulent rooms then dingy rooms in the Ritz
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Old 30th July 2018, 08:03 AM   #33
SOdhner
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Originally Posted by Dave Rogers View Post
Yes, we can. There is an infinite number of positive integers. What is the probability that a randomly chosen positive integer will be divisible by ten?
Well sure, but... um. Hang on I got my foggily-remembered stuff mixed up. I was thinking about calculating probability of picking a specific number or item rather than a category (divisible by ten), which... is that still...

Anyway regardless of exactly how wrong I am (anywhere from 50% to 100%) I really like your example. So, to map it on to the example in the OP:

There are infinite rooms in two hotels in town (and also the Hilbert but it's full and while they insist they can accommodate more we don't want to inconvenience an infinite amount of guests). Every billionth room at the Ritz is shabby, and every billionth room at the Sunshine Suites is fancy. Since they have room numbers it's easy to talk about them using just numbers, so just like you can ask about the odds of something being divisible by ten we can ask if it's divisible by a billion.

So the odds of a randomly chosen room being an outlier at each hotel WOULD still be different, so it WOULD be something to base your guess off of. Cool.

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Old 30th July 2018, 08:05 AM   #34
Dave Rogers
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Originally Posted by Robin View Post
Well there are the same number divisible by ten as there are not divisible by ten, so I guess 0.5.

Again, how is the calculation being made?
Seriously, do you think the probability of picking a random integer divisible by ten is equal to the probability of picking one not divisible by ten, and do you think there's more than one way of calculating those probabilities?

Dave
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Old 30th July 2018, 08:06 AM   #35
Robin
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Originally Posted by phunk View Post
Robin, by your logic, picking a random number out of all possible integers has a 50/50 chance of it being divisible by 9342193475892734. Because there are an infinite number of integers divisible by that number, and an infinite number not divisible...

Does that seem right to you?
I am not sure how you are even proposing to select numbers randomly from an infinite set.

If you picked 100 numbers randomly from the set of natural numbers, what would be the average of them all?

But again, show me how you are deriving this probability? Do you suggest that there are more numbers not divisible by 9342193475892734 than there are numbers divisible by 9342193475892734?
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Old 30th July 2018, 08:10 AM   #36
Robin
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Originally Posted by Dave Rogers View Post
Seriously, do you think the probability of picking a random integer divisible by ten is equal to the probability of picking one not divisible by ten, and do you think there's more than one way of calculating those probabilities?

Dave
That is what I am asking, show me your calculation.

Are there more numbers not divisible by ten than numbers divisible by ten?
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Old 30th July 2018, 08:13 AM   #37
SOdhner
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Originally Posted by Robin View Post
Are there more numbers not divisible by ten than numbers divisible by ten?
There are nine times as many.
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Old 30th July 2018, 08:14 AM   #38
Robin
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Originally Posted by SOdhner View Post
There are nine times as many.
No there aren't.
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Old 30th July 2018, 08:15 AM   #39
SOdhner
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Originally Posted by Robin View Post
No there aren't.
Yeah huh.
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Old 30th July 2018, 08:21 AM   #40
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Originally Posted by Robin View Post
I am not sure how you are even proposing to select numbers randomly from an infinite set.

If you picked 100 numbers randomly from the set of natural numbers, what would be the average of them all?

But again, show me how you are deriving this probability? Do you suggest that there are more numbers not divisible by 9342193475892734 than there are numbers divisible by 9342193475892734?
You seem to be relying on the fact that, with an infinite set, there are infinite occurrences of any given number or 'type' of number. While this is true, it doesn't mean the ratio of certain types of number (divisible/not divisible by 9342193475892734 in this case) is the same. Probability calculations depend on that ratio, not the absolute numbers.
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