Is The Answer 42?

[Rev. Green]

An infinite number of points A are placed at random on an infinitely long line. A second infinity of points B are also placed at random on the line. How many B's coincide with one or more A's on average?

[Old_Geezer]

The way I see it, there are infinite numbers of 'A' points and of 'B' points so there's an infinite number of chances where they'll coincide on the line, and an infinite number of chances where they won't. The idea that on an infinite line there are zero places they coincidence suggests a diety deliberately ensuring that they never do.

[fiction-factory]

Hi O-G. I think you are using the second interpretation of the question which is where we just need some point on B to match any of the points on line A.
If the question is interpreted as requiring A1 to match B1 or A2 to match B2, etc then might the answer be different? I can see a case for zero still (in the sense of being a limit- an infinitely small number- just as 1/x tends to zero but never actually reaches 0). Even in an infinitely small range between two points there is an infinite number of points between them.
My head hurts these days trying to think about things like this.
I probably did proofs like this and probably knew once what the meaning was of an aleph to which Rev Green referred.

[jim360]

I don't think it makes that much difference if A1 can be matched by only B1 or by any of B1, B2, etc. Having countably infinitely many chances to hit isn't enough if you have an uncountably infinite target.

I'm not satisfied by the "0" answer, though. I feel there's got to be a way, that's beyond me at the moment, to make the calculation more rigorous.

[Zacs-Master]

‘ there's an infinite number of chances where they'll coincide on the line, and an infinite number of chances where they won't’

Which is exactly why I came up with the 50% answer, as it’s the only reasonable prediction in the absence of a concrete answer.

[fiction-factory]

I'm trying to break this down. If the lines were not infinitely long- say they were 1 metre long -and you put an infinite number of points on each line, would the answer be zero or something else?

[jim360]

You can fit countable infinities infinitely often into any uncountable infinity. In that sense I'd say that a one-metre line v. an infinitely long line wouldn't make a difference.

There's no way that 50% is the most sensible prior just because "it either will or it won't".

[Zacs-Master]

I don’t see a better one.

[THECORBYLOON]

There's no way that 50% is the most sensible prior just because "it either will or it won't".

I suggested a 0.5 probability just as ZACS said 50%.

You suggested it was zero but if there's an infinite number of A and B points, are you saying the points would never overlap?

Is that more reasonable then 50/50?

[jim360]

That's a fair point. It's worth clarifying that probability zero isn't the same, in continuous probability, as something never happening. Eg the probability of tossing a coin and never getting a tail no matter how often you try is zero (= (1/2)^n as n tends to infinity), but it clearly "could" happen in principle.

The point here, then, is that uncountable infinity is just so vast that you will never exhaust the number of available points if you are choosing only countably many. It's counter-intuitive, I'll admit, but there we are.

[Old_Geezer]

Ok, but we aren't seeking a percentage, we're seeking the number of coincidences.

[jim360]

Yes, there's something missing in my full understanding of the problem in order to work out what the answer is -- but that doesn't stop anybody from commenting on what it isn't.

[royfromaus]

Any more good guesses yet?

[Old_Geezer]

Would it matter whether the instinctive version of the question is considered, or the more detailed one ? It's still infinite, plenty of occasions in which to get an infinite sized subset of coincidences. The percentage, if calculatable, may change, probably approach (but maybe not hit ?) zero, but the number of hits will still be infinite.

[tomus42]

Zero doesn't make sense to me. Infinity dictates that the coincidence of points must inevitably occur. Anything that can happen, will happen if you wait long enough.

[Old_Geezer]

Yes but infinities come in different sizes. If you have an infinite number of occurrences in an infinitely larger infinity set, then your infinite number of occurrences amount to 0%.

[jim360]

// Infinity dictates that the coincidence of points must inevitably occur. //

Except that's not how infinity works, because that's the entire point I am making. There are (at least) two types of infinity, and one of them is literally bigger than the other.

// Anything that can happen, will happen if you wait long enough. //

Also no, for the same reasons. A fair coin toss isn't bound to turn up a tail eventually.

[fiction-factory]

But if it happens once couldn't it happen again.... and again... if we repeat the process ad infinitum.
Having said that I'm still attracted to an answer of zero on average )on the basis that even a large number divided by a number approaching infinity will tend to zero

[royfromaus]

If the answer is zero, then the monkeys have no chance of knocking out the complete works of Shakespeare.

[fiction-factory]

Too slow- my response was to Tomus

[fiction-factory]

I think there's an infinite number of infinities. There's an infinite number of odd numbers, and an infinite number of even numbers. There's an even bigger infinite number of whole numbers. And an even bigger infinite number of non-integer decimals/fractions. An there's infinitely small = eg 1/n as n tends to infinity.