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?

[Peter Pedant]

er I cant believe the answer is zero =nor a finite number such as 42 - soI tend toward infinity as an average

it of course depends on how you generate ( or which algorithm you use ) the infinite humbers and whether they are are countable or uncountable

Taking a set of odd numbers and an infinite one of evens you have got zero

and so if you have an infinite number of tries
with different algorithms...
then I suggest there are an infinite number of zero s and an infinite number of (infinite number of overlays)

[jim360]

There are, in fact, as many even numbers as there are whole numbers. As many primes as there are cubes and squares. They are all countable and the same size as each other, and they are all strictly smaller than the size of the real numbers.

[jim360]

Similarly there are as many rational numbers as there are whole numbers.

[Atheist]

I really don't think that AB is the right forum for this sort of mathematics. Best stick to 'is god real?'

[tomus42]

//
// 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.
//

It is though. If you haven't tossed a tail yet, it isn't because it won't happen. It's because you haven't been going long enough.

[Peter Pedant]

// There are, in fact, as many even numbers as there are whole numbers. //

er yeah and countable - I intended
I just wanted to show that it was possible depending on how you to construct a line that never has another point overlying it
if you feel an even humber line isnt random then pick each elet randomly

[Peter Pedant]

// I really don't think that AB is the right forum for this sort of mathematics. Best stick to 'is god real?'//

no I think the best threads are where the usual 100 ABers demand that the police go round and kneel on the necks of all those AWFUL bristol fun seekers !

[Rev. Green]

I think the problem is similar to: chose any number, say pi, then how many of an infinity of random numbers will be the same as pi? The number of digits in pi is commensurate with the number of test numbers, so the answer is close to one.

[jim360]

// If you haven't tossed a tail yet, it isn't because it won't happen. It's because you haven't been going long enough. //

If you go infinitely long it still doesn't "have" to happen at some point. It's mistaken to assume that something will happen if you carry on long enough. That's not how probability works. When you get to infinity, an individual event with probability 1 doesn't necessarily have to happen, and an individual event with probability 0 still could happen.

[fiction-factory]

> and an individual event with probability 0 still could happen

So if I roll a fair six sided die an infinite times could I eventually throw a 7?

I must confess that I'm still not fully understanding the original question or the pi one.

[jim360]

"So if I roll a fair six sided die an infinite times could I eventually throw a 7?"

No. But if you roll a fair six-sided die infinitely many times you may never get a six. An event still has to be in the space of possibilities in order to happen, and an event outside the space can never happen -- but in between these two, there are events which have a probability zero without being "impossible".

[jim360]

I confess I can't parse the question about pi either.

[royfromaus]

Me neither, it sounded bonkers to me. What is the 'average' bit all about?

[tomus42]

I get the probability thing Jim, but I think you're underestimating the size of infinity. The point of it is that until the possible event happens, you've got to keep going until it does. If you ever get to the end of infinity, (which obviously you can't) and it still hasn't happened, then it wasn't possible in the first place.

[jim360]

You've already explained the flaw in your own logic. You can't get to the end of infinity, so you can't reach the conclusion that an event was impossible "after" finishing.

[jim360]

I mean, it's true that if you kept tossing a coin and never got a head then you'd question the assumption that it was a fair coin. But the fact remains that the event TTTTT... ad infinitum is possible, albeit with probability zero.

[Rev. Green]

What I meant was: Pick a number. It can be any number. Let's assume you've chosen 3.14159265358979....
Now start picking numbers at random. Perhaps the first is 4.6692016025... This isn't the same as your chosen number, so don't count it. Pick another. Keep picking random numbers and comparing then with your chosen number. Count those that match exactly. After you have picked an infinite number of random numbers, you have picked approximately as many numbers as there are digits in pi, or whatever your chosen number was. The number of numbers you have chosen, and the number of digits of pi, are both infinite, but they are similar sizes of infinity - what is known as Aleph-null or a countable infinity. Because you had the same number of comparisons as the number of digits in your chosen number, you should very, very roughly have an even chance of matching that number exactly with one of the random numbers.

[jim360]

I can follow that more or less up until the last step, which seems to just drop "even chance" out of nowhere.

[Peter Pedant]

// I think the problem is similar to: chose any number, pi, //

er pi isnt any number - transcendental for a start
and you cant enumerate it ( finitely - like you can 0.5)

and how are you going to select your second random - it is via an algorithm ( as they are - pseudo random) then it is in the set of countable numbers and Pi is in the uncountables

so no - I dont think it is the same at all

[Peter Pedant]

by picking a random you are selecting a countable set
and so missing out on a few than might fit ....