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?

[fiction-factory]

I'd say an infinite number

[mushroom25]

//Is the answer 42?//

sadly we'll never know. the program that was set to deliver the question was corrupted by the arrival of the Golgafrinchans…..

[dannyk13]

Yes, according to this:-
https://www.independent.co.uk/life-style/history/42-the-answer-to-life-the-universe-and-everything-2205734.html

[Old_Geezer]

And an infinite number missed.

If they are points I'm not sure how a member of one set can coincide with more than one member of the other; unless one was allowed to place two points of the same set at the same point on the line.

[jim360]

My gut feeling is that the answer should be zero but I am sure there's a better way to answer this than relying on points/line = countable/uncountable -> 0

[Prudie]

I'd also suggest an infinite number but am at 6s and 7s trying to prove it :-)

[Old_Geezer]

Yeah, but what deep insight was it that caused Douglas' subconscious to send up the right answer to his consciousness ?

[weecalf]

I’d go for none as the infinitely of line A would be random Need I go on

[Prudie]

It would also depend if they are placed at the same time or set A first. Also if a point can only used once then I see the zero answer.

[fiction-factory]

Yes, I don't get the "or more" bit. Or the "on average". Lunch beckons so will think later

[THECORBYLOON]

Would the extremes not be they all coincide or none coincide? If so, is there not a probability of 0.5 they will coincide?

I'm not good with probabilities so my thinking might be wide of the mark.

[jim360]

It's rarely so neat as all that, especially when you have to deal with the issue of countable points on an uncountable line. An example of the paradoxes involved would be to consider the following problem: take a perfect circle of radius 1 metre, start at the top, and then keep walking around its circumference in steps of exactly 1 metre. No matter how many steps you take you'll never run out of new places to visit (assuming your shoe size is 0), or to put it another way you'll never be standing in exactly the same place twice.

Depending on how you choose points A and B in this case, there is a halfway decent change that you'll never pick the same point twice -- indeed, it's virtually certain. So the probability that no points overlap is, at the very least, far greater than the probability that they all overlap.

[Zacs-Master]

Uncountable points, surely?

[jim360]

Well, possibly, but if you are "placing" the points then it sort of feels to me like you'd do that in a sequence, which makes them countable. I may have misunderstood and, in any case, I don't know the answer.

[woofgang]

an infinite number of course

[Zacs-Master]

Isn the ‘logical’ answer 50%? It’s like asking ‘how long is a piece of string’. The only logical answer is ‘twice as long as half it’s length’. It’s not conclusive but it’s the best fit.

[JJ109]

If there is a probability that just one point overlaps, surely in an infinite line the number has to be infinity. Even if the probability is one in a trillion, in an infinite line, infinity will overlap.

[Cloverjo]

I posted this to my clever son and he says there’s not enough information to answer it properly.

[jim360]

Thanks, clover. Ask if there's enough info if you assume uniform distributions.

I can't accept that 50% is even remotely the most logical answer, but I'm open to being persuaded otherwise by a coherent mathematical argument.

[Cloverjo]

My son:

Could analyze a finite line and then tend to infinity