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?

[THECORBYLOON]

Regarding the random numbers, if the number 1 is to be matched, the probability of the number 1 being selected or generated from an infinite amount of numbers is virtually zero.

The previously generated numbers could be repeated but even if only unique numbers were allowed, there is no end to real numbers.

No matter how many have been selected previously, there is an infinite amount of real numbers remaining.

[Rev. Green]

If you pick a number in the range 0 to 99, it has two digits and I'll have to guess fifty numbers, on average, to find your number. If you pick a number with n digits, I'll need half of 10 to the power n guesses, on average, to guess it. That's many more guesses, but not so many more to make it a different order of infinity more. Now pick a number with a countable infinity of digits. I'll need a huge number of guesses to find it, but the number of guesses will still be a countable infinity. So the chances of choosing one certain number with an infinite number of guesses is the ratio of two countable infinities. The chance is pretty close to zero, so "very, very roughly evens" exaggerated the odds, but I was trying to emphasise that the odds were finite, not zero. If the number of digits and the number of necessary comparisons had been different orders of infinity, then the odds would have been zero.

[Rev. Green]

My post is wrong. I thought I could map all numbers onto the integers by taking alternate digits from either side of the decimal point. So 5 would map to five. 5.6 would map to sixty five. 12.34 would map to 4132, and 123.4567.would map to 70615243. This can't work for real numbers, though I can't see why it doesn't.