Counterpoint:
Pick a random number between 0 and Infinity.
What set of infinite numbers are you more likely to wind up with?
Do this a million times.
Therefore 1 to Infinity is significantly larger than 0 to 1
What set of infinite numbers are you more likely to wind up with?
Do this a million times.
Therefore 1 to Infinity is significantly larger than 0 to 1
Here's one hand-wavy proof of why the cardinality of the real interval P=[0,1] is the same as the cardinality of the real interval Q=[0, infinity]: The function f(x) = 1/x-1 is a bijective function that maps the P interval onto the Q interval, which also proves the cardinality of the two sets is equal. (https://en.wikipedia.org/wiki/Bijection#Cardinality).
If you're not comfortable with 1/0 = infinity as a general matter, then simply replace the f(x) I gave with an explicit piecewise function f(x) = { 0 if x = ∞, else (1/x-1) } and the proof still works.
To judge what set of numbers you're more likely to end up with, you need to specify a probability distribution. Without any specific information to prefer one number over another, you want the highest entropy distribution. It turns out that the most "natural" probability distribution for x extending between 0 and infinity is not uniform over x, but over its logarithm; therefore, by symmetry arguments, the probability is actually more like 50% that you draw a number between 0 and 1.
And it turns out that floating point numbers more or less respect this property.
You sound like you want a uniform distribution (i.e., P(x in [a, b]) = b - a / total support of D), but when you have an infinite support, the denominator is infinity - 0 = infinity, so the probability that x is in any finite interval in that set is 0. I've never taken real analysis, so my knowledge here is quite shaky, but I'm not even certain that the resulting probability distribution function is well-defined as a probability distribution function in the first place.
Real analysis, aka, real numbers (and consequently infinite sets) are far weirder than you ever expected them to be.
Even ignoring the realities of floating point, your argument depends on sampling uniformly from an infinite interval, which is not possible: https://math.stackexchange.com/questions/14777/why-isnt-ther...
Consider the tan function. When you give it a number between 0 and pi/2, it gives you a number between 0 and infinity, and it does so in a bijective way. Therefore there are equal numbers between 0 and pi/2 as compared to 0 to infinity. Now consider a simple linear function that multiplies its input by pi/2. From here we know that there are equal numbers between 0 and 1 as compared to 0 and pi/2.
You may select any real number between zero and infinity. We will call this R.
I will give you 1/R, which is between zero and one.
QED.
I believe you're conflating range, where it is trivially true that [0,∞] is of greater extent than (0,1), with quantity, where it is not the case that there exists a greater quantity of values in the former range than in the latter.
Should that have been “between one and infinity”? Otherwise you cannot claim that 1/R is between zero and one.
I would prefer, given the domain, to amend to "given an R between 0 and infinity, I will return R for all R < 1, or 1/R otherwise". But yes, the proof was flawed.