$$P(N=i) = {1\over{2^{i+1}}}$$

when $i$ is an integer with $0<= i < k$, $P(N=k) = 1/2^k$, and $P(N=i) = 0$ otherwise. I didn’t take care of the edge cases in my “proof” because I did not think it through carefully.

Pr(N = i) = 1/2^{i}

but my sum goes from i = 1 to k, so it all boils down to the same thing. I think we’ve beaten this thing to death. It’s nice to know that this works and it’s all very cute.

If $I$ and $J$ are independent random integers from the set $\{ 1, \dots, 2^k \}$, $M$ is the largest integer such that $2^M$ divides $I$, and $N$ is the largest integer such that $2^N$ divides $J$, then

P(N is less than M), P(N equals M), and P(N is greater than M) are all about 1/3.

“Proof”

It is fairly obvious that P(N is less than M) = P(N is greater than M). So now we compute $P(N=M)$.

$P(N=M)$

$\ \ = \sum_{i=0}^k P(N = i \ and \ M= i) $

$\ \ = \sum_{i=0}^k P(N = i) P(M= i)$

$\ \ = \sum_{i=0}^k 2^{-i-1} 2^{-i-1}$

$\ \ = \sum_{i=0}^k 4^{-i-1}$

$\ \ = {{1/4- 1/4^{k+2}}\over{1 – 1/4}}$

$\ \ \approx {{1/4}\over{1 – 1/4}}$

$\ \ = 1/3.$

To prove this statement, I suspect we require Carl's insights. I'd also note, there is something interesting and counter-intuitive happening here viz. the uniform distribution over the integers.

What seems to be interesting is that the factorization is twisting the uniform distribution in a cute way [in response to Vance's comment by e-mail that looking at the prime factorization causes us to look at an entirely different distribution.

OK, here's the little experiment in Maple:

with(RandomTools);

TwoPower := proc (N::integer)::integer;
local M, count;
M := N; count := 0;
while modp(M, 2) = 0 do
count := count+1;
M := (1/2)*M
end do;
count end proc;

Simulate := proc (N::integer)::list;
local eCount, oCount, nlCount, nhCount, neCount, tpn, tpd, i, num, deno, fra, dd;
eCount := 0;
oCount := 0;
nlCount := 0;
nhCount := 0;
neCount := 0;
for i to N do
num := RandomTools:-Generate(integer);
deno := RandomTools:-Generate(integer);
fra := num/deno; dd := denom(fra);
if modp(dd, 2) = 0 then
eCount := eCount+1
else oCount := oCount+1
end if;
tpn := TwoPower(num);
tpd := TwoPower(deno);
if tpn < tpd then
nlCount := nlCount+1
elif tpn = tpd then
neCount := neCount+1
else nhCount := nhCount+1
end if end do;
[eCount, oCount, nhCount, neCount, nlCount]
end proc;

evalf((1/10000)*Simulate(10000));
[.3335000000, .6665000000, .3331000000, .3334000000, .3335000000]

# The previous example illustrates that for the appropriate measures, fractions with
#even numerators show up 1/3 of the time and illustrates Hein's conjecture that 1/3 of the
#two integers are chosen with the 2's in their prime factorizations having equal powers.

#
# Proving that will likely require the post Carl made.

#In order of output, we have the proportion of times an even number is the denominator;
#the proportion of times an odd number is the denominator;
#the proportion of time the random integers have a higher power of two in the prime
# factorization on the (non-reduced) numerator; an equal power of two in the prime
# factorizations of the two integers and; a lower power of two in the prime factorization
# of the (non-reduced) denominator.

For example, each structure has it’s own natural “absolute values”. The natural absolute value for the integers considered as an additive group is the standard Archimedean absolute value: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$. There are many natural absolute values for the integers considered as a multiplicative group, however: $|x|_p = 1/p^{v_p(x)}$ for each prime number $p$, where $v_p(x)$ is the highest power of $p$ that divides $x$. ($v_3(9) = 2$ and $v_7(1/49) = -2$, for example.)

Before I continue, a few comments. First, the absolute values $|\cdot|_p$ are called "non-Archimedean" because they satisfy a stronger triangle inequality: $|x + y|_p \le \max(|x|_p, |y|_p)$. Second, the standard Archimedean absolute value is sometimes called the "absolute value at infinity" and denoted $|\cdot|_\infty$. Finally, with these conventions we have the identity (for any rational $x$):
$$\prod_p |x|_p = 1$$
where the product is over all "finite primes" and the "prime at infinity". Number theorists sometimes joke that real analysis is "number theory at infinity!"

Now, consider the harmonic series
$$\sum_{n \ge 1} {1 \over n} = 1 + {1 \over 2} + {1 \over 3} + \ldots$$
and the identity
$${1 \over 1 - x} = 1 + x + x^2 + \ldots$$
Setting $x = 1/p$ this identity becomes
$${1 \over 1 - 1/p} = 1 + {1 \over p} + {1 \over p^2} + \ldots$$
and then by the Fundamental Theorem of Arithmetic we have
$$\sum_{n \ge 1} {1 \over n} = \prod_{p {\text { prime}}} {1 \over 1 - 1/p}$$
as formal expressions.

To make the expressions converge we can do one of two things. First, we can add an exponent to the denominators, in which case we have
$$\sum_{n \ge 1} {1 \over n^s} = \prod_{p {\text { prime}}} {1 \over 1 - 1/p^s}$$
and both sides converge and are analytic for $\Re s > 1$. This, of course, is the Riemann zeta function $\zeta(s)$, and the product on the right-hand side is why it’s important.

The other way to make the expressions converge is to limit the number of terms (or factors):
$$\sum_{n < N} {1 \over n}$$
and
$$\prod_{p < N} {1 \over 1 – 1/p}$$

It is easy to see that $\sum_{n < N} {1 \over n} \sim \log N$; what about $\prod_{p < N} {1 \over 1 – 1/p}$? Obviously the two expressions aren’t equal, but do we have
$$\sum_{n < N} {1 \over n} \sim \prod_{p < N} {1 \over 1 – 1/p}$$
as $N \to \infty$? The answer turns out to be ‘no’ (although they are of the same order as $N \to \infty$), but the point I’m trying to get to is that the sum is limiting the size of the denominators of the fractions additively, while the product is limiting the size of the denominators (in the multiplied-out sum) multiplicatively, and therefore different fractions are included in each sum.

In essence, we have defined two different measures to measure the size of fractions. And likewise two different distributions for “random” fractions.

Now the punch line: given the first distribution, fractions with even denominators show up $1/3$ of the time; given the second, even denominators show up $1/2$ the time. But since both distributions are natural, and since they are similar to each other, it is easy to miss that they are actually different!

pCount := 0;
qCount := 0;
N := 100000;
for i to N do
v := Generate(integer(range = 0 .. 10000));
u := Generate(integer(range = 0 .. 10000));
if v > u then
pCount := pCount+1
else
qCount := qCount+1
end if
end do;
unassign(‘i’, ‘v’, ‘u’);

evalf(qCount/N);
0.4983500000
evalf(pCount/N);
0.5016500000

Maybe I’m missing something deeper that everyone else is saying.