I have seen Collatz conjecture at http://en.wikipedia.org/wiki/Collatz_conjecture . I present my probabilistic approach: Let $x$ denotes the random natural number. Let denote by $P(x)$ the probability that Collatz sequence starting at $x$ will return at 1. If $x$ is divided to $2$ then $x$ is sending to $x/2$ with probability $1/2$ and now the probability that Collatz sequence starting at $x/2$ will return at 1 will be equal to $P(x/2)$ or
If $x$ is not divided to $2$ then $x$ is sending to $3x+1$ with probability $1/2$ and now the probability that Collatz sequence starting at $3x+1$ will return at 1 will be equal to $P(3x+1)$.
Using total probability formula we get the following functional equation:$P(x)=1/2P(x/2)+1/2P(3x+1)$. Notice that each constant function satisfies this equation, i.e. $P(x)=c $ for each natural number $x$. Taking into account that $P(2)=1$ we deduce $c=1$.
Since the probability that Collatz sequence starting at an arbitrary natural number will return to 1 is equal to 1 and there exists only one Collatz sequence starting at each natural number I deduce that Collatz Conjecture is valid.
Question. Where is mistake in this proof?
P.S. By probability measure constructed in [ Martin Sleziak, and Miloš Ziman. Lévy group and density measures, J. Number Theory, 128 (12), (2008), 3005–3012. MR2464850 (2009j:11019)] can be obtained equation mentioned above.
"If $x$ divided to 2 then $x$ is sending to $x/2$ with probability 1/2." This statement does not capture the Collatz conjecture at all: if $x$ is even, it is always divided by 2. Whereas you're saying that with probability 1/2 it is divided by 2.
It looks like what you've written is: "pick a random number $x$ (nevermind that there's no uniform distribution on $\mathbb{N}$), then with probability 1/2 it is even and 1/2 that it's odd.
Now, you're right that $P(x)=\frac{1}{2}P(x/2)+\frac{1}{2}P(3x+1)$. But it's fallacious to conclude $P(x)$ is constant. Similar reasoning: $P(x)=P(x+1)$ implies $P(x)$ is a constant. False, $P(x)=\sin(2\pi x)$ works.