What is wrong with this argument?
- Let $X$ be a random positive integer in base 2
- The probability that the number of digits of X is even is $1/2$ (Since number of digits can be any positive integer)
- Let $Y$ be $X$ but in base 4
- The first digit of $Y$ can be $1$,$2$, or $3$ so the first digit of $Y$ being $2$ or $3$ is $2/3$
- If the first digit of $Y$ is $2$ or $3$, then $X$ will have an even number of digits
- So, the probability of $X$ having an even number of digits is $2/3$, but this contradicts the second statement
The number of $k$-bit binary numbers is $2^{k-1}$. So if $X$ is chosen uniformly from $1,\ldots,N$, then the probability that the binary representation of $X$ has an even number of bits is
$$\frac{2^{2-1} + 2^{4-1} + \cdots + 2^{n-1}}{2^n-1} = \frac{2}{3} \text{ if $N=2^n-1$ where $n$ is even}$$ and $$\frac{2^{2-1} + 2^{4-1} + \cdots + 2^{n-2}}{2^n-1} = \frac{2 (2^{n-1} - 1)}{3(2^n-1)} \approx \frac{1}{3} \text{ if $N=2^n-1$ where $n$ is odd}.$$
So as $N \to \infty$, the probability oscillates between $1/3$ and $2/3$, and isn't actually $1/2$.