For a sequence $X_1, X_2, \ldots X_n$ with $E X_n = a$ I proved that $X_n \to_p a$, i.e. $$ \lim_{n \to \infty}P(|X_n - a|>\varepsilon)=0 $$ Does this type of convergence imply a probabilistic $ordering$ on the sequence, i.e. $$ P(|X_1-a|>\varepsilon) > P(|X_2-a|>\varepsilon)> \ldots > P(|X_n-a|>\varepsilon) $$ This would be useful for deriving a lower bound, e.g. for proof of convergence a.s., e.g. if $P(|X_1-a|>\varepsilon) = \pi$, the sum of probabilities is lower-bounded by $n \pi $ and hence diverges.
Intuitively it makes sense, but it sounds a bit too general and simplistic, hence probably wrong, but I couldn't find a good explanation anywhere.
Certainly false. In such questions it is always advisable to look at constant random vaiables. If something is not true for a sequence of real numbers it cannot be true for a sequence of random variables.
Take $X_n=\frac1 n$ if $n$ is even and $X_n =\frac 1 {4n}$ if $n$ is odd. Then $a=0$. Take any $\epsilon $ between $\frac 1 4$ and $\frac 1 2$. Then $P(|X_1-a| >\epsilon)=0$ and $P(|X_2-a| >\epsilon)=1$.