Say you have positive $\{a\_i\}\_{i=1}^n$ and you have $p\_i=\frac{a\_i}{\sum\_{i=1}^na\_i}$, then assume you have a $C$ such that $C<2a\_n\ll\sum\_{i=1}^na\_i$ (that is $C$ is not very large), then define $q\_i=\frac{a\_i}{C+\sum\_{i=1}^na\_i}$ and $q\_{n+1}=\frac{C}{C+\sum\_{i=1}^na\_i}$. Does Shannon entropy of $q$ dominate entropy of $p$?


Take $C=a\_n+\log\_2^ka\_n$ for any fixed $k$. Now does Shannon entropy of $q$ dominate entropy of $p$ after certain $n$?


There are two cases to consider. Case $(1)$ $a\_{i+1}=a\_i+O(\log^ka\_i)$ Case $(2)$ $a\_{i+1}=a\_i+O(a\_i)$.


When can one expect Shannon entropy to be dominant in sequences of these types? Clearly in this post [Entropy dominance](https://mathoverflow.net/questions/191391/entropy-dominance), one criteria for negative result is given.



Entropy dominance of certain restricted sequenes
I'll show that **Yes**, if you rule out an event of sufficiently small probability then the entropy decreases.


I'll change the notation around a bit so that my $p\_1,\dots,p\_n$ correspond to your $q\_{n+1},\dots,q\_1$.


Suppose $\sum\_{i=1}^n p\_i=1$ and we are given that the event corresponding to $p\_1$ did not occur.
Then the new probability of the event corresponding to $p\_k$, $k\ne 1$, is
$$
\hat p\_k = \frac{p\_k}{1-p\_1}.
$$
So the new entropy $H(\hat p)$ is (where $S(p)=-\log\_2(p)$ and $q$ is the experiment that only determines whether $p\_1$ occurs)
$$
\sum\_{k=2}^n S\left(\frac{p\_k}{1-p\_1}\right)\frac{p\_k}{1-p\_1}
= \frac1{1-p\_1}\sum\_{k=2}^n [S(p\_k)-S(1-p\_1)]{p\_k}
$$
$$
= \frac1{1-p\_1}\left(\sum\_{k=2}^n S(p\_k)p\_k - (1-p\_1)S(1-p\_1)\right)
= \frac1{1-p\_1}\left(H(p) - S(p\_1)p\_1 - (1-p\_1)S(1-p\_1)\right)
$$
$$
= \frac1{1-p\_1}\left(H(p) - H(q)\right) \le H(p)\quad\text{(as we want) iff }
$$
$$
\frac{H(q)}{p\_1}\ge H(p).
$$
The function $p\_1\mapsto H(q)/p\_1$ is
$$
p\_1\mapsto -\log\_2(p\_1) - \frac{1-p\_1}{p\_1}\log\_2(1-p\_1)
$$
which goes to infinity as $p\_1\rightarrow 0$, whereas $H(p)$ is bounded by $\log\_2 n $,
so it is enough to take $p\_1\le 1/n$.


