Eccoci.

A secret integer $t$ is selected at random within the range $1\le t\le n$.

The goal is to guess the value of $t$ by making repeated guesses, via integer $g$. After a guess is made, there are three possible outcomes, in which it will be revealed that either $g<t$, $g=t$, or $g>t$. Then the process can repeat as necessary.

Normally, the number of guesses required on average can be minimized with a binary search: Given a lower bound $L$ and upper bound $H$ (initialized to $L=1$ and $H=n$), let $g=\lfloor (L+H)/2\rfloor$ where $\lfloor \cdot \rfloor$ is the integer floor function. If $g=t$, the process ends. Otherwise, if $g<t$, set $L=g+1$, but if $g>t$ instead, set $H=g-1$. After setting the new bounds, the search process repeats, and ultimately ends once $t$ is found. Even if $t$ can be deduced without searching, assume that a search will be required anyway to confirm the value.

Your friend Bob believes that the standard binary search is not that much better than his randomized variant: Instead of setting $g=\lfloor (L+H)/2\rfloor$, simply let $g$ be a random integer between $L$ and $H$, inclusive. The rest of the algorithm is the same as the standard binary search. This new search routine will be referred to as a random binary search.

Given that $1\le t\le n$ for random $t$, let $B(n)$ be the expected number of guesses needed to find $t$ using the standard binary search, and let $R(n)$ be the expected number of guesses needed to find $t$ using the random binary search. For example, $B(6)=2.33333333$ and $R(6)=2.71666667$ when rounded to $8$ decimal places.

Find $R({10}^{10})-B({10}^{10})$ rounded to $8$ decimal places.