The highest number of coins that Stanlio can guarantee for himself is:

>!46

Proof:

>! Define $\rm{best}(i,j,k)$ to be the highest number of coins that Stanlio can guarantee for himself with $i$ coins remaining to be given out and $j$ further chances of receiving remaining for Stanlio, and $k$ chances remaining for Ollio. What we want to find is $\rm{best}(100,9,9)$<br/>
>! Finding a recurrence relation is straightforward. At this step, suppose that Stanlio offers $l$ coins. Ollio will choose to assign this set to Stanlio or himself based on which gives less payout to Stanlio. Of all possible values of $l$, Stanlio chooses whichever gives him the best payout. Thus we have:<br/>
>!  $$ \rm{best}(i,j,k) = max_l\;min[\rm{best}(i-l,j-1,k)+l,\rm{best}(i-l,j,k-1)]$$
>! The base cases for the recurrence are $\rm{best}(0,j,k) = \rm{best}(i,0,k) = 0$, $\rm{best}(i,j,0) = i$, $\rm{best}(i,0,0) = -\inf$ for $i$,$j$,$k>0$<br/>
> ! Bottom up dynamic programming using the recurrence gives $\rm{best}(100,9,9)=46$