Assume we have an closed interval $I = [a,b]$ where $a,b\in\mathbb{R}_+$ ($a,b\geq0$). Three persons pick a number each in the interval, lets call the numbers $A$, $B$ and $C$.
We then look at 1/3rd of the average of the numbers. So $M = (A+B+C)/9$. If your bet is closest to $M$, you win the amount of money you bet.
In case of a tie (where the tie is the closest to $M$) then the participants will draw who wins. This essentially divides the price in two or three so ties are to be avoided.
Assume that you know the numbers $A$ and $B$, how should you pick $C$ in order to be closest, and win the most amount of money?
Note you can always bet $C = (A+B)/8$, to ble closest. But this will rarely make the most money. Say the interval is $[0,100]$, $A=1$ and $B=50$. Then you can bet $C=51/8 \approx 6.375$ and win. Since $M = (6.375+1+50)/9 = 6.375$. However you can also bet $C = 13.2857$, which makes you win more.
Assume that $0\le A\le B$. Our main competitor is now $A$, for any sensible bid. We tie with $A$ if $$\left|\frac{1}{9}(A+B+C)-A\right|=\left|\frac{1}{9}(A+B+C)-C\right|$$ or $$B+C-8A=\pm(A+B-8C)$$ In the case of $+$, we have $C=A$, and in the case of $-$ we have $C=\frac{2}{7}B-A$.
These two options form an interval, on the interior of which $C$ wins. We want to pick the largest $C$ strictly inside this interval (to avoid sharing the prize with $A$) that is still inside $[a,b]$.