I am only guessing that what you are asking about is arbitrage betting. (Although I am not sure, considering the odds 8, 9 and 13, which seem rather unusual.) If you really want to apply this, keep in mind that arbitrage betting bears some risk (despite being called risk-free). Some of them are described in the Wikipedia article linked above.
You can also find some other posts about this topic: Arbitrage opportunity, Arbitrage sports betting, How is potential profit from an arbitrage bet (a.k.a. surebet) calculated? (The first two links are from this site, the third one is from Sports Stack Exchange.)
If we have 3 complementary events, such as home/draw/away; we can
look for an arbitrage as follows:
$$\begin{array}{|c|c|c|c|}
\hline
& \text{odds} & \text{stake} & \text{winnings} \\ \hline
1 & o_1 & b_1 & o_1b_1 \\ \hline
\text{X} & o_2 & b_2 & o_2b_2 \\ \hline
2 & o_3 & b_3 & o_3b_3 \\ \hline
\end{array}$$
We would like to maximize the winning in the case of any outcome,
hence
$$W=o_1b_1=o_2b_2=o_3b_3$$
which yields
\begin{gather*}
b_1=\frac{W}{o_1}\\
b_2=\frac{W}{o_2}\\
b_3=\frac{W}{o_3}
\end{gather*}
If we bet the amount of $B$ altogether, then
$$B=b_1+b_2+b_3=W\left(\frac1{o_1}+\frac1{o_2}+\frac1{o_3}\right).$$
To obtain an arbitrage, we need $W>B$, i.e.,
$$\frac1{o_1}+\frac1{o_2}+\frac1{o_3}<1.$$
For the sake of simplicity we assume that $B=b_1+b_2+b_3=1$. (Thus
$b_{1,2,3}$ represent the percentage of total stake wagered on one
of the three outcomes.)
This leads us to the following equalities
\begin{gather*}
b_1=\frac{\frac1{o_1}}{\frac1{o_1}+\frac1{o_2}+\frac1{o_3}},\\
b_2=\frac{\frac1{o_2}}{\frac1{o_1}+\frac1{o_2}+\frac1{o_3}},\\
b_3=\frac{\frac1{o_3}}{\frac1{o_1}+\frac1{o_2}+\frac1{o_3}}.
\end{gather*}
For example, if we assume that we found (at different bookmakers) odds for the recent Barcelona - PSG like this.
(This is perhaps not completely unrealistic. For the purpose of illustration I have improved a bit the odds taken from here. Of course, whether you can really find odds such that you can make profit is entirely different question - let's just concentrate on the mathematical side of this here.) I should probably also mention that I am using decimal odds in the example.
$$
\begin{array}{|c|c|}
\hline
\text{Barcelona} & 1.6 \\\hline
\text{draw} & 5 \\\hline
\text{PSG} & 6 \\\hline
\end{array}
$$
Now we have $1/o_1=0.625$, $1/o_2=0.2$, $1/o_3=0.167$ which means
\begin{align*}
\frac1{o_1}+\frac1{o_2}+\frac1{o_3}&=0.992\\
\frac1{\frac1{o_1}+\frac1{o_2}+\frac1{o_3}}&=1.008
\end{align*}
So we could make a bet with a very small profit of $0.8\%$.
The ratios between $1/o_1$, $1/o_2$ and $1/o_3$ tell us how to divide the wagered amount. In the other words, $$b_i=\frac{1/o_i}{\frac1{o_1}+\frac1{o_2}+\frac1{o_3}}$$
is the fraction of the amount to put on $i$-th outcome.
For the odds mentioned above we would get (approximately):
$$\begin{array}{|c|c|c|}
\hline
o_i & \frac1{o_i} & b_i \\\hline
1.6 & 0.625 & 0.63 \\\hline
5 & 0.2 & 0.202 \\\hline
6 & 0.167 & 0.168 \\\hline
\end{array}
$$
For example, if we divide $10$USD into three bets like this
$$\begin{array}{|c|c|c|}
\hline
\text{bet} & \text{odds} & \text{win} \\\hline
6.3 & 1.6 & 10.08 \\\hline
2.02 & 5 & 10.08 \\\hline
1.68 & 6 & 10.08 \\\hline
\end{array}$$
we would win $10.08$USD (making a very minor profit of $8$ cents) independently of the outcome of the game.
Virtually nothing changes if you have two possible outcomes (as in one of the questions I linked above), or four possible outcomes or any number of possible outcomes. You only have different number of summands in the above computations.
