Say in general we have $K$ outcomes, and each with probability $p_k$. And say we also have a chance of $p_0$ that none of them occurs.
Say we will split the bet each time according to $q_k$, $\sum_{k=1}^K q_k=1-q_{0}$, and $q_0$ is the fraction that we don't bet on anything. If we have $W$ to begin with, after a single bet, we will assume the return if outcome $k$ happens can be written in a form $ Wo_k(q_0, q_k)$. Note that $o_k(\cdot,\cdot)$, $k=1,\cdots, K$, is a function of $q_0$ and $q_k$ only and generally should be monotonically increasing with respect to both of them. And $o_0$ should just be $q_0$ (we don't gain or lose if we don't bet).
Now, say we have one dollar to start with and we bet for $N$ times, the approximate final net-worth will be approximately
$W = q_{0}^{Np_{0}}\prod_{k=1}^K (o_k(q_{0,}q_k))^{N p_k} = \left( q_{0}^{p_{0}}\prod_{k=1}^K (o_k(q_{0,}q_k))^{p_k}\right)^N.$
To maximize the net-worth, we should maximize $J=q_{0}^{p_{0}}\prod_{k=1}^K (o_k(q_{0},q_k))^{p_k}$ with respect to ${\bf q}=q_{0},q_1,\cdots,q_K$. Let's try to maximize the log value instead, we have to maximize
$\ln J =p_{0 } \ln q_{0}\sum_{k=1}^K p_k \ln o_k(q_0,q_k)$ such that $\sum q_k=1$ and $q_k \ge0$, $\forall k$.
Use the Lagrange multiplier method, let
$$ L = p_0 \ln q_0+\sum_{k=1}^K p_k \ln o_k(q_0,q_k) + \lambda
\left(\sum_{k=0}^{K}q_k- 1\right)+\sum_{k=0}^K \mu_kq_k,
$$
The optimal solution should satisfy the KKT conditions as below
\begin{align*}
\frac{\partial L}{\partial q_k}&=0, \forall k,
\\
\mu_k &\ge 0, \forall k\\
\mu_k q_k &=0, \forall k\\
q_k &\ge 0, \forall k,\\
\sum_{k=0}^K q_k &=1.
\end{align*}
Consider the first equation, $\frac{\partial L}{\partial q_k}=0$, finally we will get
\begin{align*}
\frac{p_k}{o_k}\frac{\partial o_k}{\partial q_k} + \mu_k &=\frac{p_0}{q_0}+\sum_{k'=1}^K\frac{p_{k'}}{o_{k'}}\frac{\partial o_{k'}}{\partial q_0}+ \mu_0, \forall k\ge 1 \\
\mu_k &\ge 0, \forall k\\
\mu_k q_k &=0, \forall k\\
q_k &\ge 0, \forall k,\\
\sum_{k=0}^K q_k &=1.
\end{align*}
=================================
For sanity check, take the original Kelly's criterion as an example, say with probability $p$, the net odd is $b$. So we have
$p_1 = p$, $p_0 = 1-p$, and
$o_1(q_1,q_{0})=b q_1 + 1$.
For $q_0,q_1 > 0$, we have $\mu_0=\mu_1=0$, thus we have optimum allocation if
$\frac{p_1}{o_1}\frac{\partial o_1}{\partial q_1}= \frac{p_0}{q_0}+ \frac{p_1}{o_1}\frac{\partial o_1}{\partial q_0}\Rightarrow \frac{bp}{b q_1 +1}= \frac{1-p}{q_0}+0\Rightarrow \frac{bp}{b q_1 +1}= \frac{1-p}{1-q_1}\Rightarrow q_1 =\frac{b p-(1-p)}{b}$ as in the original formula.
=========================
Now, back to the question, say we will split our bets into
$q_1$ for Team 1, $q_2$ for Team 2, $q_3$ for draw, and $q_0$ for leaving behind.
we have
$p_0=0, p_1=0.65$, $p_2=0.25$, and $p_3=0.1$. And
$o_1=r_1 q_1 + q_0, o_2 = r_2 q_2 + q_0$, and $o_3 = r_3 q_3 +q_4$, where
$r_1=1.54$, $r_{2}=4$, and $r_3=10$.
From above discussion, let's try to find a solution that satisfies all the above KKT conditions. Let's first assume $q_1,q_2,q_3 > 0$, this will mean $\mu_1=\mu_2=\mu_3 =0$. And hence we have
$$
\frac{p_1 r_1}{r_1 q_1+q_{0} } = \frac{p_2 r_2}{r_2 q_2 + q_0} = \frac{p_3 r_3}{r_3 q_3 + q_0}=\sum_{k=1}^3 \frac{p_k}{r_kq_k +q_0}+\mu_0
$$
Note that here $r_1 p_1 =r_2 p_2 =r_3 p_3 \triangleq R $, which is rather common in practice. Consequently,
$$ r_1q_{1}=r_2 q_2 = r_3 q_3$$
or
$$ \frac{q_1}{p_1}=\frac{q_2}{p_2}=\frac{q_3}{p_3}$$
and
\begin{align}
\frac{R}{r_3 q_3 + q_0}= \frac{1}{r_3q_3 +q_0}+\mu_0 \tag{1}
\label{eq:qall}
\end{align}
Note that when $R<1$, $\mu_0 <0$ and thus we don't have a valid solution. And when $R>1$, $\mu_0>0$ and $q_0=0$ from the KKT conditions (more precisely, the complementary slackness condition $\mu_0q_0=0$) and so we should all in. And when $R=1$, that is what we have in this problem, $\mu_0=0$, and there is no constraint on $q_0$ as long as $0 \le q_0 \le 1$. This is because there is no expected gain or loss from the bet. It does not matter how much one puts in. But if we ever put in anything (say all-in), the money should be split proportional to $p$'s values. That is,
\begin{align*}
q_1=p_1=0.65\\
q_2=p_2=0.15\\
q_3=p_3=0.1
\end{align*}
==============================
One may wonder why we will have all-in in this case. This is precisely because there is no chance of losing in the example ($p_0=0$). Say, if there is a chance the dealer will run away $(p_0>0)$, then \eqref{eq:qall} will become
\begin{align}
\frac{R}{r_3 q_3 + q_0}= \frac{p_0}{q_0}+\frac{1-p_0}{r_3q_3 +q_0}+\mu_0
\end{align}
Note that $q_3=(1-q_0)\frac{p_3}{p_1+p_2+p_3}=\frac{p_3(1-q_0)}{1-p_0}$, since we expect
$q_0$ should not be $0$, we expect $\mu_0=0$. Thus, we can solve out
$q_0$ as
$$
q_0 = \frac{R p_0 }{(R-1)(1-p_0)+R p_0}
$$
and
$$
q_k =p_k\frac{1- q_0}{1-p_0}, \forall k \ge 1.
$$
