A question of odds 
Consider an experiment with four possible outcomes, and suppose that the quoted odds for the first three of these outcomes are as follows.

What must be the odds against outcome 4 if there is to be no possible arbitrage when one is allowed to bet both for and against any of the outcomes?

This is exercise 6.2 on p. 102 of Sheldon M. Ross's "An Elementary Introduction to Mathematical Finance", 3rd edition, Cambridge University Press, 2011.
Following is my solution. I'd appreciate it if someone could either confirm that this is in fact the answer and that the way I solved the problem is correct, or else explain why my solution fails.
Thanks.

My attempt at solving the problem
We solve under the assumption that the rate of interest is $0$.
Denote by $S^k$, $k \in \{1, 2, 3, 4\}$ a bet on outcome $k$. Suppose there is no arbitrage opportunity in the market. Then there's an EMM distribution $p$. So, for every $k \in \{1, 2, 3, 4\}$,
$$
0 = E_p\left(S^k_\text{tomorrow} - S^k_\text{today}\right) = E_p\left(\Delta S^k\right) = \sum_{j = 1}^4\Delta S^k(\omega_j) p_j
$$
Substituting $k = 1, 2, 3$ we obtain the following linear system of equation:
$$
\begin{align}
0 & = 2 \cdot p_1 + (-1) \cdot p_2 + (-1) \cdot p_3 + (-1) \cdot \left(1 - p_1 - p_2 - p_2\right) \\
0 & = (-1) \cdot p_1 + 3 \cdot p_2 + (-1) \cdot p_3 + (-1) \cdot \left(1 - p_1 - p_2 - p_2\right) \\
0 & = (-1) \cdot p_1 + (-1) \cdot p_2 + 4 \cdot p_3 + (-1) \cdot \left(1 - p_1 - p_2 - p_2\right)
\end{align}
$$
The unique solution to this system of equations is
$$
\begin{align}
p_1 &= \frac{1}{3} \\
p_2 &= \frac{1}{4} \\
p_3 &= \frac{1}{5}
\end{align}
$$
Hence
$$
p_4 = 1 - \frac{1}{3} - \frac{1}{4} - \frac{1}{5} = \frac{13}{60}
$$
By the same rationale, denoting the odds against outcome $4$ by $c$ we have
$$
0 = (-1) \cdot p_1 + (-1) \cdot p_2 + (-1) \cdot p_3 + c \cdot p_4 = - \frac{1}{3} - \frac{1}{4} - \frac{1}{5} + \frac{13}{60} c = -\frac{47}{60} + \frac{13}{60} c
$$
Solving for $c$, we obtain $c = 47/13$. In other words, the odds against outcome $4$ are $1 : 3\frac{8}{13}$.
 A: A very similar situation is analyzed in Example 6.1a in the same chapter. Here is a quote of a relevant parts. (Of course, it is ideal to read the chapter in full.)

The return from such a bet is often quoted in terms of odds. If the odds against outcome $i$ are $o_i$ (often expressed as "$o_i$ to $1$"), then a one-unit bet will return either $o_i$ if $i$ is the outcome of the experiment or $-1$ if $i$ is not the outcome. That is, a one-unit bet on i will either win $o_i$ or lose $1$. 
...
Since the $p_i$ must sum to $1$, this means that the condition for there not to be an arbitrage is that $$\sum_{i=1}^m\frac1{1+o_i}=1.$$
  That is, if $\sum_{i=1}^m (1+o_i)^{-1}\ne1$, then a sure win is possible.

I will stress that this part of the text of exercise is important: "one is allowed to bet both for and against any of the outcomes". It means that we can either bet that outcome $i$ happens, but we can also bet that it does not happen. (Which in the book you refer to is modeled by betting negative amount on the $i$-th outcome.) 
If we use the formula quoted from the analysis in Example 6.1a, we get that there is no possible arbitrage if and only is
\begin{align*}
\frac1{1+o_1}+\frac1{1+o_2}+\frac1{1+o_3}+\frac1{1+o_4}&=1\\
\frac13+\frac14+\frac15+\frac1{1+o_4}&=1\\
\frac1{1+o_4}&=\frac{47}{60}\\
o_4&=\frac{13}{47}
\end{align*}
