Given odds $o_i$ for $i=1,2,\dots,n$ and the possibility to bet the amount $b_i \in \mathbb{R}$ on each event such that if event $i$ occurs you receive $b_i o_i$ and if it doesn't you receive $-b_i$. I am trying to find out the condition for arbitrage. My immediate thoughts are that $1/o_i$ represents probability, and since these events are independent then $1/o_1+1/o_2+\dots+1/o_n=1$ has some significance and that if $$\sum\limits_{i=1}^n o_i^{-1}\neq1$$ then arbitrage is possible.
In a specific example, $n=3$, $o_1=1,o_2=2,o_3=3$ I have worked out that if $b_1=-5,b_2=-4,b_3=-3$, then the profit is always greater or equal to zero and positive with finite probability. (i.e. $\{1\}\rightarrow 2,\{2\}\rightarrow 0,\{3\}\rightarrow 0$)
How do I show this without trial and error? I want a way to find the bids, maybe even for general $n$? Also, I get the feeling that if the sum is less than unity you need to back all possibilities for a sure profit ($b_i>0$) and if it is greater than unity you need to lay all possibilities ($b_i<0$) How would I show this?
Your odds are what I would call odds against.
If you want a Dutch Book (a guaranteed outcome) then you can bet $\dfrac{k}{1+o_i}$ on each option $i$. Then if possibility $j$ wins you will receive $\dfrac{k}{1+o_j}o_j-\displaystyle\sum_{i\not =j} \dfrac{k}{1+o_i}$, so whatever happens you will end up with $k\left(1-\displaystyle\sum_{i=1}^n\dfrac{1}{1+o_i}\right)$.
If you want to come out ahead, make sure that $k$ has the same sign as $\left(1-\displaystyle\sum_i\dfrac{1}{1+o_i}\right)$.
In your example $\displaystyle\sum_i\dfrac{1}{1+o_i}=\dfrac{13}{12}$, you could let $k=-120$ and make bets of $-60,-40,-30$ for a guaranteed outcome of $+10$.