# Arbitrage opportunity

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?

• What you mean by odds is unclear. If the $o_i$ were fair mathematical odds then the associated probabilities would be $p_i = \dfrac{o_i}{o_i+1}$, i.e. odds of $2$ would mean a probability of $2/3$. But I think your odds may be the inverse of this, i.e. odds against, so fair odds of $2$ would imply a probability of $1/3$ which would make $p_i = \dfrac{1}{o_i+1}$. You want to look up Dutch book May 27, 2013 at 17:56
• The odds $o_i$ are defined as: if you bet $x$ and $i$ occurs, then you receive $o_i x$, positive or negative. If it doesn't occur then you get $-x$ May 28, 2013 at 12:06
• @RodrigodeAzevedo I have made a post on meta, so that these two tags can be discussed: Do we need separate tags for (betting) and (gambling)? Jan 15, 2017 at 11:55
• Related Dec 4, 2022 at 12:18

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 $+10$.