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Given odds $o_i$ for $i=1,2,\ldots,n$ and the possibility to bet the amount $b_i\in \mathbb{R}$ on each event such that if event $i$ occurs you receive $b_io_i$ and if it doesn't you recieve $-b_i$. 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 $1/o_1+1/o_2\ldots+1/o_n=1$ has some significance and that if $\displaystyle\sum\limits_{i=1}^n o_i^{-1}\neq1$ 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?

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  • $\begingroup$ 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 $\endgroup$ – Henry May 27 '13 at 17:56
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    $\begingroup$ 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 $-x$ $\endgroup$ – shilov May 28 '13 at 12:06
  • $\begingroup$ @RodrigodeAzevedo I see that you have created new tag called (betting). Is there some reason why the already existing tag (gambling) cannot be used for such questions? If needed, we can continue this discussion in chat or make a post on meta to discuss this. $\endgroup$ – Martin Sleziak Jan 15 '17 at 11:29
  • $\begingroup$ @MartinSleziak I didn't know that the "gambling" tag existed. Also, "gambling" is somewhat derogatory, whereas "betting" is somewhat neutral. More importantly, in arbitrage betting there is no chance of losing. Hence, it's not gambling. $\endgroup$ – Rodrigo de Azevedo Jan 15 '17 at 11:41
  • $\begingroup$ @RodrigodeAzevedo Honestly, I do not see how gambling is derogatory. As I said, I would prefer not to pile more comments here - this is not related to this question, it only happens to be the first question where you used the new tag. As I mentioned above, there are more appropriate places to continue this discussion. $\endgroup$ – Martin Sleziak Jan 15 '17 at 11:45
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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 $+10$.

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