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Recently I came across this article about sports betting arbitrage. The article gives formulas for calculating arbitrage profit and individual bet amounts for a two-outcome event. But it doesn't prove that those formulas will always yield the optimal arbitrage profit. Those formulas are reproduced below.

Arbitrage Profit

Let P(A) and P(B) be probabilities of the only two possible outcomes of an event. These probabilities are simply inverse of decimal odds. Let I be the total investment we are willing to make. Also let P(T) = P(A) + P(B). Then

Arbitrage Profit = [I / P(T)] - I

Individual Bets

Amount to bet on outcome A = I * P(A) / P(T)

Amount to bet on outcome B = I * P(B) / P(T)

Two things I can't understand are:

  1. Why would the formulas above always yield the optimum profit for the given probabilities and investment?
  2. Are these formulas extendable to more than two outcomes?

I'm not a mathematician and trying to prove the above is doing my head in. Your help will be much appreciated!

Thanks in advance :)

Clarification - What is meant by 'Event'

Example of an event here would be a tennis match between Djokovic and Murray, the two outcomes being Djokovic and Murray. P(Djokovic) is obtained by taking reciprocal of decimal odds offered by the bookmaker for Djokovic. So if the odds for Djokovic win are 1.9 then P(Djokovic) = 0.526. In this case it is possible to have P(Djokovic) + P(Murray) = P(T) <> 1.0. When P(T) >= 1.0 there is no possibility of arbitrage. When P(T) < 1.0, then there is arbitrage profit. The above equations relate to the latter situation, i.e. when P(T) < 1.0.

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  • $\begingroup$ a) I asssume you're using the term "event" with its everyday meaning here, not in its technical sense? b) If $A$ and $B$ are the only possible outcomes, we should have $P(T)=P(A)+P(B)=1$. Since this is apparently not the case, as it would make the profit zero, there must be some third possibilitiy and it must be this third possibility that creates the opportunity for profit? $\endgroup$ – joriki Feb 4 '13 at 8:26
  • $\begingroup$ Hi, sorry about the ambiguity. You're right that it's not an event in the technical sense. I have updated the question, could you have a look please? $\endgroup$ – bytefire Feb 4 '13 at 15:16
  • $\begingroup$ The question is rather hard to understand because you're using technical terms with meanings other than their technical meanings. If I understand your edit correctly, what you're calling "probabilities" aren't actual probabilities, but hypothetical probabilities at which the odds offered by the bookmaker would be fair. As Ross shows in his answer, the opportunity for arbitrage is independent of actual probabilities; in fact, if your ability to turn a profit would rely on knowledge about the probabilities, one wouldn't call it arbitrage but expertise or good guessing or the like. $\endgroup$ – joriki Feb 4 '13 at 16:11
  • $\begingroup$ If these quantities were actual probabilities of mutually exclusive events that exhaust all possibilities, they'd add up to $1$. My advice would be not to mention those quantities at all, or, if you need to mention them, to call them something like "reciprocal odds" or the like that doesn't allow them to be confused with actual probabilities. $\endgroup$ – joriki Feb 4 '13 at 16:14
  • $\begingroup$ I read somewhere that the reciprocal odds represent implied probability of that outcome so that's why I ended up calling it probability. Obviously it was quite misleading, but that wasn't the intention. Thanks for your input. I'll keep that in mind the next time :) $\endgroup$ – bytefire Feb 4 '13 at 19:40
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First, by your definition $P(T)=1$ as $A,B$ are the only possibilities. Now you need the return of betting on each of $A,B$. The idea of arbitrage is that if the payoffs are too large, there is a risk free profit to be had. If the payoffs are too small (think the lottery) the optimum is not to bet at all.

So let the payoff from betting one unit and winning on $A$ be $a$ and the payoff on $B$ be $b$. The payoff of betting one unit and losing is $-1$. If $a+1 \gt \frac 1{P(A)}$ we have a winning bet, as the expected value is $aP(A)-(1-P(A))=P(A)(1+a)-1$. We may not know $P(A)$ however-that is one explanation why people bet on sports, that they disagree on $P(A)$.

The point of the calculation is that if $a,b$ are high enough, we can find a bet that guarantees a profit independent of $P(A), P(B)$. We can even find a bet that gives the same profit no matter which occurs. If I bet $x$ on $A$ and $y$ on $B$ and $A$ occurs, my payoff is $ax-y$, while if $B$ occurs my payoff is $by-x$. If I want to be indifferent which happens these should be equal. So $ax-y=by-x$ and $y=\frac {a+1}{b+1}x$. My total investment is $x+y=I$ and you can solve the two equations to find $x=\frac {b+1}{a+b+2}I, y=\frac {a+1}{a+b+2}I$.

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  • $\begingroup$ Thanks! Marked as answer, although there is one bit I still need help on. $\endgroup$ – bytefire Feb 4 '13 at 16:51
  • $\begingroup$ Say we use brute force method to divide up the investment I into amounts to place on outcomes A and B. E.g. if I = 100 units, we start with 1 unit on A and 99 units on B, then 2 units on A and 98 units on B, so on (total 99 iterations). In each iteration we take the worse of the two possible payoffs, thus ending up with a collection of 99 min payoffs. Then we take the max of those 99 min payoffs and compare it with the payoff given by the equations above. Will the payoff from the equations above always be greater than or equal to the max-of-the-mins payoff obtained from brute force method? $\endgroup$ – bytefire Feb 4 '13 at 16:51
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    $\begingroup$ @bytefire: Yes. Let's say $a=b=2.25$. We would be told to bet the same amount on each, so we will be paid $2.25$ for betting $2$ If you are going to bet some other ratio, you can divide the bet into the part that is equally split and the part that is just bet on the larger. The worst case of the second part will be negative, so you want to reduce it to zero. $\endgroup$ – Ross Millikan Feb 4 '13 at 16:55
  • $\begingroup$ Got it. Brilliant explanation. I can't up vote your comment (may be I don't have enough reputation yet) but thanks again for clarifying the whole muddle! $\endgroup$ – bytefire Feb 4 '13 at 19:41

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