Need help with proof for arbitrage betting 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:


*

*Why would the formulas above always yield the optimum profit for the given probabilities and investment?

*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.
 A: 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 $x=\frac {b+1}{a+b+2}I, y=\frac {a+1}{a+b+2}I$.
