Suppose I have teams A and B playing each other. The odds on A to win are $2/5$, odds on B to win are $6/1$ and odds for a draw are $11/4$.
Suppose I have $x$ total amount of money. In what proportion should I divide $x$ in bets on every possible outcome (team A winning, team B winning, and a draw), in order to ensure that I $\textbf{always}$ obtain an overall profit?
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Based on how these odds are given, it appears that there's a substantial "vig" protecting the house from losing money. Therefore, it's mathematically impossible to bet in a way that guarantees you profit. Now, the only way that changes is if you know what the true probabilities of win, lose and draw are. Of course, no one ever knows these probabilities but they can be estimated. The house usually does this better than 99% of gamblers though.
If you were really confident that you knew the probabilities better than the house, then you could devise an optimal betting strategy in some cases but not all. Again, the vig protects the casino from losing money to "sharps" who do, on occasion, have more information to work with.
If you bet $\frac{5x}7$ on $A$ and the rest on $B$ or draw, and $A$ wins, then if $A$ wins you break even. If you bet $\frac x7$ on $B$ and the rest on $A$ or draw, and $B$ wins, you break even. If you bet $\frac{4x}{15}$ on draw and the rest on $A$ and $B$, and there is a draw, you break even.
That means, if you want to be guaranteed to walk away with $x$ money no matter what, you need to bet a total of $\frac{5x}7 + \frac{x}{7} + \frac{4x}{15} = \frac{118x}{105}$ money.