Suppose four horses – $A, B, C$, and $D$ – are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $\$1$ on $A$, then you receive $\$6$ if $A$ wins, or you realize a net gain of $\$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?
I found this question from Horse Betting Odds – But Guaranteed Win! and made some revision.
I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.
However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?
Best Answer
Each of the horses has a certain return when they win, A is a $6\times$ return, B is $5\times$, C $4\times$, and D $3\times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i \ge \frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i \ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when $$B_i = \frac{\frac{B}{r_i}}{\sum \frac{1}{r_i}}$$ and your guaranteed profit is $$\frac{B}{\sum \frac{1}{r_i}} - B$$ This is why in a real horse race $\sum \frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!