# Bet to place for profit

Suppose there are $$3$$ runners $$R1, R2$$ and $$R3$$. The betting odds paid on each runner are $$r1:1$$, $$r2:1$$ and $$r3:1$$. This means that if, for example, you bet $$\2$$ for runner 1 winning, you will make $$2((r1+1)/1)-2$$ if runner 1 ends up winning.

The actual probabilities of each runner winning are unspecified. How much should you bet on each runner so that you always end up with a profit?

Is it simply a matter of solving the following system of inequalities (let $$a,b$$ and $$c$$ be the amount you bet for each of the runners) for $$a,b$$ and $$c$$:

$$a(r1+1)-a-b-c >0 \\ b(r2+1)-b-a-c >0 \\ c(r3+1)-c-b-a >0$$

## 1 Answer

Your equations are fundamentally correct, although it doesn't give much intuition into exactly how the solution works.

This is an arbitrage betting situation ("riskless" profit). The idea is simply to maximize the minimum payout you get, regardless of which actual choice wins.

In other words, you should bet an amount on each choice proportional to $$\frac{1}{r_i+1}$$, and you now guarantee the exact payout amount (per original unit of bets) of $$\frac{1}{\frac{1}{r_1+1}+\frac{1}{r_2+1}+\frac{1}{r_3+1}}$$ regardless of which actual choice wins.

Arbitrage exists when the above is greater than 1, that is, when the $$r_i+1$$s average more than 3, where the harmonic mean is appropriate.