# Arbitrage bet on $3$-way horse race

There are $$3$$ horses in this week's race. When we say "the stated odds against a horse winning are $$r$$-to-$$s$$", we mean if you wager $$1$$ dollar on the horse, you will lose $$1$$ dollar if the horse loses and win $$r/s$$ dollars if the horse wins. The stated odds against the horse $$A$$ winning are $$2$$-to-$$1$$. The stated odds against $$B$$ winning are $$3$$-to-$$1$$. If the odds against horse $$C$$ are $$4$$-to-$$1$$, how can you get to ensure you will win money?

I did some guesswork: Intuitively since one of the three has to win, we should pile the most money onto horse $$A$$ since it has the lowest payout if it wins. $$2$$-$$1$$-$$1$$ didn't work, so then I tried $$3$$-$$2$$-$$2$$ and it magically worked.

But I am wondering how one could deduct this more systematically. Here's my attempt. Let $$X$$ be the total amount of money I bet, $$X_A$$, $$X_B$$, $$X_C$$ the amounts I bet on each horse. We want:

$$2X_A > X_A + X_B + X_C, \qquad 3 X_B > X_A + X_B + X_C, \qquad 4X_C > X_A + X_B + X_C.$$

• Does 3-2-2 work? $3+2+2=7>2\times 3$ and so even if horse A wins you don’t get your money back Jul 14, 2021 at 7:46
• How are you sure this question has a solution - maybe I’m just tired, but I’ve tried fiddling with the inequalities and reaching only paradoxes Jul 14, 2021 at 7:56
• You have the inequalities incorrect. You only need $2X_A>X_B+X_C$, for example. Jul 14, 2021 at 12:48
• @IdioticShrike $3-2-2$ does work. If you win the $3$ bet and lose the other two bets, then your winnings are $3\times 2-2-2>0$. You do not lose your stake if the bet wins. Jul 14, 2021 at 19:05
• @MikeEarnest I considered this, and considered your answer, and even considered editing my answer to include the correct inequalities, but I was completely thrown by my lack of understanding with betting systems - if something is on 1 to 1 odds, and I bet 5 dollars, do I get back just my five dollars or do I get back my five dollars plus another five? I feel stupid for this.. Jul 14, 2021 at 20:05

Letting $$a,b,c$$ be the amounts of money you make on each bet, the correct inequalities are $$2a>b+c\qquad 3b>a+c\qquad4c>a+b$$ since a winning bet only needs to recoup the losses from the other bets.

A way to simplify this is to note that all the inequalities are scale invariant; if $$a,b,c$$ satisfy the above, then so will $$ra,rb$$ and $$rc$$, for any $$r>0$$. To eliminate this redundancy, let $$x=b/a,\qquad y=c/a$$ and notice the inequalities become $$2>x+y,\qquad 3x>1+y,\qquad 4y>1+x.$$ It is simple to graphically solve these inequalities, and determine the set of $$(x,y)$$ which works is the interior of a certain triangle: https://www.desmos.com/calculator/e1vjvdpwh6. For example, it is pretty clear that $$(2/3,2/3)$$ is inside that triangle, which leads to the solution $$(a,b,c)=(3,2,2)$$.

• However, with this solution the amount won is not constant for all odds. If you bet $20$ on $A$, $15$ on $B$ and $12$ on $C$ you will get a net win of $13$ in any case. I believe this can be solved analytically, but I wasn't able to prove it. See this related question still open, where differently from the present formulation, $1$ must be subtracted to the winning quote too. Thus to adapt this question to that one we need to add $1$ to all quotes here. Nov 30, 2022 at 20:13
• If I understand correctly, all points inside that triangle are feasible solutions. However, how can I decide the optimal solution? (i.e. largest profit) Feb 17, 2023 at 20:19
• @user21 Lourran's answer gives the optimal distribution, in terms of the largest guaranteed profit. Feb 17, 2023 at 20:21

Let $$a$$, $$b$$ and $$c$$ be the amounts bet on horses $$A$$, $$B$$ and $$C$$, respectively. You pay $$a+b+c$$, and you win $$3a$$, $$4b$$ or $$5c$$, depending on which horse wins. I say $$3, 4, 5$$ instead of $$2,3,4$$ to have the same data as you, because I want to separate phase 1 (what you bet) from phase 2 (what you win).

You want to win $$\100$$ in all options. So, you have to bet $$\100/3 = \33.33$$ on $$a$$, plus $$\100/4 = \25$$ on $$b$$ and $$\100/5=\20$$ on $$c$$. Total, you bet $$33.33+25+20=78.33$$, and you are sure to win $$100$$. The profit is $$\21.67$$, no matter the winner.

In real bets, you will never have this configuration, you need to pay more than $$\100$$ if you want to be sure to win $$\100$$.

• You could mention that there's an arbitrage to be exploited because $\frac13 + \frac14 + \frac15 < 1$. Nov 30, 2022 at 19:34