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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:

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

However, I am not sure what to do from here. Can anyone please help?

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    $\begingroup$ 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 $\endgroup$
    – FShrike
    Jul 14, 2021 at 7:46
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    $\begingroup$ 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 $\endgroup$
    – FShrike
    Jul 14, 2021 at 7:56
  • $\begingroup$ You have the inequalities incorrect. You only need $2X_A>X_B+X_C$, for example. $\endgroup$ Jul 14, 2021 at 12:48
  • $\begingroup$ @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. $\endgroup$ Jul 14, 2021 at 19:05
  • $\begingroup$ @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.. $\endgroup$
    – FShrike
    Jul 14, 2021 at 20:05

2 Answers 2

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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 $(a,b,c)=(3,2,2)$.

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  • $\begingroup$ 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. $\endgroup$ Nov 30, 2022 at 20:13
  • $\begingroup$ If I understand correctly, all points inside that triangle are feasible solutions. However, how can I decide the optimal solution? (i.e. largest profit) $\endgroup$
    – user21
    Feb 17, 2023 at 20:19
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    $\begingroup$ @user21 Lourran's answer gives the optimal distribution, in terms of the largest guaranteed profit. $\endgroup$ Feb 17, 2023 at 20:21
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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 $\$100$.

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

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