Designing arbitrage bet on horses 
We have $3$ horses: A, B, C. For every dollar I wager, I get $\$2$, $\$4$, or $\$6$ if A, B, or C wins the race, respectively. Design a strategy that never loses money.


I'm stuck on this question. I know the winning strategy is to bet \$6 on A, \$3 on B and \$2 on C so that you always end up paying \$11 but winning \$12, but I was wondering what the general strategy to approach this question is. I have seen something similar to this question on reddit but I don't think I understood it quite enough, unfortunately. Thank you.
 A: Suppose the initial bet is $\$P$, and that we are betting on three horses $A,B,C$ with payouts $\$a, \$b, \$c$, and I buy $\alpha, \beta, \gamma$ tickets of each respectively.  Assume $a<b<c$.
Then, my total bet is simply $\$(\alpha+\beta+\gamma)P$, and we want that $$a\alpha >(\alpha+\beta+\gamma)P\implies\frac{\alpha}{\beta+\gamma}>\frac{P}{a-P}$$
$$b\beta >(\alpha+\beta+\gamma)P\implies\frac{\beta}{\alpha+\gamma}>\frac{P}{b-P}$$
$$c\gamma >(\alpha+\beta+\gamma)P\implies\frac{\gamma}{\alpha+\beta}>\frac{P}{c-P}$$
The triple $(\alpha,\beta,\gamma)$ which solves this triad of inequalities meets the criteria. Let's apply this to your scenario, where we have $P=1, a=2, b=4, c=6$.  We get:
$$\frac{\alpha}{\beta+\gamma}>1\implies \alpha>\beta+\gamma\tag1$$
$$\frac{\beta}{\alpha+\gamma}>\frac13\implies \alpha<3\beta-\gamma\tag2$$
$$\frac{\gamma}{\alpha+\beta}>\frac 15\implies \alpha<5\gamma-\beta\tag3$$
The third equation makes resolving this remarkably easy, since we know $\alpha\geq\beta\geq\gamma> 0$. Let me explain why it is $>0$ not $\geq 0$:
Set $\gamma=0$ and we get $\alpha<-\beta$ which is impossible. No solution like this exists (and it shouldn't, because if you didn't bet on horse C at all and it won, you'd have lost money).
Let's now test $\gamma=1$. $(3)$ tells us $5-\beta>\beta\to\beta<\frac52\to \beta=1,2$. Note also that $\alpha<\frac 52$ is necessary, I used that $\alpha\geq \beta$ to say $\beta<5-\beta$.
For each, $(2)$ gives $\alpha<2$ (i.e $\alpha=1$) and $\alpha<5$, while $(1)$ yields $\alpha>2$ and $\alpha>3$ respectively. Both cases are complete contradictions, and neither work.
We now test $\gamma=2$. Comparing $(1)$ and $(3)$ gives us $\beta+2<10-\beta\to\beta<4$, so $\beta=2,3$. With $\beta=2$ we get the set: $\alpha>4, \alpha<4, \alpha<8$ which is complete nonsense, but with $\beta=3$ we get the set: $\alpha>5, \alpha<7, \alpha<7$ which is satisfied by $\alpha=6$ and gives us $(6,3,2)$
