Given the following bets you can make:
(1) A bet of $1$ wins $2$ if event A occurs
(2) a bet of $1$ wins $2$ if event B occurs
(3) a bet of $1$ wins $4$ if event C occurs.
In each case, you lose $1$ if the appropriate event does not win. Suppose that you know that event $A$ has a $90\%$ chance of occuring, event $B$ has a $9\%$ chance of occuring, and there is a $1\%$ chance of event $C$ occurs. Find a set of bets such that you are guaranteed to make money, regardless of the outcome.
My attempt: Based on my understanding, "guaranteed to make money" means profit is positive at the end. But we could easily see that if we bet $a$ dollars on event $A$ occurs, $b$ dollars on event $B$ occurs, and $c$ dollars on event C occurs, then our outcome is either $2a - b-c$ or $2b-a-c$ or $4c-a-b$. We need all these three $>0$ to guarantee we would make money. So if we solve for the intersection points of these 3 planes, we get: $a=b = 2c$.
My question: I got stuck on how to pick a good pair $(a,b,c)$ such that the overlapped regions of these $3$ planes give us the $3$ inequalities above. Plus, the information about the probability of each event occurring is not used at all, so I am skeptical whether my solution above is correct. Could someone please help me with this last step?