You are correct that you want to find a bet so that your return in every possible situation is positve. If you put amounts (a, b, c) on each outcome then your possible profits, as you correctly deduced, are 2a - b - c, 2b - a - c and 4c - a - b.
One way of proceeding is to look for a solution where you make the same amount of money in each scenario, which means setting the three quantities above equal to each other. In this case you discover that you require a = b and 3a = 5c. You can satisfy these constraints by picking a = b = 5 and c = 3, in which case the possibilities are
- A wins - you make 10 - 5 - 3 = 2
- B wins - you make 10 - 5 - 3 = 2
- C wins - you make 12 - 5 - 5 = 2
In each case you have a profit of 2, so this is your arbitrage. Indeed, the real-world probabilities are completely irrelevant.
The key is the idea of implied probabilities. Since you lose 1 if A doesn't happen, but win 2 if it does happen, then the implied probability $p$ (defined to be the probability that would give you zero expected return on this bet) satisfies
$$
(1-p)\times (-1) + p \times 2 = 0 \quad\Rightarrow\quad p = 1/3
$$
Similarly the implied probability of B is 1/3 and the implied probability of C is 1/5. These sum to less than one (in bookmaking this is called an under-round) so there is the possibility of arbitrage. The arbitrage exists because the odds given are incoherent (they don't sum to one).
To take advantage, you make bets which are proportional to the inverse of the implied probability.
In the real world, a book keeper will always set odds that sum to more than one, meaning that any combination of bets which guarantees a payout, will always guarantee a negative payout (so there are no arbitrages).
You intuition about choosing the bookmaker with the best odds for each outcome is certainly correct. How could it make sense to do otherwise?
So you bet $b_1$ with $B_1$ that $Y$ wins, $b_2$ with $B_2$ that $X$ wins, and $B_3$ with $B_3$ on a draw. Presumably, you have a predetermined amount to bet, so we assume $b_1+b_2+b_3=1$ and interpret the $b_i$ as the fraction bet with each book maker.
Now you want to analyze the profit or loss in the case of each of the three outcomes, and choose the $B_i$ to maximize the profit in the worst case. This is predicated on my understanding of the goal is to make money no matter what the outcome.
EDIT
Out of curiosity, I worked out the details, so I might as well post them. We have the following odds $$\begin{matrix}
&\text{X}&\text{Y}&\text{Draw}\\
\text{B}_1&4.2&\mathbf{2}&3.4\\
\text{B}_2&\mathbf{4.7}&1.9&3.1\\
\text{B}_3&4.1&1.6&\mathbf{3.7}
\end{matrix}$$
where the best odds on each outcome have been bolded.
If we bet $b_i$ with bookmaker $\text{B}_i,\ i=1,2,3$ where we assume $b_1+b_2+b_3=1$ then in order that we not lose money whatever happens, the following conditions must hold. $$\begin{align}
-b_1 +4.7b_2 -b_3 &\geq0\tag{1}\\
2b_1 -b_2 -b_3 &\geq0\tag{2}\\
-b_1 -b_2 +3.7b_3 &\geq0\tag{3}
\end{align}$$
Substituting $b_3=1-b_1-b_2$ in $(1)$ and simplifying gives $$b_2\geq{1\over5.7}\approx .1754$$Similarly, substituting in $(2)$ gives $$b_1=\frac13\approx .3333$$ and substituting in $(4)$ gives $$b_1+b_2\leq{37\over47}\iff b_3\geq{10\over47}\approx.2127$$
The sum of the lower bounds comes to $\approx.7215$ so the problem is feasible. If we lay out about $72\%$ of our capital as computed above, we will never lose, and so the problem is how to bet the remaining $28\%$ to best advantage.
Since this is arbitrage, we want our profit to be the same no matter what the outcome. In order to have the same profit if X wins or there is a draw, we equate the left-hand sides of $(1)$ and $(3)$ and we find$$5.7b_2=4.7b_3\tag{4}$$ To ensure the same profit if there is a draw of if Y wins, we equate the left-hand sides of $(2)$ and $(3)$ and find $$3b_1=4.7b_3\tag{5}$$ Comparing $(4)$ and $(5)$ we see that we must have $$b_1=1.9b_2\tag{6}$$ $(4)$ and $(6)$ now give$$b_2\left(1.9+1+{57\over47}\right)=1\implies b_2={470\over1933}\approx.243145 $$so we have $$\boxed{b_1\approx.461976\\b_2\approx.243145\\b_3\approx.294878}$$
You can, and should, verify that no matter what the outcome, the profit is about $ 38.59\%$ of the amount wagered.
In retrospect, I could have simply solved for the bets that would make the profit the same for all outcomes, and then tested whether it was positive. You can recast the solution that way, if you like.
Best Answer
First, by your definition $P(T)=1$ as $A,B$ are the only possibilities. Now you need the return of betting on each of $A,B$. The idea of arbitrage is that if the payoffs are too large, there is a risk free profit to be had. If the payoffs are too small (think the lottery) the optimum is not to bet at all.
So let the payoff from betting one unit and winning on $A$ be $a$ and the payoff on $B$ be $b$. The payoff of betting one unit and losing is $-1$. If $a+1 \gt \frac 1{P(A)}$ we have a winning bet, as the expected value is $aP(A)-(1-P(A))=P(A)(1+a)-1$. We may not know $P(A)$ however-that is one explanation why people bet on sports, that they disagree on $P(A)$.
The point of the calculation is that if $a,b$ are high enough, we can find a bet that guarantees a profit independent of $P(A), P(B)$. We can even find a bet that gives the same profit no matter which occurs. If I bet $x$ on $A$ and $y$ on $B$ and $A$ occurs, my payoff is $ax-y$, while if $B$ occurs my payoff is $by-x$. If I want to be indifferent which happens these should be equal. So $ax-y=by-x$ and $y=\frac {a+1}{b+1}x$. My total investment is $x+y=I$ and you can solve the two equations to find $x=\frac {b+1}{a+b+2}I, y=\frac {a+1}{a+b+2}I$.