Let’s name the three opponents as 1, 2, and 3, and let \(p_i\) be the probability of winning against opponent \(i\). Let’s also take, without loss of generality, \(p_1 \leq p_2 \leq p_3\), meaning that opponent 3 is the weakest. For a given order of opponents \(o_1, o_2 , o_3\), you win the tournament if you win against \(o_2\), and you win against either \(o_1\) or \(0_3\). Therefore, the probability that you win the tournament is:

\[\begin{align*} & p_{o_2} \left( p_{o_1} + (1-p_{o_1}) p_{o_3} \right) \\ = & p_{o_2} \left( p_{o_1} + p_{o_3} - p_{o_1} p_{o_3} \right) \\ = & p_{o_2} p_{o_1} + p_{o_2} p_{o_3} - p_{o_1}p_{o_2} p_{o_3} \\ \end{align*}\]

Now, we need to compare three probabilities corresponding to three order decisions we can make, and as the subtraction term will be common to all three orders, we can ignore it. Here are the three alternatives:

\[\begin{align*} & p_1 p_2 + p_2 p_3 \\ & p_1 p_3 + p_2 p_3 \\ & p_1 p_2 + p_1 p_3 \\ \end{align*}\]

A pairwise comparison will quickly show that the highest probability is provided by the middle row, where the weakest opponent (3) is played second.