(Slightly amended from Hulett)
Let \(S\) be a set of competitors partitioned into three components \(C_1\), \(C_2\), \(C_3\) such that every member of component \(C_i\) beats every member of component \(C_{(i \mod 3) + 1}\). Fix probabilities \(p_i^{(0)}\) summing to 1, and let \(p_i^{(k)}\) denote the probability that a member of component \(C_i\) wins a balanced upper-bracket of height \(k\) where each leaf is labeled independently according to \(p_i^{(0)}\).
Let \(q_i^{(k)}\) denote the probability that a member of component \(C_{i}\) wins the lower bracket that corresponds to the upper bracket of height \(k\). Let \(o_i^{(k)}\) denote the probability that the winner of a lower bracket game between the upper bracket loser and prior lower bracket winner is from component \(i\). Let \(z_i^{(k)}\) denote the probability that a member of component \(C_i\) loses at the top level of a balanced upper-bracket of height \(k\) where each leaf is labeled independently according to \(p_i^{(0)}\).
Since we have labeled each leaf independently, the two children of a node are independent, so we can easily calculate \(p_i^{(k)}\) recursively. For all \(k \ge 0\),
\[p_i^{(k + 1)} = \left( p_i^{(k)} \right)^2 + 2 p_i^{(k)} p_{(i \mod 3) + 1}^{(k)} = p_i^{(k)} \left( p_i^{(k)} + 2 p_{(i \mod 3) + 1} \right)^{(k)}\].
Similarly, we have
\[\begin{align*} z_i^{(k)} &= p_i^{(k)} \left( p_i^{(k)} + 2p_{(i + 1) \mod 3 + 1}^{(k)} \right) \\ q_i^{(0)} &= p_i^{(0)} \left( p_i^{(0)} + 2p_{(i + 1) \mod 3 + 1}^{(0)} \right) = z_i^{(0)} \\ o_i^{(k + 1)} &= q_i^{(k)} \left(z_i^{(k + 1)} + z_{i \mod 3 + 1}^{(k + 1)}\right) + q_{i \mod 3 + 1}^{(k)} \left(z_i^{(k + 1)} \right) \\ q_i^{(k + 1)} &= o_i^{(k)} \left( o_i^{(k)} + 2o_{i \mod 3 + 1}^{(k)} \right) \\ \end{align*}\]
Suppose \(p_1^{(0)} = p_3^{(0)} = \epsilon \lt 2^{-10}\). Then, \(p_2^{(k)} = 1 - 2*\epsilon\).
Phase 1 will be the time during which either \(p_1^{(k)} + p_3^{(k)}\) grows to be larger than \(\sqrt 2 - 1 - 2^{-6}\) (TODO) or \(q_1^{(k)} + q_3^{(k)}\) grows to be larger than \(\sqrt 2 - 1 - 2^{-6}\).
First, notice that this step must exist since while \(p_1^{(k)} + p_3^{(k)} \lt \sqrt 2 - 1 - 2^{-6}\), \(p_2^{(k)} \gt \frac{1}{2}\), so
\[p_3^{(k + 1)} = p_3^{(k)} \left( p_3^{(k)} + 2 p_1^{(k)} \right) \lt p_3^{(k)} \le \epsilon\],
i.e., \(p_3^{(k)}\) is (Hulett has weakly) decreasing on this interval. Thus also,
\[p_1^{(k + 1)} = p_1^{(k)} \left( p_1^{(k)} + 2p_2^{(k)} \right) = p_1^{(k)} \left( 1 - p_3^{(k)} + p_2^{(k)} \right) \ge p_1^{(k)} \left( 1.5 - \epsilon \right) \ge p_1^{(k)} \sqrt 2\]
for \(k\) in this interval since \(p_1^{(k)} + p_3^{(k)} \le \sqrt 2 - 1\) implies \(p_2 \gt 1/2\), and since \(\epsilon \le 2^{-10} < 1.5 - \sqrt 2\). Therefore, \(p_1^{(k)}\) is increasing by at least a constant factor every step, and so eventually \(p_1^{(k)} + p_3^{(k)}\) will exceed 1/2. Note also that \(p_i^{(k + 1)} \le 2p_i^{(k)}\) for all \(i\), \(k\), so \(p_1^{(k)}\) is increasing by at least a factor of \(\sqrt 2\) and at most a factor of \(2\) on this interval.
Now, note that
\[\begin{align*} q_1^{(0)} &= 3 ε^{2} \\ q_2^{(0)} &= 1 - 2 ε \\ q_3^{(0)} &= - 3 ε^{2} + 2 ε \end{align*}\]
So, \[\begin{align*} o_1^{(1)} &= 9 ε^{4} - 12 ε^{3} + 6 ε^{2} \\ o_2^{(1)} &= 12 ε^{3} - 10 ε^{2} + 1 \\ o_3^{(1)} &= - 9 ε^{4} + 4 ε^{2} \end{align*}\]
Thus,
\[\begin{align*} q_1^{(1)} &= - 81 ε^{8} + 108 ε^{7} - 162 ε^{6} + 192 ε^{5} - 76 ε^{4} - 12 ε^{3} + 10 ε^{2} \\ q_3^{(1)} &= 81 ε^{8} - 108 ε^{7} + 162 ε^{6} - 192 ε^{5} + 76 ε^{4} \\ \end{align*}\]
Note that since \(\epsilon \le 2^{-10}\), \(q_3^{(1)} \lt \epsilon\). Then, notice that while \(1 \le k \lt K_1\), \(q_3^{(k)}, o_3^{(k)} \lt \epsilon\).
On this interval,
\[\begin{align*} z_3^{(k)} &= p_3^{(k)} \left( p_3^{(k)} + 2 p_2^{(k)} \right) \\ &= p_3^{(k)} \left( 1 - p_1^{(k)} + p_2^{(k)} \right) \\ &\le \epsilon \left( 1 - \epsilon + 1 - 2 \epsilon \right) \\ &= 2\epsilon - 3\epsilon^2 \end{align*}\]
(TODO: verify that \(1 - 2\epsilon\) is correct)
For temporary (TODO) notational efficiency, let \(x = \sqrt 2 - 1 - 2^{-6}\).
Additionally,
\[\begin{align*} z_1^{(k)} &= p_1^{(k)} \left( p_1^{(k)} + 2p_3^{(k)} \right) \\ &\le x \left( x + 2 \epsilon \right) \\ &\le x^2 + 2 x \epsilon \end{align*}\]
Thus,
\[\begin{align*} o_3^{(k)} &= q_3^{(k - 1)} \left( z_3^{(k)} + z_1^{(k)} \right) + q_1^{(k-1)} \left( z_3^{(k)} \right) \\ &\le \epsilon \left( 2\epsilon - 3\epsilon^2 + x^2 + 2 x \epsilon \right) + x \left( 2\epsilon - 3\epsilon^2 \right) \\ &= -3 \epsilon^3 + (2 - x) \epsilon^2 + (x^2 + 2x) \epsilon \\ &\lt \epsilon \qquad \text{(Since $\epsilon \le 2^{-10}$; also, TODO)} \end{align*}\]
Thus, we also know that \(q_3^{(k)} \lt \epsilon\) since \(q_3^{(k)} \lt o_3^{(k)} \lt \epsilon\) (same argument as for \(p\)s)