Two-player Variant of Collatz procedure?

Question

Recently i got interested at a Relaxation of Collatz Conjecture. This goes just like the normal Collatz procedure, except when $x$ is even, you have the choice of applying either $x \rightarrow 3x+1$ or $x \rightarrow x/2$. The idea of adding the ability to chose to Collatz procedure intrigues me. This relaxed Collatz procedure can be treated as a game where the goal is to reach $1$ as fast as possible. It turns it into a $1$-player game.

Here, i went one-step further, and created a two-player game based on Relaxed Collatz Procedure. The game has two player, the minimizer and the maximizer. The game went as follow

1. Start with a initial number $x_1$, and the first move goes to the minimizer, and the game alternates between the maximizer and the minimizer's turn
2. At each timestep $t < T$, the current player can either set $x_{t+1} = 3x_t+1$ or $x_{t+1} = \frac{x_t}{2^k} $, where $k$ is natural number, and the player can choose any $k$ as long as we end up with integer.
3. At the timestep $T$, the game terminates.

The goal of the minimizer is to, as the name suggest, minimize $x_T$, while the maximizer's goal is to maximize $x_T$. Let $X(x_1, T)$ be the value of $x_T$ given a stopping time of $T$ and a starting of $x_0$ assuming both the minimizer and maximizer plays optimally. Do we have any approximation of $X(x_1, T)$? Does the game favors the minimizer or the maximizer?

[EDIT] In the original game, the maximizer have so much power, so i add a rule, that the maximizer could only set $x_{t+1} = 3x_t + 1$ if $x_t$ is odd

Answer 1

(I'm ignoring the bound on the number of turns in this answer.)

Well, if the minimizer ever gives the maximizer an even number, clearly the maximizer can always win by just always choosing the $3x+1$ option. (That way, the minimizer always obtains an odd number $x$ and has no choice but to give back $3x+1$, which will again be even.) In this situation, the game state will roughly triple in size at each turn from the first point at which the maximizer obtained an even number.

Thus, the only hope for the minimizer is if the initial number is even and they always give back an odd number (so they have to divide by the largest $2$-power possible). Hence, such a starting condition just leads to the usual Collatz sequence (where successive turns of dividing out $2$ are condensed to a single turn). In particular, the dynamics of this case are not easier than the usual Collatz problem.