Conjecture on the optimal strategy for dependent Bernoulli variables

The proposed rule fails in both directions, even for three Bernoulli variables with full joint support. Moreover, identical future conditional success probabilities can require opposite optimal decisions.

Take $n=3$, let $I_1$ be an independent fair Bernoulli variable, and condition on $I_1=1$. Write

$$p_{ab}=\mathbb P(I_2=a,I_3=b\mid I_1=1), \qquad a,b\in\{0,1\}.$$

If the current player retains the turn, optimal backward induction gives the winning probability

$$V=p_{01}+\max\{p_{10},p_{11}\}. \tag{1}$$

Indeed, on $I_2=0$, the player cannot pass and wins exactly when $I_3=1$, contributing $p_{01}$. On $I_2=1$, retaining wins on outcome $11$, whereas passing wins on outcome $10$, contributing the larger of $p_{11}$ and $p_{10}$. Passing immediately after $I_1=1$ instead wins with probability $1-V$.

First, take

$$(p_{00},p_{01},p_{10},p_{11}) = \frac1{10}(3,3,3,1).$$

Then

$$\mathbb P(I_2=1\mid I_1=1) = \mathbb P(I_3=1\mid I_1=1) = \frac25<\frac12.$$

The conjecture therefore prescribes passing. But

$$V=\frac3{10}+\max\left\{\frac3{10},\frac1{10}\right\} =\frac35,$$

so retaining wins with probability $3/5$, while passing wins with probability $2/5$.

Conversely, take

$$(p_{00},p_{01},p_{10},p_{11}) = \frac1{20}(7,1,7,5).$$

Here

$$\mathbb P(I_2=1\mid I_1=1)=\frac35>\frac12,$$

so the conjecture prescribes retaining. Nevertheless,

$$V=\frac1{20}+ \max\left\{\frac7{20},\frac5{20}\right\} = \frac25,$$

and passing is strictly optimal, with winning probability $3/5$.

Finally, taking

$$(p_{00},p_{01},p_{10},p_{11}) = \frac15(2,1,1,1)$$

gives the same two conditional success probabilities $(2/5,2/5)$ as the first example, but now

$$V=\frac25,$$

so the optimal decision is the opposite. Thus even the complete list of future one-coordinate conditional success probabilities does not determine the optimal action. All eight joint outcomes of $(I_1,I_2,I_3)$ have strictly positive probability in every example.