Conjecture on the optimal strategy for dependent Bernoulli variables

At least 7 years old · documented by 2018

Let $I_{1},I_{2},...,I_{n}$ be $n$ dependent Bernoulli random variables, and let

p_{i,k}:=P(I_{i}=1\mid I_{k}=1).

Optimal-strategy conjecture. After observing $I_{k}=1$ , the player whose turn it is should give up his turn to his opponent if and only if $p_{i,k}<\frac{1}{2}$ for all $i>k$ . This predicts that the adversarial Last-Success-Problem remains simple for dependent variables, with the decision determined by the conditional success probabilities of the remaining variables. The conjecture is presented as a prediction in the source; no resolution is given.

References

Primary source

José María Grau Ribas, “Concerning an adversarial version of the Last-Success-Problem”, arXiv:1812.05381 (2019).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

An unverified reader-written attempt claims a three-variable counterexample that would disprove the conjecture in both directions, while the published source gives no resolution.

Grau Ribas (2018) formulates the conjecture that, after observing $I_k=1$ , passing is optimal exactly when every later conditional success probability $p_{i,k}$ is below $\frac{1}{2}$ . The paper presents this only as a prediction for dependent Bernoulli variables.

Known results

For independent Bernoulli variables, the analogous strategy is proved: pass exactly when no remaining success probability is at least $\frac{1}{2}$ (Grau Ribas, 2018).

Posted attempt

A reader-written attempt claims a complete counterexample using three Bernoulli variables with full joint support: it gives examples where the conjecture prescribes passing but retaining is optimal, and vice versa, and examples showing that one-coordinate conditional probabilities do not determine the optimal action. The calculation has not been independently verified.

Current status (as of August 2026): The conjecture has an unverified claimed counterexample and therefore is not settled; absent verification, no published proof or disproof is recorded.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.