Constructing valid counterexamples

Build finite probability models that satisfy a premise and violate a proposed conclusion, then audit exactly what the witness establishes.

20 minutes · advanced · Original Poker4Quant lesson

What you will learn

Construct and verify a counterexample to a universal implication while distinguishing irrelevant candidates, confirming examples and unsupported reverse claims.

Prerequisites: Expected value as a weighted average · Variance and standard deviation · Testing event independence · Approximation error and decision margins

A universal implication has two obligations for a witness

Suppose a claim says that every model satisfying premise P also satisfies conclusion Q. A counterexample must be an allowed model with P true and Q false. A case with a failed premise is irrelevant to that implication, even if its conclusion is also false. A case with both P and Q true merely confirms one instance. The construction needs the stated domain and all assumptions, not just an eye-catching numerical result.

One valid counterexample refutes the universal implication. It does not show that every model fails Q, prove the reverse implication, or estimate the fraction of real situations that fail. Searching many cases without finding a witness is not an unrestricted proof. Exhaustive verification can establish a claim over a genuinely complete finite domain, but the domain coverage itself must be justified. Our bounded integer editor is a search tool, not a claim that every real-valued model was checked.

Positive expectation is compatible with an individual loss

Consider four equally likely payoff cases. The proposed claim is that positive mean forces every case payoff to be nonnegative. The candidate(−2,2,2,2)has mean 1, so it satisfies the premise. One case is a loss, with probability1/4, and the minimum payoff is−2. The conclusion therefore fails. Every arithmetic step is exact for this specified finite distribution.

A candidate such as(−2,0,0,0)also contains a loss but has mean−0.5. It does not satisfy the premise and is not a counterexample to this particular claim. Likewise,(1,1,1,1)has positive mean but no loss, so it is a confirming case, not a proof that the universal claim is true. Fake-chip ledgers in the activity settle each alternative from the same wallet/treasury baseline; outcomes are not added together as repeated realized profit.

Matching one summary does not identify a distribution

The second claim says that any four-case distribution with mean 1 has the same variance as reference B = (−1, 1, 1, 3). The reference deviations from mean 1are(−2,0,0,2), with mean squared deviation2. Construct A=(1,1,1,1). Its mean is also1but variance is0. Equal means therefore do not determine equal dispersion. A model with a different mean would not meet the premise and would not answer this claim.

A counterexample can be deliberately simple: a constant distribution isolates the freedom left by a mean constraint. It is not an empirical claim about typical bets or people. The numerical construction also does not imply the opposite universal rule that equal means always require different variances: choosing A equal to B gives the same mean and variance. Valid refutation and the reverse overgeneralization must stay separate.

Zero covariance is weaker than independence

For four equally likely atoms, fix X=(−1,0,0,1)and construct binary Y=(1,0,0,1). Then E[X]=0, E[Y]=1/2and E[XY]=(−1+0+0+1)/4=0, so covariance is0. Yet Y is determined by whether X is nonzero. Specifically P(X=0,Y=1)=0, while P(X=0)P(Y=1)=(1/2)(1/2)=1/4. One failed joint-product equality disproves independence.

This is a valid witness against the claim that covariance 0forces independence in this allowed model class. It is not a claim that independence never occurs: constant Y is independent of X here and also has covariance 0. No bivariate-normal assumption is present; adding distributional restrictions would create a different proposition. We construct an exact joint model rather than estimate a covariance from a sample.

Counterexample construction and proof by contradiction are related logical tools but are not interchangeable labels for every argument. A counterexample supplies one qualifying instance that violates a universal conclusion. A contradiction proof assumes a proposition and derives a false consequence by valid steps. The linked MIT video provides additional proof-method context; its algebra examples are not the activity’s payoff or joint-law constructions.

Worked example

Evaluate three candidates against “every positive-mean four-case payoff has no negative case”: A=(−2,0,0,0), B=(1,1,1,1), C=(−2,2,2,2).

A has mean−0.5, so the premise fails. Its negative case is irrelevant to this implication.
B has mean 1and no negative case. It satisfies both premise and conclusion; it confirms only this instance.
C has mean 1but one negative case, probability1/4. It satisfies the premise while violating the conclusion.
Ctherefore refutes the universal claim. AandBhelp distinguish a failed premise from a valid witness and from a confirming example.

Verify domain, every premise and the failed conclusion. One valid witness is enough for refutation, but it is not a universal reverse theorem or an observed failure frequency.

Hands-on: Construct a qualifying counterexample and audit its scope

Edit four-case payoff or binary joint models, inspect every case after final edits, test distinct candidates and verify a true premise, failed conclusion and correctly scoped refutation.

Reading counterexample workshop…

Common mistakes

Presenting a loss without checking the positive-mean premise.
Changing an assumption to manufacture a witness outside the claim’s domain.
Treating many confirming examples as a proof over an unrestricted domain.
Concluding that all models fail from one counterexample.
Reading exact constructed case weights as estimated real-world frequencies.
Assuming zero covariance proves independence without checking distributional assumptions.

Keep these ideas

A valid witness has premise true and conclusion false.
An irrelevant candidate and a confirming example do not refute the implication.
One summary statistic can leave important distributional structure undetermined.
State the precise scope of a refutation and distinguish search from exhaustive proof.