Condorcet Jury Theorem

Condorcet Jury Theorem

Setup: From One Juror to Many

In Lectures 11 and 12 we analyzed a single juror's decision. She follows her signal (under the assumption $1-p < q \leq p$), achieves $\text{Prob}(\text{correct}) = p$, and makes mistakes with probability $1-p$. Now we extend the framework to a panel of $N$ jurors voting collectively.

The assumptions are:

• $N$ is odd.
• Each juror has prior $\text{Prob}(G) = \text{Prob}(I) = \frac{1}{2}$.
• Each juror receives a binary signal with accuracy $p > \frac{1}{2}$.
• Signals are independent across jurors conditional on the true state.
• Each juror follows her signal: convicts after $g$, acquits after $i$.
• There is no conflict of interest: all jurors share the objective of convicting the guilty and acquitting the innocent.
• The jury reaches a verdict by a social choice function applied to the $N$ individual votes.

We analyze two voting rules: simple majority and unanimity.

The $N=3$ Case Under Simple Majority

With 3 jurors and simple majority, the jury convicts if at least 2 of the 3 jurors vote to convict — that is, if at least 2 receive signal $g$.

There are $2^3 = 8$ possible signal profiles, each with a computable probability given the true state. Let us work through the case $\theta = G$ first.

When the defendant is guilty, each juror independently gets signal $g$ with probability $p$ and signal $i$ with probability $1-p$. The eight profiles and their probabilities are:

• $ggg$: probability $p^3$. All three convict — majority convicts. Correct.
• $ggi$: probability $p^2(1-p)$. Two convict — majority convicts. Correct.
• $gig$: probability $p^2(1-p)$. Two convict — majority convicts. Correct.
• $igg$: probability $p^2(1-p)$. Two convict — majority convicts. Correct.
• $gii$: probability $p(1-p)^2$. One convicts, two acquit — majority acquits. Mistake.
• $igi$: probability $p(1-p)^2$. Mistake.
• $iig$: probability $p(1-p)^2$. Mistake.
• $iii$: probability $(1-p)^3$. None convicts — majority acquits. Mistake.

The jury convicts (correctly) in exactly the four profiles with $\geq 2$ signals of $g$:

$\text{Prob}(\text{convict} \mid G) = p^3 + 3p^2(1-p) = p^2\bigl[p + 3(1-p)\bigr] = p^2(3 - 2p) = 3p^2 - 2p^3$

When the defendant is innocent, by the symmetry of the signal distribution (swap $p$ and $1-p$, and $g$ and $i$), the jury acquits correctly when $\geq 2$ signals are $i$:

$\text{Prob}(\text{acquit} \mid I) = p^3 + 3p^2(1-p) = 3p^2 - 2p^3$

The overall probabilities are:

$\text{Prob}(\text{correct}) = \frac{1}{2}(3p^2 - 2p^3) + \frac{1}{2}(3p^2 - 2p^3) = 3p^2 - 2p^3$

$\text{Prob}(\text{mistake}) = 1 - (3p^2 - 2p^3) = 1 - 3p^2 + 2p^3$

Is Three Jurors Better Than One?

We have:

• $N=1$: $\text{Prob}(\text{correct}) = p$
• $N=3$: $\text{Prob}(\text{correct}) = 3p^2 - 2p^3$

Claim: $3p^2 - 2p^3 > p$ for all $p \in \left(\frac{1}{2}, 1\right)$.

Proof: Consider the difference:

$3p^2 - 2p^3 - p = p(3p - 2p^2 - 1)$

Factor the quadratic $-2p^2 + 3p - 1 = -(2p^2 - 3p + 1) = -(2p-1)(p-1) = (2p-1)(1-p)$. Therefore:

$3p^2 - 2p^3 - p = p(2p-1)(1-p)$

When $p \in \left(\frac{1}{2}, 1\right)$: $p > 0$, $(2p-1) > 0$, and $(1-p) > 0$. The product of three positive numbers is positive. Therefore $3p^2 - 2p^3 > p$.

A three-person majority always outperforms a single juror (as long as $p > \frac{1}{2}$). The gain is exactly $p(2p-1)(1-p)$, which is zero at $p = \frac{1}{2}$ (no signal information) and at $p = 1$ (perfect information), and positive everywhere in between.

The Condorcet Jury Theorem

This single comparison generalizes dramatically. Let $Q(N)$ denote the probability of a correct collective decision with an odd jury of size $N$ under simple majority, with all jurors following their signals.

Condorcet Jury Theorem (Condorcet, 1785): For any $p \in \left(\frac{1}{2}, 1\right)$ and any odd $N \geq 1$:

Part 1: $Q(N) > p$ for all odd $N \geq 3$. Any jury of at least three members under simple majority outperforms a single decision-maker.

Part 2: $Q(N+2) > Q(N)$ for all odd $N$. Larger juries (odd) make better decisions — adding two jurors always strictly improves accuracy.

Part 3: $Q(N) \to 1$ as $N \to \infty$. As the jury size grows without bound, the probability of a correct decision approaches certainty.

The third part is the most striking. No single juror, however well-informed ($p < 1$), can achieve certainty. But a sufficiently large jury can get arbitrarily close to it, even if each member is only marginally better than random ($p$ only slightly above $\frac{1}{2}$). This is a form of the Law of Large Numbers applied to collective decision-making: with enough independent votes, the signal of the majority overwhelms any individual noise.

The theorem justifies the use of collective decision-making over individual dictatorships not merely on grounds of fairness, but on grounds of accuracy. Groups can be wiser than individuals, not because any individual is especially wise, but because aggregation pools information.

Connection to Hypothesis Testing

The jury's decision problem has a direct parallel in statistics. Think of the jury as performing a hypothesis test:

Null hypothesis: the defendant is innocent ($I$).

Alternative hypothesis: the defendant is guilty ($G$).

Type I error (false positive): convicting an innocent defendant — $\text{Prob}(\text{convict} \mid I)$.

Type II error (false negative): acquitting a guilty defendant — $\text{Prob}(\text{acquit} \mid G)$.

Different social choice functions correspond to different tradeoffs between Type I and Type II errors. Simple majority minimizes the total error probability (it gives the highest $Q(N)$), but it treats the two types of error symmetrically. Unanimity, as we see next, dramatically reduces Type I error at the cost of dramatically increasing Type II error.

The analogy is useful because it connects jury design to a well-developed statistical theory of testing and provides vocabulary — Type I and Type II errors — that is standard in legal and policy discussions.

Unanimity Rule with $N=3$

Under unanimity, the jury convicts only if all three jurors vote to convict — that is, only if all three signals are $g$.

Probability of conviction given guilt: Only profile $ggg$ leads to conviction. $\text{Prob}(\text{convict} \mid G) = p^3$

Probability of acquittal given innocence: All profiles except $ggg$ lead to acquittal when $\theta = I$. But from the innocent defendant's perspective, the signals are $g$ with probability $1-p$ and $i$ with probability $p$. So all three being $g$ has probability $(1-p)^3$, and at least one being $i$ has probability $1-(1-p)^3$. $\text{Prob}(\text{acquit} \mid I) = 1-(1-p)^3$

Type I error (wrongful conviction): $\text{Prob}(\text{convict} \mid I) = (1-p)^3$

Type II error (wrongful acquittal): $\text{Prob}(\text{acquit} \mid G) = 1-p^3$

Overall accuracy: $\text{Prob}(\text{correct}) = \frac{1}{2} \cdot p^3 + \frac{1}{2} \cdot \bigl[1-(1-p)^3\bigr]$

Overall mistake probability: $\text{Prob}(\text{mistake}) = \frac{1}{2}(1-p)^3 + \frac{1}{2}(1-p^3)$

Comparing the Two Rules at $p = 0.6$

Let us evaluate both rules numerically at the specific accuracy $p = 0.6$.

Simple majority, $N=3$: $\text{Prob}(\text{correct}) = 3(0.6)^2 - 2(0.6)^3 = 3(0.36) - 2(0.216) = 1.08 - 0.432 = 0.648$ $\text{Prob}(\text{mistake}) = 1 - 0.648 = 0.352$

Breaking this down by error type: $\text{Prob}(\text{convict} \mid G) = 0.648$, $\text{Prob}(\text{acquit} \mid I) = 0.648$, $\text{Prob}(\text{convict} \mid I) = 0.352$, $\text{Prob}(\text{acquit} \mid G) = 0.352$.

Both types of error occur with equal probability $35.2\%$ under simple majority.

Unanimity, $N=3$: $\text{Prob}(\text{convict} \mid I) = (1-0.6)^3 = (0.4)^3 = 0.064$ $\text{Prob}(\text{acquit} \mid G) = 1-(0.6)^3 = 1-0.216 = 0.784$ $\text{Prob}(\text{convict} \mid G) = (0.6)^3 = 0.216$ $\text{Prob}(\text{acquit} \mid I) = 1-(0.4)^3 = 1-0.064 = 0.936$ $\text{Prob}(\text{correct}) = \frac{1}{2}(0.216) + \frac{1}{2}(0.936) = 0.108 + 0.468 = 0.576$ $\text{Prob}(\text{mistake}) = 1 - 0.576 = 0.424$

The comparison reveals a stark tradeoff. Unanimity reduces Type I error from $35.2\%$ to $6.4\%$ — an enormous improvement in protecting the innocent. But it increases Type II error from $35.2\%$ to $78.4\%$ — most guilty defendants go free. Overall accuracy falls from $64.8\%$ to $57.6\%$: unanimity is less accurate in total, but it offers much stronger protection against the specific error of convicting the innocent.

Which Rule Is Better?

There is no objective answer to this question. The answer depends on how society values the two types of error — convicting the innocent versus acquitting the guilty.

If the two errors are considered equally bad — if wrongfully convicting an innocent person is exactly as bad as wrongfully acquitting a guilty one — then simple majority is better: it achieves higher overall accuracy.

If wrongful conviction is considered catastrophically worse than wrongful acquittal — if convicting an innocent person is a far graver injustice than letting a guilty one go free — then unanimity is better: it keeps the Type I error rate very low.

The legal principle "innocent until proven guilty" and the standard of proof "beyond a reasonable doubt" encode a strong asymmetric preference against Type I errors. The US criminal justice system requires unanimous consent of 12 jurors precisely because it takes this asymmetry seriously. Lecture 14 develops this connection further.