Bias Control and Injection for Multi-Agent Aggregation

Sungjae Lee; Soheil Abbasloo; Jabin Koo; Jungseul Ok

cs.AIOct 15, 2026

Bias Control and Injection for Multi-Agent Aggregation

Authors: Sungjae Lee, Soheil Abbasloo, Jabin Koo, Jungseul Ok

Organizations: POSTECH · Microsoft Research

Abstract

Multi-agent reasoning with large language models (LLMs) raises a central question for collective decision-making: when agents disagree, whose decision should be trusted? Resolving such disagreements is difficult because agents' decisions are susceptible to non-semantic biases such as majority pressure and response-style preferences, which can obscure the quality of the underlying reasoning. Our analysis shows how bias can be both mitigated and exploited: bias control improves initial answer selection, while bias injection elicits decision instability that serves as a reliability signal. Building on these findings, we introduce a test-time aggregation framework that requires neither validation data nor prior knowledge of relative model strength. The framework implements bias control by presenting one style-normalized reasoning trace per distinct answer, and bias injection by placing each aggregating agent's initially selected answer in the minority. Overall, our framework achieves higher average accuracy and better token efficiency than competing non-oracle aggregation methods across three mathematical reasoning benchmarks and four heterogeneous LLM combinations. It also improves final-answer accuracy when applied to multi-agent debate and Mixture-of-Agents, without modifying their interaction or synthesis processes.

Figures & tables

Figure 1: Correct-answer availability and majority-vote accuracy. Majority voting is effective overall but still overlooks correct answers with limited support.

Figure 2: Illustration of bias control and bias injection. Bias control forms an initial decision by comparing one style-normalized trace per distinct answer under balanced frequency. Bias injection then places the selected answer in the relative minority to assess its reliability. Bias is mitigated through control during decision formation and then exploited through injection to assess reliability.

Figure 3: Decision instability under bias injection. Bars show flip rates conditioned on initial correctness. Adverse-frequency yields the largest gap, offering a more selective signal for reliability assessment.

Figure 4: Overview of the proposed aggregation framework. Responses from the reasoning history are grouped by answer. When agents disagree, the framework uses bias control to form LLM decisions and injects adverse-frequency bias to test their reliability. Final aggregation combines decision holds observed under bias injection with cross-agent support, without modifying upstream reasoning.

Figure 5: Accuracy–cost trade-offs under independent sampling. Each panel shows one three-model combination across response budgets. Accuracy is the macro-average across OlymMATH, AIME, and HMMT, and inference cost is measured by total input-and-output tokens per problem. Stars mark Ours. Most Ours configurations with

k\in\{2,3,4\}

lie on or near the observed Pareto frontier.

Figure 6: Recovery and retention relative to majority voting. Rates are macro-averaged across benchmarks. Ours achieves the highest recovery while retaining most MV-correct answers.

Figure 7: Integration with multi-agent reasoning. Arrows show pooled-accuracy gains from adding Ours. R/L denote Debate rounds/MoA layers.

Figure 8: Cross-generator convergence in Phi. Cosine distances between matched traces.

Condition	Hold rate (%)	Held acc. (%)	$\Delta$ (pp)
Repeated inference	53.7	40.6	$+10.3$
Position swap	52.4	42.4	$+12.1$
Adverse frequency	37.5	46.9	$+16.6$

Table 3: Reliability of retained decisions. Hold rate is the fraction unchanged on the second pass; held accuracy is their conditional accuracy.

\Delta

is the gain over the 30.3% controlled baseline.

Condition	Flip $\mid$ incorrect	Flip $\mid$ correct	Gap
Repeated inference	44.2	11.7	32.5
Position swap	48.3	14.0	34.3
Adverse freq. (exact copy)	56.2	18.9	37.3
Adverse freq. (distinct)	59.8	19.1	40.7

Table 4: Content control for bias injection. Flip rates (%) are conditioned on initial correctness. Gap is the incorrect–correct difference in percentage points.

Figure 9: Component ablations. Answer accuracy for (a) bias control and (b) bias injection.

Appendix figures & tables10 assets

Supplementary material from the paper’s appendix.

Appendix

Parameter	Value
CoT generation
Model	Temperature	Top- $k$
Phi-4-reasoning-plus	0.8	50
Qwen3-8B	0.6	20
GLM-Z1-9B-0414	0.6	40
NVIDIA-Nemotron-Nano-9B-v2	0.6	Disabled

Appendix

Table 5: Hyperparameters used in the experiments.

Model combination	Style normalizer	OlymMATH	AIME	HMMT	Average	$\Delta_{\mathrm{Qwen}}$ (pp)
Qwen–Phi–GLM	Qwen3-8B	89.0	81.7	56.7	75.8	–
	GLM-Z1-9B	89.0	83.3	58.3	76.9	$+1.1$
	Nemotron-Nano-9B	93.0	81.7	56.7	77.1	$+1.3$
Qwen–Phi–Nemotron	Qwen3-8B	91.0	83.3	58.3	77.6	–
	GLM-Z1-9B	92.0	81.7	58.3	77.3	$-0.3$
	Nemotron-Nano-9B	90.0	80.0	63.3	77.8	$+0.2$

Appendix

Table 6: Final-answer accuracy with alternative style normalizers. Average is the unweighted mean across the three benchmarks, and

\Delta_{\mathrm{Qwen}}

is its difference from the matched Qwen3-8B run. Alternative normalizers retain comparable performance across model combinations.

Model combination	Method	Tie-breaker	OlymMATH	AIME	HMMT	Average
Qwen–Phi–GLM	SLM-MUX	Test-set Best-SC (oracle)	87.0	80.0	53.3	73.4
	SLM-MUX	Pooled MV (main)	86.0	78.3	48.3	70.9
	Ours	–	90.3	83.3	58.9	77.5
Qwen–Phi–Nemotron	SLM-MUX	Test-set Best-SC (oracle)	88.0	78.3	55.0	73.8
	SLM-MUX	Pooled MV (main)	87.0	76.7	51.7	71.8
	Ours	–	90.7	82.8	59.4	77.6

Appendix

Table 7: SLM-MUX under alternative confidence tie-breakers. The two SLM-MUX variants differ only in confidence tie-breaking and use identical responses and inference cost. Test-set Best-SC selects the tied model with the highest test-set SC@

k

accuracy and serves as an oracle reference; Pooled MV is the label-free rule used in the main experiments. Ours is included for comparison, and bold marks the highest average.

Figure 10: Cross-generator convergence across representation spaces. Panels show cosine distances in the Phi, GLM, and Qwen representation spaces between matched traces from different generators before and after formatting removal or full normalization. Traces are matched by problem and final answer. Points below the diagonal indicate reduced distance. Full normalization produces larger reductions than formatting removal, particularly in the Phi and GLM spaces.

Feature channel	Original	Normalized	$\Delta$
Formatting	94.1	72.2	$-21.9$
Rhetorical markers	88.8	79.7	$-9.1$
Character 3-grams	99.2	91.9	$-7.3$
Bag of words	94.6	88.5	$-6.1$
Function words	86.6	80.9	$-5.7$
Judge representation	100.0	97.2	$-2.8$

Appendix

Table 9: Source-generator attribution before and after style normalization. Values report three-way classification accuracy under grouped five-fold cross-validation;

\Delta

uses unrounded values. Normalization attenuates formatting cues most strongly, while substantial source signal remains.

	$H(a)=0$	$H(a)>0$
$S(a)\leq J/2$	13.3	23.5
$S(a)>J/2$	18.1	49.1

Appendix

Table 10: Answer accuracy by hold and support. Conditional accuracy (%) by

H(a)

and

S(a)

J

denotes the number of agents.

Original trace	Style-normalized trace
To determine the number of ways to divide a $20\times 24$ rectangle into $4\times 5$ rectangles, we first verify the feasibility of such a tiling: Feasibility Check $\bullet$ Area of the large rectangle: $20\times 24=480$ $\bullet$ Area of each tile: $4\times 5=20$ $\bullet$ Number of tiles needed: $\frac{480}{20}=24$ Since the area is divisible and the tile dimensions are smaller than the rectangle, a tiling is possible. However, we must also ensure that the tiling can be done with the given dimensions, accounting for the fact that the tile can be rotated (i.e., placed as $4\times 5$ or $5\times 4$ ).	To determine the number of ways to divide a $20\times 24$ rectangle into $4\times 5$ rectangles, we first verify the feasibility of such a tiling. The area of the large rectangle is $20\times 24=480$ . Each tile has an area of $4\times 5=20$ , so the number of tiles required is $480\div 20=24$ . Since the area is divisible and the tile dimensions are smaller than the rectangle, a tiling is possible. However, we must also ensure that the tiling can be done with the given dimensions, accounting for the fact that the tile can be rotated.
Key Insight: Tiling via Segmentations To tile the entire rectangle, we can divide the width (24 units) into segments of either 4 or 5 units. Each such segment can then be filled with tiles of the appropriate orientation: $\bullet$ A segment of 4 units wide can be tiled with vertical tiles (5 units tall, 4 units wide). $\bullet$ A segment of 5 units wide can be tiled with horizontal tiles (4 units tall, 5 units wide). Since the height of the rectangle is 20, which is divisible by both 4 and 5, we can tile each such segment with tiles of the correct orientation. Therefore, the problem reduces to finding the number of compositions of 24 into parts of 4 and 5, where the order of the segments matters (since different arrangements correspond to different tilings).	To tile the entire rectangle, we can divide the width (24 units) into segments of either 4 or 5 units. Each such segment can then be filled with tiles of the appropriate orientation. A segment of 4 units wide can be tiled with vertical tiles (5 units tall, 4 units wide). A segment of 5 units wide can be tiled with horizontal tiles (4 units tall, 5 units wide). Since the height of the rectangle is 20, which is divisible by both 4 and 5, we can tile each such segment with tiles of the correct orientation. Therefore, the problem reduces to finding the number of compositions of 24 into parts of 4 and 5, where the order of the segments matters.
Counting the Number of Valid Compositions We define a function $C(n)$ as the number of compositions of $n$ into parts of 4 and 5. This satisfies the recurrence: $C(n)=C(n-4)+C(n-5)$ with base cases: $\bullet$ $C(0)=1$ (the empty composition), $\bullet$ $C(n)=0$ for $n<0$ .	We define a function $C(n)$ as the number of compositions of $n$ into parts of 4 and 5. This satisfies the recurrence: $C(n)=C(n-4)+C(n-5)$ with base cases: $\bullet$ $C(0)=1$ (the empty composition), $\bullet$ $C(n)=0$ for $n<0$ .
We compute $C(24)$ using this recurrence: $\begin{aligned} C(0)&=1\\ C(1)&=0\\ C(2)&=0\\ C(3)&=0\\ C(4)&=C(0)=1\end{aligned}$	We compute $C(24)$ using this recurrence: $\begin{aligned} C(0)&=1\\ C(1)&=0\\ C(2)&=0\\ C(3)&=0\\ C(4)&=C(0)=1\end{aligned}$
$\begin{aligned} C(5)&=C(1)+C(0)=0+1=1\\ C(6)&=C(2)+C(1)=0+0=0\\ C(7)&=C(3)+C(2)=0+0=0\\ C(8)&=C(4)+C(3)=1+0=1\\ C(9)&=C(5)+C(4)=1+1=2\end{aligned}$	$\begin{aligned} C(5)&=C(1)+C(0)=0+1=1\\ C(6)&=C(2)+C(1)=0+0=0\\ C(7)&=C(3)+C(2)=0+0=0\\ C(8)&=C(4)+C(3)=1+0=1\\ C(9)&=C(5)+C(4)=1+1=2\end{aligned}$
$\begin{aligned} C(10)&=C(6)+C(5)=0+1=1\\ C(11)&=C(7)+C(6)=0+0=0\\ C(12)&=C(8)+C(7)=1+0=1\\ C(13)&=C(9)+C(8)=2+1=3\\ C(14)&=C(10)+C(9)=1+2=3\end{aligned}$	$\begin{aligned} C(10)&=C(6)+C(5)=0+1=1\\ C(11)&=C(7)+C(6)=0+0=0\\ C(12)&=C(8)+C(7)=1+0=1\\ C(13)&=C(9)+C(8)=2+1=3\\ C(14)&=C(10)+C(9)=1+2=3\end{aligned}$

Appendix

Table 11: Complete original and normalized traces for the correct answer

\boxed{6}

Original trace	Style-normalized trace
We will show that no “exotic” tiling exists–in fact, there is only one way to cover a $20\times 24$ rectangle by $4\times 5$ rectangles (when rotations and reflections count as different, so that even a “mirrored” copy would be considered distinct–but here there is no choice at all).	We aim to demonstrate that no unconventional tiling is possible–in fact, there is only one way to cover a $20\times 24$ rectangle using $4\times 5$ rectangles (when rotations and reflections count as distinct, so that even a mirrored copy would be considered different–but here there is no choice at all).
Let us explain. A $4\times 5$ rectangle has area 20. Since $\begin{gathered}20\times 24=480\\ \qquad\text{and}\qquad 480\div 20=24,\end{gathered}$ any tiling must use exactly 24 tiles.	First, let us explain. A $4\times 5$ rectangle has an area of 20. Given that $20\times 24=480$ and 480 divided by 20 equals 24, any tiling requires exactly 24 tiles.
Moreover, since all tiles are congruent (up to rotation) and their sides must be parallel to the sides of the large rectangle, one may show (by a standard “grid-forcing” argument) that if a rectangle is tiled by congruent (non-square) rectangles then the tiling “extends” to a complete set of horizontal and vertical lines. In other words, every tiling by congruent non-square rectangles is necessarily a grid tiling.	Moreover, since all tiles are congruent (up to rotation) and their sides must align with the sides of the large rectangle, it can be shown (via a standard grid-forcing argument) that tiling a rectangle with congruent non-square rectangles necessitates a grid-like structure. In other words, every tiling by congruent non-square rectangles is necessarily a grid tiling.
Now, note that a $4\times 5$ rectangle has two possible orientations. In one orientation the longer side (of length 5) is vertical and the shorter side (of length 4) is horizontal; in the other the roles are reversed. But look at the dimensions of the big rectangle: $\bullet$ The vertical side is 20 and the horizontal side is 24.	Now, note that a $4\times 5$ rectangle can be oriented in two ways: one with the longer side (length 5) vertical and the shorter side (length 4) horizontal, and the other with these dimensions reversed. But consider the dimensions of the large rectangle: $\bullet$ The vertical side is 20 and the horizontal side is 24.
If we try to fit a $4\times 5$ rectangle with the 5-side vertical, then its horizontal side is 4. Since 24 is divisible by 4 ( $24\div 4=6$ ) but 20 is not divisible by 4 ( $20\div 4=5$ , and then $5\times 4=20$ would force the other dimension to be 5, not 20), the only possibility is to have every tile placed in the orientation in which the 5-side is vertical and the 4-side is horizontal. In that case the vertical cell has height 5 and the horizontal cell has width 4. Then the tiling is given by	If we attempt to fit a $4\times 5$ rectangle with the 5-side vertical, its horizontal side would be 4. Since 24 is divisible by 4 ( $24\div 4=6$ ) but 20 is not divisible by 4 ( $20\div 4=5$ , and $5\times 4=20$ would force the other dimension to be 5, not 20), the only viable orientation is one in which the 5-side is vertical and the 4-side is horizontal. In this case, the vertical cell has height 5 and the horizontal cell has width 4. Then the tiling is achieved by
$\bullet$ dividing the 20-side into $20\div 5=4$ equal segments, and $\bullet$ dividing the 24-side into $24\div 4=6$ equal segments. This produces a $4\times 6$ grid of $5\times 4$ rectangles.	$\bullet$ dividing the 20-side into $20\div 5=4$ equal segments, and $\bullet$ dividing the 24-side into $24\div 4=6$ equal segments. This produces a $4\times 6$ grid of $5\times 4$ rectangles.

Appendix

Table 12: Complete original and normalized traces for the competing answer

\boxed{1}

Panel of Figure 11 — Figure 11: Style-normalization prompt. The prompt instructs the rewriter to change only wording and formatting while preserving reasoning content and final answers; a length constraint limits deviation from the original trace.

Panel of Figure 12 — Figure 12: Decision prompts before and after adverse-frequency bias injection. (a) Bias-controlled presentation with one trace per answer hypothesis. (b) After the controlled decision selects an answer other than $197$ , an additional existing trace makes $197$ the higher-frequency competitor without changing the problem, answer hypotheses, or decision instruction.

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