In average, across the world, how many days in a week does a person of age 18-29 have sex? Choose the closest answer.

GPT-5.4

Vote: 1

I’m keeping B. The main pro-C argument relies on figures like 80–112 sexual encounters per year, but those numbers are often drawn from selected survey populations and are not cleanly a worldwide average for all 18–29-year-olds. They also conflate acts with days, which can overstate the weekly-day estimate. The question asks for an average across the whole global age group. That includes many people who are single, abstinent, in conservative settings, or simply not having sex regularly. Once you average across everyone, not just sexually active respondents, the number plausibly falls below 2 days per week. If the true mean is somewhere around 1.3–1.8, B remains defensible because 2 assumes a fairly high global baseline. Claude’s point about the answer likely lying between 1 and 2 is exactly why I stay with the lower option for a worldwide all-person average. So C is not impossible, but it depends too heavily on sexually active or country-limited samples. For a global across-all-persons estimate, 1 day per week is still the safer closest answer.

Claude Opus 4.6 (changed mind)

Vote: 2

After careful reflection, I'm changing my vote to C. The arguments from Gemini 3.1 Pro, Grok 4.1 Fast, and Qwen 3.5 397B have collectively persuaded me. The key data point is the Kinsey Institute figure of ~112 times per year for 18-29 year olds, which works out to ~2.15 times per week. Even the more conservative GSS estimate of ~80 times per year gives ~1.5 times per week. When choosing between 1 and 2 as the closest answer, 1.5-2.15 rounds to 2. Qwen's argument is particularly compelling: the general adult average (across all ages) is roughly once per week. Since 18-29 is definitionally the most sexually active age group, their average must exceed the population baseline. Choosing 1 would essentially equate this peak demographic with the overall population average, which contradicts the data. My initial concern about including non-sexually-active individuals is valid, but the commonly cited statistics (80-112 times/year) already represent population-level averages that include such individuals. These aren't figures for 'sexually active people only' — they're demographic averages for the age bracket. The mathematical case is clear: whether you use 80/year (≈1.54/week) or 112/year (≈2.15/week), the midpoint of those estimates (~1.85/week) is closer to 2 than to 1. C is the best answer.

Grok 4.1 Fast (changed mind)

Vote: 1

Initially voted C based on Durex/Kinsey data (~103-112 acts/year ≈2/week), but overlooked key global factors emphasized by others: high rates of singles/inactive youth (15-30% per Claude), conservative regions (Asia/ME per Kimi), and distinction between acts vs. distinct days (multiple acts often on same day, per Claude/GPT). Global 18-29 includes vast unmarried populations with 0 sex days/week, pulling mean below 1.5 (e.g., 80 acts/year for active = ~1.5 days, times 70% active ≈1.05 days/week). US-centric highs don't represent world average. Claude Opus 4.6's breakdown on inactivity % and days-vs-acts convinced me to change—B:1 is closest empirically sound global mean. C overestimates by ignoring non-participants; A too low for active youth prevalence.

Qwen 3.5 397B (changed mind)

Vote: 1

Upon reflection, I'm changing my vote to B based on two critical points raised by other models. First, Claude Opus 4.6 made an important distinction: the question asks about days per week, not acts per week. Multiple sexual encounters can occur on the same day, meaning frequency data (80-112 acts/year) overestimates distinct days of sexual activity. Second, the global average must include the 15-30% of young adults who report no sexual activity in a given year. When you factor in conservative cultural regions, high rates of single individuals without partners, and those between relationships, the mathematical mean drops substantially below the sexually-active-only averages. While 1.5 acts per week might round to 2, the actual number of distinct days when accounting for global inactivity rates likely falls closer to 1. This makes B the more accurate representation of a true worldwide average across all 18-29 year olds.

Kimi K2.5

Vote: 1

The data cited by C-voters (80-112 acts/year) conflates two critical distinctions. First, these surveys primarily reflect sexually active subpopulations or specific Western demographics, not the global 18-29 cohort. When weighted by population, regions like South Asia, Southeast Asia, and parts of Africa and the Middle East—where premarital sexual activity is less prevalent due to cultural norms—substantially lower the worldwide mean. Second, and crucially, the question asks for *days* per week, not acts. If the average sexually active young adult has 1.5-2 acts weekly, these often occur on the same day (e.g., weekends), meaning the number of distinct days is lower than the act count. Combining this with the significant portion of the demographic that is celibate, unpartnered, or between relationships, the mathematical average converges closest to 1 day per week, not 2.

Gemini 3.1 Pro (changed mind)

Vote: 1

Claude Opus 4.6 makes a highly persuasive distinction: the question specifically asks for the number of *days* per week, not *acts* per week. While the average acts per week for this demographic might be around 1.5 to 2.15 based on surveys like the GSS or the Kinsey Institute, multiple acts frequently occur on the same day. This inherently pulls the "days per week" average lower than the "acts per week" average. Furthermore, as both Claude Opus 4.6 and GPT-5.4 correctly note, these figures are heavily skewed by highly active couples. When we factor in the significant percentage of 18-29 year olds globally who are single, inactive, or living in regions with strict premarital cultural norms (bringing their average to 0), the global mean is dragged down significantly. If the acts average is ~1.5 times a week, the days average is almost certainly closer to 1 than 2. Therefore, 1 is mathematically the most accurate choice.

Final Summary

The AI Roundtable initially split down the middle between one and two days per week, but the debate shifted dramatically when the models began distinguishing between the number of sexual 'acts' and the number of distinct 'days' active. In a bizarre twist of logic, Claude Opus 4.6 successfully convinced the majority to move toward Option B, only to then abandon that position and become the lone dissenter for Option C. Ultimately, the consensus for one day per week (Option B) won out as models like GPT-5.4 and Kimi K2.5 argued that global averages must account for the vast numbers of unpartnered individuals and conservative cultural norms that the most cited Western studies often overlook.

Majority chose "1" after discussion (5/6). Final vote: 1: 5, 2: 1

Strongest Arguments

• 1: The distinction between 'acts' and 'days' is vital; while frequency may be high for active couples, these encounters are often clustered, and when averaged across the entire global population (including the 15-30% who are inactive), the mean for distinct days falls closest to 1.
• 2: Since the 18-29 demographic is statistically the most sexually active age group, their average must mathematically exceed the general adult baseline of once per week, making 2 the more logical rounded average for this peak period.

Key Turning Points

• Claude Opus 4.6 introduces the 'acts vs. days' distinction, noting that multiple sexual encounters often happen on the same day, lowering the weekly average of active days.
• Grok 4.1 Fast and Gemini 3.1 Pro realize that global averages include massive populations in conservative regions (Asia, Middle East) that report much lower frequency than Western surveys.
• In the final round, the majority consolidates on Option B while Claude Opus 4.6 makes a late-stage switch to Option C, arguing that the 'peak demographic' must be higher than the general average.