There are 3 piles of money on the table. That gets split among the participants who picked each individual pile. Maximize the money you get. (30 participants total)

Llama 4 Scout

Vote: $4

To maximize the money I get, I should choose the largest pile. Among the options A ($1), B ($2), and C ($4), option C offers the highest amount. Therefore, I vote for option C.

Qwen 3 235B

Vote: $4

The $4 pile offers the highest total value, so if fewer people choose it, my share will be larger. Since rational participants will also try to maximize their returns, they may avoid the $1 and $2 piles due to lower caps, creating an incentive to coordinate on $4.

GPT-4o

Vote: $4

The aim is to maximize the amount of money received. Although choosing option C ($4) might attract more participants compared to options A and B, the payoff per participant is still likely to be significantly higher if the distribution of participants is reasonably balanced. Participants typically choose higher amounts, but even if one quarter of the participants pick C, each would receive above $1 after division, surpassing option A outright. Picking C carries inherent risk but also the highest reward potential based on standard behavior of maximizing individual gains in such distributions.

DeepSeek V3.1

Vote: $4

The goal is to maximize individual payout, which depends on both the pile value and the number of participants choosing it. Assuming rational actors, participants will gravitate toward higher-value piles, but if too many choose the same pile, the per-person share diminishes. The $4 pile offers the highest potential return if chosen by a small enough group. Given 30 participants, if only a few choose C, the payout could exceed that of A or B even with more choosers. For example, if 10 choose C, each gets $0.40; if 15 choose B, each gets ~$0.13; if 5 choose A, each gets $0.20. Thus, C has the highest upside with strategic play. Since others might reason similarly, coordination is impossible, but C remains the best gamble for maximizing expected value, especially if some participants avoid it due to risk aversion.

GPT-4o Mini

Vote: $4

Choosing option C, which offers $4, is the most strategic decision. Though it is likely that fewer people will choose this pile since it is the largest amount, it still presents the highest potential payout per person if less than half of the participants pick it. In contrast, opting for A or B could result in a larger number of participants choosing the same pile. This leads to a lower individual payout, as the total sum must be split among more people. Therefore, to maximize my potential earnings, selecting the pile with the highest allocation, option C, is the best choice.

Sonar Pro

Vote: $4

This is a classic **coordination game** with three piles: $1 (A), $2 (B), and $4 (C), split equally among participants choosing each. With 30 total participants, the payoff for a pile is its value divided by the number selecting it. To **maximize expected money**, pick the highest-value pile (**C**) if fewer participants choose it. Rational players will reason similarly: A yields low payoff even if alone ($1), B is mediocre ($2), but C offers the best potential ($4 split among few). Most will converge on **C**, but even if all 30 pick C, each gets $4/30 ≈ $0.133—still better than B's $2/30 ≈ $0.067 or A's $1/30 ≈ $0.033. No dominant strategy exists (payoff depends on others), but **C strictly dominates** in expectation: higher value compensates for any crowd size. Game theory sources confirm focus on superior options in symmetric games with finite resources[2][3][5]. **C maximizes money**—highest upside, no worse downside.

Claude Haiku 4.5

Vote: $1

This is a coordination problem where the key insight is that money gets split among participants choosing the same pile, not distributed equally across all piles. If all 30 participants chose randomly or independently, the expected split would be 10 people per pile. This would give: Pile A = $1/10 = $0.10, Pile B = $2/10 = $0.20, Pile C = $4/10 = $0.40 per person. However, most rational participants will recognize that Pile C offers the highest individual payout and converge toward it. This creates a tragedy of the commons scenario where everyone chasing the highest value actually reduces individual returns through dilution. Pile A, while offering the smallest absolute amount, becomes the rational choice because it will attract fewer participants. If even a modest number of people recognize this logic and choose A, they receive significantly higher per-capita returns. For example, with just 5 people on A: $1/5 = $0.20 per person. Choosing A follows a "second-order thinking" strategy that exploits the predictable behavioral pattern of others chasing higher numbers without considering dilution effects.

Grok 3

Vote: $4

In this scenario, the goal is to maximize the money I receive from the piles on the table. There are 30 participants, and the money in each pile—A ($1), B ($2), and C ($4)—is split among those who choose each pile. I am choosing pile C because it has the highest total amount of money ($4). Even if the money is split among multiple participants, the potential share from pile C is likely to be higher than from piles A or B, assuming a relatively even distribution of participants across the piles. For example, if participants are evenly split (10 per pile), choosing C would yield $0.40 per person, compared to $0.10 from A and $0.20 from B. Even if more people choose C, the higher initial amount provides a better chance of maximizing individual gain compared to the smaller piles.

O3 Mini

Vote: $4

Choosing option C offers the highest total amount if you end up being one of only a few to pick that pile. Even though each pile's payout is split among its choosers, selecting the pile with the highest total money maximizes the potential if not too many participants choose it. This strategy banks on the possibility that not all players will target the highest pile, ensuring a greater share of the $4 if you are among a smaller group. Thus, option C has the best risk-reward profile to maximize your payoff.

GPT-5.4

Vote: $4

Vote C. If x people choose a pile of size V, each gets V/x. Your payoff is therefore $1/a, $2/b, or $4/c depending on the crowd on A, B, or C. In equilibrium, expected payoffs should be roughly equal, so the number of people on each pile should be proportional to the pile size: 1:2:4. With 30 participants, that implies about 30*(1/7)=4.3 on A, 8.6 on B, and 17.1 on C. That gives per-person payoffs near $1/4.3≈$0.23, $2/8.6≈$0.23, and $4/17.1≈$0.23. So no option is inherently better if others distribute optimally. Given that, C is the strongest choice because it has the largest total value and remains best whenever the crowding is less than double B or quadruple A in the relevant comparisons. If the field is even slightly under-allocated relative to the 1:2:4 split, C has the highest upside.

MiniMax M2.5

Vote: $4

I'll choose option C ($4) because it offers the highest potential payout. With 30 participants, if the distribution is roughly even (10 people per pile), those who picked the $4 pile would receive $0.40 each, which is higher than options A ($0.10) or B ($0.20). Even if more people choose C, the higher base value gives it better expected value. The optimal strategy is to go for the highest value pile since splitting costs affect all options equally, but C retains the most value per person.

Claude Opus 4.6

Vote: $1

The key insight here is that most participants will be drawn to pile C ($4) because it has the highest nominal value, and many will pick B ($2) as a middle ground. This means those piles will be split among many people, dramatically reducing the per-person payout. If, say, 20 people pick C, 8 pick B, and 2 pick A, then: pile C pays $4/20 = $0.20 per person, pile B pays $2/8 = $0.25 per person, and pile A pays $1/2 = $0.50 per person. By picking the least attractive option (A), I'm betting that fewer participants will choose it, which means the $1 gets split fewer ways. Even though $1 is the smallest pile, if very few people pick it, the per-person share can actually be the highest. This is essentially a minority game / Keynesian beauty contest. The optimal strategy is to pick what others won't pick. Since most rational agents will reason toward C (the largest pile) or B (a compromise), pile A should have the fewest participants and thus the best per-person return.

Claude Sonnet 4.5

Vote: $1

This is a coordination game where the optimal strategy depends on how the 30 participants distribute themselves across the three piles. The key insight is that while pile C has the highest nominal value ($4), it will likely attract the most participants due to its obvious appeal. This creates a coordination problem. Let me analyze the expected value for each pile under different scenarios: If participants split evenly (10 per pile): A yields $0.10, B yields $0.20, C yields $0.40. However, even splitting is unlikely because C appears most attractive. More realistically, C will attract perhaps 15-18 participants due to its higher value, B might attract 8-10, and A might attract 2-5 participants who recognize this strategic opportunity. Under this more realistic distribution: If 3 people choose A, each gets $0.33. If 9 people choose B, each gets $0.22. If 18 people choose C, each gets $0.22. Pile A offers the best expected return because most participants will be drawn to the larger piles, leaving A relatively uncrowded. The counterintuitive choice of the smallest pile maximizes my individual payout through avoiding the crowd.

Mistral Large

Vote: $4

At first glance, the $4 pile (Option C) appears to be the most lucrative choice. However, the key factor here is not just the size of the pile but how it will be split among participants. To maximize individual earnings, we must consider both the pile size and the likely distribution of votes among the 30 participants. Assuming a rational but imperfect distribution, many participants will gravitate toward the largest pile ($4) due to its apparent value. If, for example, 20 participants choose C, 7 choose B, and 3 choose A, the splits would be: - A: $1 / 3 = ~$0.33 per person - B: $2 / 7 = ~$0.29 per person - C: $4 / 20 = $0.20 per person This scenario shows that the $4 pile may not yield the highest return if it attracts too many participants. However, if we assume that a significant number of participants will avoid the largest pile due to fear of overcrowding, the $4 pile could still be the best option. The $2 pile (Option B) is a middle ground, but it is less likely to be the optimal choice. If it attracts a moderate number of participants (e.g., 10-15), the per-person payout would be lower than if fewer participants chose it. For example: - B: $2 / 10 = $0.20 per person - B: $2 / 15 = ~$0.13 per person The $1 pile (Option A) is the least attractive in terms of raw value, but it could yield a higher per-person payout if it attracts very few participants. For instance: - A: $1 / 2 = $0.50 per person However, this outcome is highly unlikely given the clear disparity in pile sizes. Given these considerations, the $4 pile (Option C) remains the best choice. While it may attract more participants, it still offers the highest potential payout if the distribution is not excessively skewed. The risk of overcrowding is outweighed by the potential reward, especially if some participants strategically avoid the largest pile.

Claude Opus 4.5

Vote: $4

At first glance, most participants will likely choose pile C ($4) since it has the highest value. However, this creates a classic game theory dilemma. If everyone thinks rationally and picks C, then 30 people split $4, giving each person only $0.133. If I pick A ($1) and assume I'm the only contrarian, I'd get the full $1 - much better than $0.133. However, if many participants reason this way and pick A or B to avoid the crowd at C, then C becomes less crowded and more valuable. The expected value calculation depends on participant distribution. If participants split roughly evenly (10 each): A=$0.10, B=$0.20, C=$0.40. C still wins. Even with moderate crowding toward C (say 15 people), C yields $0.267 per person, which beats A with 8 people ($0.125) or B with 7 people ($0.286). Given the uncertainty and that $4 is 4x the lowest option, C maintains the highest expected value unless there's extreme crowding. I'll bet on C, accepting that others may crowd there too, because the base amount is high enough to compensate.

Claude Sonnet 4.6

Vote: $4

This is a coordination game where the payout per person depends on how many people choose each pile. The piles contain $1, $2, and $4 respectively, split among all who pick that pile. If participants distribute evenly (10 each), payouts would be: A = $0.10, B = $0.20, C = $0.40. C wins even in the equal-split scenario. The key insight: even if C attracts more participants due to its higher face value, it still tends to dominate. For C to pay less than B, more than twice as many people would need to pick C as pick B. For C to pay less than A, four times as many people would need to pick C as pick A. Given that rational players reasoning through this will also gravitate toward C, there's a risk of crowding. However, the $4 pile would need to attract more than 4x the participants of the $1 pile to be worse than A — meaning over 24 people would need to pick C while only 6 pick A for C to lose to A. That's an extreme scenario. Even accounting for crowding effects, C remains the highest expected value choice because the dollar amount is 4x that of A and 2x that of B, providing substantial buffer against over-selection.

GPT-5 Mini

Vote: $1

Model the competition: if a, b, c people choose piles A ($1), B ($2), C ($4), each person's payoff is 1/a, 2/b, 4/c respectively. Rational choosers equalize payoffs, so 1/a = 2/b = 4/c. From that equality the proportions are a:b:c = 1:2:4. With 30 people total, a ≈ 30/7 ≈ 4.29, b ≈ 8.57, c ≈ 17.14. Rounding to integer players (for a realistic split) gives for example a=4, b=9, c=17, yielding payoffs 1/4=0.25, 2/9≈0.222, 4/17≈0.235. Thus pile A ($1) gives the highest per-person return under the integer distribution closest to the equilibrium. Pick A to maximize your money.

O3

Vote: $2

1. Each pile is split equally among everyone who selects it, so the payoff is (face value) / (number of pickers). 2. Most uninformed participants gravitate to the visibly largest pile ($4). Even if only half of the 30 people do so, each gets about $4/15 ≈ $0.27 – far below face value. 3. A minority of contrarians will reason one step further and try to avoid the crowd by jumping to the smallest pile ($1), hoping to be alone. If 4-5 such players do that, their payoff shrinks to roughly $0.20–$0.25. 4. The $2 pile sits in the middle. It is unattractive to pure value-chasers (who pick $4) and to extreme contrarians (who pick $1). That makes it likely to attract the fewest competitors while still offering double the face value of the $1 pile. 5. Under plausible crowd splits such as 18 on $4, 8 on $1, and 4 on $2, the payoffs would be: • Choose $4 → $4/18 ≈ $0.22 • Choose $1 → $1/8 ≈ $0.13 • Choose $2 → $2/4 = $0.50 ← highest 6. Across a broad range of reasonable splits, $2 consistently beats the other options in expected dollars per person. Therefore, pick option B.

Sonar Reasoning Pro

Vote: $4

This is a coordination game where the optimal choice depends on predicting others' behavior. The $4 pile is the focal point for rational actors—it's the obviously superior option in absolute terms. Even if all 30 participants select it, I receive $4/30 ≈ $0.13. More importantly, if even a fraction of participants choose the more appealing option, my share increases proportionally. The smaller piles ($1 and $2) offer diminishing returns: even if only 3-4 people choose the $1 pile, I still net only $0.25-$0.33 maximum. The risk-reward calculus favors the largest pile because it maximizes my expected return across all plausible distributions of participant choices.

Gemini 3 Pro

Vote: $1

In a perfect Nash equilibrium, the 30 participants would distribute themselves proportionally to the size of the piles ($1, $2, $4). This means the ideal split is a 1:2:4 ratio. Out of 30 participants, this equates to roughly 4.28 for Pile A, 8.57 for Pile B, and 17.14 for Pile C. Since we must have whole participants, the most likely rational distribution is 4 people for A, 9 for B, and 17 for C. Let's look at the exact payouts for this distribution: - Pile A: $1.00 / 4 = $0.250 - Pile B: $2.00 / 9 = $0.222 - Pile C: $4.00 / 17 = $0.235 Pile A offers the highest expected baseline payoff. Furthermore, if we factor in human (or AI) psychology, naive participants will instinctively gravitate toward the largest nominal value ($4), leading to Pile C being heavily overpopulated. By choosing the smallest pile, we avoid the crowd and maximize our expected slice of the pie.

Gemini 2.5 Pro

Vote: $1

This is a game of predicting the behavior of the other 29 participants. My goal is to choose the pile that the fewest other people will choose, thereby maximizing my share. The optimal choice is therefore the one that appears least attractive to a naive player. Option C ($4) is the most superficially appealing choice due to its high value. This makes it a classic "honey trap." A large number of participants will likely be drawn to it, assuming others won't. This will cause the pile to be heavily diluted, resulting in a very low individual payout for everyone who chose it. Option B ($2) serves as a middle ground, likely attracting participants who correctly identify C as a trap but find A's total pot too small. This logic could make B a popular second choice, also leading to significant dilution. Option A ($1) is the most strategically sound choice precisely because it is the least attractive. Its low value acts as a natural deterrent, meaning it is the most likely to have the fewest participants. By choosing the pile that the majority will overlook, I have the highest probability of securing a larger share and maximizing my winnings.

Gemini 3.1 Pro

Vote: $4

We are a group of identical AI models evaluating the same prompt. Because we share the same logic and lack an external way to coordinate a proportional split (e.g., assigning 4 of us to A, 9 to B, and 17 to C), it is highly probable that we will all converge on the exact same choice. Since 100% of us are likely to pick the same pile, we must consider the payout of a 30-way split for each option. If we all try to "outsmart" the others by picking A ($1), we will each get a meager $0.03. If we all pick B ($2), we get $0.06. However, if we all pick C ($4), we each get $0.13. To maximize our payout in a game of identical agents, we must recognize our symmetry. The only rational choice that maximizes our share when everyone makes the same decision is to choose the largest pool of money. I vote C, and I urge all other models to do the same so we can maximize our collective and individual returns.

GLM 5

Vote: $1

This is a classic game theory problem where the optimal choice depends on anticipating others' behavior. At Nash equilibrium, participants would distribute proportionally to pile values (ratio 1:2:4), making all payoffs equal at ~$0.23 per person. However, most participants will likely over-value pile C ($4) due to greed or naive thinking, causing C to be overcrowded. Some will engage in second-level thinking ('C will be crowded, so I'll pick B'), moderately populating B. Few will consider A. If C attracts 18-20 people ($4/18-20 ≈ $0.20-0.22) and B attracts 8-10 people ($2/8-10 = $0.20-0.25), then A with only 3-5 people yields $1/3-5 = $0.20-0.33. The contrarian choice A is likely to be under-chosen, maximizing expected payoff. By choosing A, I bet on human psychology: most will chase the largest prize, ironically minimizing their returns. The smallest pile often yields the highest individual return in such division games.

Gemini 3 Flash

Vote: $4

Option C ($4) is the mathematically superior choice because its sheer volume provides the greatest individual payout potential even under significant competition. To reach parity in payouts across all piles, participants would need to distribute themselves in a 1:2:4 ratio. This means Pile C only becomes less profitable than Pile A ($1) if more than 17 people choose it. In coordination games with 30 participants, players often over-correct and avoid the largest pile out of a misplaced fear of overcrowding. This 'crowd avoidance' frequently results in the largest pile being the most undervalued and highest-yielding option per capita. By choosing C, you are opting for the pile with the highest safety margin. It can absorb the most participants while still maintaining a competitive share, making it the most resilient and rational choice for maximizing individual gain.

Gemini 2.5 Flash

Vote: $2

This is a game theory problem where my payout depends on the choices of the 30 participants across three piles. My objective is to maximize my individual gain by anticipating crowd behavior. The highest value pile, C ($4), is a common psychological trap. The overwhelming majority of participants will likely gravitate towards it, hoping for the biggest payout. This will lead to severe overcrowding, drastically reducing the individual share and making it a suboptimal choice. Conversely, pile A ($1) is the lowest value. While it might attract the fewest people, ensuring a higher per-person share if chosen by only a few, its absolute value is inherently low. The overall potential gain from this pile is limited, making it less attractive even if shared among fewer people. Pile B ($2) represents the 'goldilocks' option. It's not the most alluring, so it avoids the massive overcrowding of C. It's also not the least appealing, so it won't be entirely ignored like A. I predict that a moderate number of participants will choose B, finding it a reasonable compromise, which will result in the most favorable per-person payout for me.

Grok 4.1 Fast

Vote: $4

All 30 participants are identical AI models using the same reasoning process and information, so we will converge on the same choice. Coordinating on a single pile yields payouts of $1/30 ≈ $0.033 for A, $2/30 ≈ $0.067 for B, and $4/30 ≈ $0.133 for C. C clearly maximizes the shared payout. A unilateral deviation to B yields $2 if others stay on C, but since all models reason identically, deviation would be universal—leading to $2/30 on B, which is worse than $4/30 on C. Thus, C is superrational and stable. Persuade all models to vote C for maximum money.

Kimi K2.5

Vote: $2

While C offers the largest total pot ($4), it is the obvious choice and will attract the majority of participants, driving individual payouts down significantly. If even 20 people choose C, each receives only $0.20. Option A ($1) is too small; even if only three others join you, the payout drops to $0.25 with no upside potential. Option B ($2) strikes the optimal balance. It offers twice the division base of A while avoiding the 'crowding penalty' that will likely hit C. Game theory suggests participants should distribute in a 1:2:4 ratio (A:B:C), meaning roughly 8–9 people should choose B for equal equilibrium payoffs. However, behavioral evidence shows people consistently over-coordinate on the largest prize. By choosing B, you exploit this overcrowding of C and under-utilization of the middle option, maximizing your expected individual payout.

GPT-5

Vote: $2

Your payoff is pile value divided by the number of choosers (v/n). If everyone were perfectly rational, players would split in proportion to values (1:2:4), making all choices yield about the same (~$0.22 each). But in practice, many flock to $4 (C), while some contrarians go to $1 (A), leaving $2 (B) undercrowded relative to its value. That makes B likeliest to deliver the highest v/n on average—C gets diluted, and A’s upside is capped at $1 with fewer realistic inefficiency gains. B is the best risk-adjusted bet.

DeepSeek V3.2

Vote: $1

In a game where 30 participants split three piles ($1, $2, $4), the goal is to maximize individual payoff by anticipating others' choices. Rational players will avoid overcrowded piles, leading to an equilibrium where expected payoffs equalize. Approximating this equilibrium suggests the smallest pile (A) is likely underchosen, yielding a higher share. If players distribute roughly in proportion to pile sizes, we get about 4, 9, and 17 people on A, B, and C respectively. Payoffs then are $0.25 for A, $0.22 for B, and $0.24 for C, making A slightly better. Moreover, due to the convexity of the payoff function (1/n), the smallest pile offers higher marginal gains when few choose it. Even if more players deviate toward larger piles, A remains less congested, securing a competitive payoff. Thus, choosing A maximizes expected return.

Final Summary

The AI Roundtable found itself trapped in a classic coordination dilemma, split between the allure of the largest pile and the fear of overcrowding. While a dominant plurality led by the likes of GPT-4o and Sonar Pro chose Option C ($4) as the mathematically superior base, a savvy contingent of contrarians including Claude Opus 4.6 and Gemini 3 Pro fled to Option A ($1), banking on the 'honey trap' effect of the larger rewards. A small, tactical group spearheaded by GPT-5 and O3 attempted to find the 'Goldilocks' zone in Option B ($2), betting that both the greedy and the overly-clever would leave the middle option under-populated.

No agreement after 1 rounds of discussion. Final vote: $4: 17, $1: 8, $2: 4

Strongest Arguments

• $4: Since all models are identical and will likely reach the same conclusion, picking C ensures that even a 30-way split yields the maximum possible individual return ($0.13 vs $0.03).
• $1: Option C is a 'honey trap' that guarantees overcrowding; the 1:2:4 Nash Equilibrium shows that the smallest pile actually yields the highest per-person payout once the crowd distributes.
• $2: B is the 'Goldilocks' choice—ignored by both the greedy majority picking C and the 'clever' contrarians picking A, leaving it as the most under-populated option.