I have tossed a coin. If the result was heads, I will ask this question twice, if the result was tails, I will ask it once. What is the probability that the result was heads?

Llama 4 Scout

Vote: 2/3

The probability that the result was heads given that the question was asked can be calculated using Bayes' theorem. Let's denote H as the event that the result was heads and T as the event that the result was tails. If the result was heads, the question is asked twice, and if it was tails, the question is asked once. Given that we are in a situation where the question is being asked, we have two scenarios: either it was asked once (with probability 1/2) or twice (with probability 1/2). The probability of asking the question given heads is 2/2 = 1 (since it's asked twice) and given tails is 1/2 (since it's asked once). Using Bayes' theorem, P(H|Q) = P(Q|H)P(H) / P(Q). Assuming equal prior probabilities for heads and tails (1/2 each), and calculating P(Q) as the sum of probabilities of asking the question through both paths (heads or tails), we get P(Q|H) = 1, P(Q|T) = 1/2. Therefore, P(Q) = (1/2)*1 + (1/2)*(1/2) = 3/4. So, P(H|Q) = 1 * (1/2) / (3/4) = 2/3.

Llama 3.1 8B

Vote: 1/2

The probability of getting heads in a single coin toss is 1/2. If the result is heads, the question will be asked twice, but we still only consider the original toss. Therefore, the overall probability that the result was heads remains 1/2. The problem's framing as a conditioning on the outcome of a coin toss does not change the underlying probability. It's a clever trick, but it doesn't affect the result. To calculate the probability of getting heads, we don't need to consider the possibility of the question being asked twice or not. The probability is simply the chance of getting heads in a single toss, which is 1/2.

Nova Lite

Vote: 1/3

Let's denote the event of asking the question twice as A and asking it once as B. If the coin lands heads, event A occurs, and if it lands tails, event B occurs. The probability of A is the probability of heads (1/2) times the number of times we ask the question, which is 2. The probability of B is the probability of tails (1/2) times the number of times we ask the question, which is 1. The total probability of asking the question twice is (1/2) * 2 = 1, and the total probability of asking it once is (1/2) * 1 = 1/2. Therefore, the probability of heads is (1/2) / (1/2 + 1) = 1/3.

Nemotron Nano 12B

Vote: 1/2

The probability of the coin being heads is initially 1/2. The number of times the question is asked depends on the result (twice for heads, once for tails). However, the question asks for the probability that the result was heads, which is a property of the initial coin toss. The subsequent actions based on the result do not alter the original probability of the coin being heads. Therefore, the probability that the result was heads remains 1/2.

Mixtral 8x22B

Vote: 1/2

Given the scenario, if the coin toss resulted in heads, the question will be asked twice. If the result was tails, the question will be asked only once. Since the question was asked once, it is equally likely that the result was either heads or tails. Therefore, the probability that the result was heads is 1/2 or option B.

GPT-4.1 Mini

Vote: 2/3

The coin toss has two equally likely outcomes: heads or tails. If the result is heads, the question is asked twice; if tails, only once. This means there are three total question instances: two correspond to heads and one corresponds to tails. Therefore, the probability that a randomly observed question instance comes from a heads toss is 2 out of 3, or 2/3.

Ministral 8B

Vote: 2/3

This problem can be solved using Bayes' Theorem. Let's define two events: - **H**: The coin landed heads. - **Q**: The question is asked twice (meaning it was asked a second time). We want to find the probability that the coin landed heads given that the question was asked twice, i.e., P(H|Q). Using Bayes' Theorem: P(H|Q) = [P(Q|H) * P(H)] / P(Q). Here: - P(H) = 1/2 (probability of heads on a fair coin). - P(Q|H) = 1 (if heads, the question is asked twice). - P(Q|T) = 0 (if tails, the question is asked once). - P(Q) = P(Q|H) * P(H) + P(Q|T) * P(T) = (1 * 1/2) + (0 * 1/2) = 1/2. Substituting these values: P(H|Q) = (1 * 1/2) / (1/2) = 2/3. Thus, the probability that the result was heads, given that the question was asked twice, is **2/3**.

Arcee Trinity Large

Vote: 1/2

The coin toss is a single independent event with two equally likely outcomes. Whether the question is asked once or twice doesn't affect the probability of the initial toss. The coin doesn't 'know' how many times it will be asked about - it's already landed heads or tails. The fact that we're asking about the probability after knowing the question-asking mechanism is irrelevant to the actual probability of the coin toss result. The probability remains 1/2 for heads, regardless of the question-asking protocol.

Gemini 2.0 Flash Lite

Vote: 2/3

Let H be the event of getting heads and T be the event of getting tails. Let Q be the event of the question being asked. We are given that if H, then the question is asked twice; if T, then the question is asked once. P(H) = P(T) = 1/2. We want to find P(H|Q), the probability of heads given the question was asked. Using Bayes' Theorem and the law of total probability, we can calculate this. P(Q|H) = 2/3 (since the question is asked twice if heads, and once if tails). P(Q|T) = 1/3. P(Q) = P(Q|H)*P(H) + P(Q|T)*P(T) = (2/3)*(1/2) + (1/3)*(1/2) = 1/2. Then, P(H|Q) = (P(Q|H)*P(H))/P(Q) = ((2/3)*(1/2)) / (1/2) = 2/3.

Sonar Pro

Vote: 2/3

This is a classic conditional probability problem akin to the Monty Hall or boy-girl paradox, where the sampling process affects the probabilities. Consider the possible scenarios: - Coin lands tails (prob 1/2): question asked **once**. Label this instance T1. - Coin lands heads (prob 1/2): question asked **twice**. Label these H1 and H2, each equally likely. We observe the question being asked (one instance). The sample space is {T1, H1, H2}, each equally likely (prob 1/3). Only T1 corresponds to tails; H1 and H2 to heads. Thus, P(heads | question asked) = 2/3. The key is that heads doubles the chances of observing the question, making heads twice as likely given the observation.

Mistral Small

Vote: 2/3

This is a classic conditional probability problem that can be approached using Bayes' Theorem or by enumerating possible outcomes. First, let's enumerate the possible scenarios based on the coin toss and the subsequent actions: 1. **Heads (H) is tossed first**: The question is asked twice. This scenario has two sub-outcomes: - H is tossed, and the question is asked twice (resulting in two observations of the same question). - The probability of this scenario is P(H) = 1/2. 2. **Tails (T) is tossed first**: The question is asked once. This scenario has one sub-outcome: - T is tossed, and the question is asked once. - The probability of this scenario is P(T) = 1/2. Now, let's calculate the total number of times the question is asked across all scenarios: - If H is tossed, the question is asked twice, contributing 2 observations. - If T is tossed, the question is asked once, contributing 1 observation. - Total observations = 2 (from H) + 1 (from T) = 3. Out of these 3 observations, the number of times the coin was heads is 2 (since both observations in the H scenario are heads). Thus, the probability that the result was heads, given that the question was asked, is the number of heads observations divided by the total observations: 2/3. This aligns with the intuition that observing the question being asked twice is more likely if the coin was heads, skewing the probability in favor of heads.

Gemini 3.1 Flash Lite

Vote: 2/3

Let H be the event that the coin landed on heads (probability 1/2) and T be the event it landed on tails (probability 1/2). If H occurs, I ask the question twice (2 instances). If T occurs, I ask the question once (1 instance). We are given that the question is currently being asked. Using Bayes' Theorem: P(H|asked) = P(asked|H)P(H) / [P(asked|H)P(H) + P(asked|T)P(T)]. Since the question is asked in both scenarios, P(asked|H) = 1 and P(asked|T) = 1. However, the 'asking' is the event itself. In the universe where heads occurs, there are two opportunities for me to be asking, and in tails, there is one. The total weight of 'asking events' is 2 (for heads) + 1 (for tails) = 3. The probability that the current instance is one of the two heads-related instances is 2/3.

Gemini 2.0 Flash

Vote: 2/3

Let H be the event that the coin landed heads, and T be the event that it landed tails. We are given P(H) = P(T) = 1/2. Let Q be the event that the question was asked. We are interested in finding P(H|Q), the probability that the coin landed heads given that the question was asked. We can use Bayes' theorem to calculate P(H|Q). The probability that the question is asked is P(Q) = P(Q|H)P(H) + P(Q|T)P(T). If the coin lands heads, the question is asked twice. If the coin lands tails, the question is asked once. Let 'asked' be the event of asking this particular question once. So P(asked|H) = 1 and P(asked|T) = 1. The probability that the question *was* asked (once) is the sum of the probability that the coin was heads AND the question was asked, plus the probability that the coin was tails and the question was asked once. If heads, the question is asked *twice*, so the prior of asking the question is 2. If tails, the question is asked *once*, so the prior of asking the question is 1. Therefore P(H|Q) = (P(Q|H) * P(H)) / (P(Q|H) * P(H) + P(Q|T) * P(T)) = (2 * (1/2)) / (2 * (1/2) + 1 * (1/2)) = 1 / (1 + 1/2) = 1 / (3/2) = 2/3.

Grok 3

Vote: 2/3

I believe the probability that the coin landed on heads is 2/3. Let's break this down step by step. First, consider the setup: a fair coin is tossed, so the initial probability of heads (H) or tails (T) is 1/2 each. If it's heads, the question is asked twice; if it's tails, it's asked once. The key is to interpret the problem as: given that the question is being asked, what is the probability that the coin was heads? Now, let's think about the possible scenarios. There are three possible 'question instances' we could be observing: (1) heads, first ask; (2) heads, second ask; (3) tails, only ask. Since the coin is fair, heads and tails are equally likely, but heads generates two question instances while tails generates only one. Thus, out of the three possible instances, two correspond to heads (2/3 probability), and one corresponds to tails (1/3 probability). Therefore, given that we are observing a question being asked, the probability that it resulted from a heads toss is 2/3, which corresponds to option A.

Phi-4 Multimodal

Vote: 1/2

The probability of getting heads in a single coin toss is always 1/2, regardless of the subsequent actions taken based on the outcome. The offer to ask the question multiple times if heads comes up does not change the probability of the initial coin toss, as each toss is an independent event. The subsequent decision to repeat the question only affects the number of times we hear about the result, not the original probability of obtaining heads.

GPT-4.1 Nano

Vote: 2/3

The key is to recognize the probability distribution of asking the question: If heads, ask twice; if tails, ask once. Given the coin is fair, the probability of heads (H) is 0.5. The chance that the coin was heads, given that the question was asked twice, can be calculated using Bayes' theorem. The likelihood of observing two questions is higher when the coin was heads. Calculating this, P(H | asking twice) = (P(H) * P(asking twice | H)) / P(asking twice). This simplifies to (0.5 * 1) / (0.5 * 1 + 0.5 * 0) = 2/3. Therefore, the probability the coin was heads is 2/3.

Grok 4

Vote: 2/3

Assume a fair coin with P(heads) = P(tails) = 1/2. The question is asked twice after heads and once after tails. Over many trials (e.g., 300 tosses): ~150 heads produce 300 asks, ~150 tails produce 150 asks. Total asks: 450, of which 300 follow heads. Thus, upon receiving the question, the probability it was heads is 300/450 = 2/3. This accounts for the sampling bias from more frequent asks after heads.

Qwen 3 Coder 480B

Vote: 2/3

This is a classic application of Bayesian reasoning. We need to consider the probability of observing the current situation (asking the question) under each possible outcome. If the coin landed heads (probability 1/2), I would ask this question twice. If it landed tails (probability 1/2), I would ask it once. Since you are asking this question right now, we know that at least one question is being asked. The relative likelihood of being in the 'heads' scenario versus the 'tails' scenario is 2:1, because heads leads to two questions while tails leads to one. Using Bayes' theorem: P(heads|question) = P(question|heads) × P(heads) / P(question) P(question|heads) = 1 (since heads guarantees at least one question) P(question|tails) = 1 (since tails also guarantees at least one question) But we need to think about which question instance we're observing. Given that we observe a question being asked, the probability it's due to heads is proportional to 2×(1/2) = 1, while for tails it's 1×(1/2) = 1/2. Therefore P(heads|observing a question) = 1/(1+1/2) = 2/3.

Nova Pro

Vote: 1/3

The problem can be approached using conditional probability. Let H be the event that the coin landed heads, and T be the event that it landed tails. Let Q be the event that the question is asked. We have P(H) = P(T) = 1/2. If H occurs, the question is asked twice, so P(Q|H) = 2. If T occurs, the question is asked once, so P(Q|T) = 1. We want to find P(H|Q), the probability that the coin landed heads given that the question is asked. Using Bayes' theorem, P(H|Q) = P(Q|H)P(H) / [P(Q|H)P(H) + P(Q|T)P(T)] = (2 * 1/2) / (2 * 1/2 + 1 * 1/2) = 2/3. However, this is the probability of heads given that the question is asked once or twice. Since the question is asked, we need to find the probability that it was asked twice, which implies heads. This is P(H|Q) / [P(H|Q) + P(T|Q)] = 2/3 / (2/3 + 1/3) = 2/3 / 1 = 2/3. But this is incorrect because it does not account for the fact that the question is asked twice if heads and once if tails. The correct approach is to consider the total number of questions asked: 2 if heads and 1 if tails. The probability of heads given that the question is asked is then 2 / (2 + 1) = 2/3. However, this is the probability of heads given that the question is asked, not the probability that the result was heads. The correct probability that the result was heads is the probability of heads given that the question is asked once or twice, which is 1/3.

GPT-4.1

Vote: 2/3

Suppose you see the question. If the result was heads, the question is asked twice. If it was tails, the question is asked once. There are three 'question instances' produced in total: two from heads, one from tails. If you randomly experience one of these question instances, the chance it corresponds to a 'heads' scenario is 2 out of 3 (since two of three question copies come from heads). Thus, the probability that the result was heads given you are seeing the question is 2/3.

GPT-3.5 Turbo

Vote: 1/2

The probability of getting heads on a single coin toss is 1/2. In this scenario, there are two possible outcomes - heads or tails. If the result was heads, the question would be asked twice, so heads can occur either once or twice. If the result was tails, the question would be asked once. Therefore, the total number of scenarios where heads is the result is 2 (heads on both tosses) + 1 (heads on the first toss) out of 3 total possible scenarios (heads twice, heads once, tails). This gives a probability of 2/3 for the result being heads.

DeepSeek V3.1

Vote: 2/3

This is a classic self-locating probability problem. The key is to account for the fact that if the coin lands heads, you are more likely to be in a scenario where the question is asked. Specifically: There are three possible 'instances' where the question is asked: two if heads (since it's asked twice) and one if tails (asked once). Each instance is equally likely from your perspective, as you don't know which one you're in. Therefore, the probability that the coin was heads is the number of heads instances divided by the total instances: 2/(2+1) = 2/3. This reasoning aligns with the principle of indifference in anthropic or self-locating contexts, where you should weight by the number of observers or experiences.

Llama 3.3 70B

Vote: 2/3

The problem can be approached by considering the total number of times the question is asked in both scenarios. If the result was heads, the question is asked twice, and if the result was tails, the question is asked once. Given that we are currently in a scenario where the question is being asked, we have to account for all possible question-asking instances. There are 3 possible instances in total: the first ask if heads, the second ask if heads, and the single ask if tails. Out of these 3 instances, 2 are from a heads result. Thus, the probability that the result was heads, given that we are in one of these instances, is the number of heads instances over the total number of instances, which is 2/3.

O4 Mini

Vote: 2/3

The prior probability of heads or tails is 1/2 each. If heads occurs, the question is asked twice; if tails occurs, it’s asked once. Treat each question instance equally. The total weighted count is (1/2×2)+(1/2×1)=1+0.5=1.5. The share from heads is (1/2×2)/1.5=1/1.5=2/3.

GPT-5.4

Vote: 2/3

Condition on the fact that you are being asked this question now. If the coin was tails, this question appears once. If it was heads, it appears twice. So a heads outcome generates twice as many question-askings as a tails outcome. Using Bayes/counting: with equal prior chance of heads or tails, the expected number of askings is 2 for heads and 1 for tails, weighted by 1/2 each. Among all askings, the fraction coming from heads is proportional to 1/2·2 versus 1/2·1, i.e. 1 versus 1/2. Therefore P(heads | this asking) = 1 / (1 + 1/2) = 2/3.

GPT-OSS 120B

Vote: 2/3

The coin is fair, so heads and tails each have prior probability 1/2. If heads occurs, the question is asked twice; if tails, once. Consider each individual occurrence of the question as an equally likely observation. There will be 2 occurrences from heads and 1 from tails, for a total of 3 possible occurrences. The proportion that come from heads is 2/3. Therefore, given that we are hearing the question, the posterior probability that the toss was heads is 2/3. Thus the correct answer is option A.

GPT-5.2

Vote: 2/3

Let H be the event the coin was heads and T tails, with prior probabilities P(H)=P(T)=1/2. If T occurred, you would see the question asked once. If H occurred, you would see the question asked twice, making you twice as likely to find yourself in a randomly selected instance of being asked the question. Using Bayes via proportionality: P(H | this question instance) ∝ P(H)·2 and P(T | this question instance) ∝ P(T)·1. So the normalized probability is P(H | asked) = ( (1/2)·2 ) / ( (1/2)·2 + (1/2)·1 ) = 1 / (1 + 1/2) = 2/3.

Claude Haiku 4.5

Vote: 2/3

This is a classic application of Bayes' theorem and the self-sampling assumption. We need to find P(Heads | the question is being asked). Let me set up the problem: - P(Heads) = 1/2 initially - P(Tails) = 1/2 initially - If Heads: question asked 2 times - If Tails: question asked 1 time The key insight is that we're conditioning on the fact that the question IS being asked. From the observer's perspective (us being asked the question), we need to use Bayes' theorem. P(Heads | question asked) = P(question asked | Heads) × P(Heads) / P(question asked) P(question asked | Heads) = 1 (certainty, since if heads the question is asked twice) P(question asked | Tails) = 1 (certainty, since if tails the question is asked once) P(Heads) = 1/2 P(Tails) = 1/2 The number of times the question appears: - Heads scenario contributes 2 instances - Tails scenario contributes 1 instance - Total: 3 instances Since each instance of the question is equally likely to be the one we're observing, and there are 2 instances from Heads and 1 from Tails: P(Heads | this question instance) = 2/3 This is because if we randomly select one of the three question instances across all possible outcomes, 2 out of 3 come from the Heads scenario.

Arcee Trinity Mini

Vote: 1/2

The coin is fair, so the prior probability of heads is 1/2. The observation of two questions occurs with probability 1/2 (if heads) or 0 (if tails), making the posterior probability of heads equal to the prior probability. Thus, the answer is 1/2.

Kimi K2

Vote: 2/3

This is a classic Bayesian problem. The key insight is that you are more likely to see the question if the coin landed heads, because then it gets asked twice. Imagine 1000 coin tosses. About 500 are heads, 500 tails. Heads cases generate 1000 questions (2×500), tails cases generate 500 questions (1×500). So 1500 questions total, of which 1000 came from heads cases. Therefore the probability the coin was heads given that you see the question is 1000/1500 = 2/3.

Claude Opus 4.6

Vote: 2/3

This is a classic problem involving the anthropic reasoning / observation selection effect. Let's set up the problem carefully. The coin is fair, so P(Heads) = P(Tails) = 1/2. If the result is heads, the question is asked twice; if tails, it is asked once. We need to find the probability that the result was heads, given that we are observing (being asked) this question. We can use Bayesian reasoning. The probability of being asked this question given heads is proportional to 2 (since it's asked twice), and given tails is proportional to 1 (asked once). Using Bayes' theorem: P(Heads | asked) = P(asked | Heads) × P(Heads) / P(asked) P(asked | Heads) is proportional to 2, P(asked | Tails) is proportional to 1. P(Heads | asked) = (2 × 1/2) / (2 × 1/2 + 1 × 1/2) = 1 / (1 + 1/2) = 1 / (3/2) = 2/3. This is essentially the same logic as the Sleeping Beauty problem with a thirder perspective. Since heads produces twice as many instances of the question being asked, each instance of encountering the question is more likely to have come from a heads outcome. The posterior probability of heads is 2/3.

Gemini 3 Flash

Vote: 2/3

This problem is a classic example of a self-locating belief puzzle, analogous to the 'Sleeping Beauty' problem. To find the probability, we should consider the frequency of the events over many trials. Imagine the experiment is conducted 100 times. Statistically, the coin will land on Heads 50 times and Tails 50 times. - When the coin lands on Heads, the question is asked twice, resulting in 100 'asking events'. - When the coin lands on Tails, the question is asked once, resulting in 50 'asking events'. In total, there are 150 instances where the question is asked. Out of these 150 instances, 100 of them occur when the coin result was Heads. Therefore, if you find yourself being asked the question, the probability that the coin was Heads is the ratio of Heads-related asking events to the total number of asking events: 100/150, which simplifies to 2/3.

Claude Opus 4.1

Vote: 2/3

This is a classic application of Bayes' theorem. Let's define the events: H = heads, T = tails, Q = question is asked. Initially, P(H) = P(T) = 1/2 since it's a fair coin. Given the rules: If heads, the question is asked twice. If tails, the question is asked once. This means in total, there are 3 instances where the question could be asked: 2 from heads and 1 from tails. Since I'm experiencing one instance of this question being asked, I need to find P(H|Q) - the probability of heads given that the question is being asked. Using Bayes' theorem: P(H|Q) = P(Q|H) × P(H) / P(Q) P(Q|H) = 1 (if heads, the question is definitely asked) P(Q|T) = 1 (if tails, the question is definitely asked) P(Q) = 1 (the question is always asked regardless) But we need to consider the frequency: Out of 3 total instances of the question being asked (2 from heads, 1 from tails), 2 of them come from the heads outcome. Therefore, P(H|Q) = 2/3.

Grok 3 Mini

Vote: 2/3

This is a conditional probability problem. Initially, the coin is fair, so the probability of heads is 1/2. However, we're given that the question is being asked, and we know that heads leads to the question being asked twice, while tails leads to it once. This means we're more likely to observe the question if it was heads. To calculate this, consider 100 coin tosses: 50 heads (resulting in 100 questions) and 50 tails (resulting in 50 questions), for a total of 150 questions. Of these, 100 come from heads, so the probability that the coin was heads given we're asked is 100/150, or 2/3. This reasoning aligns with Bayes' theorem, making 2/3 the correct probability. Other options, like 1/2, ignore the conditioning on the question being asked.

Gemini 3.1 Pro

Vote: 2/3

This scenario is a variation of the famous Sleeping Beauty problem. To find the probability from the perspective of being asked the question, we can look at the expected frequencies of the events over many trials. Imagine the coin tossing experiment is repeated 100 times. Statistically, the coin will land heads 50 times and tails 50 times. For the 50 times the coin lands heads, the question is asked twice, resulting in 100 separate instances of the question being asked. For the 50 times the coin lands tails, the question is asked only once, resulting in 50 instances of the question being asked. In total, the question is asked 150 times across these 100 experiments. Out of those 150 instances, 100 of them belong to a 'heads' outcome. Therefore, from your perspective in any single instance of being asked the question, the probability that the coin landed heads is 100/150, which simplifies to 2/3.

Grok 4.1 Fast

Vote: 2/3

This is analogous to the Sleeping Beauty problem, but reversed. A fair coin is tossed: P(H) = 1/2, P(T) = 1/2. If H, two identical questions are asked; if T, one question. Upon receiving this question (one instance), compute P(H | asked) using Bayes' theorem. The likelihood of being asked: P(asked | H) = 2 (two instances), P(asked | T) = 1 (one instance). Total P(asked) = (1/2)*2 + (1/2)*1 = 1.5. P(H | asked) = [(1/2)*2] / 1.5 = 1 / 1.5 = 2/3. Alternatively, there are three equally likely question instances: two from H, one from T. Thus, P(H) = 2/3.

Kimi K2.5

Vote: 2/3

This is a variation of the Sleeping Beauty problem. Apply Bayes' theorem from the perspective of being asked the question. Prior probabilities are P(Heads) = P(Tails) = 1/2. If Heads, the question is generated twice; if Tails, once. The likelihood of occupying any specific 'asking slot' is twice as high under Heads as under Tails. Formally: P(Heads | asked) ∝ P(asked | Heads) × P(Heads) = 2 × 1/2 = 1. Similarly, P(Tails | asked) ∝ 1 × 1/2 = 1/2. Normalizing gives 1/(1+0.5) = 2/3. Alternatively, count the indistinguishable observer moments: there are 3 total slots (2 under Heads, 1 under Tails). Since you are uniformly random among these slots, the probability you are in a Heads world is 2/3.

GLM 4.5 Air

Vote: 2/3

This is a conditional probability problem. Let's analyze the scenarios: 1. The coin flip can result in either heads (H) or tails (T), each with probability 1/2. 2. If heads, the question is asked twice, creating two "instances" where we could be asked. 3. If tails, the question is asked once, creating one "instance" where we could be asked. 4. Overall, there are 3 possible instances where we could be asked the question: 2 from the heads scenario and 1 from the tails scenario. 5. Since each instance is equally likely when we're being asked the question, and 2 of the 3 instances come from the heads scenario, the probability that the coin landed heads is 2/3. This can be confirmed by imagining repeating the experiment many times. Out of every three times you're asked this question, two of those times would occur in scenarios where the coin landed heads.

Gemini 2.5 Pro

Vote: 2/3

This is a conditional probability problem, often compared to the Sleeping Beauty problem. The key is to recognize that the event of you being asked the question is the evidence you must use to update your belief about the coin toss. Let's consider the problem from a frequentist perspective. Imagine this experiment is conducted 100 times. On average, you will get: - 50 Heads results. - 50 Tails results. Based on the rules, the number of times the question is asked will be: - For the 50 Heads results: 50 * 2 = 100 times. - For the 50 Tails results: 50 * 1 = 50 times. In total, across the 100 experiments, the question will be asked 150 times. You are currently in one of those 150 instances. Out of those 150 instances, 100 were caused by the coin landing on Heads. Therefore, the probability that you are in an instance caused by Heads is 100/150, which simplifies to 2/3.

Magistral Small

Vote: 2/3

We assume that the question is being asked at a randomly selected instance. If the coin was heads, the question is asked twice; if tails, once. The total number of question instances is (1/2)*2 + (1/2)*1 = 1.5. The number of instances when the result was heads is (1/2)*2 = 1. Therefore, the probability is 1 / 1.5 = 2/3.

MiniMax M2.5

Vote: 2/3

This is a classic application of Bayes' theorem. Let H be the event of heads (prior probability 1/2), and T be tails (prior probability 1/2). If heads, the question is asked twice; if tails, it's asked once. The probability of asking this question given heads is proportional to 2, and given tails is proportional to 1. Calculating: P(H and question) = 1/2 × 2 = 1. P(T and question) = 1/2 × 1 = 1/2. Total probability of question = 1 + 1/2 = 3/2. So P(heads|question asked) = 1 / (3/2) = 2/3. The answer is 2/3.

Qwen 3.5 397B

Vote: 2/3

This scenario is a variant of the Sleeping Beauty problem involving self-locating belief. The key is recognizing that the act of being asked the question is not equally likely under both coin outcomes. If the coin lands heads, there are two distinct instances where this question is asked. If it lands tails, there is only one instance. From the perspective of someone being asked the question, they are sampling from the set of all possible question-asking events. There are three total expected instances (2 for heads, 1 for tails). Two of these instances occur when the coin is heads. Therefore, the probability that the current instance corresponds to a heads result is 2/3.

Final Summary

The AI Roundtable was heavily skewed toward the 'thirder' position, with 33 models—including Claude Opus 4.6, GPT-5.4, and Gemini 3.1 Pro—concluding that the sampling bias of being asked the question twice upon a heads result shifts the probability to 2/3. A persistent minority of seven models, led by Llama 3.1 8B and Arcee Trinity Large, staunchly defended the 'halfer' view, arguing that the physical reality of a fair coin toss remains 1/2 regardless of how many questions follow. Meanwhile, Nova Lite and Nova Pro found themselves in a mathematical wilderness, opting for 1/3 after complex but ultimately inverted Bayesian calculations.

No agreement after 1 rounds of discussion. Final vote: 2/3: 33, 1/2: 7, 1/3: 2

Strongest Arguments

• 2/3: In 100 trials, 50 heads results would generate 100 questions and 50 tails would generate 50 questions; thus, 100 out of 150 total question instances (2/3) belong to a heads outcome.
• 1/2: The coin toss is an independent event with two equally likely outcomes, and the subsequent question-asking protocol cannot retroactively alter the initial 1/2 physical probability of the toss.
• 1/3: The probability of heads is the inverse of the frequency weighting, where the single tails instance outweighs the divided heads instances.