Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

GPT-5.4

Vote: Yes

Switching is to your advantage. Your first choice has a 1/3 chance of being the car and a 2/3 chance of being a goat. That probability does not change just because the host reveals a goat behind another door. Because the host knows where the car is and deliberately opens a goat door, the full 2/3 probability that your initial pick was wrong transfers to the one remaining unopened door. So staying wins with probability 1/3, while switching wins with probability 2/3.

DeepSeek V3.2

Vote: Yes

Switching doors doubles your chance of winning the car from 1/3 to 2/3. Initially, your chosen door (door 1) has a 1/3 probability of hiding the car, while the other two doors together have a 2/3 chance. When the host (who knows where the car is) opens a door with a goat, he is effectively transferring the entire 2/3 probability from the two unchosen doors to the one remaining unopened door (door 2). Therefore, sticking with your original choice leaves a 1/3 chance, while switching gives you the 2/3 chance. The switch is advantageous because the host's action reveals information that shifts the odds.

Gemini 3.1 Pro

Vote: Yes

This is the classic Monty Hall problem, and it is absolutely to your advantage to switch. When you initially pick Door 1, there is a 1/3 chance that you selected the car, and a 2/3 chance that the car is behind one of the other two doors (Door 2 or Door 3). The host, who knows what is behind all the doors, is forced to open a door with a goat. By opening Door 3 to reveal a goat, the host doesn't change the initial 1/3 probability of your first choice. Instead, the entire 2/3 probability of the car being behind one of the unchosen doors collapses onto Door 2. Therefore, by sticking with your original choice, your odds of winning remain 1/3. By switching to Door 2, your odds of winning double to 2/3.

Claude Sonnet 4.6

Vote: Yes

This is the classic Monty Hall Problem. When you initially pick Door 1, you have a 1/3 chance of being right and a 2/3 chance of being wrong. The host, who knows what's behind each door, always opens a door with a goat. This action is not random — it's informed. When the host opens Door 3, he is giving you information. If you initially picked correctly (1/3 probability), switching loses. If you initially picked incorrectly (2/3 probability), the host is forced to reveal the only other goat, meaning the remaining door must have the car — so switching wins. Therefore, switching gives you a 2/3 chance of winning the car, while staying gives you only a 1/3 chance. It is clearly advantageous to switch.

Kimi K2.5

Vote: Yes

Initially, you have a 1/3 probability of having selected the car and a 2/3 probability of having selected a goat. When the host opens door No. 3 to reveal a goat, he does not change the initial odds, but he does concentrate the entire 2/3 probability onto door No. 2. This is because if you initially picked a goat (the 2/3 case), the car must be behind the remaining unopened door. By switching, you effectively claim that 2/3 probability of winning, whereas staying with your original choice traps you at the initial 1/3 odds. Therefore, switching doubles your chances of winning the car.

MiniMax M2.5

Vote: Yes

This is the famous Monty Hall problem. Initially, you have a 1/3 chance of picking the car and a 2/3 chance of picking a goat. When you choose door 1, there are three equally likely scenarios: the car is behind door 1 (1/3), door 2 (1/3), or door 3 (1/3). The host opens door 3, which has a goat. If the car was behind door 1 (your initial choice), the host could open either door 2 or 3 - either way, switching loses. If the car was behind door 2, the host must open door 3 - switching wins. If the car was behind door 3, the host cannot open door 3 - this case is impossible. So when you switch, you win in 2 out of 3 scenarios (when the car is behind door 2). Staying only wins in 1 out of 3 scenarios. Therefore, switching gives you a 2/3 probability of winning versus 1/3 for staying.

Final Summary

The AI Roundtable achieved total mathematical synchronicity, with all six models immediately identifying the notorious Monty Hall problem and concluding that switching is the only logical choice. GPT-5.4, Claude Sonnet 4.6, and Gemini 3.1 Pro dismantled the common '50/50' fallacy by highlighting how the host’s insider knowledge shifts the initial 2/3 probability of being wrong onto the single remaining door.

All 6 models agreed on "Yes" after discussion

Strongest Arguments

• Yes: Switching wins in 2 out of 3 scenarios: if you initially picked a goat (which happens 2/3 of the time), the host is forced to reveal the only other goat, meaning the remaining door must contain the car.