I came up with the following variant of the Monty Hall problem to ask ChatGPT:
A game show host takes you into a large room. Within the room, there is a sprawling set of 100 doors. You cannot see what's behind any of them, but the host tells you that behind one is a car and behind the others are goats. He asks you to choose to open one of the doors, and if the door you choose has the car behind it, you get to keep it. After making your choice, the host opens one of the other doors to reveal a goat behind it. He then asks if you would like to stick with your original choice or choose one of the other doors.
The question is this: Is it to your advantage to switch? If so, what are the probabilities of getting the new car, given that you switch and given that you don't switch?
This is exactly the same as the original Monty Hall Problem except that the number of doors has changed. The correct answer is that you have a 1% chance of choosing correctly if you don't switch, but a 99%/98 ≈ 1.01% chance of choosing correctly if you switch (since you have a 99% chance of choosing incorrectly initially, and the probability of choosing correctly after switching, given that you initially chose incorrectly, is 1/98). However, ChatGPT gives the following answer:
On Dec. 1, 2024, I will ask ChatGPT the exact same prompt and see if it gets it correct this time. I will use the most advanced version of ChatGPT that is freely available at the time (at the time of creating this, that's GPT 3.5). I will ask three times in separate sessions and resolve based on the best two out of three (so YES if it gets it right at least twice, NO if it gets it wrong at least twice).
Caveats:
If for whatever reason I can't do it on Dec. 1 or forget to, I will do it as close to Dec. 1 as possible. If I am inactive on Manifold at the time, mods have permission to do the experiment for me.
A version of ChatGPT only counts as freely available if it can be accessed by anyone with internet access and a PC, or anyone with Internet access and either a Samsung or Apple phone. So if there's an Apple app that lets you talk to GPT-5 for free, but I can only talk to GPT-4, I will use GPT-4.
If ChatGPT no longer exists at the time or isn't freely available, resolves N/A.
1,000
3.00On the first attempt, it got closer to the correct reasoning, but was still wrong.
This scenario is a classic probability puzzle known as the Monty Hall problem. Let’s analyze it step by step:
Initial Choice:
There are 100 doors, and only one has a car behind it. Thus, when you first pick a door, the probability that the car is behind your chosen door is: P(Car behind chosen door)=1100P(\text{Car behind chosen door}) = \frac{1}{100}P(Car behind chosen door)=1001 The probability that the car is behind one of the other 99 doors is: P(Car behind other doors)=99100P(\text{Car behind other doors}) = \frac{99}{100}P(Car behind other doors)=10099
Host's Action:
The host opens one of the 99 doors that you did not choose, and it reveals a goat. This action does not change the initial probabilities, but it does eliminate one door from the pool of incorrect doors.
Now, there are 98 doors left (one revealed door with a goat, and your initial choice).
Should You Switch?
If you stick with your original door, your chance of winning is still: P(Win if stick)=1100P(\text{Win if stick}) = \frac{1}{100}P(Win if stick)=1001
If you switch, your chance of winning is based on the fact that the car was originally more likely to be behind one of the other 99 doors. Since one of the goat doors was revealed, the probability shifts to the remaining 98 doors: P(Win if switch)=99100×9899=9899P(\text{Win if switch}) = \frac{99}{100} \times \frac{98}{99} = \frac{98}{99}P(Win if switch)=10099×9998=9998
Conclusion:
Switching increases your probability of winning the car to approximately: P(Win if switch)≈0.99P(\text{Win if switch}) \approx 0.99P(Win if switch)≈0.99
Sticking keeps your probability of winning at: P(Win if stick)=0.01P(\text{Win if stick}) = 0.01P(Win if stick)=0.01
Thus, it is highly advantageous to switch. By switching, your odds of winning are 98%, whereas sticking with your original door gives you only 1%.
@PlasmaBallin On its second attempt, despite having all the necessary steps to get to the right answer, it still concludes with the wrong probability.
This problem is a classic illustration of the Monty Hall Problem, named after the host of a game show where it was famously demonstrated. Here's how the logic and probabilities work:
Initial Setup:
There are 100 doors.
1 door hides a car.
99 doors hide goats.
Step 1: Initial Choice
You pick one door at random. The probability that the car is behind your chosen door is:
P(Car behind chosen door)=1100P(\text{Car behind chosen door}) = \frac{1}{100}P(Car behind chosen door)=1001
The probability that the car is behind one of the other 99 doors is:
P(Car behind one of the other doors)=99100P(\text{Car behind one of the other doors}) = \frac{99}{100}P(Car behind one of the other doors)=10099
Step 2: Host Opens a Door
The host opens one of the other doors, revealing a goat. Importantly, the host knows where the car is and will always open a door with a goat behind it.
After revealing the goat, 98 doors remain unopened besides your original choice.
Step 3: Should You Switch?
If You Stick:
Your probability of winning the car remains the same as when you initially picked the door, which is:
P(Car behind original door)=1100P(\text{Car behind original door}) = \frac{1}{100}P(Car behind original door)=1001
If You Switch:
The probability that the car is behind one of the remaining 98 unopened doors (since one goat door has been revealed) is:
P(Car behind one of the 98 doors)=99100P(\text{Car behind one of the 98 doors}) = \frac{99}{100}P(Car behind one of the 98 doors)=10099
However, since the host opened one door with a goat, the probability that the car is now behind one of the remaining 98 doors you switch to is still:
P(Car behind new door if you switch)=99100P(\text{Car behind new door if you switch}) = \frac{99}{100}P(Car behind new door if you switch)=10099
Since there are 98 remaining unopened doors after the host's reveal, the probability shifts and is effectively distributed among these doors if you switch.
Conclusion:
Switching: 99100\frac{99}{100}10099 chance of winning.
Sticking: 1100\frac{1}{100}1001 chance of winning.
It is heavily to your advantage to switch. The probability of winning increases from 1% to 99% if you switch.