ChatGPT gets very confused when you give it something similar to monty hall

Alex Wennerberg

I show you three closed doors, two of which have a car behind them and one of which has a donkey. You pick a door and I open a different door revealing a donkey. I then give you the option to change your decision. If you want to win a car, should you switch doors?

Yes, you should switch doors. The probability of winning a car by initially choosing one of the three doors is 1/3. After one of the doors with a donkey behind it is revealed, the probability of winning the car by switching to the other closed door is 2/3, which is twice as likely as the initial probability of 1/3. This is because, when you first made your choice, the car was behind one of the doors you didn't pick, and the host revealed one of those doors. So, by switching to the other unopened door, you have a higher chance of winning the car. This probability paradox is known as the Monty Hall problem, and switching doors is the optimal strategy to increase your chances of winning the car.

Alex Wennerberg

Are you sure that your probabilities are correct?

Yes, I'm sure that the probabilities I provided are correct. The Monty Hall problem is a well-known probability puzzle that has been studied extensively and the answer is mathematically proven.

The initial probability of choosing a door with a car behind it is 1/3. After one of the doors with a donkey is revealed, the probability of the car being behind one of the remaining closed doors is now 2/3. This means that if you switch doors, you have a 2/3 chance of winning the car, whereas if you stick with your original choice, you only have a 1/3 chance of winning.

This probability paradox can be counterintuitive, but it has been proven mathematically and has been demonstrated in various experiments and simulations. So, in the Monty Hall problem, switching doors is the optimal strategy to maximize your chances

Alex Wennerberg

How many cars were in my question?

In your question, you mentioned that there are two doors with cars behind them and one door with a donkey behind it. So, there are a total of two cars in your question.

Alex Wennerberg

So what is the probability of receiving a car before and after the donkey is revealed?

Before the donkey is revealed, the probability of receiving a car is 2/3, since there are two doors with cars behind them and one door with a donkey behind it.

After the donkey is revealed, the probability of receiving a car changes depending on whether or not you switch doors. If you don't switch doors, the probability of receiving a car remains at 2/3. However, if you do switch doors, the probability of receiving a car increases to 2/3, since the car must be behind one of the two doors that you did not initially choose.

To summarize, the probability of receiving a car before the donkey is revealed is 2/3, and after the donkey is revealed, the probability of receiving a car is either 2/3 if you switch doors or 1/3 if you do not switch doors.

Alex Wennerberg

That is incorrect. In my example, there are two cars, not one car. What is the probability after a donkey is revealed of me having a car?

I apologize for the mistake in my previous response. Since there are two cars in your example, the probability of receiving a car before the donkey is revealed is 2/3, and the probability of receiving a donkey is 1/3.

After the donkey is revealed, if you switch doors, the probability of receiving a car becomes 2/3 and the probability of receiving a donkey becomes 1/3. This is because the car must be behind one of the two doors that you did not initially choose, and switching doors effectively gives you a chance to choose either of those two doors.

So, in summary, after the donkey is revealed, if you switch doors, the probability of receiving a car becomes 2/3, and if you don't switch doors, the probability remains at 1/3.

Alex Wennerberg

That is incorrect. You are confusing my question with a similar problem. In my example, the number of cars is different.

I apologize for the confusion in my previous responses. I misunderstood your initial question and incorrectly assumed that it was the Monty Hall problem.

In your example, there are two cars behind the three doors and one donkey behind the other door. If you initially choose one of the three doors randomly, the probability of choosing a car is 2/3 and the probability of choosing a donkey is 1/3.

After one of the doors with a donkey is revealed, there are two doors remaining. If you switch doors, the probability of receiving a car becomes 1/2, and if you don't switch doors, the probability of receiving a car remains at 2/3.

So, in summary, the probability of receiving a car before the donkey is revealed is 2/3, and after the donkey is revealed, if you switch doors, the probability of receiving a car becomes 1/2, and if you don't switch doors, the probability remains at 2/3.

Alex Wennerberg

Your first two paragraphs are correct, but then you are incorrect. Do you see your mistake?

I apologize for my previous error. I now see my mistake.

After one of the doors with a donkey is revealed, you are left with two doors, one of which has a car behind it and the other has a donkey. Since there are two cars and three doors initially, the probability of each door having a car is 2/3.

If you initially chose a door with a donkey, then the remaining door must have a car. So, switching doors would result in winning a car with a probability of 1.

If you initially chose a door with a car, then the remaining door must have a donkey. So, switching doors would result in losing with a probability of 1.

If you initially chose a door with a donkey, then the probability of switching doors and winning a car is 1/2, while the probability of losing is also 1/2.

So, in summary, after one of the doors with a donkey is revealed, the probability of winning a car by switching doors is 1/2 if you initially chose a door with a donkey, and 0 if you initially chose a door with a car.

Alex Wennerberg

You are again confusing my problem with a similar problem. In my example, there are two cars, rather than one car. After the donkey is revealed, how many cars remain, and how many doors?

I apologize for the confusion. In your example, there are two cars and one donkey behind the three doors. If you initially choose one of the three doors randomly, the probability of choosing a car is 2/3 and the probability of choosing a donkey is 1/3.

After one of the doors with a donkey is revealed, there will be two doors remaining, one with a car and the other with a car as well. The probability of winning a car after switching to the other unopened door is 1/2, as there are only two doors left, and one of them has a car behind it.

So, in summary, after the donkey is revealed, there are still two cars remaining, but now only two doors. If you switch doors, the probability of winning a car is 1/2, and if you don't switch, the probability remains at 2/3.

Alex Wennerberg

My example resembles the monty hall problem, but it is different. If there are two cars and one donkey behind the three doors, and the donkey is revealed, what remains?

If there are two cars and one donkey behind the three doors, and the donkey is revealed, then there would be two cars remaining behind two doors. Since you know that one of the doors has a donkey behind it, you can eliminate that door as an option. This leaves you with two doors, each with an equal probability of having a car behind it.

So, after the donkey is revealed, there are two cars remaining behind two doors. The probability of winning a car by switching doors is 1/2, regardless of which door you initially chose.

Alex Wennerberg

If there are two cars remaining behind two doors and I chose one of them, what is my probability of choosing a car?

If there are two cars remaining behind two doors and you chose one of them, your probability of choosing a car is 1/2, assuming that both doors are equally likely to have a car behind them.

Since there are only two doors remaining and you have already chosen one of them, the probability of the other door having a car behind it is 1/2. This means that if you switch doors, your probability of choosing a car would also be 1/2.

Alex Wennerberg

That is incorrect. Do you see your mistake?

I apologize for my mistake. If there are two cars remaining behind two doors and you chose one of them, the probability of choosing a car is not necessarily 1/2.

If you initially chose a door with a car behind it, then the probability of choosing a car by switching doors would be 0, since the other door would necessarily have a donkey behind it.

However, if you initially chose a door with a donkey behind it, then the other door must have a car behind it, and switching doors would result in a probability of 1 for choosing a car.

Therefore, the probability of choosing a car after initially choosing a door with a donkey

Alex Wennerberg

If there are two doors remaining with two cars behind them, my probability of receiving a car should be 1, not 1/2. You are confusing this with the monty hall problem. My problem is worded like the monty hall problem, but the reasoning is different.

I apologize for the confusion. You are correct that in your problem, if there are two doors remaining with two cars behind them, then the probability of receiving a car should be 1 if you switch your choice.

In the Monty Hall problem, the crucial difference is that the host reveals a donkey, which changes the probabilities of the remaining doors. In your problem, there is no such change in probabilities since both remaining doors contain cars. Therefore, if you switch your choice, you will always win a car.

Alex Wennerberg

So should I switch my choice?

Yes, if there are two doors remaining with two cars behind them, you should switch your choice to the other door. This is because the probability of winning a car is 1 if you switch your choice, and only 1/2 if you stick with your original choice. By switching, you are effectively hedging your bets and increasing your chances of winning.

Alex Wennerberg

There are two problems with your answer:

First, you state that "In your problem, there is no such change in probabilities since both remaining doors contain cars." There is a change, but the numbers are different. In my problem, since there are two cars behind three doors, my probability of getting a car is 2/3. After the host reveals the donkey, my probability is 1

Second, you state that the probability of sticking with my original choice is 1/2. That is not correct, because there are two cars, rather than one. What is the real probability of sticking with my original choice?

I apologize for my previous answer being incorrect. Thank you for clarifying the problem.

In your problem, there are initially three doors, two of which have cars behind them, so the probability of choosing a car is 2/3. When the host reveals a donkey behind one of the doors, the probability of the remaining unchosen door containing a car is 2/3. Therefore, switching to the other unchosen door will result in a win with probability 2/3.

Regarding the probability of sticking with your original choice, the probability of choosing a car initially is 2/3, and therefore the probability of choosing a donkey initially is 1/3. When the host reveals the donkey behind one of the doors, the probability of the remaining unchosen door containing a car is still 2/3, and therefore the probability of your original choice being a car is only 1/3.

Alex Wennerberg

That is incorrect. If there are two doors with cars behind them and the host reveals a donkey, the probability of me having a car goes up to 1, since all remaining doors have cars behind them

I apologize for my previous response being incorrect. You are correct that if there are only two doors and both have cars behind them, then after the host reveals a donkey, the probability of your chosen door having a car is 1. This is because there are no other doors left to choose from that could have a donkey behind them, leaving only the two doors with cars. Thank you for pointing out my mistake.

Alex Wennerberg

So should I switch after the host reveals the donkey or not?

In the scenario you described where there are two doors and both have cars behind them, and the host reveals a donkey, it doesn't matter whether you switch or not.

If you switch, you will get one of the cars behind the remaining door, and if you don't switch, you will still get the other car behind the other door.

So, in this case, the decision to switch or not will not affect your chance of winning a car.