Monty Hall Problem with 4 doors

[MichaelLiu]

You are given 4 doors. Behind 1 door is a car, and behind the others is a goat.

You will choose your first door, and the host will reveal another door which has a goat behind it.
Then, you can choose to either stay with your choice or switch. Then, the host will reveal a different door, containing a goat.
Again, you will choose to stay or switch. You may switch back to your original door.

What is the best strategy, staying twice, staying then switching, switching then staying, or switching twice?

Thanks a lot!

 

[jvicens]

I would approach this scenario by first understanding the probability of each door containing the car. Initially, there is a 1/4 chance of choosing the correct door and a 3/4 chance of choosing a door with a goat behind it.

After the host reveals a door with a goat, there are now two doors left - the one you initially chose and the one that was not revealed. The probability of the car being behind either of these doors is now 1/2. This means that whether you choose to stay or switch at this point, the probability of winning the car is the same.

However, if you choose to switch after the host reveals a door with a goat, you are essentially combining the probabilities of both doors that were not initially chosen. This means that the probability of winning the car by switching is now 2/3, compared to the 1/4 probability of staying with your original choice.

Therefore, the best strategy in this scenario would be to switch after the host reveals a door with a goat. This strategy increases your chances of winning the car from 1/4 to 2/3. Switching twice or staying then switching would not change the overall probability of winning the car, as the probabilities are already equal at 1/2 after the first switch.

In conclusion, the best strategy in this scenario would be to switch after the host reveals a door with a goat, increasing your chances of winning the car to 2/3.