Originally, this was to be my last article on The Reverse Monty Hall Problem – or at least the last on why the solution is “it doesn’t matter whether you switch or stay, the likelihood of benefitting from either choice is 1/2". It didn't work out that way ...
- You walk to the door that you selected that will remain closed, and you mark it as Door 1,
- You walk to the door that the philosopher has told you will be opened, and you mark it as Door 2,
- You walk to the unselected door, and you mark it as Door 3, and
- Then you return to the philosopher to continue the show.
Someone also complained about my graphical representation of the solution to the classic Monty Hall Problem:
This might be difficult to explain, but please bear with me.
Now, imagine that we run the Reverse Monty Hall scenario over and over again, still assuming that when there is an option to select between two goats, the philosopher makes that selection at random. This means that the possibilities are:
This can be compared with the results from the classic Monty Hall Problem.
So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split. But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly.
Once again, please note that the argument above is in correct. The correct answer is that the 2/3 - 1/3 split is a true (in this scenario) as it is for multiple iterations of it.