So, Let's say we have three doors and three guests in our show. A car is behind one door. The other two doors have goats . Please note.. I am **not** asking about the original Monty Hall problem here. This is another problem (but eventually it might be equivalent to the original one).

Each guest picks a different door. and the host opens one losing door and says goodbye to one contestant.Then He asks each of the remaining guests if they would like to switch.

**Q1** What is winning probability of each of the remaining guests if they decide to switch?

**Q2** If both have to switch doesn't that mean that the computed probabilities are useless?

**NOTE**I am aware that we have two events which**aren't**disjoint here

Event A : Car is behind the losing door or first remaining guest's choice

Event B : Car is behind the losing door or second remaining guest's choice

and each has 2/3 chance to win. So, the sum here isn't 1

So, eventually.. If the answer is to **Switch** or **NOT** , doesn't it mean that we have same probability for the remaining doors.
Thanks.. please **give time for discussion** and don't down vote for the question if it's against your mathematical beliefs. I am suggesting a new problem here, so be open to other's ideas :)