Optimizing Your Chances: Solving the Monty Hall Variation | Homework Help

[superconduct]

Homework Statement

Original Monty Hall problem: There are 3 doors, 1 of them contains a car and the other 2 goats. You choose 1 door, the host opens a door that is not chosen by you and does not contain the car. Then you can change to the other closed door, or keep your own chosen door at first. Most people agree with changing as there is 2/3 chance of getting a car.

Now add conditions to the original problem, that is the host will have 80% chance of opening a door closer to himself and can stand only on the 2 ends of the stage.

Which decision should you make now and why?

Homework Equations

I found on the net that 'If the host opens the rightmost door, switching wins with probability 1/(1+q) where he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q=1−p.'

The Attempt at a Solution

1. I get 2/3 because the new condition will only take in effect when your first choice is the car.
2. From b2 I get 5/6 but I don't understand why b2 is true and why not 2/3

 

[HallsofIvy]

I'm not sure I understand "the host will have 80% chance of opening a door closer to himself and can stand only on the 2 ends of the stage." Are you saying that there is now a chance that the host will open the door with the car behind it or are you saying that the 80% or 20% chance applies only to the doors with goats behind them? If it is the latter then the probability remains the same- as long as the host only opens a door that does NOT have the prize behind it, which of the two doors he opens is irrelevant.

 

[superconduct]

what if it's not the latter? in which condition would you get the 5/6 answer?