New Monty Hall Puzzle(s)

Based upon some comments in the other Monty Hall puzzle, I propose the following scenarios:

Three doors which the host knows the location of a car and two goats. Two contestants get to choose SEPARATE doors.

• Scenario 1: The host always opens the non-chosen door, regardless of car or goat. A car shown means both contestants lose immediately. No choice offered.
  
• Scenario 2: The host always opens either goat door that does not declare a winner, but can declare a loser (random goat 1/2)
  
• Scenario 3: The host always opens the non-chosen door if it's a goat (2/3), but chooses a random contestant's door if the non-chosen door is a car.
  
• Scenario 3a: The contestants do not know this rule
  
• Scenario 3b: The contestants do know this rule
  

In each scenario, the option to switch is given, not forced. In what cases is it advantageous to switch?


The billion door equivalencies:

• Scenario 1: The host always opens the 1 billion minus 2 remaining doors
  
• Scenario 1a: There is one remaining door and 1 billion minus 1 contestants
  

As I'm running through the permutations of this in my head, I realize that adding multiple contestants greatly convolutes the problem. With two contestants head to head, the math is indicative that at least one of the contestants is likely to have chosen the car (2 contestants out of 3 choices). More contestants means it's more likely that one of the contestants have already won the car. With multiple contestants already knowing that one of their set has the car, there appears to be little incentive to change even if the host shows a goat. However, if the host starts showing known losers the set of unopened doors plus shown losers increases the probability that the subset of unopened doors more likely contains the car than the original probability of the set of original choice of doors versus the set of total doors.