[armourall]

Not that familiar with Hold 'em, so I'll sit this hand out. Just to finish up on Monty, here is what I consider to be the correct rundown of the probabilities: (My previous "flawed" argument conveniently excluded possibilities #4 and #6.)

1. The player originally picked the car. The door with goat 1 is revealed. The remaining door has goat 2. (1/6)
2. The player originally picked the car. The door with goat 2 is revealed. The remaining door has goat 1. (1/6)
3. The player originally picked goat 1. The door with goat 2 is revealed. The remaining door has the car. (2/6)
4. The player originally picked goat 1. The door with the car is revealed. The remaining door has goat 2. (0/6)
   (Rules don’t allow this option, so probability of #3 outcome is doubled.)
5. The player originally picked goat 2. The door with goat 1 is revealed. The remaining door has the car. (2/6)
6. The player originally picked goat 2. The door with the car is revealed. The remaining door has goat 1. (0/6)
   (Rules don’t allow this option, so probability of #5 outcome is doubled.)