Panda writes:
I think the reasoning is:
Since there is now a chance of Monty winning the car, your chance of winning the car (on the right hand side of your diagram) must be lower.
This reasoning cannot be correct.
The question asks what one should do when Monty shows a goat. In other words, Monty's possibility of showing the car should be removed from the considered outcomes, and probabilities would be calculated based on the remaining possible outcomes. It doesn't matter whether Monty picked the goat randomly, by ESP, or by being told which doors hide goats.
Slevesque's overlooks that when you initially pick a car, there are
two ways for Monty to randomly pick the goat. If you switch your pick in either of those two situations, you lose.
Edited by NoNukes, : Address slevesque's post