So I agree that Rich is short-sighted, though I think that's separate from his reactionary nature. Alas, you are 1000% wrong about the Monty Hall problem. It is proven theoretically, and plays itself out in simulation. That last real challenges fizzled out the 90's when computer simulation demonstrated it to be true to those who had previously refused to accept it. It's what Quine calls a veridical paradox, the outcome seems bogus and it is true nonetheless.
Your problem is that you are skipping over a key piece of information. After revealing a tiger, the host is not the only one with knowledge, the player has additional knowledge as well. In fact, not only is it better to switch doors than to keep your door, it's also better to switch doors than to choose one of the remaining doors at random. I get that with 3 doors, this is a bit hard to intuit, so let's do the counter-intuitive and scale this problem up to make it more intuitive!
The scenario is much the same, but this time there are 100 doors - for simplicity of typing, we'll say 1/100 doors wins, 99/100 doors lose. After choosing a door, the host will open 98 doors that lose, leaving only your chosen door and one other.
When you initially choose a door, you have a 1/100 chance of winning. After the host opens 98 losing doors, that has not changed. Even though there are only 2 doors remaining, you still have a 1/100 chance. Why?
Because the host is guaranteed to open only losing doors. If the host chooses doors at random and may open the winning door (but not your door), then things change: If the host opens the winning door, you now have a 0% chance of winning. If the host does not open the winning door, then you now have a very good chance of winning with your current door. Why?
If the host opens randomly, then he is effectively playing the same game as you, except with 99 doors (since you've put one out of play). By opening all but one of the remaining doors at random, he has a 98/99 chance of opening the winning door if you had not chosen it, and a 0% chance of opening the winning door if you had. So, there is almost a 100% chance that the host would open the winning door if you had not chosen it, and of course no chance at all if you had.
So as the player, you are watching the host open doors randomly, and the winning door does not get opened. Whereas you previously had a 1/100 chance of having the correct door, you would now have a 98/99 chance of having the correct door, so you would of course keep it. Of course, if the winning door had been opened by the host, you would go home sad.
But that's not the scenario. The scenario is that the host knows the contents, and is guaranteed to only open losing doors. However, the previous example illustrates a point, which is that the player gains knowledge as the host opens doors. When the host knowingly opens a losing door, he is telling you that this particular door is not a winning door. More importantly, he is telling you that the winning door is still in the remaining set.
So you choose 1 of the 100 doors which has a 1/100 chance of winning. That means there is a 99/100 chance that the winning door is in the set of doors that you did not pick - the set of doors which the host must narrow down to only 1 (let's call it the hosts' set). So if there is a 99% chance that the host's set contains the winning door, and the host knows which door is the winning door and is contractually obligated to not remove it from the set, the single door he leaves in his set at the end still has a 99% chance of winning.
So now you get to change your mind, do you want the door that still has a 1% chance of winning, or the one with the 99% chance of winning?
This scales no matter the number of doors. If you have 3 doors, and you choose 1, it is a 1/3 chance of winning. The host's set then has a 2/3 chance of containing the winning door. Since the host must keep the winning door if his set contains it, the remaining door that he leaves also has a 2/3 chance of winning. So you are indeed better of switching doors.
Your problem is that you are skipping over a key piece of information. After revealing a tiger, the host is not the only one with knowledge, the player has additional knowledge as well. In fact, not only is it better to switch doors than to keep your door, it's also better to switch doors than to choose one of the remaining doors at random. I get that with 3 doors, this is a bit hard to intuit, so let's do the counter-intuitive and scale this problem up to make it more intuitive!
The scenario is much the same, but this time there are 100 doors - for simplicity of typing, we'll say 1/100 doors wins, 99/100 doors lose. After choosing a door, the host will open 98 doors that lose, leaving only your chosen door and one other.
When you initially choose a door, you have a 1/100 chance of winning. After the host opens 98 losing doors, that has not changed. Even though there are only 2 doors remaining, you still have a 1/100 chance. Why?
Because the host is guaranteed to open only losing doors. If the host chooses doors at random and may open the winning door (but not your door), then things change: If the host opens the winning door, you now have a 0% chance of winning. If the host does not open the winning door, then you now have a very good chance of winning with your current door. Why?
If the host opens randomly, then he is effectively playing the same game as you, except with 99 doors (since you've put one out of play). By opening all but one of the remaining doors at random, he has a 98/99 chance of opening the winning door if you had not chosen it, and a 0% chance of opening the winning door if you had. So, there is almost a 100% chance that the host would open the winning door if you had not chosen it, and of course no chance at all if you had.
So as the player, you are watching the host open doors randomly, and the winning door does not get opened. Whereas you previously had a 1/100 chance of having the correct door, you would now have a 98/99 chance of having the correct door, so you would of course keep it. Of course, if the winning door had been opened by the host, you would go home sad.
But that's not the scenario. The scenario is that the host knows the contents, and is guaranteed to only open losing doors. However, the previous example illustrates a point, which is that the player gains knowledge as the host opens doors. When the host knowingly opens a losing door, he is telling you that this particular door is not a winning door. More importantly, he is telling you that the winning door is still in the remaining set.
So you choose 1 of the 100 doors which has a 1/100 chance of winning. That means there is a 99/100 chance that the winning door is in the set of doors that you did not pick - the set of doors which the host must narrow down to only 1 (let's call it the hosts' set). So if there is a 99% chance that the host's set contains the winning door, and the host knows which door is the winning door and is contractually obligated to not remove it from the set, the single door he leaves in his set at the end still has a 99% chance of winning.
So now you get to change your mind, do you want the door that still has a 1% chance of winning, or the one with the 99% chance of winning?
This scales no matter the number of doors. If you have 3 doors, and you choose 1, it is a 1/3 chance of winning. The host's set then has a 2/3 chance of containing the winning door. Since the host must keep the winning door if his set contains it, the remaining door that he leaves also has a 2/3 chance of winning. So you are indeed better of switching doors.
