Time to discuss the Monty Hall problem (strategy and game show)

66.6% > 33.3%

Switching.

Switching, but I'll admit it took me forever to accept this.

Was talking about this the other day. Switch every time

It's amazing how even mathematicians didn't grasp it.

Your initial odds of picking a goat are 2/3. If you picked a goat, the host is guaranteed to open another goat-door (it's important to remember the host knows which is which), meaning you should switch.

Let's not discuss Monty Hall, it's been done before. Let's discuss how to find the door that hides this fine specimen.

1,1,1,1,1,1....

I have to disagree with the "always switch" conclusion. The correct answer about the probability is that it is 50/50.

The problem is not in the math--it is that the setup and explanation of the problem always given is a cheat. The problem is changed in the middle, but the original probabilities are not. That is why it is so counterintuitive.

Those who discuss the probability in terms of 33% and 66% after the first door is eliminated are carrying forward into the new problem (where only two doors exist) information from the original problem, and thus not establishing anything more earth shattering than the fact that we originally had 66% goats and 33% cars to choose from.

In fact, the reason people commonly say the answer is 50/50 is that most people intuitively accept the initial premise that a door is simply removed from the problem--leaving two doors, known to hide a goat and a car. This is clearly a 50/50 choice.

You can appreciate this truth if you consider the person who comes into the game at the point where the contestant is told that there is goat behind one of the doors he didn't pick. That door is effectively removed from the game, as if it never existed. The new person faces simply two doors, with no knowledge about what has gone on before, and the certainty that there is one goat and one car. It does not matter which of the two doors he picks (or whether he "picks" one then "switches", or doesn't switch.) In the end, he chooses one door--and has a 50/50 shot at getting it right. Thus, for him, it doesn't matter if he switches or not--his probability is always 50/50.

Again, the problem with this riddle is that the mathematical explanations always start with the premise that you must carry forward the old 33/66 probabilities from the first problem into the second. That is the cheat--you don't. The Monty Hall problem really is a 33/66 probability problem changed into a 50/50 problem, but discussed mathematically (incorrectly)after the basic premise has been changed as if it is still a 33/66 problem.

We have a non-believer.

20+ pages.

Thanks Broseph Barksdale for proving my point.

In your explanation, you are absolutely correct--but only because you consider it all one game.

However, the problem is always described by asking what the odds are of choosing the car once one door is eliminated (and is always a goat). That IS a new game, and the probabilities are 50/50. If mathematicians don't like the fact that people perceive it correctly to be a new game, that is fine, they can continue to run it all together as one multi-stage probability exercise. But that is mere slight of hand, not math.