re: Marilyn vos Savant and the history of the Montel Hall question

I have a better idea.

Take any two doors, where you know that there is a car behind one and a goat behind the other. Pick one. Switch or don't switch, I don't care. Record your results over 100 tries. (Hint: if you flip a coin this will be the same, and easier).

The point is: If you accept carrying the initial 3-door probabilities forward, the mathematical proofs are correct. But that is a lie; the problem has been changed to eliminate one non-car door. That is what people intuitively understand, and why they say that it is a 50/50 shot at that point.

quote:

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But you are basically choosing 2 doors when you switch (2/3 chance). So it is technically two choices, but the odds are still 2/3 to 1/3.

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Yes, which is why I clarified my agreement "conceptually" versus reality.

But, in the end, you are not actually being given 2 chances to pick the car. Remember, the Host will ALWAYS drop a goat for you so it is simply impossible to pick 2 goats. So, it was always one car and one goat to pick from.

If you picked the car correctly (your 1/3 pick), you will now be given yet another chance to screw up and change to the wrong pick but this time you have a 50/50 chance instead of the "initial" 1/3 chance.

With all that said, the explanation of the solution is provided in the article and there is nothing, that I know of, that any of us can do to refute the results.