The Monty Hall Problem- Revision and extensive analysis

My understanding is that NO NEW INFORMATION is given - which is why the probability is 2/3 if you DO switch.

If you have 100 doors, a car behind one of them and goats behind 99 of them - you KNOW ALREADY that of the 99 doors you didn't select, at least 98 of them will be goats.

So by opening 98 doors and revealing goats you are not revealing any new information.

So in the 3-door problem, it is because no new information is given that you should swap to the door you didn't initially select.
The door you initially picked has 1/3 probability.
Since no new info is given, the probability of that door winning is always going to stay 1/3.

 

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This is confusing to me. I don't see how you have a greater chance of winning if you switch.

Let's take the same example and assume that instead of one, there are two contestants. If one contestant chooses one unopened door and the other contestestant chooses the other, and they both switch, how can they both have a 2/3 chance of winning? Doesn't that add up to over 100%?

 

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The probability of winning under the always-switch strategy is 1/2, not 2/3.

 

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The probability of winning under the always-switch strategy is indeed 2/3. There are three outcomes for the initial choice: the car, or either of the two goats. Choosing the car initially and switching is a losing proposition, but this only happens 1 out of 3 times. If you happen to pick one of the goats on the first choice, Monte has no choice but to show you the other goat, and the third door must hide the car. Switching is a winning proposition in the 2 out of 3 cases where the door chosen originally hid a goat. Putting it all together,

P(car | switch) = 1/3*0 + 2/3*1 = 2/3.

 

Whoops, my mistake...

Getting back to the scenario Prosoothus raised: it's not possible to play the Monte Hall game with two players like that. You assume that they choose different doors, but if they both choose doors with goats, Monte won't have any doors with goats left to open. So you'd have to modify the game somewhat, and the analysis would be different..

 

There are a couple of things wrong here.

First, probabilities are not in general additive. The probability of either or both of two events happening is the sum of the two individual events only if those events are mutually exclusive.

It appears that these are indeed mutually exclusive events by construction: Only one of the two contestants, if any, will win the car. The second problem is: What if both contestants chose doors hiding goats? Monte better not show a third goat or the game is rigged.