For Nerds Only - How To Calculate Odds

[Crispy]

You can assume that he picks randomly among the doors he knows do not contain the prize.

I've never seen that assumption made explicit though, although it certainly is a reasonable one under the circumstances.

 

Here's the proof I promised earlier. I didn't take notes at the time the lecturer was explaining, so I recreated the proof based on this one from Wikipedia: http://en.wikipedia.org/wiki/Monty_Hall_pr...ayes.27_theorem

 

Notice that the assumption that the host makes his choice randomly has to be stated explicitly there.

 

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Let P1 be the event "the prize is behind door 1".

Let P2 be the event "the prize is behind door 2".

Let P3 be the event "the prize is behind door 3".

Let O1 be the event "the host opens door 1".

Let O2 be the event "the host opens door 2".

Let O3 be the event "the host opens door 3".

 

Assume the prize is distributed with uniform probability: P(P1) = P(P2) = P(P3) = 1/3.

 

Assume, without loss of generality, that you have chosen door #1.

 

Assume that if the host has a choice of doors to open (i.e. the prize is behind door #1), then he will always open door #2. That is, P(O2 | P1) = 1, and P(O3 | P1) = 0. [usually, everyone assumes that these values are both 1/2.]

 

If the prize is behind door #2, then the host will always open door #3. i.e. P(O3 | P2) = 1. Similarly, P(O2 | P3) = 1.

 

Of course, P(O2 | P2) = 0 and P(O3 | P3) = 0 (the host will never open the door which conceals the prize).

 

Using a law of conditional probability, P(O2) = P(P1)*P(O2 | P1) + P(P2)*P(O2 | P2) + P(P3)*P(O2 | P3)

-> P(O2) = (1/3) * (1 + 0 + 1) = 2/3

 

Using Bayes' Theorem, P(P3 | O2) = P(O2 | P3)*P(P3) / P(O2)

-> P(P3 | O2) = 1 * (1/3) / (2/3) = 1/2

 

Also, P(P2 | O2) = 0 and P(P1 | O2) + P(P2 | O2) + P(P3 | O2) = 1; therefore P(P1 | O2) = 1/2.

 

In other words, if the host opens door #2, then the probability that the prize is behind door #3 is 1/2; and the probability that the prize is behind door #1 is also 1/2. So in this situation there is no advantage in switching.

 

On the other hand, if the host opens #3, then of course the prize must be behind door #2 (since he would have opened door #2 if he was able), so switching guarantees a win. You can also calculate this using a method similar to the above.

 

Granted, this is somewhat artificial, since it relies on you knowing that the host will always choose a certain door when he is able (which in a real-world gameshow you probably wouldn't know), but it does demonstrate my point.

 

Note that always switching is still an optimal strategy (one of two, the other being "always switch unless the host chooses his preferred door"), so that piece of advice holds true no matter what the probabilities involved are.

[OrbWeaver]

I get a bit lost in the details, but I understand your explanation via the concept of "information leakage".

 

As in, if you know that the host will choose a specific door if possible, you can gain information by observing whether he does so or not.

[Crispy]

Yep. That's the simplified intuitive explanation.