# One variation of Monty Hall problem

Here once Monty has a choice to open any of two doors each containing a goat (i.e. when our initial choice is a car) then he chooses the rightmost one. We need to prove that here also the "overall" probability of winning by switching is $\frac{2}{3}$ using the rule of total probability. I am able to calculate the probabilities for each of the configurations (it is sure wining or 50% wining after switching). I want to claculate the "total" proability using them.

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I think this needs a bit more context. People might not know which role Monty has in the game (I didn't). The strategy isn't fully specified by "switching", since the doors are no longer equivalent so the strategy needs to include which door to pick originally. Also, it's not entirely clear what you mean by a "configuration", and it wouldn't hurt to tell us the probabilities that you were able to calculate. –  joriki Jan 23 '13 at 12:01

As in the ordinary Monty Hall problem, the simplest way to calculate the probability of winning with various strategies is simply to enumerate all the possible occurrences and their probabilities.

First, let's summarize what we know. As in the usual Monty Hall problem, we assume that there are three doors (which we'll arbitrarily label A, B, and C from left to right), behind one of which is a car (which the contestant wants) and behind the other two is a goat (which the contestant doesn't want). We assume that the car is, a priori, equally likely to be behind any of the doors, and that, after the contestant has chosen one of the doors, Monty (who knows which door the car is behind) always opens one door which has not been chosen by the contestant and has a goat behind it, and then offers the contestant the option to switch to the other unopened door. In this variation, we'll also assume that, if Monty has two possible doors to choose from, he will always open the rightmost one of them.

Then the possibilities are:

• Car is behind door A (probability 1/3):
• Contestant chooses door A ⇒ Monty opens door C (stay ⇒ win, switch ⇒ lose).
• Contestant chooses door B ⇒ Monty opens door C (stay ⇒ lose, switch ⇒ win).
• Contestant chooses door C ⇒ Monty opens door B (stay ⇒ lose, switch ⇒ win).
• Car is behind door B (probability 1/3):
• Contestant chooses door A ⇒ Monty opens door C (stay ⇒ lose, switch ⇒ win).
• Contestant chooses door B ⇒ Monty opens door C (stay ⇒ win, switch ⇒ lose).
• Contestant chooses door C ⇒ Monty opens door A (stay ⇒ lose, switch ⇒ win).
• Car is behind door C (probability 1/3):
• Contestant chooses door A ⇒ Monty opens door B (stay ⇒ lose, switch ⇒ win).
• Contestant chooses door B ⇒ Monty opens door A (stay ⇒ lose, switch ⇒ win).
• Contestant chooses door C ⇒ Monty opens door B (stay ⇒ win, switch ⇒ lose).

By summing over all possible locations of the car, we can see that, as in the usual Monty Hall problem, the strategy "always stay" wins 1/3 of the time while the strategy "always switch" wins 2/3 of the time, regardless of which door the contestant initially chooses.

However, we could also consider strategies where the decision to switch depends on which door Monty opens. For example, we can easily see that, if we choose door A and Monty opens door B, we'll know for sure that the car must be behind door C, whereas if Monty opens door C, the car is equally likely to be behind either door A or B. Thus, the strategy "choose door A, switch if door B is opened" also wins with probability 2/3, as do the strategies "choose B, switch if A is opened" and "choose C, switch if A is opened".

Unfortunately, we can't do any better than that: regardless of which door we choose initially, there's always a 2/3 chance that Monty's action will not tell us which of the two unopened doors the car is behind, and so we'll always have a 1/2 · 2/3 = 1/3 chance of guessing wrong. Thus, the "always switch" strategy, with its 2/3 chance of winning, still remains optimal even in this variant of the problem, even though it's no longer the unique optimal strategy.

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