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The Monty Hall problem (https://en.wikipedia.org/wiki/Monty_Hall_problem) is about a game show in which the player is given the choice of three doors. There's a car behind one of the doors and a goat behind the others. The players is asked to pick a door, and he picks door number 1 which remains closed for now. The host, who knows where the car is, then opens door number 3, which has a goat, and then asks the player if he wishes to stay on door 1 or change to door 2.

Now, since the host knows where the car is - then the odds for door number 2 are obviously higher, and the player should switch from door 1 to door 2 (.. because it has 2/3 chance to have the car).

My question is: If the host doesn't know where the car is, and he randomly picks door 3 - are the odds the same as in the above scenario (i.e. door 2 has 2/3 chance to be the one with the car), or is it a different case in which doors 1 and 2 have 50% chance each?

Thanks! Dan

user1124556
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  • You can find answer here: https://math.stackexchange.com/questions/3085994/in-the-basic-monty-hall-problem-why-not-the-probability-are-50-50/3086008#3086008 – Rutger Moody Dec 14 '22 at 08:48
  • There are many versions of Monty Hall although the original one was just designed to support understanding conditional probabilities. It turned out that this backfired because the people that did not understand the original version constructed versions where switching is not only bad, but fatal. The original version is clearly defined : 1. A door is always opened 2. It is never the door with the car 3. If there is a choice which wrong door is opened, a random door is opened. – Peter Dec 14 '22 at 08:48
  • In the original version , switching wins iff the original choice is not the door with the car and not switching wins iff the original choice is the door with the car. That's it. – Peter Dec 14 '22 at 08:50
  • @trueblueanil I think you're wrong: if the host opens the door with the goat by chance (he knows nothing about the location of the car) then it's exactly like another player.. in that case his choice is not "a smart choice", but purely random, and it says nothing about door 2. In case that the host knows where the car is then his choice might be a deliberate one (if the car is behind door 2, and he knows that). hence the option of door 2 is stronger.
    This is just my intuition :-) , I'm just looking for a more concrete (mathematical?) proof.     – user1124556 Dec 14 '22 at 09:40
  • You are right, I have deleted my comment. The crux of the matter is that if I initially picked a goat, by switching I will necessarily win because the host must open the door for the other goat, whereas if he is randomly opening a door, he will do so only 1/2 the time. – true blue anil Dec 14 '22 at 10:58
  • Also, in your version, it is not clear as to how the game will proceed if the host opens the door with the car by chance ! – true blue anil Dec 14 '22 at 14:03

1 Answers1

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Your intuition is correct: It becomes a toss-up whether you stay or switch.

We make the following assumptions:

  • The three doors conceal (in some order) two goats and a car.
  • All three positions for the car are equally likely.
  • You choose one of the three doors.
  • Monty opens one of the remaining two doors, irrespective of where the car is.

Assume, without loss of generality, that you choose door $1$ and Monty opens door $2$ to reveal a goat. Before he did so, there were three equally likely positions for the car. Afterwards, he has eliminated one of the options, but because he effectively does not know where the car is, he has given you no additional information with respect to doors $1$ and $3$. Therefore, they remain equally likely as they did before.

If we like, we can bring the machinery of Bayes's Law to bear. Again, assume you open door $1$. Then let $C_k$ ($k = 1, 2, 3$) be the event that the car is behind door $k$, and let $D_k$ ($k = 1, 2, 3$) be the event that Monty opens door $k$. Then the probability that the car is behind door $1$, given that Monty opened door $2$ and the car wasn't behind door $2$, is

\begin{align} P(C_1 \mid D_2, \neg C_2) & = \frac{P(D_2 \mid C_1, \neg C_2)P(C_1 \mid \neg C_2)} {P(D_2 \mid \neg C_2)} \\ & = P(C_1 \mid \neg C_2) \\ & = \frac12 \end{align}

where the middle equality obtains because we assume that $D_2$ is independent of the location of the car. The difference between this situation and the usual interpretation is that in the latter case, we assume that Monty never opens the door where the car is. So that middle equality doesn't obtain.

Brian Tung
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