The Monty Hall problem

Q250, Monty Hall
18 November 2011
1 The Game
The Monty Hall game goes as follows. There are three doors; behind one of the doors is a car,
and behind the other two are goats. First you pick one door and stand in front of it. Then Monty
(the gameshow host) will open one of the other two doors and show what’s behind it. And then
finally you have to choose whether to stay with the door you first chose or whether to switch to
another door. In either case, you get to keep what’s behind the door you pick the second time (and
we assume you want the car).
2 Formalizing the Game
Let D be the set {a,b,c} of doors. Let C be a D-valued random variable, namely which door has
the car. And let M be a D-valued random variable, namely the door that Monty opens. Assume the
game is fair, so C will be a uniform distribution over D: Pr(C≡a)=Pr(C≡b)=Pr(C≡c)=1/3.
Because the game is fair, we can assume you always pick door a the first time (because the problem
is symmetric, so the cases where you pick door b or c are the same modulo renaming the doors).
If we assume that Monty’s choice is symmetric, then we can say that Monty always chooses
door b (because choosing door c is symmetric). So the general question of whether to switch or
not is to solve:
argmax Pr(C≡x|M ≡ b)
x∈D
i.e., to determine whether Pr(C ≡ a|M ≡ b), Pr(C ≡ b|M ≡ b), or Pr(C ≡ c|M ≡ b) is largest,
and then you move to the corresponding door. However, in specific instances of the game we may
have a more restricted set of givens, since we’ll know more than we do in the general case.
From what we know so far we can start with the following joint probability table.
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Q250, Monty Hall
C≡a C≡b C≡c
M ≡ a 0 0 0 0
M ≡ b 1/2
M ≡ c 1/2
1/3 1/3 1/3 1
That the margins for C are all 1/3 come from the fact that we assumed that C is a uniform
distribution over D. And that the margins for M are 1/2 come from our assumption that the game
is symmetric. We can relax the assumtion that doors b and c are symmetric, but that means we’ll
have to rephrase the problem (below) to marginalize over whether Monty chooses b or c.
In order to know anything more, we need to know something about the distribution of M. And
herein lies the danger of making assumptions.
3 The Problem
Let’s say that when we’re playing the game, Monty opens door b and there’s a goat behind it (thus
C≢b). Now, the question is, do you stay before door a or do you switch to door c?
Because we know M ≡ b and C ≢ b, the problem is argmax x∈D Pr(C ≡ x | M ≡ b,C ≢ b);
namely, is Pr(C≡a|M ≡ b,C≢b) or Pr(C≡c|M ≡ b,C≢b) bigger?
4 Situation 1
Let us assume that Monty doesn’t care whether we win or loose, so he just picks a door uniformly
at random without regard for where the car is. Thus Pr(M ≡ b)=Pr(M ≡ c)=1/2.
Therefore, the joint probability is Pr(M≡ x,C≡ y)=Pr(M≡ x)∗Pr(C≡ y)=Pr(M≡ x)∗1/3.
For x=a we have Pr(M ≡ x)∗1/3=0∗1/3=0, and for x=b or x=c we have Pr(M ≡ x)∗1/3=
1/2∗1/3=1/6. Thus, we can fill in the table as follows.
C≡a C≡b C≡c
M ≡ a 0 0 0 0
M ≡ b 1/6 1/6 1/6 1/2
M ≡ c 1/6 1/6 1/6 1/2
1/3 1/3 1/3 1
To solve the problem we compute both probabilities. We’ll do the first one in excruciating
detail, and the second one more briefly.
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Q250, Monty Hall
Pr(C≡a|M ≡ b,C≢b)=
Pr(C≡a,M ≡ b,C≢b)
Pr(M ≡ b,C≢b)
defn. conditional probability
=
Pr(C≡a∧M ≡ b∧(C≡a∨C≡c))
Pr(M ≡ b∧(C≡a∨C≡c))
partitioning of C≢b
=
Pr((C≡a∧M ≡ b)∨(C≡a∧M ≡ b∧C≡c))
Pr((M ≡ b∧C≡a)∨(M ≡ b∧C≡c))
distributivity
=
Pr(C≡a,M ≡ b)+Pr(C≡a,M ≡ b,C≡c)
Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)
by disjointness
=
Pr(C≡a,M ≡ b)+Pr(⊥)
Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)
a≠c
=(1/6+0)÷(1/6+1/6) from the table
=(1/6)÷(2/6) arithmetic
= 1/2 arithmetic
Pr(C≡c|M ≡ b,C≢b)=
Pr(C≡c,M ≡ b,C≢b)
Pr(M ≡ b,C≢b)
defn. conditional probability
=
Pr(C≡c,M ≡ b)
Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)
partitioning, distributivity, disjointness
=(1/6)÷(1/6+1/6) from the table
=(1/6)÷(2/6) arithmetic
= 1/2 arithmetic
So in the case where Monty choses to open a door uniformly at random, it doesn’t matter
whether we switch or not. This happens because in this case the random variables C and M are
independent, so knowing that M ≡ b doesn’t give us any special information (beyond that C≢b).
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Q250, Monty Hall
5 Situation 2
Now, more realistically, let’s assume that Monty wants you to lose so that he can save money for
his show. Since he wants us to lose, rather than picking a door uniformly at random (which might
mean picking a door with the car), he will always pick a door with a goat.
In this case we know that Pr(M≡ b| C≡ b)=Pr(M≡ c| C≡ c)=0. Without the assumption
that Monty’s choice is symmetric, we can fill in the table as follows:
C≡a C≡b C≡c
M ≡ a 0 0 0 0
M ≡ b 0 1/3
M ≡ c 1/3 0
1/3 1/3 1/3 1
If we assume Monty’s choice is symmetric, then that will resolve the empty cells. Notably,
the marginal probabilities of Monty’s choice are the same as in the first situation, even though the
joint and conditional probabilities are quite different.
C≡a C≡b C≡c
M ≡ a 0 0 0 0
M ≡ b 1/6 0 1/3 1/2
M ≡ c 1/6 1/3 0 1/2
1/3 1/3 1/3 1
So now we can do the same computations as before, but using this new table.
Pr(C≡a|M ≡ b,C≢b)=Pr(M ≡ b,C≡a)÷(Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)) see situation 1
=(1/6)÷(1/6+1/3) from the table
=(1/6)÷(3/6) arithmetic
= 1/3 arithmetic
Pr(C≡c|M ≡ b,C≢b)=Pr(M ≡ b,C≡c)÷(Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)) see situation 1
=(1/3)÷(1/6+1/3) from the table
=(2/6)÷(3/6) arithmetic
= 2/3 arithmetic
So in the case where Monty always chooses to open a door with a goat, we should always
switch since that doubles our chances of winning. Ironically, in trying to make us lose, Monty gave
us information which will help us win! This happens because in this case the random variables
C and M are not independent, so knowing that M ≡ b gives us special information (beyond that
C≢b).
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Q250, Monty Hall
6 Situation 3
We could imagine another case in which Monty wants us to win. (Maybe he’s trying to bankrupt
the show for some reason.) In this case, whenever he can choose the door with the car he will. If
we assume his choice is symmetric in the case where the car is behind door a, then we can fill in
the table as follows.
C≡a C≡b C≡c
M ≡ a 0 0 0 0
M ≡ b 1/6 1/3 0 1/2
M ≡ c 1/6 0 1/3 1/2
1/3 1/3 1/3 1
But what about the case where he picks door b and there’s no car there? We can do the same
computations as before, but using this new table.
Pr(C≡a|M ≡ b,C≢b)=Pr(M ≡ b,C≡a)÷(Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)) see situation 1
=(1/6)÷(1/6+0) from the table
= 1 arithmetic
Pr(C≡c|M ≡ b,C≢b)=Pr(M ≡ b,C≡c)÷(Pr(M ≡ b,C≡a)+Pr(M ≡ b,C≡c)) see situation 1
= 0÷(1/6+0) from the table
= 0 arithmetic
If Monty will always show us the car when he can, we know that if he does not show the car
then it must be behind door a, so we should always stay. Again, this is a case where C and M are
not independent, so knowing M ≡ b gives special information (beyond that C≢b).
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