The Monty Hall example demonstrates the use
of a small Bayesian network to solve the Monty Hall Puzzle.
Describing the Monty Hall Puzzle
The Monty Hall puzzle gets its name from an
American TV game show,
"Let's make a deal", hosted by Monty Hall. In
this show, you have the chance to win some prize if you are lucky
enough to find the prize behind one of three doors. The game goes
like this:
- You are asked to select one of the
three doors
- Monty Hall (who knows where the prize
is) opens another door which does not contain the prize
- You are asked if you want to redo your
selection (select one of the two closed doors)
- You get the prize if it is behind the
door you selected
The problem of
the puzzle is: What should you do at your second selection? Some
would say that it does not matter because it is equally likely that
the prize is behind the two remaining doors. This, however, is not
quite true. The right answer is that you have the best chance of
winning if you redo your selection - odds 2/3 in stead of 1/3. In
the following section, the solution is calculated through the use
of a simple Bayesian network (BN).
Finding the Solution Using a BN
The Monty Hall puzzle can be modeled in three
random variables:
Prize,
First Selection, and
Monty Opens.
-
Prize represents the information about which door contains
the prize. This means that it has three states: "Door
1", "Door 2" and "Door 3"
-
First Selection represents your first selection. This
variable also has the three states: "Door 1",
"Door 2" and "Door 3"
-
Monty Opens represents Monty Halls choice of door when you
have made your first selection. Again, we have the three states:
"Door 1", "Door 2" and "Door
3"
The door
containing the prize is known to Monty and thus
Prize has impact on
Monty Opens. Monty will never choose to open the door of
your first selection so also
First Selection has impact on
Monty Opens. This give us the BN shown in figure 1.
|
Figure
1: BN of the Monty Hall puzzle.
The causal arrows describes that both
Prize and First Selection have impact on
Monty Opens. |
The conditional probability table (CPT) of
Prize is shown in table 1.
|
Prize="Door 1" |
Prize="Door 2" |
Prize="Door 3" |
| 0.33 |
0.33 |
0.33 |
|
| Table 1:
P(Prize). |
The CPT of
First Selection is shown in table 2 (this table is really
not important since you will always select a specific state of this
variable when you use the BN).
First Selection
= "Door 1" |
First Selection
= "Door 2" |
First Selection
= "Door 3" |
| 0.33 |
0.33 |
0.33 |
|
| Table 2: P(First
Selection). |
Table 3 shows the CPT of
Monty Opens. This table states that if the prize is behind
door 1 and you have chosen door 3, then Monty will open door 2 for
sure since this door is the only possible door he can open which
does not contain the prize. If you have selected the right door in
your first selection, he will randomly choose one of the remaining
doors.
|
Prize
|
First Selection
|
Monty Opens
= "Door 1" |
Monty Opens
= "Door 2" |
Monty Opens
= "Door 3" |
| "Door 1" |
"Door 1" |
0 |
0.5 |
0.5 |
| "Door 1" |
"Door 2" |
0 |
0 |
1 |
| "Door 1" |
"Door 3" |
0 |
1 |
0 |
| "Door 2" |
"Door 1" |
0 |
0 |
1 |
| "Door 2" |
"Door 2" |
0.5 |
0 |
0.5 |
| "Door 2" |
"Door 3" |
1 |
0 |
0 |
| "Door 3" |
"Door 1" |
0 |
1 |
0 |
| "Door 3" |
"Door 2" |
1 |
0 |
0 |
| "Door 3" |
"Door 3" |
0.5 |
0.5 |
0 |
|
| Table 3: P(Monty
Opens | Prize, First Selection). |
The BN shown in
figure 1 with nodes having the CPTs of table 1, 2, and 3 can be
constructed in the Hugin GUI in less than 10 minutes. Right after
compilation, the node list pane of the Hugin GUI will look as in
figure 2.
This network has been installed on your
computer with the Hugin software. Open the network in the Hugin
GUI. You can find the network in the directory where you installed
Hugin (eg. C:/Program Files/Hugin/Hugin Lite/Samples).
Testing the
network shows that it will always be best for you to redo your
selection when Monty has opened a door (gives you 66.67% chance of
winning).
An easy way to convince yourself that this is
true goes like this: At first you have 33.33% chance of choosing
the right door and there is 66.67% chance of the prize being
somewhere else. You know that Monty is going to open an empty door
so when he does, this should not change a thing about your belief
of your door being the right one. You still have 33.33% chance of
having selected the right door. Thus there must be 66.67% chance of
the prize being somewhere else (behind the last
door).