# 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 belief network (Bbn).

## Finding the Solution Using a Bbn

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 choise 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 Bbn shown in figure 1.

Figure 1: Bbn of the Monty Hall puzzle. The causal arrows describes that both Prize and First Selection has 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 Bbn).

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 Bbn shown in figure 1 with nodes having the cpts of table 1, 2, and 3 can be constructed in HUGIN Runtime in less than 10 minutes. Right after compilation, the node list pane of HUGIN Runtime will look as in figure 2. Figure 2 is an interactive Java applet allowing you to select first your own first selection - then the door opened by Monty. After this, the solution can be retrieved from the probabilities of Prize. The functionality of this applet is very similar to the functionality of HUGIN Runtime.

Figure 2: Interactive figure showing the list of nodes (variables) of the Monty Hall puzzle domain. Use the mouse to select your own first selection and then the door Monty opens.

Figure 2 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).

Back