Let us recall the Monty Hall Game from its statement in the Wikipedia
article ^{2}
on the subject:

`"Suppose you're on a game show. Three doors *A*, *B* and *C* are in front of you. Behind one door is a brand new car, and behind the two others, a goat. You are asked to pick one of the three doors. Then the host of the game, who knows what's behind the doors, opens one of the two remaining doors and shows a goat. He then asks you: "Do you want to switch doors or keep your initial choice?" `

Say you have chosen the door *A* and the host has opened door
*B*. The question now is: Is it to your advantage to switch your
choice from door *A* to door *C*?”

**Some notation to begin with**

For each door *A, B, C*, consider the same frame of
discernment (Fod) *F* with three possible values: \[F = \{car, goat 1, goat 2\}.\] I use
(0,1)-vectors to identify each element of the frame. Hence, the element
“car” is identified by the vector \((1, 0,
0)\), goat 1 by the vector \((0, 1,
0)\) and goat 2 by the vector \((0, 0,
1)\).

With this notation, any subset of *F* has a unique (0,1)
representation. For example, the subset \(\{goat 1, goat 2\}\) is represented by the
vector \((0, 1, 1)\).

We have three things to consider:

- How the three doors are linked;
- Evidence pertaining to door
*A*(choice of the contestant); - Evidence pertaining to door
*B*(the action of the host).

There are six possible combinations of the car and the two goats
behind the three doors *A, B, C*:

\(\{car, goat 1, goat 2\}\) \(\{car, goat 2, goat 1\}\) \(\{goat 1, car, goat 2\}\) \(\{goat 1, goat 2, car\}\) \(\{goat 2, car, goat 1\}\) \(\{goat 2, goat 1, car\}\).

These combinations are elements of the product space \(F_{ABC} = \prod(A, B, C)\). The number of
elements of \(F_{ABC}\) is \(3^3 = 27\). The six possible dispositions
of the car and goats determine a subset *S* of the Fod \(F_{ABC}\). A mass of 1 is allotted to this
subset *S*.

I use the function *bcaRel* to code the desired relation
between the doors.

```
# 1. define the tt matrix MHABC_tt, which encodes the subset S
#
MHABC_tt <- matrix(c(1,0,0,0,1,0,0,0,1,
1,0,0,0,0,1,0,1,0,
0,1,0,1,0,0,0,0,1,
0,1,0,0,0,1,1,0,0,
0,0,1,1,0,0,0,1,0,
0,0,1,0,1,0,1,0,0), ncol=9, byrow=TRUE)
colnames(MHABC_tt) <- rep(c("car", "goat1", "goat2"), 3)
#
# 2. define the spec matrix.
# Here we have one subset of six elements
#
MHABC_spec = matrix(rep(1,12), ncol = 2, dimnames = list(NULL, c("specnb", "mass")))
#
# 3. define the info matrix.
# for each variable, we attribute a number and give the size of the frame
#
MHABC_info =matrix(c(1:3, rep(3,3)), ncol = 2, dimnames = list(NULL, c("varnb", "size")) )
#
# 4. call of the function with the name of the variables and the numbering of the relation
#
MHABC_rel <- bcaRel(tt = MHABC_tt, spec = MHABC_spec, infovar = MHABC_info, varnames = c("MHA", "MHB", "MHC"), relnb = 1)
#
# Relation vetween the three doors A, B and C
bcaPrint(MHABC_rel)
#> MHABC_rel
#> 1 car goat1 goat2 + car goat2 goat1 + goat1 car goat2 + goat1 goat2 car + goat2 car goat1 + goat2 goat1 car
#> specnb mass
#> 1 1 1
# Note that row labels can become pretty long. If it is the case, the prmatrix function can be used to view results, for example:
#
# prmatrix(t(MHABC_rel$tt), collab = "")
#
# Another way to check the tt matrix is:
#
# which(MHABC_rel$tt[1,] == TRUE)
```

You have chosen door *A*. At this point, the problem is quite
simple. Your belief is equally divided between 3 possible outcomes: car,
goat 1 or goat 2: \(m({car}) = m({goat1}) =
m({goat2}) = 1/3\). Let’s encode this evidence with function
*bca*.

```
# Evidence related to choice of door A
MHA_E <- bca(tt= diag(1,3,3), m= rep(1/3, 3), cnames =c("car", "goat1", "goat2"), varnames = "MHA", idvar = 1)
# Evidence of the contestant (function MHA_E attached to variable A)
bcaPrint(MHA_E)
#> MHA_E specnb mass
#> 1 car 1 0.333333333333333
#> 2 goat1 2 0.333333333333333
#> 3 goat2 3 0.333333333333333
```

But the host wanted to add some thrill to the game. He has opened door B and revealed a goat. The host has given us a small piece of evidence: Goat 1 or goat 2 was behind door B. Since the host knows what is behind each door, the mass value of this piece of evidence is: \(m({goat1, goat2}) = 1\).

Let’s translate this in R with function *bca*:

We now have all the elements of a small belief network made of one
relation (MHABC_rel) between three variables: Door *A*, Door
*B*, Door *C*, and two pieces of evidence coming from the
Contestant (MHA_E) and the Host two (MHB_E). Variables A, B and C
(doors) are the nodes of the graph. The edges (hyperedges) are the
evidences MHA_E (named ev_A on the graph) and MHB_E (ev_B on the graph)
and the relation MHABC_rel (r_ABC on the graph). We use the package
igraph ^{3}
to produce a bipartite graph corresponding to the desired
hypergraph.

Since we want to know if there is advantage to switch doors, our goal
is the calculation of a belief function *MHC* attached to the
variable of interest *C* (Door C). To obtain this belief
function, we need to combine evidence for doors A (Contestant) and B
(Host) with the relation linking the three doors in the product space
\(F_{ABC}\). We will use a process of
successive elimination of variables until only variable *C*
remains.

The calculations involved follow the principles of the valuation
language of Shenoy ^{4}; see also ^{5}. The variables are
linked to functions (called valuations). A function can be a piece of
evidence attached to a variable or a relation between two or more
variables.

Three kinds of operations are involved in the calculations: a) the minimal (vacuous) extension of a mass function to a larger Fod; b) the combination of two mass functions by Dempster’s rule; c) the marginalization of a mass function, i.e. eliminating a variable to reduce the function to a smaller Fod.

Using function *extmin*, we extend the mass function MHA_E to
the space \(\prod(A, B, C)\); then we
combine MHA_E extended with MHABC_rel, using functions *dsrwon*
and *nzdsr* (normalization); finally, we use function
*elim* to eliminate *A* by marginalizing to \(\prod(B, C)\). The mass function obtained
is named MHBC. This gives a reduced network with *B* and
*C*.

```
# 1. Extend MHA to the product space A x B x C
MHA_ext <- extmin(MHA_E, MHABC_rel )
"Evidence of Contestant extended to the product space A x B x C"
#> [1] "Evidence of Contestant extended to the product space A x B x C"
bcaPrint(MHA_ext)
#> MHA_ext
#> 1 car car car + car car goat1 + car car goat2 + car goat1 car + car goat1 goat1 + car goat1 goat2 + car goat2 car + car goat2 goat1 + car goat2 goat2
#> 2 goat1 car car + goat1 car goat1 + goat1 car goat2 + goat1 goat1 car + goat1 goat1 goat1 + goat1 goat1 goat2 + goat1 goat2 car + goat1 goat2 goat1 + goat1 goat2 goat2
#> 3 goat2 car car + goat2 car goat1 + goat2 car goat2 + goat2 goat1 car + goat2 goat1 goat1 + goat2 goat1 goat2 + goat2 goat2 car + goat2 goat2 goat1 + goat2 goat2 goat2
#> specnb mass
#> 1 1 0.333333333333333
#> 2 2 0.333333333333333
#> 3 3 0.333333333333333
#
# 2. Combine MHA_ext and MHABC_rel
MHA_ABC_comb <- dsrwon(MHA_ext,MHABC_rel)
# since the measure of contradiction is 0, no need to normalize
MHA_ABC_comb$con
#> [1] 0
# "Subsets resulting from the combination of Expert 1 extended and r1"
bcaPrint(MHA_ABC_comb)
#> MHA_ABC_comb specnb mass
#> 1 car goat1 goat2 + car goat2 goat1 1 0.333333333333333
#> 2 goat1 car goat2 + goat1 goat2 car 2 0.333333333333333
#> 3 goat2 car goat1 + goat2 goat1 car 3 0.333333333333333
#
# 3. Eliminate variable A
MHBC <- elim(MHA_ABC_comb, xnb = 1)
bcaPrint(MHBC)
#> MHBC specnb mass
#> 1 goat1 goat2 + goat2 goat1 1 0.333333333333333
#> 2 car goat2 + goat2 car 2 0.333333333333333
#> 3 car goat1 + goat1 car 3 0.333333333333333
```

After this first step, the graph is updated.

```
#> + 4/4 vertices, named, from 0d61498:
#> [1] B C r_BC ev_B
```

Similarly, we extend evidence on B in the product space \(\prod(B, C)\), combine the extended evidence with the relation MHBC, then marginalize to C. This will give the final result.

```
# 1. Extend MHB_E to the space B x C
MHB_ext <- extmin(MHB_E, MHBC )
# Evidence of Host extended to the product space B x C"
bcaPrint(MHB_ext)
#> MHB_ext mass
#> 1 1 1
#
# 2. combination of MHB_ext and MHBC
MHB_BC_comb <- dsrwon(MHB_ext, MHBC)
# "Subsets of the space B x C resulting from the combination of Host extended and MHBC"
bcaPrint(MHB_BC_comb)
#> MHB_BC_comb specnb mass
#> 1 goat2 car 1 0.333333333333333
#> 2 goat1 car 2 0.333333333333333
#> 3 goat1 goat2 + goat2 goat1 3 0.333333333333333
# MHA_BC_comb$con = 0, no need to normalize)
MHB_BC_comb$con
#> [1] 0
#
# 3. Eliminate variable B
MHC <- elim(MHB_BC_comb, xnb = 2)
# Final result: the belief function MHC attached to variable C
round(belplau(MHC), digits = 2 )
#> bel disbel unc plau rplau
#> car 0.67 0.33 0 0.67 2.0
#> goat1 + goat2 0.33 0.67 0 0.33 0.5
```

As we can see, we double our chances of winning the car if we switch from door A to door C.

Note that there is no loss of generality by fixing the choices in the analysis (door A for the contestant, door B for the host).

To be more specific and make a bridge with probability theory, we can add to our result all the elementary events that have 0 mass, so that we can see their measure of plausibility.

The function *addTobca* serves this purpose.

```
MHC_plus_singl <- addTobca(MHC, tt = matrix(c(0,1,0,0,0,1), ncol = 3, byrow = TRUE))
result <- tabresul(MHC_plus_singl)
round(result[[1]], digits = 2)
#> car goat1 goat2 mass bel disbel unc plau rplau
#> car 1 0 0 0.67 0.67 0.33 0.00 0.67 2.00
#> goat1 0 1 0 0.00 0.00 0.67 0.33 0.33 0.33
#> goat2 0 0 1 0.00 0.00 0.67 0.33 0.33 0.33
#> goat1 + goat2 0 1 1 0.33 0.33 0.67 0.00 0.33 0.50
cat("\n", " conflict:", result[[2]] )
#>
#> conflict: 0
```

Retired Statistician, Stat.ASSQ↩︎

https://en.wikipedia.org/w/index.php?title=Monty_Hall_problem&oldid=829292640↩︎

Csardi G, Nepusz T: The igraph software package for complex network research, InterJournal, Complex Systems 1695. 2006. https://igraph.org↩︎

P. P. Shenoy. A Valuation-Based Language for Expert systems. International Journal of Approximate Reasoning 1989, 3 383–411↩︎

P. P. Shenoy. Valuation-Based Systems. Third School on Belief Functions and Their Applications, Stella Plage, France. September 30, 2015↩︎