In a four-door Monty Hall game, the contestant guesses door A, Monty opens empty door B, and the contestant switches to C.

decision/monty.four.doors New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

In a four-door Monty Hall game, the contestant guesses door A, Monty
opens empty door B, and the contestant switches to C.  Monty then opens
another empty door.  What are the odds that the prize is behind C if
Monty now opens A, and if he opens D?

Statement of problem:
1. 4 doors, prize hidden equally randomly behind one of them.
2. Contestant guesses a door.
3. Monty, who knows where the prize is, always shows an empty door.
4. Contestant can switch doors if desired.
5. Monty now opens another empty door.

Question: if contestant guesses A, Monty opens B, contestant switches
to C, what are the odds of the prize being behind door C if (i) Monty
now opens A; (ii) Monty now opens D?

-- Scott Brown <sbrown@symcom.math.uiuc.edu>

[ Editor: the first paragraph is the editor's summary. ]

Show solution

Solutions: (i)  behind door C 1/3 of the time;
           (ii) behind door C 3/7 of the time.

Strategy: At this point, if you can, switch from C to the remaining
door.

                 -=o=-     -=o=-

Step 1: calculate odds of prize being behind one of the three doors
        left after Monty opens the first empty door, which we are told
        was door B.

Note that, without loss of generality, by symmetry we can assume the
contestant has guessed door A on the first guess and analyze the game
from there.

Since Monty's behavior is assumed(!) to be: never open the door with
the prize, and never open the door guessed by the contestant, and
choose equally randomly between remaining doors, we have (under
assumption contestant has guessed A) the following possible sequences
of events in the game:

    Prize Behind:    Monty Opens:    Odds:  Did Monty open door B?
         A                B          1/12      Yes (2 times/24)
                          C          1/12
                          D          1/12
         B                C          1/8
                          D          1/8
         C                B          1/8       Yes (3 times/24)
                          D          1/8
         D                B          1/8       Yes (3 times/24)
                          C          1/8

(Odds are: 1/4 odds of prize behind given door, times the odds of monty
selecting to open either 1 of 2 or 1 of 3 available doors)

If we assume we are in the case posed in the problem above, Monty has
opened door B.  Inspection of the options and odds above leads to the
conclusion that in 24 plays of the game this will occur 8 times, with
the prize behind either door A, C or D as indicated above.  Thus:

After step 1, the odds of the prize being behind door A are 2/8; behind
C, 3/8; behind D, 3/8.

                 -=o=-     -=o=-

Step 2: Calculate odds of the prize being behind a given door after
        Monty opens either door A or door D.

Note that, again, Monty is assumed to choose a door that has neither
been guessed nor contains the prize.  Thus:

    Prize Behind:     Monty Now Opens:    Odds:
         A  (2/8)           D             2/8  = 4/16
         C  (3/8)           A             3/16
                            D             3/16
         D  (3/8)           A             3/8  = 6/16

(Here again the final odds are the odds of the prize being behind the
given door, as calculated in step 1, times the odds of Monty choosing
to open the indicated door.  Note that Monty often has no choice as to
which door to open)

                 -=o=-     -=o=-

Thus if Monty opens door B, then after the contestant switches to door
C Monty opens door A, the prize is behind door C 1/3 of the time and
behind D 2/3 of the time;

while if Monty opens door B, then door D after the contestant switches,
the prize is behind door A 4/7 of the time and behind door C 3/7 of the
time.

-- Scott Brown <sbrown@symcom.math.uiuc.edu>

[ Editor: see also decision/monty.hall. ]