Monty Hall Problem Redux
The Monty Hall Dilemma continues to fascinate lay and professional mathematicians. In the References section below I cite a few of the latest papers and books.
The original formulation that was popularized by Marylin vos Savant contained a degree of ambiguity.
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
In case you picked a door that conceals a goat, Monty has no choice but to open the only other door with the goat behind it. The case where you pick the prize door is the source of controversy. All that was said was that Monty knows what he's doing and that he always opens the door with a goat. But how does he choose that door? When I first wrote about the Monty Hall Dilemma (MHD), my interpretation was that Monty makes a random selection and chooses one of the two remaining doors each with probability 1/2. This is not the only possibility that may govern Monty's behavior, although I do believe that this interpretation is rather common among mathematicians, see, for example [Havil, 59-63, Stein, 121-124].
Before I continue with examples let me give what appears to be an unassailably unambiguous (albeit wordy) formulation of the problem [Lucas et al] - an article from which I also draw some examples.
You are a player on a game show and are shown three identical doors. You are told that behind one of the doors there is a car, behind the other two are goats. Monty Hall, the host of the show, asks you to choose one of the doors, but not open it just yet. Monty, who knows where the car is, now opens one of the doors. He does so in accordance with the following rules:
- Monty always opens the door that conceals a goat.
- Monty never opens the door you initially chose.
- If Monty can open more than one door without violating the two previous rules, then he opens a random door among the available ones.
After Monty opens a door, he gives you a choice of sticking with your original selection or switching to the other unopened door. What should you do to maximize your chances of winning the car? Switch or not to switch?
As is well known, in this formulation, you will win the car with probability 2/3 if you switch and with probability 1/3 if you stay with your original choice.
Most people come up with the 50/50 argument: "Once Monty opens his door, there are only two doors remain to choose from. Since nothing is known about the objects these two doors conceal, the car is as likely to be behind one as it is to be behind the other. It does not matter whether I switch or not."
The modern psychology explains the affinity for this argument by a well-known human disposition. Negative consequences incurred by inaction hurt less than same negative consequences incurred through some sort of action. A failed action elicits a feeling of "I brought this on myself" much more than inaction with the same consequences.
Let me note that one simulation available elsewhere at the site does, in a sense, away with ambiguity by simply opening two remaining doors at once, thus letting one verify his or her own strategy, whatever it is.
One way to diverge from the consensus interpretation (by which I mean mine and that of many others) is to claim that Monty's knowledge of the prize layout, although known to him, plays no role in his selection of a door to open. It just so happens that the door he opens contains a goat. The probability of a win in this case is 1/2.
Another possibility is to assume that Monty does act on his knowledge but does it in a peculiar manner. For example, Monty may follow the rule, according to which, when there is a choice, he chooses the door with the highest number. If you are aware of the rule and chose door #1 then Monty's choice of door #2 would tell you that the car is behind door #3. And this is with probability 1. However, if Monty opens door #3, the car may be either behind door #1 or door #2, both with probabilities of 1/2. This is rather obvious but is also supported by Bayes' Theorem.
Let C_{j} denote the event that the car is behind door #j, and D_{j} the event that Monty opens door #j. Without loss of generality we also assume that you open door #1 and Monty opened door #3. The probability P(C_{1}|D_{3}) that your door conceals the car given that Monty opened door #3 is found from Bayes' Theorem:
(1) | P(C_{1}|D_{3}) = P(C_{1})P(D_{3}|C_{1}) / P(D_{3}). |
Expnading the denominator,
(2) | P(C_{1}|D_{3}) = P(C_{1})P(D_{3}|C_{1}) / (P(C_{1})P(D_{3}|C_{1}) + P(C_{2})P(D_{3}|C_{2}) + P(C_{3})P(D_{3}|C_{3})). |
Now, at the outset P(C_{1}) = P(C_{2}) = P(C_{3}) = 1/3 and, of course,
(3) | P(C_{1}|D_{3}) = P(D_{3}|C_{1}) / (P(D_{3}|C_{1}) + P(D_{3}|C_{2})). |
Under our assumptions that Monty is required to open the door with the highest number,
(4) | P(C_{1}|D_{3}) = 1 / (1 + 1) = 1/2. |
References
- R. Deaves, The Monty Hall Problem: Beyond Closed Doors, lulu.com, 2007
- J. Havil, Impossible?, Princeton University Press, 2008
- S. Lucas, J. Rosenhouse, A. Schepler, The Monty Hall Problem, Reconsidered, Mathematics Magazine, Vol. 82, No. 5, December 2009, 332-342
- J. Rosenhouse, The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser, Oxford University Press, USA (June 4, 2009)
- J. S. Rosenthal, Monty Hall, Monty Fall, Monty Crawl, Math Horizons, Vol. 16, No. 1, September 2008, 5-7
- M. vos Savant, The Power of Logical Thinking, St. Martin's Press, NY 1996
- S. K. Stein, Strength in Numbers, John Wiley & Sons, 1996
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