The Monty Hall Conundra Revisited
Kam Hon Chu Associate Professor Department of Economics Memorial University of Newfoundland St. John's, Newfoundland & Labrador CANADA A1C 5S7 Tel: (709) 737-8102 Fax: (709) 737-2094 E-mail:
[email protected]
April 18, 2009
Acknowledgments: I would like to thank John Nachbar for his comments and suggestions on an earlier version of this paper. All remaining errors are, however, mine.
Electronic copy available at: http://ssrn.com/abstract=1391943
The Monty Hall Conundra Revisited Abstract The purpose of this note is to enhance the teaching of probability theory by reexamining the original Monty Hall game (Conundrum # 1) and its variation (Conundrum # 2, Raiffa et al. 2002, p. 40). The solution to the original Monty Hall game, commonly known as Marilyn’s solution, has that the posterior probability of winning the game by switching door is 2/3 in Conundrum # 1. But Marilyn's solution does not apply to Conundrum # 2, for which the right answer is ½. This note offers a formal solution to Conundrum # 2 by applying the Bayes formula like Marilyn’s solution does. However, the posterior probability has to be computed conditional on not only the door opened by Monty Hall but also the door chosen by the contestant.
JEL Classifications: A20, C44, D81 Key Words: Bayes Theorem, Conditional Expectations, Monty Hall Problem, Statistical Decision Theory
Electronic copy available at: http://ssrn.com/abstract=1391943
The Monty Hall Conundra Revisited Experimental results and anecdotes reveal that the human mind seems to have a bit of problem in understanding probability concepts, let alone estimating correct probabilities. The best known examples among economists are probably Allais’s Paradox (Allais 1953), Ellsberg’s Paradox (Ellsberg 1961), and the Prospect Theory of Kahneman and Tversky (1979), to name just a few. In brief, these studies have documented “choice anomalies” as individuals’ decisions are found to be inconsistent with rational choice theory. However, Savage (1972) argues that those who succumb to the paradoxes simply make mistakes in probability reasoning and calculations because they are uneducated in probability theory; and they will not react as the paradoxes tempt them to react once we have educated them by pointing out their errors clearly and perspicuously. Whether Savage’s argument is a convincing resolution to those paradoxes or not, there are laboratory results indicating that individuals make the right choice more often through learning and as a result choice anomalies declined substantially (see, for example, Friedman 1998, and Palacios-Huerta 2003). Following the popular presentation by Nalebuff (1987), the Monty Hall game is widely used in economics, statistics and game theory classes (see, for example, Becker and Greene 2001, Brokaw and Merz 2004, among many others). This note aims at enhancing the teaching of probability by demonstrating how the Bayes formula can be correctly applied to compute the posterior probability of winning the prize in a variation of the original Monty Hall game. For pedagogical reasons, the original game is also reconsidered here. Consider the famous TV game show “Let’s make a Deal,” in which a contestant is given a choice of three closed doors. Behind one of the doors is an attractive prize, say, a car, and the other
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two doors are empty. After the contestant has picked a door, say, Door 1, which remains closed, the host Mr. Monty Hall, who knows which door hides the prize, opens one of the remaining two doors, say, Door 3, to reveal an empty door. The host then offers the contestant the option of switching to Door 2. Should the contestant stick to Door 1 or switch to Door 2? This famous Monty Hall game problem, also known as the three-door game problem or the car-goat problem in the literature, was the focus of much discussion more than a decade ago because of its underlying probability paradox. See, for example, Morgan et al. (1991a, 1991b, 1991c), Seymann (1991), Tierney (1991), vos Savant (1990a, 1990b, 1991a, 1991b), among others. This game remains durable even after years of controversy, as evidenced by the recent applications of game-theoretic concepts and techniques in an attempt to solve the puzzle. See, for example, Bailey (2000), Chu (2003), Fernandez and Piron (1999), and Gintis (2000). Following Raiffa et al. (2002, p.39), I call the game described above the Monty Hall Conundrum # 1. This note considers this conundrum and also its variation called the Monty Hall Conundrum # 2 (see below or Raiffa et al. 2002, p.40). Many solutions to the Monty Hall Conundrum # 1 have been offered, and they vary depending on the assumptions made and the analytical framework used. See, for example, Falk (1993, pp.184-9), Gillman (1992), Isaac (1995, pp. 8-10, pp. 26-7), Selvin (1975a, 1975b), and the works cited above for details. The prevailing view, commonly known as Marilyn's solution (vos Savant 1990), has that the conditional probability of winning the car by switching door is 2/3 in the Monty Hall Conundrum # 1. In contrast, the chance of winning in the Monty Hall Conundrum # 2 is ½, which appeals to the intuition of most people including many statisticians and mathematicians. In fact, ½ had been widely accepted as the correct answer to Conundrum # 1 before Marilyn’s solution appeared. Instead -2-
of using game-theoretic concepts and techniques as in some recent studies, the analysis here adheres closely to the original formulation by vos Savant (1990). It shows that the Bayes formula can be applied to both conundra. More specifically, the formula can be applied to obtain a posterior probability of 2/3 in Conundrum #1 but a posterior probability of ½ for Conundrum # 2 when we identify precisely and accurately the events conditional upon which the posterior probability is computed. To make this note self-contained and, more importantly, to contrast the differences in the computations and results of the two conundra, let us first reexamine the Monty Hall Conundrum # 1. To avoid unnecessary confusion and potential controversies, the analysis below assumes the following: (i) the host always offers to the contestant the option to switch or not after the contestant has chosen a door and the host has opened a door showing no prize behind it; (ii) the host does not have biased preferences in favour of which door to open; (iii) the host does not use any psychological factor or strategic behaviour to influence the contestant’s choice to switch or not; and finally (iv) when the contestant or the host has no reason or information to favour one door over any other doors, she or he will choose at random (i.e., a uniform distribution is assigned to all equally likely outcomes). The counterintuitive solution that the contestant should always switch because the probability of winning the prize is 2/3 is based on applying the Bayes formula to compute the conditional probability of winning. For expository purpose, let us re-derive the result here. First, the probability of having the prize behind Door i is P(A = i) = 1/3, for i = 1, 2, and 3, because of equal priors. Here A = i represents the event that the prize is behind Door i. Assume the contestant picks Door 1. Now let H = j denote the event that the host opens Door j, j
1 by assumption. By the rule of the game that the host cannot open a door to reveal the prize, the probability of having -3-
the host to open Door 2 when the prize is behind Door 2 is P(H = 2 | A = 2) = 0, whereas P(H = 3 | A = 2) = 1. By the same token, P(H = 3 | A = 3) = 0, and P(H = 2 | A = 3) = 1. But if the prize is behind Door 1, then by assumptions (ii)-(iv) above, P(H = 2 | A = 1) = P(H = 3 | A = 1) = ½. Now suppose the host opens Door 3. Then by the Bayes formula, the posterior probability that the prize is behind Door 1 is given as:
(1)
Accordingly, P(A = 2 | H = 3) = 2/3 and P(A = 3 | H = 3) = 0. Therefore, the contestant should switch to Door 2 because of a higher posterior probability of winning the prize. This is the famous Marilyn’s solution (vos Savant 1990) in the literature. Now let us turn to an interesting variation of the above game, coined by Raiffa et al. (2002, p. 40) as the Monty Hall Conundrum # 2. The setup and the rules of this game are exactly the same as in the original game we have discussed above, except that three contestants line up in front of each door. After opening a door not containing the car, say, Door 3, Monty Hall asks the contestant in front of Door 1 if she wants to swap places with the contestant in front of Door 2. In this game, Marilyn’s solution, i.e., switching door for the probability of winning is 2/3, is definitely not the right answer. If one contestant has a higher chance of winning by switching door, it should also be the case for the other contestant because the two remaining contestants are in symmetrical positions. Hence Marilyn’s solution is apparently ludicrous in this case. The contestant in front of Door 1 should infer that her chance of winning the car is the same as that of the other contestant, i.e., ½, -4-
simply because both of them are exactly in the same position. Raiffa et al. (2002, p.40) convincingly explain the right answer in an intuitive way without giving a formal solution using the Bayes formula. So why can’t we apply Equation (1) or Marilyn’s solution to Conundrum # 2? Or what is missing here? The intuitive answer is that the setup and rules of the two games, though largely similar, are not exactly the same. In Conundrum #1, Monty Hall has to open a non-prize door out of the two doors not chosen by the contestant. In Conundrum # 2, Monty Hall can open any nonprize door. Therefore, the different rules about which door Monty Hall can open alter the probability calculations. As a formal analysis, this note shows from basic principles that the Bayes formula (not Equation (1), however) can still be applied in the case of Conundrum # 2 to obtain the correct answer. To begin with, it should be pointed out that in deriving Equation (1) it is assumed throughout that the contestant has chosen Door 1. The posterior probability is then computed from the Bayes formula conditional on that Monty Hall opens Door 3. Assuming symmetry to hold, the result is then generalized to the other cases in which the contestant chooses a door other than Door 1 and Monty Hall opens a door other than Door 3. In Conundrum # 2, however, in applying the Bayes formula a contestant cannot assume that she has chosen Door 1 and then computes the posterior probability of winning based on which door Monty Hall opens. The posterior probability should be computed conditional on which door Monty Hall opens and on which door the contestant chooses. More explicitly, for the contestant who chooses Door 1 and observes that Monty Hall has opened Door 3, her posterior probability of winning the prize should be based on P(A = 1 | C = 1 1 H = 3) but not simply P(A = 1 | H = 3) as in -5-
Marilyn's solution. Here C = k denotes the event that the contestant chooses Door k, and C = k 1 H = j represents the joint event that the contestant chooses Door k and Monty Hall opens Door j. The other notation remains the same as previously defined. Based on the rules of the game and the standard assumptions, the event space and the associated conditional probabilities are tabulated as Table 1. Take the case in which the prize is behind Door 1 as an example for illustration. Since the prize is behind Door 1, Monty Hall will not open Door 1. So any event with H = 1 conditional on A = 1 will have a conditional probability of zero regardless of what C is. That is to say, P(C = k 1 H = 1 | A = 1) = 0 for any k = 1,2,3. On the other hand, from the perspective of a contestant who is offered a chance of switching door, Monty Hall has not opened the door chosen by her. Therefore, P(C = k 1 H = j | A = 1) = 0 if j = k, for any j, k = 1,2,3. After assigning zero probability to the above events, the remaining (four) possible events will each carry an equal probability, i.e., 1/4, because of the assumption that a uniform distribution is assigned to all equally likely outcomes. The probabilities for the events associated with A= 2 and 3 can be assigned in a similar fashion. Now for the contestant who chooses Door 1 and observes Monty Hall opens Door 3, the posterior probability of winning the prize for her should therefore be computed according to the Bayes formula as follows:
(2)
In other words, she should be indifferent to switching or not. By the same token, we can repeat this exercise to compute the posterior probability of winning for the contestant who chooses Door 2 and -6-
to obtain the same answer of ½. For the contestant who chooses Door 3, the posterior probability of winning is of course zero as can be seen from Table 1 that P(C = 3 1 H = 3 | A = i ) = 0 for i = 1,2,3. Or simply put, this contestant has no chance of winning at all because Monty Hall has opened her door and eliminated her from the game. Therefore, we have provided a right answer for the Monty Hall Conundrum # 2 based on applying the Bayes formula. To conclude, this note has offered a formal solution to the Monty Hall Conundrum # 2 by a direct application of the Bayes formula with a careful consideration of what events should be conditional on in the computation. The importance of grasping the correct probability concept and computation should be well recognized. The original Monty Hall game has already shown us that the posterior probability of winning the prize by switching door, i.e. 2/3, is counter-intuitive when compared with the commonly chosen answer of ½. It clearly demonstrates that we should rely on analysis and logic rather than intuition in order to make the right decision. In Palacios-Huerta’s (2003) Monty Hall game experiment, for example, students majoring in Applied Math Economics, economics majors, and students who have prior knowledge about the Monty Hall game tend to switch door more often than other students. These findings may partly reflect the positive effect of relevant education on making right decisions. Besides decision making, the theory and computation of conditional probability have wide applications and interesting results in many other areas, like empirical studies in economics, economic theories and other social sciences. To support this view, take the famous Siegel Paradox in international finance as an example. In a nutshell, the paradox can be phrased as follows: if the forward rate is an unbiased predictor of the future spot exchange rate in the home country, then it cannot simultaneously hold in the foreign country (see Siegel 1972 for details). This paradoxical result is based on the well-known statistical theorem known as -7-
Jensen’s Inequality. However, the paradox can be resolved if the correct conditional probability is applied (see Chu 2005 for details). All in all, both teaching and learning of probability should be taken seriously.
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References Allais, M. 1953. Le comportement de l’homme rationnel devant le risque: critiques de postulates et axiomes de l’ecole Americane. Econometrica 21: 503-46. Bailey, H. 2000 Monty Hall Uses a Mixed Strategy. Mathematics Magazine 73 (April): 135-41. Becker, W.E., and W. Greene. 2001. Teaching Statistics and Econometrics to Undergraduates. Journal of Economic Perspectives 15: 169-82. Brokaw, A.J., and T.E. Merz. 2004. Active Learning with Monty Hall in a Game Theory Class. Journal of Economic Education 35: 259-68. Chu, K.H. 2003. To Switch or Not To Switch? InterStat: Statistics on the Internet, at http://interstat.statjournals.net, (February) first article. ________. 2005. Solution to the Siegel Paradox. Open Economies Review 16: 399-405. Ellsberg, D. 1961. Risk, Ambiguity and the Savage Axioms. Quarterly Journal of Economics 75: 643-9. Falk, R. 1993. Understanding Probability and Statistics: A Book of Problems. Wellesley, MA: A K Peters. Fernandez, L., and R. Piron. 1999. Should She Switch? A Game-Theoretic Analysis of the Monty Hall Game. Mathematics Magazine 72 (June): 214-7. Friedman, D. 1998. Monty Hall’s Three Doors: Construction and Destruction of a Choice Anomaly. American Economic Review 88: 933-46. Gillman, L. 1992.) The Car and the Goats. American Mathematical Monthly 99: 3-7. Gintis, H. 2000. Game Theory Evolving: A Problem-Centered Introduction to Modeling Strategic Behavior. Princeton University Press. -9-
Isaac, R. 1995. The Pleasure of Probability. New York: Springer-Verlag. Kahneman, D., and A. Tversky. 1979. Prospect Theory: An Analysis of Decision under Risk. Econometrica 47: 263-91. Morgan, J.P. N.R. Chaganty, R.C. Dahiya, and M.J. Doviak. 1991a. Let’s Make a Deal: The Player’s Dilemma. The American Statistician 45: 284-7. ________ 1991b. Rejoinder. The American Statistician 45: 289-8. ________ 1991c. Rejoinder to vos Savant. The American Statistician 45: 347-8. Nalebuff, B. 1987. Puzzles. Journal of Economic Perspectives 1: 157-63. Palacios-Huerta, I.. 2003. Learning to Open Monty Hall’s Doors. Experimental Economics 6: 23551. Raiffa, H., J. Richardson, and D. Metcalfe. 2002. Negotiation Analysis: The Science and Art of Collaborative Decision Making. Cambridge, M.A.: Belknap Press of Harvard University Press. Savage, L. 1972. The Foundations of Statistics, 2nd ed. New York: Dover. Siegel, J. 1972. Risk, Interest Rates and the Forward Exchange. Quarterly Journal of Economics 86: 303-9. Selvin, S. 1975a. A Problem in Probability. The American Statistician 29: 67. ________1975b. On the Monty Hall Problem. The American Statistician 29: 134. Seymann, R.G. 1991. Comment. The American Statistician 45: 287-8. Stigler, S.M. 1999. Statistics on the Table: the History of Statistical Concepts and Methods. Cambridge: Harvard University Press. Tierney, J. 1991. Behind Monty Hall’s Doors: Puzzle, Debate, and Answer? New York Times, July -10-
21, 1991. vos Savant, M. 1990a. Ask Marilyn. Parade Magazine. September 9, 1990, p.15. ________ 1990b. Ask Marilyn. Parade Magazine. December 2, 1990, p. 25. ________ 1991a. Ask Marilyn. Parade Magazine. February 17, 1991, p.12. ________ 1991b. Marilyn vos Savant’s Reply. The American Statistician 45: 347.
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Table 1: Event Space and Probabilities Award Behind Door
P(A=1)
P(A=2)
P(A=3)
Prior Pro bability
1/3
1/3
1/3
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Event
Cond itional Probability
(C=1 1 H =1 | A=1)
0
(C=1 1 H =2 | A=1)
1/4
(C=1 1 H =3 | A=1)
1/4
(C=2 1 H =1 | A=1)
0
(C=2 1 H =2 | A=1)
0
(C=2 1 H =3 | A=1)
1/4
(C=3 1 H =1 | A=1)
0
(C=3 1 H =2 | A=1)
1/4
(C=3 1 H =3 | A=1)
0
(C=1 1 H =1 | A=2)
0
(C=1 1 H =2 | A=2)
0
(C=1 1 H =3 | A=2)
1/4
(C=2 1 H =1 | A=2)
1/4
(C=2 1 H =2 | A=2)
0
(C=2 1 H =3 | A=2)
1/4
(C=3 1 H =1 | A=2)
1/4
(C=3 1 H =2 | A=2)
0
(C=3 1 H =3 | A=2)
0
(C=1 1 H =1 | A=3)
0
(C=1 1 H =2 | A=3)
1/4
(C=1 1 H =3 | A=3)
0
(C=2 1 H =1 | A=3)
1/4
(C=2 1 H =2 | A=3)
0
(C=2 1 H =3 | A=3)
0
(C=3 1 H =1 | A=3)
1/4
(C=3 1 H =2 | A=3)
1/4
(C=3 1 H =3 | A=3)
0
