The Monty Hall Problem
From Math Images
Line 5: | Line 5: | ||
:When the Monty Hall problem was published in Parade Magazine in 1990, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. It remains one of the most disputed mathematical puzzles of all time. | :When the Monty Hall problem was published in Parade Magazine in 1990, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. It remains one of the most disputed mathematical puzzles of all time. | ||
- | |ImageDescElem=<nowiki></nowiki> | + | |ImageDescElem= |
+ | <nowiki></nowiki> | ||
===The Problem=== | ===The Problem=== | ||
<blockquote>The show's host, Monty Hall, asks a contestant to pick one of three doors. One door leads to a brand new car, but the other two lead to goats. Once the contestant has picked a door, Monty opens one of the two remaining doors. He is careful never to open the door hiding the car. After Monty has opened one of these other two doors, he offers the contestant the chance to switch doors. Is it to his advantage to stay with his original choice, switch to the other unopened door, or does it not matter?</blockquote> | <blockquote>The show's host, Monty Hall, asks a contestant to pick one of three doors. One door leads to a brand new car, but the other two lead to goats. Once the contestant has picked a door, Monty opens one of the two remaining doors. He is careful never to open the door hiding the car. After Monty has opened one of these other two doors, he offers the contestant the chance to switch doors. Is it to his advantage to stay with his original choice, switch to the other unopened door, or does it not matter?</blockquote> | ||
Line 78: | Line 79: | ||
Let the door picked by the contestant be called door ''a'' and the other two doors be called ''b'' and ''c''. Also, V<sub>''a''</sub>, V<sub>''b''</sub>, and V<sub>''c''</sub>, are the events that the car is actually behind door ''a'', ''b'', and ''c'' respectively. We begin by looking at a scenario that leads to Monty opening door ''b'', so let O<sub>''b''</sub> be the event that Monty Hall opens curtain ''b''. | Let the door picked by the contestant be called door ''a'' and the other two doors be called ''b'' and ''c''. Also, V<sub>''a''</sub>, V<sub>''b''</sub>, and V<sub>''c''</sub>, are the events that the car is actually behind door ''a'', ''b'', and ''c'' respectively. We begin by looking at a scenario that leads to Monty opening door ''b'', so let O<sub>''b''</sub> be the event that Monty Hall opens curtain ''b''. | ||
- | Then, the problem can be restated as follows: Is <math>P(V_a | + | Then, the problem can be restated as follows: Is <math>P(V_a|O_b) = P(V_c|O_b)</math>? <math>P(V_a|O_b)</math> is the probability that door ''a'' hides the car given that Monty opens door ''b''. Similarly, <math>P(V_c|O_b)</math> is the probability that door ''c'' hides the car given that monty opens door ''b''. So, when <math>P(V_a|O_b) = P(V_c|O_b)</math> the probability that the car is behind one unopened door the same as the probability that the car is behind the other unopened door. If this is the case, it won't matter if the contestant stays or switches. |
+ | |||
+ | |||
+ | Using Bayes' Theorem, we know that | ||
+ | |||
+ | <math>P(V_a|O_b)=\frac{P(V_a)*P(O_b|V_a)}{P(O_b)}</math> | ||
+ | |||
+ | <math>P(V_c|O_b)=\frac{P(V_c)*P(O_b|V_c)}{P(O_b)}</math> | ||
+ | |||
+ | Also, we can assume that the prize is randomly placed behind the curtains, so | ||
+ | |||
+ | <math>P(V_a) = P(V_b) = P(V_c) = \frac{1}{3}</math> | ||
+ | |||
+ | Then we can calculate the conditional probabilities for the event ''O''<sub>''b''</sub>, which we can then use to calculate the probability of event ''O''<sub>''b''</sub>. | ||
+ | |||
+ | First, we can calculate the conditional probability that Monty opens door ''b'' if the car is hidden behind door ''a''. | ||
+ | |||
+ | <math>P(O_b|V_a) = 1/2 </math> because if the prize is behind ''a'', Monty can open either ''b'' or ''c''. | ||
+ | |||
+ | <math>P(O_b|V_b) = 0 </math> because if the prize is behind door ''b'', Monty can't open door ''b''. | ||
+ | |||
+ | <math>P(O_b|V_c) = 1 </math> because if the prize is behind door ''c'', Monty can only open door ''b''. | ||
+ | |||
+ | Each of these probabilities is conditional on the fact that the prize is hidden behind a specific door, but we are assuming that each of these probabilities is mutually exclusive since the car can only be hidden behind one door. As a result, we know that P(''O''<sub>''b''</sub>) is equal to | ||
+ | |||
+ | <math> P(O_b) = P(O_b \cap V_a) + P(O_b \cap V_b) + P(O_b \cap V_c) </math> | ||
+ | |||
+ | Using the equation for the probability of non-independent events, we can say | ||
+ | |||
+ | <math> P(O_b)= P(V_a)P(O_b|V_a) + P(V_b)P(O_b|V_b) +P(V_c)P(O_b|V_c) </math> | ||
+ | |||
+ | ::<math>= \frac{1}{3} * \frac{1}{2} + \frac{1}{3} * 0 + \frac{1}{3} * 1 </math> | ||
+ | |||
+ | ::<math>= \frac{1}{2} </math> | ||
+ | |||
+ | Then, we can use <math>P(O_b)</math>, <math>P(O_b|V_a)</math>, and <math>P(V_a)</math> to calculated <math>P(V_a|O_b)</math>. | ||
+ | |||
+ | <math> P(V_a|O_b) = \frac {P(V_a)*P(O_b|V_a)}{P(O_b)} </math> | ||
+ | |||
+ | :::<math>= \frac {\frac{1}{3} * \frac{1}{2}} {\frac{1}{2}} </math> | ||
+ | |||
+ | :::<math>= \frac {1}{3} </math> | ||
+ | |||
+ | Similarly, | ||
+ | |||
+ | <math> P(V_c|O_b) = \frac {P(V_c)*P(O_b|V_c)}{P(O_b)} </math> | ||
+ | |||
+ | :::<math> = \frac {\frac{1}{3} * 1} {\frac{1}{2}} </math> | ||
+ | |||
+ | :::<math> = \frac {2}{3} </math> | ||
+ | |||
+ | The probability of ''V''<sub>''c''</sub> (the event that car is hidden behind door ''c'') in this case is ''not'' equal to the probability of ''V''<sub>''a''</sub> (the case where the car is hidden behind the door that Monty hasn't opened and the contestant hasn't selected). The contestant is offered an opportunity to switch to door ''c''. We have calculated that the probability of winning when door ''c'' is selected is 2/3 and the probability of winning with the contestant's original choice, door ''a'' is 1/3. | ||
+ | |||
+ | Since Monty is equally likely to open any of the three doors, we can generalize this strategy for any door that he opens. The probability that the car is hidden behind the contestant's original choice is 1/3, but the probability that the car is hidden behind the unopened and unselected door is 2/3. If the contestant switches, he doubles his chance of winning. | ||
+ | |||
|AuthorName=Grand Illusions | |AuthorName=Grand Illusions | ||
|Field=Algebra | |Field=Algebra | ||
Line 146: | Line 201: | ||
http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf | http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf | ||
- | |||
- | |||
- | |||
|InProgress=No | |InProgress=No | ||
}} | }} |
Revision as of 13:45, 13 July 2010
Let's Make a Deal |
---|
Let's Make a Deal
- The Monty Hall problem is a probability puzzle based on the 1960's game show Let's Make a Deal.
- When the Monty Hall problem was published in Parade Magazine in 1990, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. It remains one of the most disputed mathematical puzzles of all time.
Contents |
Basic Description
The Problem
The show's host, Monty Hall, asks a contestant to pick one of three doors. One door leads to a brand new car, but the other two lead to goats. Once the contestant has picked a door, Monty opens one of the two remaining doors. He is careful never to open the door hiding the car. After Monty has opened one of these other two doors, he offers the contestant the chance to switch doors. Is it to his advantage to stay with his original choice, switch to the other unopened door, or does it not matter?
The Solution
If you answered that the contestant's decision doesn't matter, then you are among about 90% of respondents who were quickly able to determine that the two remaining doors must be equally likely to hide the car. You are also wrong. The answer to the Monty Hall Problem is viewed by most people—including mathematicians—as extremely counter–intuitive.
It is actually to the contestant's advantage to switch: the probability of winning if the contestant doesn't switch is 1/3, but if the contestant switches, the probability becomes 2/3. To see why this is true, we examine each possible scenario below.
We can first imagine the case where the car is behind door 1.
In the diagram on the below, we can see what prize the contestant will win if he always stays with his initial pick after Monty opens a door. If the contestant uses the strategy of always staying, he will only win if he originally picked door 1.
If the contestant always switches doors when Monty shows him a goat, then he will win if he originally picked door 2 or door 3.
A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. Since we know that the car is equally likely to be behind each of the three doors, we can generalize our strategy for the case where the car is behind door 1 to any placement of the car. The probability of winning by staying with the initial choice is 1/3, while the probability of winning by switching is 2/3. The contestant's best strategy is to always switch doors so he can drive home happy and goat-free.
Aids to Comprehension
The Monty Hall problem has the distinction of being one of the rare math problems that has gained recognition on the front page of the Sunday New York Times. On July 21, 1991, the Times published a story that explained a heated argument between a Parade columnist, Marilyn vos Savant, and numerous angry readers. Many of these readers held distinguished degrees in mathematics, and the problem seemed far too elementary to warrant such difficulty in solving.
Further explanation of readers' debate with vos Savant's can be found in the Why It's Interesting section. However, if you aren't completely convinced that switching doors is the best strategy, be aware that the Monty Hall problem has been called "math's most contentious brain teaser." The following explanations are alternative approaches to the problem that may help clarify that the best strategy is, in fact, switching doors.
Why the Probability is not 1/2
The most common misconception is that the odds of winning are 50-50 no matter which door a contestant chooses. Most people assume that each door is equally likely to contain the car since the probability was originally distributed evenly between the three doors. They believe that they have no reason to prefer one door, so it does not matter whether they switch or stick with their original choice.
This reasoning seems logical until we realize that the two doors cannot be equally likely to hide the car. The critical fact is that Monty's choice of which door to open is not random, so when he opens a door, it gives the contestant new information.
Marilyn defended her answer in a subsequent column addressing this point specifically. Suppose we pause after Monty has revealed a goat and a UFO settles down onto the stage and a little green woman emerges. The host asks her to point to one of the two unopened doors. Then the chances that she'll randomly choose the one with the prize are 1/2. But, that's because she lacks the advantage the original contestant had—the help of the host.
"When you first choose door #1 from three, there's a 1/3 chance that the prize is behind that one and a 2/3 chance that it's behind one of the others. But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don't switch, you win only if the prize is behind door #1," Marilyn explained.
This is true because when Monty opens a door, he is reducing the probability that it contains a car to 0. When the contestant makes an initial pick, there is a 1/3 chance that he picked the car and a 2/3 chance that one of the other two doors has the car. When Monty shows him a goat behind one of those two doors, the 2/3 chance is only for the one unopened door because the probability must be 0 for the one that the host opened.
An Extreme Case of the Problem
Imagine that you are on Let's Make a Deal are there are now 1 million doors. You choose your door, then Monty opens all but one of the remaining doors, showing you that they hide goats. It’s clear that your first choice is unlikely to have been the right choice out of 1 million doors.
Since you know that the car must be hidden behind one of the unopened doors and it is very unlikely to be behind your door, you know that it must be behind the other door. In fact, on average in 999,999 out of 1,000,000 times the other door will contain the prize because 999,999 out of 1,000,000 times the player first picked a door with a goat. Switching to the other door is the best strategy.
Simulation
Using a simulation is another useful way to show that the probability of winning by switching is 2/3. A simulation using playing cards allows us to perform multiple rounds of the game easily.
One simulation proposed by vos Savant herself requires only two participants, a player and a host. Three cards are held by the host, one ace that represents the prize and two lower cards that represent the mules. The host holds up the three cards so only he can see their values. The contestant picks a card, and it is placed aside so that he still cannot see the value. Monty then reveals one of the remaining low cards which represents a mule. He must choose between the two lower cards if they both remain in his hand.
If the card remaining in the host's hand is an ace, then this is recorded as a round where the player would have won by switching. Contrastingly, if the host is holding a low card, the round is recorded as one where staying would have won. Performing this simulation repeatedly will reveal that a player who switches will win the prize approximately 2/3 of the time.
Play the Game
http://www.nytimes.com/2008/04/08/science/08monty.html?_r=1
A More Mathematical Explanation
The following explanation uses Bayes' Theorem to show how Monty revea [...]
The following explanation uses Bayes' Theorem to show how Monty revealing a goat changes the game.
Let the door picked by the contestant be called door a and the other two doors be called b and c. Also, V_{a}, V_{b}, and V_{c}, are the events that the car is actually behind door a, b, and c respectively. We begin by looking at a scenario that leads to Monty opening door b, so let O_{b} be the event that Monty Hall opens curtain b.
Then, the problem can be restated as follows: Is ? is the probability that door a hides the car given that Monty opens door b. Similarly, is the probability that door c hides the car given that monty opens door b. So, when the probability that the car is behind one unopened door the same as the probability that the car is behind the other unopened door. If this is the case, it won't matter if the contestant stays or switches.
Using Bayes' Theorem, we know that
Also, we can assume that the prize is randomly placed behind the curtains, so
Then we can calculate the conditional probabilities for the event O_{b}, which we can then use to calculate the probability of event O_{b}.
First, we can calculate the conditional probability that Monty opens door b if the car is hidden behind door a.
because if the prize is behind a, Monty can open either b or c.
because if the prize is behind door b, Monty can't open door b.
because if the prize is behind door c, Monty can only open door b.
Each of these probabilities is conditional on the fact that the prize is hidden behind a specific door, but we are assuming that each of these probabilities is mutually exclusive since the car can only be hidden behind one door. As a result, we know that P(O_{b}) is equal to
Using the equation for the probability of non-independent events, we can say
Then, we can use , , and to calculated .
Similarly,
The probability of V_{c} (the event that car is hidden behind door c) in this case is not equal to the probability of V_{a} (the case where the car is hidden behind the door that Monty hasn't opened and the contestant hasn't selected). The contestant is offered an opportunity to switch to door c. We have calculated that the probability of winning when door c is selected is 2/3 and the probability of winning with the contestant's original choice, door a is 1/3.
Since Monty is equally likely to open any of the three doors, we can generalize this strategy for any door that he opens. The probability that the car is hidden behind the contestant's original choice is 1/3, but the probability that the car is hidden behind the unopened and unselected door is 2/3. If the contestant switches, he doubles his chance of winning.
Why It's Interesting
Variations of the problem have been popular game teasers since the 19th century, but the "Lets Make a Deal" version is most widely known.History of the Problem
The earliest of several probability puzzles related to the Monty Hall problem is Bertrand's box paradox, posed by Joseph Bertrand in 1889. In Bertrand's puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. The player chooses one random box and draws a coin without looking. The coin happens to be gold. What is the probability that the other coin is gold as well? As in the Monty Hall problem the intuitive answer is 1/2, but the actual probability 2/3.
Ask Marilyn: A Story of Misguided Hatemail
The question was originally proposed by a reader of “Ask Marilyn”, a column in Parade Magazine in 1990. Marilyn's correct solution, that switching doors was the best strategy, caused an uproar among mathematicians. While most people responded that switching should not matter, the contestant’s chances for winning in fact double if he switches doors. Part of the controversy, however, was caused by the lack of agreement on the statement of the problem itself.
Most statements of the problem, including the one in Marilyn's column, do not match the rules of the actual game show. This was a source of great confusion when the problem was first presented. The main ambiguities in the problem arise from the fact that it does not fully specify the host's behavior.
For example, imagine a host who wasn't required to always reveal a goat. The host's strategy could be to open a door only when the contestant has selected the correct door initially. This way, the host could try to tempt the contestant to switch and lose.
When first presented with the Monty Hall problem, an overwhelming majority of people assume that switching does not change the probability of winning the car even when the problem was stated to remove all sources of ambiguity. An article by Burns and Wieth cited various studies on the Monty Hall problem that document difficulty solving the Monty Hall problem specifically. These previous articles reported 13 studies using standard versions of the Monty Hall dilemma, and reported that most people do not switch doors. Switch rates ranged from 9% to 23% with a mean of 14.5%, even when the problem was stated explicitely.
This consistency is especially remarkable given that these studies include a range of different wordings, methods of presentations, languages, and cultures.
Marilyn quotes cognitive psychologist Massimo Piattelli-Palmarini in her own book saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." When the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong.
One letter written to vos Savant by Dr. E. Ray Bobo of Georgetown University was especially critical of Marilyn's solution: "You are utterly incorrect about the game show question, and I hope this controversy will call some public attention to the serious national crisis in mathematical education. If you can admit your error, you will have contributed constructively toward the solution to a deplorable situation. How many irate mathematicians are needed to get you to change your mind?"
Monty and Monkeys
A recent article published in The New York Times uncovered an interesting relationship between the Monty Hall problem and a study on cognitive dissonance using monkeys. If the calculations of Yale economist M. Keith Chen are correct, then some of the most famous experiments in psychology might be flawed. Chen believes the researchers drew conclusions based on natural inclination to incorrectly evaluate probability.
The most famous experiment in question is the 1956 study "Postdecision changes in the desirability of alternatives" on rationalizing choices. The researchers studied which M&M colors were most preferred by monkeys. After identifying a few colors of M&Ms that were approximately equally favored by a monkey - say, red, blue, and yellow, - the researchers gave the monkey a choice between two of the colors.
In one case, imagine that a monkey chose yellow over blue. Then, the monkey would be offered the choice between blue and red M&Ms. Researchers noted that about two-thirds of the time the monkey would choose red. The 1956 study claimed that their results reinforced the theory of rationalization: Once we reject something we are convinced that we never like it anyway.
Dr. Chen reexamined the experimental procedure, and says that monkey's rejection of blue might be attributable to statistics alone. Chen says that there must be some difference in preference between the original red, blue, and yellow. If this is the case, then the monkey's choice of yellow over blue wasn't arbitrary. Like Monty Hall's decision to open a door that hid a goat, the monkey's choice between yellow and blue discloses additional information. In fact, when a monkey favors yellow over blue, there's a two-thirds chance that it also started off with a preference for red over blue- which would explain why the monkeys chose red 2/3 of the time in the Yale experiment.
To why this is true, consider Chen's conjecture that monkeys must have some slight preference between the three colors they are being offered. The table below shows all the possible combinations of ways that a monkey could possibly rank its M&Ms.
We can see that in the case where the monkey preferred yellow over blue, they monkey preferred red over blue in 2/3 of the rankings.
Although Chen agrees that the study may have still discovered useful information about preferences, but he doesn't believe it has been measured correctly yet. "The whole literature suffers from this basic problem of acting as if Monty's choice means nothing."
Monty Hall problem, the study of monkeys, and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly. Even academic studies may be littered with mistakes caused by difficulty interpreting statistics.
21
The 2008 movie 21 increased public awareness of the Monty Hall problem. 21 opens with an M.I.T. math professor using the Monty Hall Problem to explain theories to his students. The Monty Hall problem is included in the movie to show the intelligence of the main character because he is immediately able to solve such a notoriously difficult problem.
Teaching Materials
- There are currently no teaching materials for this page. Add teaching materials.
References
http://www.nytimes.com/2008/04/08/science/08tier.html
http://www.cs.dartmouth.edu/~afra/goodies/monty.pdf
Heart of Mathematics. Edward B. Burger and Michael P. Starbird.
http://en.wikipedia.org/wiki/Monty_Hall_problem
The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser. Jason Rosenhouse.
Brehm, J. W. (1956) Postdecision changes in the desirability of alternatives, Journal of Abnormal and Social Psychology, 52, 384-9
http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.