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The Monty Hall Problem

(Difference between revisions)
 Revision as of 12:12, 9 July 2010 (edit)← Previous diff Revision as of 14:44, 13 July 2010 (edit) (undo)Next diff → 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= + |ImageDescElem= - + ===The Problem=== ===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 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?
Line 79: Line 78: 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''. 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 $P(V_a|O_b) = P(V_c|O_b)$? $P(V_a|O_b)$ is the probability that door ''a'' hides the car given that Monty opens door ''b''. Similarly, $P(V_c|O_b)$ is the probability that door ''c'' hides the car given that monty opens door ''b''. So, when $P(V_a|O_b) = P(V_c|O_b)$ 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. + Then, the problem can be restated as follows: Is $P(V_a - + - + - Using Bayes' Theorem, we know that + - + - [itex]P(V_a|O_b)=\frac{P(V_a)*P(O_b|V_a)}{P(O_b)}$ + - + - $P(V_c|O_b)=\frac{P(V_c)*P(O_b|V_c)}{P(O_b)}$ + - + - Also, we can assume that the prize is randomly placed behind the curtains, so + - + - $P(V_a) = P(V_b) = P(V_c) = \frac{1}{3}$ + - + - 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''. + - + - $P(O_b|V_a) = 1/2$ because if the prize is behind ''a'', Monty can open either ''b'' or ''c''. + - + - $P(O_b|V_b) = 0$ because if the prize is behind door ''b'', Monty can't open door ''b''. + - + - $P(O_b|V_c) = 1$ 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 + - + - $P(O_b) = P(O_b \cap V_a) + P(O_b \cap V_b) + P(O_b \cap V_c)$ + - + - Using the equation for the probability of non-independent events, we can say + - + - $P(O_b)= P(V_a)P(O_b|V_a) + P(V_b)P(O_b|V_b) +P(V_c)P(O_b|V_c)$ + - + - ::$= \frac{1}{3} * \frac{1}{2} + \frac{1}{3} * 0 + \frac{1}{3} * 1$ + - + - ::$= \frac{1}{2}$ + - + - Then, we can use $P(O_b)$, $P(O_b|V_a)$, and $P(V_a)$ to calculated $P(V_a|O_b)$. + - + - $P(V_a|O_b) = \frac {P(V_a)*P(O_b|V_a)}{P(O_b)}$ + - + - :::$= \frac {\frac{1}{3} * \frac{1}{2}} {\frac{1}{2}}$ + - + - :::$= \frac {1}{3}$ + - + - Similarly, + - + - $P(V_c|O_b) = \frac {P(V_c)*P(O_b|V_c)}{P(O_b)}$ + - + - :::$= \frac {\frac{1}{3} * 1} {\frac{1}{2}}$ + - + - :::$= \frac {2}{3}$ + - + - 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. + - + |AuthorName=Grand Illusions |AuthorName=Grand Illusions |Field=Algebra |Field=Algebra Line 201: Line 146: http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf http://www.math.jmu.edu/~lucassk/Papers/MHOverview2.pdf + |ToDo=Helper Pages: + :* Bayes' Theorem + :* Probability |InProgress=No |InProgress=No }} }}

Revision as of 14:44, 13 July 2010

{{Image Description |ImageName=Let's Make a Deal |Image=Mainimage.jpg |ImageIntro=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.

|ImageDescElem=

Contents

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 |ImageDesc=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, Va, Vb, and Vc, 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 Ob be the event that Monty Hall opens curtain b.