Monty Hall problem: Difference between revisions
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The player initially chooses door number 1. Initially, the odds that the car is behind this door are 2 to 1 against (it is two times more likely that his choice is wrong than that it is right). | The player initially chooses door number 1. Initially, the odds that the car is behind this door are 2 to 1 against (it is two times more likely that his choice is wrong than that it is right). | ||
The host opens one of the other two doors, revealing a goat. Let's suppose that for the moment, the player doesn't take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind Door 1, this new information does not change the player's odds that the car is indeed behind Door 1; they are still 2 to 1 against. | The host opens one of the other two doors, revealing a goat. Let's suppose that for the moment, the player doesn't take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind Door 1, this new information does not change the player's odds that the car is indeed behind Door 1; they are still 2 to 1 against. The player should switch. | ||
Many authors essentially stop at this point. Some however add one further step in the argument. | |||
Therefore, also knowing that the host opened specifically Door 3 to reveal a goat, the | The player of course also gets informed that the specific door opened is, for instance, Door 3. Does this influence the odds that the car is behind Door 1? No: from the player's point of view, the chance that the car is behind Door 1 can't depend on whether the host opens Door 2 or Door 3 - the door numbers are arbitrary, exchangeable, mere labels. (A mathematician will here refer to the mathematical concept of <i>symmetry</>). | ||
Therefore, also knowing that the host opened specifically Door 3 to reveal a goat, the player's odds on the car being behind Door 1 are 2 to 1 against. He should switch to Door 2. |
Revision as of 06:02, 26 January 2011
The Monty Hall problem, also known as the three doors problem or the quizmaster problem, became famous in 1990 on the publication of an article in Parade magazine in the popular weekly column "Ask Marilyn". The column's author Marilyn vos Savant was reputedly, at the time, the person with the highest IQ in the world. The problem is named after the stage-name of an actual quizmaster, Monty Halperin, on a long-running 60's TV show, though it seems that the events related in the Monty Hall Problem never actually took place in reality.
Rewriting in her own words a problem posed to her by a correspondent, a Mr. Craig Whitaker, Marilyn asked the following:
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 No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Almost everyone, on first hearing the problem, has the immediate and intuitive reaction that the two doors left closed, Door 1 and Door 2, must be equally likely to hide the car, and that therefore there is no point in switching. However, some careful thought shows that this immediate reaction must be wrong. Consider many, many repetitions of the game, and suppose (for the sake of argument) that the car is hidden each time anew completely at random behind one of the three doors, while the player always starts by picking Door number 1 (perhaps, because "1" is his favourite number). In the long run, the car will be behind his initially chosen door, Door 1, on one third of the repetitions of the game. If he never switches doors, he'll take the car home with him on precisely those, one third of all, repetitions of the show, where his initial choice was actually right. On the other hand, if he always switches doors when offered the choice (and that will sometimes be to Door 2 and sometimes to Door 3, depending on which door is opened by the host each time), he'll go home with the car on two thirds of all the repetitions of the game - he'll win the car by "switching" exactly on every occasion when he would not win it by "staying".
Here is a more detailed analysis. In this solution we use "probability" in the Bayesian or subjectivist sense; that is to say, all probability statements are supposed to reflect the state of knowledge of one person, let's say, of the contestant on the show, who initially knows no more than the following: he'll choose a door; the quizmaster will thereupon open a different door revealing a goat, and make the offer that the contestant may switch to the third door. For the contestant therefore, initially all doors are equally likely to hide the car; and moreover, if he chooses any particular door, and the car happens to be behind that particular door, then for him the host is equally likely to open either of the other two doors.
The player initially chooses door number 1. Initially, the odds that the car is behind this door are 2 to 1 against (it is two times more likely that his choice is wrong than that it is right).
The host opens one of the other two doors, revealing a goat. Let's suppose that for the moment, the player doesn't take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind Door 1, this new information does not change the player's odds that the car is indeed behind Door 1; they are still 2 to 1 against. The player should switch.
Many authors essentially stop at this point. Some however add one further step in the argument.
The player of course also gets informed that the specific door opened is, for instance, Door 3. Does this influence the odds that the car is behind Door 1? No: from the player's point of view, the chance that the car is behind Door 1 can't depend on whether the host opens Door 2 or Door 3 - the door numbers are arbitrary, exchangeable, mere labels. (A mathematician will here refer to the mathematical concept of symmetry</>).
Therefore, also knowing that the host opened specifically Door 3 to reveal a goat, the player's odds on the car being behind Door 1 are 2 to 1 against. He should switch to Door 2.