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The Monty Hall Problem
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會員從Jan 26, 2011開始
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Aug 21, 2011 at 00:58
(已編輯Aug 21, 2011 at 01:04)
會員從Jan 26, 2011開始
1367帖子
<b>The 3-Door Monty Hall Problem
By Michael Shermer</b>
In nearly 100 months of writing the Skeptic column I have never received so many letters as I did in response to my October essay (“A Random Walk Through Middle Land”) on the so-called Monty Hall Problem: you are on a game show with three doors, behind one of which is a car and behind the other two are goats. You pick door #1. Monty, who knows what’s behind all three doors, reveals that behind door #2 is a goat. Before showing you what you won, Monty asks if you want to switch doors. Most people say that it doesn’t matter because it is now a 50/50 choice. I explained that you should always switch. My correspondents disagreed.
The James Madison University mathematics professor Jason Rosenhouse, who has written an entire book on the subject—The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser (Oxford University Press, 2009)—explained to me that you double your chances of winning by switching doors when three conditions are met: (1) Monty never opens the door you chose initially; (2) Monty always opens a door concealing a goat; (3) When the first two rules leave Monty with a choice of doors to open (which happens in those cases where your initial choice was correct) he makes his choice at random. “Switching turns a loss into a win and a win into a loss,” says Rosenhouse, “and since my first choice is wrong 2/3rds of the time, I will win that often by switching.”
Here’s why: At the beginning of the game you have a 1/3rd chance of picking the car and a 2/3rds chance of picking a goat. Switching doors is bad only if you initially chose the car, which happens only 1/3rd of the time. Switching doors is good if you initially chose a goat, which happens 2/3rds of the time. Thus, the probability of winning by switching is 2/3rds, or double the odds of not switching (keeping in mind the three rules above). Analogously, if there are 10 doors, initially you have a 1/10th chance of picking the car and a 9/10ths chance of picking a goat. Switching doors is bad only if you initially chose the car, which happens only 1/10th of the time. Switching doors is good if you initially chose a goat, which happens 9/10ths of the time. Thus, the probability of winning by switching is 9/10ths, again, assuming that Monty has shown you 8 other doors with goats.
Still not convinced? Google “Monty Hall Problem simulation” and try the various computer simulations yourself and you will see that you double your actual wins by switching doors. One of my correspondents, who was skeptical at first, ran his own simulation over 10,000 trials, concluding that “switching doors yields a 2/3 success rate while running without switching doors yields a 1/3 success rate.”
https://www.scientificamerican.com/article.cfm?id=the-3-door-monty-hall-problem
By Michael Shermer</b>
In nearly 100 months of writing the Skeptic column I have never received so many letters as I did in response to my October essay (“A Random Walk Through Middle Land”) on the so-called Monty Hall Problem: you are on a game show with three doors, behind one of which is a car and behind the other two are goats. You pick door #1. Monty, who knows what’s behind all three doors, reveals that behind door #2 is a goat. Before showing you what you won, Monty asks if you want to switch doors. Most people say that it doesn’t matter because it is now a 50/50 choice. I explained that you should always switch. My correspondents disagreed.
The James Madison University mathematics professor Jason Rosenhouse, who has written an entire book on the subject—The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brainteaser (Oxford University Press, 2009)—explained to me that you double your chances of winning by switching doors when three conditions are met: (1) Monty never opens the door you chose initially; (2) Monty always opens a door concealing a goat; (3) When the first two rules leave Monty with a choice of doors to open (which happens in those cases where your initial choice was correct) he makes his choice at random. “Switching turns a loss into a win and a win into a loss,” says Rosenhouse, “and since my first choice is wrong 2/3rds of the time, I will win that often by switching.”
Here’s why: At the beginning of the game you have a 1/3rd chance of picking the car and a 2/3rds chance of picking a goat. Switching doors is bad only if you initially chose the car, which happens only 1/3rd of the time. Switching doors is good if you initially chose a goat, which happens 2/3rds of the time. Thus, the probability of winning by switching is 2/3rds, or double the odds of not switching (keeping in mind the three rules above). Analogously, if there are 10 doors, initially you have a 1/10th chance of picking the car and a 9/10ths chance of picking a goat. Switching doors is bad only if you initially chose the car, which happens only 1/10th of the time. Switching doors is good if you initially chose a goat, which happens 9/10ths of the time. Thus, the probability of winning by switching is 9/10ths, again, assuming that Monty has shown you 8 other doors with goats.
Still not convinced? Google “Monty Hall Problem simulation” and try the various computer simulations yourself and you will see that you double your actual wins by switching doors. One of my correspondents, who was skeptical at first, ran his own simulation over 10,000 trials, concluding that “switching doors yields a 2/3 success rate while running without switching doors yields a 1/3 success rate.”
https://www.scientificamerican.com/article.cfm?id=the-3-door-monty-hall-problem
會員從Aug 25, 2011開始
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