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The Monty Hall Problem | Dilemmas for Thinkers |
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"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?" - Reader of "Parade" Magazine, directed towards Marilyn vos Savant in an 1990 issue.
In other words, do you have a better chance of getting a car if you switch your choice, or do you have a better chance of getting a car if you keep your initial choice? The answer? You have a 2/3 chance of winning the car if you decide to switch doors, but only a 1/3 chance of winning the car if you decide to stick to your initial door. But wait, if he opens a door, then there are only two choices remaining. Shouldn't there be a 50:50 chance of winning? No.
Receiving assistance from the picture above, observe that there is a 2/3 chance that your first choice of a door is hiding a goat. In that situation, the host, Monty Hall, MUST open a door hiding the other goat, and keep the door containing the car closed. Therefore, switching to another door will only fail to win you a car only if you initially chose the door hiding the car, which only happens 1/3 of the time.
Not getting it? Lets analyze this:
First, we'll start off with popular assumptions:
1) The host, Monty Hall, will always open a door different than what you chose. So if you chose Door 1, he will always open either Door 2 or Door 3.
2) Monty Hall knows where the car is
-2a) Monty Hall will always open a door that does NOT have the car. Meaning, if you chose Door 1, and the car was behind Door 2, he would have chosen Door 3.
"The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated." - Leonard Mlodinow, 2008
Lets go back to most well-known statement of the problem:
"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?"
(Whitaker, 1990, as quoted by vos Savant 1990a)
vos Savant replied with:
"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?"
Lets expand her response. Say that, instead of 3 doors, there are 1,000,000 doors. Out of these doors, there is only 1 door hiding a car. Thus, if you pick a door, your chance of winning would be 1 out of 1,000,000. Like in the original situation, Monty Hall opens every door and shows 999,998 goats, while skipping your door and another door. Monty Hall then offers you to switch to the door he did not open, but should you switch? Well, in 999,999 out of 1,000,000 times, the other door will contain the winning car because in 999,999 out of 1,000,000 times you picked a door containing a goat. Still not making too much sense?
Ask yourself this: how likely would it be that I picked the correct door, out of 1,000,000 to choose from. What the host, Monty Hall, is doing is giving you the opportunity to either keep your one door, or open all other 999,999 doors - he's already opened 999,998 for you while leaving, on purpose, the winning door.
Still not understanding? Don't worry, many don't.
1a | 1/6 | (Door 1)Car | (Door 2)Goat| (Door 3)Goat | 2 | Car | Goat
1b | 1/6 | (Door 1)Car | (Door 2)Goat| (Door 3)Goat | 3 | Car | Goat
2a | 1/6 | (Door 1)Goat| (Door 2)Car | (Door 3)Goat | 3 | Goat | Car
2b | 1/6 | (Door 1)Goat| (Door 2)Car | (Door 3)Goat | 3 | Goat | Car
3a | 1/6 | (Door 1)Goat| (Door 2)Goat| (Door 3)Car | 2 | Goat | Car
3b | 1/6 | (Door 1)Goat| (Door 2)Goat| (Door 3)Car | 2 | Goat | Car
Look at the diagram above, and assume you've chosen Door 1
-The number on the far left represents the cases of each door picked, remembering that Monty Hall could open either of the doors you didn't choose, thus creating two separate cases for each door(indicated by .a and .b)
-The fraction to the immediate right represent the probability of each case occurring, with each case having a 1/6 chance of happening.
-The next 3 columns represents what's behind each door (i.e in Case 1a, there was a car behind Door 1, a goat behind Door 2, and a goat behind Door 3)
-The numbers following those columns represent which door Monty Hall opened.
-The next column to the immediate right states the result if you were to stay with your initial choice of Door 1.
-The rightmost column states the result if you were to switch to the door that was being offered - the door that you did not choose, nor the one that Monty Hall opened.
By looking at the right-most column, you will notice that you will always win a car if the car is NOT behind Door 1(which only happens 2/6 of the time; 1/3 of the time), and you always switch to the other door(the car will be behind the other door 4/6 of the time; 2/3 of the time.
Math can be fun sometimes, yeah?
Special credit to "Cepheus" of Wikipedia for uploading the image used.
Read more about the "Monty Hall Problem" on its wikipedia page | http://en.wikipedia.org/wiki/Monty_Hall_problem
 oTobias · Thu Aug 01, 2013 @ 09:02pm · 0 Comments |
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