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# Suppose you're given the choice of three boxes. In one of

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Suppose you're given the choice of three boxes. In one of [#permalink]

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08 Oct 2009, 19:41
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I found the problem below extremely interesting and as this problem appeared in certain Hollywood movie, I'm not the only one who thinks so.

This problem is purely about the probability. So, would be great for those who preparing for GMAT to understand concept in it.

Suppose you're given the choice of three boxes. In one of them is the big prize (full-ride from Stanford/Harvard/Any you like); in the others, an apples. The prize and the apples were placed randomly in the boxes. The rules of the game are as follows: after you have chosen a box, the box remains closed for the time being. Then one of the two remaining boxes is opened, (by the computer which knows what is in each box) and the box contains an apple (as far as there are two boxes with apples it's always possible to open one no matter what is in the box you chose). After one box with apple is opened, you are asked to decide whether you want to stay with your first choice or to switch to the last remaining box.

A. What is the probability of winning the prize if you stay with your first choice?

B. What is the probability of winning the prize if you switch to the last remaining box?

The solution with the way of thinking welcomed.
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08 Oct 2009, 20:05
That's the famous 'Monty Hall Problem', though in the standard version of the problem, you're on a game show, and behind two of three doors are goats, and behind one is a car or some other prize. After you make your selection, the game show host, who knows what's behind each door, then opens one of the two remaining doors to reveal a goat, and asks if you want to change your selection:

en.wikipedia.org/wiki/Monty_Hall_problem

Of course, it's to your advantage to switch. One third of the time, you will have chosen the car first, and by switching you lose. Two thirds of the time, however, you've chosen a goat. The game show host then must open the remaining door hiding a goat - he can't open the door hiding the car - so when you switch, you get the car. So by switching, your probability of winning is 2/3.
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08 Oct 2009, 20:14
IanStewart wrote:
That's the famous 'Monty Hall Problem', though in the standard version of the problem, you're on a game show, and behind two of three doors are goats, and behind one is a car or some other prize. After you make your selection, the game show host, who knows what's behind each door, then opens one of the two remaining doors to reveal a goat, and asks if you want to change your selection:

en.wikipedia.org/wiki/Monty_Hall_problem

Of course, it's to your advantage to switch. One third of the time, you will have chosen the car first, and by switching you lose. Two thirds of the time, however, you've chosen a goat. The game show host then must open the remaining door hiding a goat - he can't open the door hiding the car - so when you switch, you get the car. So by switching, your probability of winning is 2/3.

Of course it's 'Monty Hall Problem', I didn't mention it because wanted to hear the way of thinking of people here.
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08 Oct 2009, 23:35
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Consider the following variation of the final round of the classic TV game show Let's Make A Deal:
There are three doors, and behind one of them is a car, while behind the other two are goats. If you choose the door with the car behind it, you win the car. Now, say you choose Door 1. The host Monty Hall then opens either Door 2 or Door 3, behind which is a goat. (He knows what is behind each door, and never opens the door with the car behind it.) Monty now gives you the choice: do you want to stick with Door 1, or switch to the other door. What should you do? Or does it matter?

A similar question is posed to Ben Campbell (played by Jim Sturgess) by Professor Micky Rosa (played by Kevin Spacey) in the movie "21". Without hesitation Ben answers this correctly, which convinces Professor Rosa that Ben would be a good addition to their "card counting team". Before reading on, try to answer this yourself.

One solves this problem by comparing the probability of choosing the car if you stick with your original choice to the probability of choosing the car if you switch after Monty opens the one door. Note that the car has an equal probability of 1/3 of being behind Door 1, Door 2, or Door 3.

First, suppose that your strategy is to stick with your original choice of Door 1. Then you only win if the car is behind Door 1, so that your probability of winning is 1/3.

Next, suppose that your strategy is to switch doors. We break this into three cases:

If the car is behind Door 1, Monty will open either Door 2 or Door 3 to reveal a goat. You switch to the other of Door 2 or Door 3, and in either case you switched to a door with a goat behind it (remember, the car is behind Door 1).

If the car is behind Door 2, Monty will open Door 3. This is because he always opens a door with a goat behind it, and he can't open Door 1 because that was your original choice. So the only door you can switch to is Door 2, which is the door with the car behind it. Ding! You win!

If the car is behind Door 3, Monty will open Door 2. This is because he always opens a door with a goat behind it, and he can't open Door 1 because that was your original choice. So the only door you can switch to is Door 3, which again is the door with the car behind it. Ding! You win!
So if your strategy is to switch doors, you win 2/3 = 1/3 + 1/3 of the time. (Remember, the probability is 1/3 that the car is behind any particular door.) Therefore, a better strategy is to switch doors - the calculated probabilities indicate that you are twice as likely to win if you do this! Ben's correct answer in the movie "21" indicates that he is a good person for "counting cards". Not only does it show that he is clever, but it also demonstrates that he realizes that it is best to go with the choice which maximizes your probability of winning. This realization is essential to the success of "counting cards" for Blackjack.

In 1990, a similar question appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (which comes in some Sunday newspapers). Marilyn gave the correct answer, but many readers (including mathematics professors) believed that this was incorrect. So don't feel too bad if you got it wrong when you answered it for yourself. But now you know!

Source : http://www.me.ucsb.edu/~moehlis/21.html
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28 Oct 2013, 17:08
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The Monty Hall Problem gets its name from the TV game show, “Let’s Make A Deal,” hosted by Monty Hall. The scenario is such: you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide “goats” (or some other such “non–prize”), or nothing at all. Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice. Here is the problem: Does it matter if you switch?
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29 Oct 2013, 01:25
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outlandish23miles wrote:
The Monty Hall Problem gets its name from the TV game show, “Let’s Make A Deal,” hosted by Monty Hall. The scenario is such: you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide “goats” (or some other such “non–prize”), or nothing at all. Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice. Here is the problem: Does it matter if you switch?

Merging similar topics.
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Re: Monty Hall Problem   [#permalink] 29 Oct 2013, 01:25
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# Suppose you're given the choice of three boxes. In one of

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