Weighing The Odds In HE Question: Monty Hall

[PokerAmateur4]

Under Extra Topics Yao writes of the Monty Hall problem. He writes:
"If you know in advance that Monty will always open up a door and show you a goat after you have already chosen a door, and then offer to let you switch doors, then the answer is yes, you should switch"
I don't get it, why should you switch?

[A_F]

One way to think about this problem is to consider the sample space, which Monty alters by opening one of the doors that has a goat behind it. In doing so, he effectively removes one of the two losing doors from the sample space.
We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:


1. The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.

2. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

3. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors. When she switches, she wins the car twice (the number of favorable outcomes) out of three possible options (the sample space). Thus the probability of winning the car is 2/3 if she switches doors, which means that she should always switch doors - unless she wants to become a goatherd.

This result of 2/3 may seem counterintuitive to many of us because we may believe that the probability of winning the car should be 1/2 once Monty has shown that the car is not behind door A or door B. Many people reason that since there are two doors left, one of which must conceal the car, the probability of winning must be 1/2. This would mean that switching doors would not make a difference. As we've shown above through the three different options, however, this is not the case.

[PokerAmateur4]

Thanks, that was very clear. Yao probably went through this too but I stopped where the quote was, as the counterintuitive nature tripped me up.

[kitaristi0]

Google is your friend.

[BigF]

[ QUOTE ]
Under Extra Topics Yao writes of the Monty Hall problem. He writes:
"If you know in advance that Monty will always open up a door and show you a goat after you have already chosen a door, and then offer to let you switch doors, then the answer is yes, you should switch"
I don't get it, why should you switch?

[/ QUOTE ]

I found a similar version of the teaser a little more interesting.

Suppose Monty doesn't know what's behind the doors and he opens a door showing you a goat, should you switch?

[disjunction]

I look at the Probability forum about once every couple of months. The Monty Hall problem is always on one of the first pages, every time I look. LOL

[Lawman]

[ QUOTE ]
[ QUOTE ]
Under Extra Topics Yao writes of the Monty Hall problem. He writes:
"If you know in advance that Monty will always open up a door and show you a goat after you have already chosen a door, and then offer to let you switch doors, then the answer is yes, you should switch"
I don't get it, why should you switch?

[/ QUOTE ]

I found a similar version of the teaser a little more interesting.

Suppose Monty doesn't know what's behind the doors and he opens a door showing you a goat, should you switch?

[/ QUOTE ]

How is this any different to the original problem? Once he has opened a door with a goat behind it his intention becomes irrelevant. The set of outcomes is exactly the same - the only way it would have been different would be if he had opened the door with a car behind it.

Or am I a retard?

[WhiteWolf]

[ QUOTE ]
[ QUOTE ]
Under Extra Topics Yao writes of the Monty Hall problem. He writes:
"If you know in advance that Monty will always open up a door and show you a goat after you have already chosen a door, and then offer to let you switch doors, then the answer is yes, you should switch"
I don't get it, why should you switch?

[/ QUOTE ]

I found a similar version of the teaser a little more interesting.

Suppose Monty doesn't know what's behind the doors and he opens a door showing you a goat, should you switch?

[/ QUOTE ]
This doesn't seem as interesting as the original version. If Monty randomly picks a door and reveals a goat, it doesn't matter if you switch or not. The intuitive answer matches the correct one in your version.

[scolapasta]

The difference is that he COULD have opened the door with the car; just because he didn't doesn't mean that it wasn't possible.

So in this case, switching is the same as not swiitching, ie 50/50 at getting the car.

[WhiteWolf]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Under Extra Topics Yao writes of the Monty Hall problem. He writes:
"If you know in advance that Monty will always open up a door and show you a goat after you have already chosen a door, and then offer to let you switch doors, then the answer is yes, you should switch"
I don't get it, why should you switch?

[/ QUOTE ]

I found a similar version of the teaser a little more interesting.

Suppose Monty doesn't know what's behind the doors and he opens a door showing you a goat, should you switch?

[/ QUOTE ]

How is this any different to the original problem? Once he has opened a door with a goat behind it his intention becomes irrelevant. The set of outcomes is exactly the same - the only way it would have been different would be if he had opened the door with a car behind it.

Or am I a retard?

[/ QUOTE ]
Hmmm, I guess the question is more interesting than I thought. I'll use A_F's approach of using the sample space to show how this is different from the original approach.

Once again, the goats are behind doors A and B, and the car is behind C. The total sample space is larger, because Monty now opens a door at random, instead of always chosing a goat. Before we start, the possible outcomes (all equally likely) are:

A1) You chose A, Monty opens B and shows a goat. Switching makes sense here.

A2) You chose A, Monty opens C and shows a car. Since the problem specifies that this did not occur, we throw out this case.

B1) You chose B, Monty opens A and shows a goat. We should switch.

B2) You chose B, Monty opens C and shows a car. Once again, this did not occur, so we can throw out this case.

C1) You chose C, Monty opens A and shows a goat. We should not switch.

C2) You chose C, Monty opens B and shows a goat. We should not switch.

Before we begin the contest, all of the above cases are equally likely. After Monty shows a goat, we now know A2 and B2 did not occur, so we are left with 2 cases where we should switch (A1 and B1), and 2 cases where we should not (C1 and C2). Since they are equally likely to occur, we are indifferrent between switching and not switching.

This differs from the original formulation because there we know, before the game starts, that A2 and B2 can never happen, and this boasts the odds of A1 and B2 happening. In the original formulation, A1 happens 1/3 of the time, B1 happens 1/3 of the time, C1 happens 1/6 of the time, and C2 happens 1/6 of the time. In this case, switching is correct 2/3 of the time, and not switching is correct 1/3 of the time, so odds dictate we switch.