The Ultimate Brainteaser [edited]

[qrac]

here's a brainteaser that some swedes didn't like the answer to:

Imagine that the set of Monty Hall's game show Let's Make a Deal has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but Monty Hall does.

The contestant picks a door and Monty opens one of the remaining doors, one he knows doesn't hide the car. If the contestant has already chosen the correct door, Monty is equally likely to open either of the two remaining doors.

After Monty has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. What should the contestant do and why?

[bobafett]

The time it took to scroll down and read the message

--Dan

[Rave E55]

That was dumb.

[ycchan]

[qrac]

yeah.. it was kind of dumb.. but i couldn't resist.. i had studied math all day and then came to the board and saw all these brainteasers.. it just became too much to me

but it's edited now

[kees]

Well...dont hold us in suspense...what should he do?

[chazzy]

Monty Hall's Paradox:
http://math.ucsd.edu/~crypto/Monty/montybg.html

[JFB]

Originally posted by chazzy
Monty Hall's Paradox:
http://math.ucsd.edu/~crypto/Monty/montybg.html

Those UCSD math department folks are darned smart!

Intuitively, however, the math department's explanation seems to over analyze the situation based on the initial set of three variables at the first decision point. To simplify, ignore the door numbers. There are three possibilities, but only two outcomes True or False:
T = Car
F= Goat
F= Goat

At the first decision point the contestant might choose any of the three doors (T-F-F) representing two possible outcomes. But the contestant cannot win in the first round.

Regardless of which door (T-F outcome) the contestant initially picks, the outcome of the first round will always be the same. Monty knows which door conceals the car (T), so he always eliminates one of the goats (F). Therefore, which door the contestant chooses in the first round is irrelevant.

In fact, the existence of the first round is irrelevant. This is true because the same two variables will always exist at the second decision point. The fact that there were 3 doors to begin with is moot, because Monty always eliminates one goat (F) choice to start the second round. Invariably one door conceals a car (T) and one a goat (F) at the second decision point.

Because the outcome of the first round is always the same, the contestant does not know at the second decision point whether his initial choice conceals the goat or the car. One of the two remaining doors conceals the car (T) and the other conceals the goat (F). There is no reason for the contestant to assume that he guessed wrongly on his first attempt -- which would be a prerequisite to changing his mind. There is equally no reason for him to assume he guessed correctly. He always has a new decision to make.

At that second decision point, therefore, the odds are 50/50 that either door contains the car. The contestant is essentially starting the game over with two doors.

What am I missing here?

(UCSD class of 1979, albeit not a mathematics or statistics major.)

[MrMan]

i remember this. If i remember correctly, there is no sure answer... this is just a probability that is supposedly 50%?

[qrac]

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:


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.

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.

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.

If it's hard to get it:
What if there were 1,000 doors? You would have a 1/1,000 chance of picking the correct door. If Monty opens 998 doors, all of them with goats behind them, the door that you chose first will still have a 1/1,000 chance of being the one that conceals the car, but the other remaining door will have a 999/1,000 probability of being the door that is concealing the car. Here switching sounds like a pretty good idea.

[qrac]

http://math.ucsd.edu/~crypto/Monty/monty.html

there's a game you can try on the site chazzy posted.

[Ben_FR]

This is a very easy problem. You always switch. The opening of the door is valuable information.

On your first pick, the odds of being correct are 1/3. Let's see the odds if you don't switch:

(1/3)(1) + (2/3)(0) = 1/3

If you do switch, 2/3 of the time you will be guaranteed winning.

(1/3)(0) + (2/3)(1) = 2/3

-Ben FR