http://www.youtube.com/watch?v=mhlc7peGlGg

I've been reading The Curious Incident of the Dog in the Night-Time by Mark Haddon and I heard of this nifty little thing for the first time.

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 is the probability of winning the car if she stays with her first choice? What if she decides to switch?

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.

One way to convince yourself that 2/3 is the correct probability is to do a simulation with a friend. Have your friend impersonate Monty Hall and you be the contestant. Keep track of how often you win the car by switching doors and by not switching doors.

Source: http://mathforum.org/dr/math/faq/faq.monty.hall.html

Crazy stuff, huh?

i didnt believe it at first, but after thinking aboujt it it's obviously true

Odds can change based upon futher information becoming revealed.  For those who get confused by the fact the odds can apparently change, consider this:

Let's say there were 100 doors to pick.  One has the prize, and you pick one.  Then they reveal all the other doors not picked but one.  Do you switch to the other unopened door?  Think of it that way and it should be clear the answer is yes.  That last door has a 99% chance of being the one door with the prize behind it, while your initial pick has a 1% chance.  The odds did change, based upon further information, for the other door not picked.  The one you picked does remain constant.

They key is that, out of the doors you haven't picked, they always open a door that doesn't have the prize. So opening it doesn't change the odds of it being behind {set of all doors you didn't choose}.

When you reduce the set of doors not picked by you to one (show the rest as mot having the prize), the odds of remaining door that has not been selected by you equals the odds of all the other doors you didn't pick.

Yeah, this was in that one movie about those college kids going to Vegas and winning a lot. Don't remember the name.

It makes logical sense, the 2/3 probability, if you consider the whole thing as one problem. But I'm not really convinced.

The way I see it is as follows:

Three doors; a car and two goats.

You choose a door. Doesn't matter which.

Host reveals a goat.

END OF PROBLEM A.

Two doors; a car and a goat (what I'm saying is that this is an entirely different puzzle. By opening a door and revealing a goat, the host has changed the situation, and the choices you can make).

You are now choosing between the car and the goat. It's a 50:50 chance.

Hm, I read the whole thing, but I don't think their explanation of probability holds much in the actual situation. The person opening the doors knows where the car is, there is absolutely no probability at work. In the end you are left to choose between two doors, and you weren't left with these two doors due to sheer chance.

It doesn't really matter how many doors, neither does your initial choice at all, so 50:50.

edit: I see the 2/3 now, pretty clever indeed

Hi, first let me start by reintroducing myself. I'm the poster formerly known as LuStaysTru. Lost my password.....I came back just for this lol

Well anyways, I believe this to be what is called the "Gambler's Fallcy". The odds in fact do not change.

From wiki:

Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only ¹⁄₃₂ (one in thirty-two), a believer in the gambler's fallacy might believe that this next flip is less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler's fallacy; the event of 5 heads in a row and the event of "first 4 heads, then a tails" are equally likely, each having probability ¹⁄₃₂. Given the first four rolls turn up heads, the probability that the next toss is a head is in fact,

While a run of five heads is only ¹⁄₃₂ = 0.03125, it is only that before the coin is first tossed. After the first four tosses the results are no longer unknown, so their probabilities are 1. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses, that a run of luck in the past somehow influences the odds in the future, is the fallacy.

Ending problem A would mean no information from it is carried forward. But knowing what door is "yours" is information. If you completely forgot which door you chose first and were now faced with two doors, you be 50-50.

You can apply this situation over different points of view and get the same result: Let's say you forget all about "problem A" and have to choose between the two doors. If someone watching it all still knows about "problem A" and hopes you'll win, he'll hope you'll select the "other" door, for that, given the information HE has is got 2/3 of chances of winning. Now, he knows you don't know anything about either door, so he knows it's 50-50 that you'll choose any door. Thus 1/2 * 1/3 1/2 * 2/3 = 1/2, making it 50-50. You can reach the same resut trhough presenter's eyes.

But you DO carry information from "problem A" onward. You know which door you chose and you know how the game works. So you can't just pretend that never happened.