Def change. before it is a one in 3 chance. Then the host removes one. The one you choose still has a one in 3 chance. But now there are only 2 choices, so the other one has a 2 in 3 chance of being the one that has the money.

well, no. Then the prob would be zero. But the original question said

Dave wrote:

after you pick, the host shows you one of the curtains you didn't pick that had nothing behind it.

He shows you an empty choice. Therefore the one you pick still has a 1 in 3 prob. (you picked your choice before knowing which the host would take away).

love faith swing wrote:

...He shows you an empty choice. Therefore the one you pick still has a 1 in 3 prob.

No. Once he shows you that “empty choice,” you have information that you didn’t have before, so the probabilities change.

...(you picked your choice before knowing which the host would take away).

Precisely the point: You made your choice when the odds were 1 in 3. However, once you learned that one of the curtains you didn’t pick was an “empty choice,” you also learned that the million dollars must be behind either the curtain you chose or the remaining curtain, so the odds changed from 1 in 3 to 1 in 2.

Martin, changing odds from 1/3 to 1/2 requires that the event (the opportunity to pick a door out of a set of doors) is independent. However, since you are given information about one of the doors before picking a new one, and this information is based on the original problem, the second event is not independent of the first. That's why the gentleman in the article linked to above gave an explanation (the second explanation) focused on the probability of the complement being the choice that pays, or opposed to discussing the probability of the door you chose first to be payday.

An experiment would be nice

I took a quick glance at the references cited; I’ll have to read them more carefully before I can offer an opinion. For now, let me explain the way I approached the problem:

Dave wrote:

Your on a gameshow. The host gives you the option to choose one of three curtains. Behind one of the curtains is a million dollars, the other two have nothing.

The way the problem is stated, initially you are presented with three equally likely configurations:

1) The cash is behind Curtain A; nothing is behind Curtain B or Curtain C.
2) The cash is behind Curtain B; nothing is behind Curtain A or Curtain C.
3) The cash is behind Curtain C; nothing is behind Curtain A or Curtain B.

So you decide to pick one of the curtains, after you pick, the host shows you one of the curtains you didn't pick that had nothing behind it. Now the host gives you the option to change your answer.

You choose a curtain—say, Curtain A. The host then chooses another curtain (in this case, B or C), and then reveals that the cash is not behind that curtain. Now you have the choice of keeping Curtain A, or switching to the remaining curtain (C or B, respectively). Let’s look at the possibilities:

The Host Chooses Curtain B.
This leaves you with two configuration possibilities:
1-B) The cash is behind Curtain A (the curtain you had chosen); nothing is behind Curtain C.
2-B) N/A
3-B) The cash is behind Curtain C; nothing is behind Curtain A (the curtain you had chosen).

The Host Chooses Curtain C.
This leaves you with two configuration possibilities:
1-C) The cash is behind Curtain A (the curtain you had chosen); nothing is behind Curtain B.
2-C) The cash is behind Curtain B; nothing is behind Curtain A (the curtain you had chosen).
3-C) N/A

What do you do and why?

Out of the four configuration possibilities 1-B), 3-B), 1-C), and 2-C), you will win if you do not change your selection in 1-B) and 1-C); and you will win if you do change your selection in 3-B) and 2-C). Hence, of the four configuration possibilities you could be faced with, a “Do Not Change Your Selection” strategy works for 50% of them, while a “Do Change Your Selection” strategy works for the other 50% of them. Each strategy, thus, is equally likely to be a winning strategy.

I’ll take a more careful look at the references cited, however, to see if I’ve made an error.

I did the experiment.

100 times I switched and found the money 70 times, or 70%

100 times I stayed and found the money 32 times or 32%

This is very close to the 2/3 and 1/3 probabilities predicted. I'm convinced.

I would immediately recognise the monty hall desicion problem, and advise her to switch.

The other possibility would be that the first door was open from the beginning, even before She picked A. Then the odds would be 50/50.

so the odds could be
A-33.33
B-66.66

or

A - 50
B - 50.

So if she switches, she either gains in chance of winning or stays the same. It would be best if she switches.

The question remains: You are offered two curtains (doors, boxes, rooms,...whatever). Which do you choose?

Chris wrote:

In the situation you described, I would tell Ms. Smith to switch.

If the host took away one door before I came in and he knew there was no prize behind it, then the odds are 2/3 on the switch, 1/3 on staying put.

If there just happened to be an open door and Ms. Smith was only given two options to begin with, then the odds or 1/2 on the switch, 1/2 on staying put. So it doesn't matter.

Nicely done!

So...specifically where is the error in the analysis (of the original problem) that I presented above?

Chris wrote:

You choose a curtain—say, Curtain A. The host then chooses another curtain (in this case, B or C), and then reveals that the cash is not behind that curtain.

You give the host the option of choosing either B or C when he/she may not have such a choice.

Huh? I did not say that the host can choose Curtain B or Curtain C without consideration of what is behind Curtain A. I simply said that if the contestant selects Curtain A (whether or not the money is behind Curtain A), then the host picks one of the two remaining curtains. At least one (possibly both) of those remaining curtains will be “empty.”

Chris wrote:

If the door you picked, A, had the money, then the host can choose between B or C. ...

Yes...and those are possibilities 1-B) and 1-C) of the four possibilities that I listed.

Now c'mon.  This does not take a physicist to figure out.  No matter which door you initially pick, the host will have a door to open.  But it's all about perspectives.  If you knew that the host was going to open one door, the whole time you would have a 1/2 chance of winning the cash because you would know that eventually you will have a choice of two.  Like Chris said, "Look from the host's perspective."  The whole time the host knows that in the end he will have a 1/2 chance of giving the prize away.  (Yet more) Now when you have 3 doors to choose from, obviously you have a 1/3 chance of winning until you find out that you have only two doors to choose from.  Then you have a 1/2 chance, just as if you had started from the beginning with two doors.  Because you know that one of the two has the prize.  I don't understand the complication.  If I am missing anything please explain in small terms.