3 Door Logic Problem

[Mid]

Well I don't care what anyone says I wouldn't change my choice of doors in this game, but then again I want the goat

 

[Lamuella]

I just re-read the initial post in this thread, and I do not see any statement that could be interpreted to indicate that the game show host ALWAYS opens a door.

This is your assumption, and it is necessary in order to answer the question. It may or may not be true, but the question itself does not provide the information.

This puzzle is based on the real life gameshow "Let's make a deal". THis isn't a hypothetical puzzle, but instead one that took place on nationally broadcast television. In that gameshow, the host ALWAYS opened a door. This isn't an assumption.

 

[StewartP]

In the French version of "Deal or No Deal" (called A Prendre ou a Lesse) the banker knows the contents of the boxes.

This makes the show more diabolical because the player know has to guess if the banker is bluffing, double bluffing or not.

The French version predates UK version by many years. In fact the reason there are 22 boxes is to represent the 22 regions of France.

 

[pgwenthold]

This puzzle is based on the real life gameshow "Let's make a deal". THis isn't a hypothetical puzzle, but instead one that took place on nationally broadcast television. In that gameshow, the host ALWAYS opened a door. This isn't an assumption.

You are making the assumption that it always behaves exactly like "Let's Make a Deal."

Let's re-read the initial statement of the problem:

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

 

[tsg]

This puzzle is based on the real life gameshow "Let's make a deal". THis isn't a hypothetical puzzle, but instead one that took place on nationally broadcast television. In that gameshow, the host ALWAYS opened a door. This isn't an assumption.

Do you have a cite for this? Because, while I know the problem is based on the gameshow, I don't believe the initial assumptions are necessarily met. I also seem to remember Monty Hall being quoted as saying that he had the option of opening another door or not, but I would have to verify that.

 

[Overman]

Wait Wait Wait...

I was just conversing with someone and they put it like this, which makes sense to me for some reason...

If the host always shows you a goat, you have 2 doors left...

Door 1 and Door 2.

Here are your scenarios:

1) you pick door 1 and you switch and the car was behind door 1. you lose.
2) you pick door 1 and you switch and the car was behind door 2. you win.
3) you pick door 1 and don't switch and the car was behind door 1. you win.
4) you pick door 1 and don't switch and the car was behind door 2. you lose.
5) you pick door 2 and you switch and the car is behind door 1. you win.
6) you pick door 2 and you switch and the car is behind door 2. you lose.
7) you pick door 2 and don't switch and the car was behind door 2. you win.
8) you pick door 2 and don't switch and the car was behind door 1. you lose.

The above appear to be the only possibilities....?

 

[tsg]

The above appear to be the only possibilities....?

Be careful. They aren't all equally likely.

 

[Beleth]

Wait Wait Wait...

I was just conversing with someone and they put it like this, which makes sense to me for some reason...

If the host always shows you a goat, you have 2 doors left...

Door 1 and Door 2.

Here are your scenarios:

1) you pick door 1 and you switch and the car was behind door 1. you lose.
2) you pick door 1 and you switch and the car was behind door 2. you win.
3) you pick door 1 and don't switch and the car was behind door 1. you win.
4) you pick door 1 and don't switch and the car was behind door 2. you lose.
5) you pick door 2 and you switch and the car is behind door 1. you win.
6) you pick door 2 and you switch and the car is behind door 2. you lose.
7) you pick door 2 and don't switch and the car was behind door 2. you win.
8) you pick door 2 and don't switch and the car was behind door 1. you lose.

The above appear to be the only possibilities....?

That's not the same problem. He is unnecessarily renumbering the doors and thus confusing the issue.

You pick  Prize behind  He shows  You switch  W/L
1         1             2 or 3    N           W
1         1             2 or 3    Y           L
1         2               3       N           L
1         2               3       Y           W
1         3               2       N           L
1         3               2       Y           W

Each of those six scenarios are equally likely. Notice that of the three cases where you don't switch, you only win once, and you lose twice. Likewise, you win two of the three cases when you switch.

The snag a lot of people get caught on is not assigning the proper probabilities to each of the cases, and thinking the case where the prize is in 1, you pick 1, and the host shows 2 is as probable as the case where the prize is in 3, you pick 1, and the host shows 2... when in fact the former is half as likely as the latter.

 

[Hellbound]

Wait Wait Wait...

I was just conversing with someone and they put it like this, which makes sense to me for some reason...

If the host always shows you a goat, you have 2 doors left...

Door 1 and Door 2.

Here are your scenarios:

1) you pick door 1 and you switch and the car was behind door 1. you lose.
2) you pick door 1 and you switch and the car was behind door 2. you win.
3) you pick door 1 and don't switch and the car was behind door 1. you win.
4) you pick door 1 and don't switch and the car was behind door 2. you lose.
5) you pick door 2 and you switch and the car is behind door 1. you win.
6) you pick door 2 and you switch and the car is behind door 2. you lose.
7) you pick door 2 and don't switch and the car was behind door 2. you win.
8) you pick door 2 and don't switch and the car was behind door 1. you lose.

The above appear to be the only possibilities....?

Only because you are starting the problem from the point where the third door is already revealed. At that point, you make your choice and you have a 1 in 2.

However, you could have picked door number three, and the host opened one of the others.

So, let's take an example with doors A, B, and C. The car is behind door C.

Initially, each door has a 1/3rd chance of being picked. We have to incorporate that into our calculation. Let's take the not switching strategy first, and look at each 1/3rd option:

1/3rd of the time you pick door A. The host opens door B. You don't switch, so you lose.

1/3rd of the time you pick door B. The host opens door A. You don't switch, so you lose.

1/3rd of the time you pick door C. The host opens either door A or B. You don't switch, so you win.

Add those up, and 1/3rd of the time you win.

Now, let's examine the switch strategy:

1/3rd of the time you pick door A. The host opens door B. You switch, so you win.

1/3rd of the time you pick door B. THe host opens door A. You switch, so you win.

1/3rd of the time you pick door C. The host opens door A or B. You switch, so you lose.

That's a win 2/3rds of the time.

The big problem here is that you are attempting to treat the host's opening of a door as a random event. IT IS NOT RANDOM. The host's choice of what door to open is dependent on your initial choice of door. IN my example, you can choose door A, B, or C. If you choose A, the host has a 100% chance to open B. If you choose B, the host has a 100% chance of opening A. Where the confusion comes in, is people calculate the odds wrong on choice C. If you choose C, the host has a 50% chance of opening door A and a 50% chance of opening door B. People calculate it, however, as if each of these is another 100% chance (they are weighted equally to the options in your initial choices of A and B), so it throws off the calculation by incorrectly increasing the denominator [2/(1+1+1+1) instead of 2/(1+1+.5+.5)].

I had to sit and figure this myself, as I started a different explanation that was wrong. It's the incorrect weighting of the host's choice that causes the confusion here.

 

[Lord Muck oGentry]

2 times in 3, you will have chosen a wrong door and have been shown the other wrong door ( Monty having no choice).
1 time in 6, you will have chosen the right door (1/3) and have been shown one of the wrong doors (1/2).
1 time in 6, you will have chosen the right door (1/3) and have been shown the other wrong door (1/2).

 

[Lucky]

First, we have to interpret the scenario in the way tsg stated:

1. The host knows where the prize is.
2. The host will always offer you a choice.
3. The host will always open another door and never show you the prize.
4. The contestant is aware of 1-3.

It’s completely pointless to discuss the maths of the situation unless we’re agreed this is the scenario, and Marilyn’s failure to make sure it was clearly stated was responsible for some of the confusion.

It can be quite difficult to explain logic problems that seem obvious, as you need to understand the other person’s incorrect point of view, but I have sometimes managed it by a two-step process.

First you have to explain the cast-iron correct way of looking at the problem. In this case it’s simple enough: Each door, including yours, had a probability of 1/3 before the host opened one of them. By the agreed scenario, your door still has a probability of 1/3 (because he was always going to open one of the other two), so the other door now has a probability of 2/3 (it has acquired the probability previously belonging to the opened door). The host’s choice can’t alter the probability of your door, as it gives no extra information about it, but it alters the probability of the one he chose not to open, as you have precisely that extra information about it: he chose not to open it.

Next, you have to explain independently why any contradictory way of looking at the problem is wrong. Like this one.

Here are your scenarios:

1) you pick door 1 and you switch and the car was behind door 1. you lose.
2) you pick door 1 and you switch and the car was behind door 2. you win.
3) you pick door 1 and don't switch and the car was behind door 1. you win.
4) you pick door 1 and don't switch and the car was behind door 2. you lose.
5) you pick door 2 and you switch and the car is behind door 1. you win.
6) you pick door 2 and you switch and the car is behind door 2. you lose.
7) you pick door 2 and don't switch and the car was behind door 2. you win.
8) you pick door 2 and don't switch and the car was behind door 1. you lose.

As tsg says, these are indeed the possibilities, but some are more probable than others. But it’s not so obvious why. The point is that the door you pick (whether 1 or 2) has a probability of 1/3 (by the cast-iron argument above), and the other door has a probability of 2/3. Before you decide whether to switch (or after you decide and before the answer is revealed), it is twice as probable that the other door is the right one. If you pick door 1, it's twice as probable that the answer is door 2 (for this reason, this is not a valid way of listing the possibilities: the probability of each door depends on your initial choice). So, for example, outcome 2) is twice as probable as 1), 4) is twice as probable as 3).

 

[Lamuella]

You are making the assumption that it always behaves exactly like "Let's Make a Deal."

Let's re-read the initial statement of the problem:

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

so despite the fact that this very well known logic puzzle is based on a specific and fixed set of rules (and is even CALLED the Monty Hall puzzle by most people), you're going to go with someone's incomplete phrasing of the puzzle?

The only assumption I'm making is that the Monty Hall puzzle is the Monty Hall puzzle.

Can someone find Articulett (the original source of the quote) and get him to clarify, please?

 

[Beleth]

You can think of it this way too:

The host showing you an empty box isn't really giving you any new information. You already knew that one of the boxes is empty. So what he's really doing is giving you a choice between the box you chose, and both of the boxes you didn't choose.

This becomes even clearer if you extend the number of boxes to more than three. Say there's a hundred boxes. You choose one. You know that at least 98 of the other 99 are empty. The host then says "You can have the contents of all of these other 99 boxes, 98 of which you know are empty, or you can keep the one box you chose."

 

[Ashles]

I love this problem.

It's so counterintuitive, and, although it involves fairly simple probability, it's darn hard to explain.

The crux is that two thirds of the time you chose the wrong door to start with so the host has to reveal the other wrong door.

So if you work from the assumption that you chose the wrong door to start with then you know exactly which door is the right one.
And you'll be correct 2/3 times (in theory).

 

[Lamuella]

Wait Wait Wait...

I was just conversing with someone and they put it like this, which makes sense to me for some reason...

If the host always shows you a goat, you have 2 doors left...

Door 1 and Door 2.

Here are your scenarios:

1) you pick door 1 and you switch and the car was behind door 1. you lose.
2) you pick door 1 and you switch and the car was behind door 2. you win.
3) you pick door 1 and don't switch and the car was behind door 1. you win.
4) you pick door 1 and don't switch and the car was behind door 2. you lose.
5) you pick door 2 and you switch and the car is behind door 1. you win.
6) you pick door 2 and you switch and the car is behind door 2. you lose.
7) you pick door 2 and don't switch and the car was behind door 2. you win.
8) you pick door 2 and don't switch and the car was behind door 1. you lose.

The above appear to be the only possibilities....?

Think of it like this:

What are the chances of you picking one door out of three and getting the prize?

1/3, correct?

Why would those odds change just because you have been told that one of the doors you didn't pick has a goat behind it? You already know that at least one of them does.

Thus, if you stick where you are, you have a one out of three chance of winning.

Can you accept that?

alternatively, let's rephrase this way:

You are on a gameshow. You have to pick one of three doors, there is a prize behind one of them. After you have picked, you are offered the chance to swap what is behind your door for what is behind both of the other doors. Would you swap?

 

[Mr. Scott]

Monty Hall Problem, a "Stupid Test" for Intellectuals

It ALWAYS pays to switch.

Well, you should word it "your chance of winning is ALWAYS better if you switch," because, being a game of chance, nothing you do will "always pay."

I'm not sure where Marylin found this problem but she popularized it and became notorious for it circa 1990 because so many intelligent people got the answer wrong. The controversy reached the front page of the New York Times IIRC. It's what I've heard is called a "stupid test," which is a question you can ask someone for which they are likely to give the wrong answer and look stupid. A grown-up version of "pull my finger." I personally know a nobel laureate in physics who got it wrong.

One argument I always hear is "if there are two choices (after one door is opened) then it's 50/50 right? No, because the probabilities are not even. It's now loaded dice. Opening one of the goat doors gives away information, but the information is so faint it flies under our intuitive radar. The reason so many get it wrong is they go by their intuition instead of by rational analysis, and it demonstrates how faulty intuition can be.

Marylin later offered a problem which is similar in its infuriating counter-intuitiveness, presented at this web site as the coin puzzle.

 

[Lamuella]

Well, you should word it "your chance of winning is ALWAYS better if you switch," because, being a game of chance, nothing you do will "always pay."

True. It doens't always pay. However it's always more likely to pay.

 

[tsg]

Marylin later offered a problem which is similar in its infuriating counter-intuitiveness, presented at this web site as the coin puzzle.

Interesting puzzle. Marylin is right. Roger is commiting the same error as the Monty Hall problem by assuming that two heads and one head-one tail are equally likely.

 

[Lamuella]

agreed. Marilyn is right on that one.

There are 4 possible results for two tossed coins:

HH HT TH TT

Person A has ruled out TH and TT
Person B has only ruled out TT

 

[tkingdoll]

I'm not sure where Marylin found this problem but she popularized it and became notorious for it circa 1990 because so many intelligent people got the answer wrong.

It was from a reader of her column, Ask Marilyn. He wrote in with the question and her answer prompted a storm of replies and eventually the publicity you describe.