Understanding the Monty Hall Problem

This is how I thought about it: imagine that door A is the one you pick. There are 3 equally likely scenarios. 1: Door A is correct. In this case, you would be better off staying. 2: Door B is correct. Monty would open C and you would be better off switching. 3: Door C is correct, Monty would open B and you would be better off switching. Two out of three times, you are better off switching.

I like the way you encourage visualizing the numbers with both shape and color. Great stuff.

Or you could you know figure out the really simple system used on this and get about 70% with out all the close examining you just made me do

The combinations of your first choice are:

That’s a 1/3 chance of winning the Car – as you would expect.

After you make your first choice the combinations behind the remaining two doors are:
Goat(1) & Goat(2)
Car & Goat(2)
Car & Goat(1)

If Monty Always picks a Goat then the remaining door contains:
Goat (1 or 2)

Hence if you switch there is a 2/3 chance of picking the Car.

Initially we make a random choice with a 33% probability that the car is behind the door. The other two doors contain 66% of the probability. Once the goat is revealed the 66% probability is condensed into a single door. No extra probability is handed to the first door until a new random choice is made. Thus if no new random choice is made the original choice maintains its 33% probability but the other door now has all of the 66% probability of its original set. That’s why switching to the other door wins 66% of the time. If a new random choice is made of those two remaining doors, then it’s truly a 50% chance.

Everything hinges on making a random choice. Those that argue it is a 50% chance problem after the goat is revealed fail to see where the original probability of 33% came from. It came from making a random choice. If there was no random choice initially then there’d be no 33% chance. If the contestant was told what to pick, they’d claim their chances were not even. So after the goat is revealed, the only way to have a 50% chance is to flip a coin and make another random from the two remaining doors. But since 66% chance of a win was held in the other two doors and condensed from those 2 doors into 1, it makes sense to just pick that other door that has the 66% chance and NOT make any more random choices.

Why are both doors examined before they are opened? I find that part confusing. So did he open both doors or only one?

We humans have a habit of being acquisitive. Once we choose something, we have a tendency to want to hang on to it. The act of choosing a door, makes it ‘our door’ and so there is a natural reluctance to part with it.

An easier way of intuitively understanding the Monty Hall game is to hold off selection until the ‘door’ has been opened, and to show the higher probability of switching choice.

Take a pack of cards - it contains a single Ace of Diamonds - this is the winning card - pick the Ace of diamonds and you win.

Shuffle the pack and take a single card out - lay it on the table. Put the rest of the pack on the table beside the single card.

Now, the choice is pick the single card or pick the rest of the pack - which do you thinks is most likely to give you the Ace of Diamonds. Clearly, most people would choose the pack with 51 chances that they have chosen the Ace.

Now we add in the ‘opening of the doors’

As before, pick a card and lay it on the table, but before putting the pack down, somebody sorts through the pack and if it contains the Ace of Diamonds, they put it on the top of the pack and then put the pack on the table.

Now you have the choice of taking the single card or the top card off the pack - again, automatically folks will take the top card from the pack.

Now we put it all together - ‘Choose’ a card and put it on the table (we now own it). Next the pack is sorted and if the Ace is present it is put on the top of the pack.

The top card is taken off the pack and placed beside your chosen card and the rest of the pack are turned face up to show that they are not the Ace of Diamonds.

Now you are offered the choice of changing the card you chose for the remaining card from the pack. Even though ‘ownership’ is now involved, most people would not have any doubt about changing, and understand that there is not the remotest chance that the first card has a 50:50 chance of being the Ace.

they will understand that they had a one in 52 chance of picking the Ace first time, and almost dead certainty of it being the other card - they would switch…

Ya right, i got 5 wins 5 losses both ways, 50 50. your strategy fails.

I’ve read a bunch of articles about this, but the multiple doors made it click and made me think of it in a way no one has mentioned.

The reality is that 66.7% of the time, I have picked a goat. This means that 66.7% of the time, Monty still has the car and the other goat. So, 66.7% of the time, Monty is revealing the 2nd goat I did not pick.

Another way that came to me is that there are only 2 versions of this game.

  1. I have picked the car and Monty is showing me one of the two goats.
    How to win: stay with my first choice because it is the car.
  2. I have picked a goat and Monty is showing me the other goat that I did not pick.
    How to win: change my choice because the car is behind the 3rd door.

I don’t know which version of the game I’m playing, but I DO know that “version 1” only happens 33.3% of the time, while “version 2” happens 66.7% of the time. Since this is true, I am going to use the winning strategy that applies to 66.7% of the games I play. I will win about 66.7% of the time.

I learned about this a while ago but never tested it until today. I tried picking a door and switching 20 times twice. Each time the score was 9 wins and 11 losses. I’ll try one more time.

One sixty seven percent win. One 50/50. It doesn’t work. The reason is because if it is predetermined that a goat will be revealed from one of the doors you didn’t pick, then you are starting out with a 50/50 chance not a 1/3 chance. No matter what you picked, it’s a 50% chance of being a goat or a car because the door you didn’t pick but will be revealed will automatically have a goat in it or be irrelevant.

Think of it like this. You have a coin which has three possible landings: heads, tails, and edge. Let’s say that landing on its edge will count it as tails and you want heads. It will never land on it’s edge, whether you know it or not, so that decision automatically starts out irrelevant. Your choice is either heads or tails because no matter what, one option will be nonexistent.

To restate it in your terms: There will be a door you will never pick, but will always be a goat. Let’s call him goat 1. There is one goat you will never choose. You will either have a goat 2 or a car.

Suppose Monty didn’t examine doors B&C first. Suppose he just opened one at random, and it just happened to have a goat behind it. (In other words, there was a chance he could have opened the car door, thereby spoiling the game.)

In that case, is it now 50:50 for us when choosing to switch or not?

@Frank: It would be no new information if Monty randomly revealed a door. But, Monty is purposefully filtering the other side, and providing the information that “The door that remains is the best of these two doors”. So, you have the choice of 1) your original door or 2) the best of the other side.

Monty provides information by picking the “winner” on the other side for you.

@Lizzy: Cool viewpoint, thanks for sharing!

@rajanKazhmin: Glad you enjoyed it.

@Domogo: I’m not sure I understand…

@Edmund: Thanks for the breakdown! I love seeing how everyone approaches this problem.

@Mike: That’s exactly it – the 66% is “collapsed” into a single door after Monty’s filter. A fresh contestant, seeing two doors, has a 50% chance. But since you saw the process, you know which door went through the filter.

@Sludgie: Good feedback, I should clarify. Monty looks behind both doors, but only fully opens one for you to see. At this point he knows where the car is since he’s seen two of the three doors and can infer the third.

There’s no reason for him to do so (in the game, he’s helping you) but it makes the question interesting :).

@DerekSmith: Great analysis into the psychology of it! I like the example too.

@Ace: As I write for @sweet…, try doing more trials, like 50… you’ll see a pattern emerge.

If you have a fair coin but only flip it twice, you could get two tails and think it was biased.

@SteaveG: Awesome, thanks for the explanation!

@sweetestsadist: You bring up a good point – how many trials do we need to be convinced of something? I need to update the article – 10 isn’t enough to really eliminate chance (in such a small sample, there is a chance that staying will be better). If you can, try doing 50 or 100 trials where you stay (just keep clicking door 1)… you should see the percentage be much closer to 1/3.

Actually, your initial guess has a 1/3 chance of being a car, not 50-50. If you have 3 choices and pick 1, that’s the best you can do, right? Monty can add and take away doors on the other side, but at the time of your initial guess, it was a 1 in 3 chance of being right.

It’s true that monty leaves you with two choices in the end, but two choices doesn’t mean a 50-50 chance between them (that’s the crux of the counter-intuitive nature of the paradox). Try giving the game a shot with 100 trials to see what happens :). Also, give SteveG’s explanation a try.

@Puzzled: Great question! Yes, if Monty randomly opened a door (and it happened to be a goat), it would indeed by 50-50. But since Monty is looking at TWO doors (and leaving you with the better one), it is an advantage to switch.

still completely do not understand this.

Ok, I think I get it. Here’s how I understand it:

There is a 1/3 chance for the door you pick to have a car, therefore there is a 2/3 chance for either of the remaining doors to have a car.

After all filtering, there is still a 1/3 chance for your door to have a car and a 2/3 chance for any other door to have a car.

Since there is now only one other door there is still a 2/3 chance for that door to have the car and still 1/3 that your door has it.

Or for 100 doors. 1/100 that you pick the car the first time and 99/100 that you didn’t. 98 doors are eliminated. It’s still 1/100 that you got it right the first time and it’s still 99/100 that you didn’t. But now, there’s only one other door, so it’s 99/100 that it’s the other door.

Awesome post.

Here’s yet another way to think about this problem.

Pretend we are dealing with a lottery that everyone in the world plays - each person is given a ticket which is equally likely to win. One person’s ticket is chosen from the world population to be the winner. Would you expect yourself to be the winner, or someone else in the world? The intuitive (and correct) answer is to expect someone else.

Now suppose you have a connection inside the agency that runs the lottery. Your “inside man” sends you a letter with the name of a person (call him James.) They inform you that if the rest of the world contains the winning ticket, then that winner is James. They then give you the opportunity to switch tickets with James. Would you?

Of course you would!

@CB: It’s a weird problem to be sure. Try playing the game above but with 10 doors instead of 3 [enter 10 for the number of doors and press reset in the game].

You’ll start to see how your initial guess is “usually” wrong (it only has a 1 in 10 chance of being right). From there, you’ll start to see that when you have 3 choices, you only have a 1 in 3 chance of being right.

@Greg: Awesome, thanks for sharing your insights! I always like seeing how other people think about the problem. I too see the “probability cloud” collapsing.

@Alex: I like the lottery example, it’s something we can all relate to!

Here’s the way I finally convinced myself that switching every time is the best strategy: if you pick a goat, Monty shows you the other goat, and when you switch to the remaining door (the car), you are ALWAYS right. If you pick the car, Monty shows you one of the goats, and when you switch to the remaining door (the other goat), you are ALWAYS wrong. So if you always switch, you will win 2/3 of the time since that is how often you will pick a goat to begin with. But if you decide to never switch, you will only win 1/3 of the time, since that is how often you will correctly pick the car to begin with. So by always switching, you will win twice as often as when you never switch!

@Bruce: Awesome, thanks for sharing!