Monty Hall Problem - Numberphile

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This video features Lisa Goldberg, an adjunct professor in the Department of Statistics at University of California, Berkeley.

NUMBERPHILE

Videos by Brady Haran

[JNDlego]

Before you comment saying its 50/50, read this.

Scenario 1: You pick the car. Monty shows a goat. You switch to the other goat. You lose.

Scenario 2: You pick goat 1. Monty shows the other goat. You switch to the car. You win.

Scenario 3: You pick goat 2. Monty shows the other goat. You switch to the car. You win.

So as you can see, you only lose by switching if you picked the car the first time around. And the chance that you picked the car first time around is 33%.

You win by switching if you picked a goat the first time around. And the chance that you picked a goat first time around is 67%.

The reason its not 50/50 after a goat is revealed is because you made your first selection BEFORE the goat was shown to you. You made your first choice when there was still 3 doors so there is a 33% chance you picked the car. Revealing where a goat is does not change the fact that you still picked your door when there was 3 doors to choose from.

4/8/20 Edit:
I’ve seen a lot of comments saying I have forgotten the other three scenarios in my explanations, which I have defined below.

Scenario 4: You pick goat 1. Monty shows goat 2. You don’t switch. You lose.

Scenario 5: You pick goat 2. Monty shows you goat 1. You don’t switch. You lose.

Scenario 6: You pick the car. Monty shows you one of the goats. You don’t switch. You win.

As you can see, Scenario’s 4-6 show that sticking with your original choice only wins 33% of the time. The reason you don’t combine the perecentages of Scenarios 1-3 with Scenarios 4-6 is because you add more then one independant variable to the experiment. The independant variable is whether you switch or not. Everything else must remain the same. There’s always 2 goats 1 car, you always get first pick, Monty always shows a goat, you are always offered the option to switch.

If you want to analyze the results of switching, there are only 3 scenarios to analyze (1-3.)
If you want to analyze the results of not switching, there are only 3 scenarios to analyze (4-6.)
But you cannot group Scenarios 1-6 together and analyze the results of all 6 percentages because you have two different independant variables that change halfway through the experiment.

Imagine you have two different plant foods, A and B, were trying to figure out which plant food makes your plant grow faster. You hve two identical plants, but lets say you also watered one twice as much as the other. You have no way of knowing whether it was the water, the plant food, or both that made the plants grow at the rates they did. If you’re thinking to yourself “Why would you water one plant twice as much as the other? It should be the same water?” Then that is perfect. That means you understand that for the Monty Hall problem, you can’t combine the switching and not switching scenarios into one large experiment.

[Viva_Reverie]

I've heard explanations for this game too many times but I've only found a few which are super easy to understand, this is my preferred "Explanation":
Say you choose 1 door out of 3. There's a 1/3 chance you get a car, and a 2/3 chance you get a zonk. This is without the switching nonsense, just barebones and I'm sure we can all agree on it.
Now let's say you chose a zonk, fair enough, it's a 2/3 chance, more likely to show up. Now Monty opens a door with a zonk in it. There are only two zonks in the game, the one you've more likely selected, and the one Monty just opened. Therefore, the third door _must_ have the car.
Let's go back a step, and go on the assumption you chose the car first try. Now Monty will open one of the other two doors, with a zonk, of course. If you switch you'll get the other zonk, no matter what. This is why it's not a foolproof strategy.
Buuut, with this logic, if you choose a car first (1/3 chance) and then switch, you'll always get a zonk. And, if you choose a zonk first (2/3 chance) and then switch, you'll always get the car. This, is why switching is the better option.
Therefore, if you stay, you'll get a 1/3 chance of getting the car, but if you switch you'll have a 2/3 chance of getting the car. That's what the video says in the end anyways, but I found this explanation far easier to comprehend xD

[maplewang]

Third option: you intentionally switch to the open door to guarantee that you get the goat

[lloydburden]

If you pick the car and switch you always lose. But if you pick a goat and switch you always win, since Monty is forced to eliminate the other goat from the game. But because you are twice more likely to pick a goat than a car in the first place, this means that you are twice more likely to be in a situation where switching wins you the game.

[criscusack]

I always sum it up as: "In order to lose by switching, you have to have picked right at first. So you will lose by switching 1/3 of the time. So switch."

[scrtmin]

How darE YOU DETECTIVE DIAZ I AM YOUR SUPERIOR OFFICER *inhales* B O N E ? ! !

[dumitruene]

The explanation with 100 doors got rid of any misunderstandings I had about the answer of this problem. Thank you!

[jayparmo]

I didnt understand it until the 1 in 100 part. That made it incredibly clear to me.

[victoriahiggs]

switching can save you 15% or more on car insurance

[caseyl]

Great explanation! However if i may add. i would postulate, math aside, that the reason the probability concentrates on the final door vs reverting back to splitting between the final and the initial door is because Monty's pick is a dependent probability and not an independent probability (very important point here). For example, if Monty were allowed to pick "your" door and show you that you picked the goat, your chances don't increase. and remain 50/50 as his choice of door is no longer dependent on your initial choice.

[Motorsheep]

Another way to look at it is from Monty's perspective:
Monty can only open a door that you didn't pick, and he can only open a door that has a Zonk behind it.
There's a 1 in 3 chance that you picked the car. So there's a 2 in 3 chance that you picked a Zonk. So 2 out of 3 times, Monty has only one door that he's allowed to open. Because one is your Zonk and the other is the car.

[cdmcfall]

The brute force method is easiest for me to visualize. With three doors and three prizes (car, goat 1, goat 2), we can get the following possible combinations:

1. *C* G1 G2
2. *C* G2 G1
3. G1 *C* G2
4. G1 G2 *C*
5. G2 *C* G1
6. G2 G1 *C*

Whether I pick door 1, door 2, or door 3, there are only two out of six possible combinations that result in a win. For each of those six possible combinations of doors and prizes, I can open one of the remaining doors and eliminate one of the remaining prizes. That means for whatever door I pick, there are 12 possible outcomes that are equally likely to occur. In this example, I have picked door one and _X_ is the door that is opened:

1a. *C* _X_ G2 or 1b. *C* G1 _X_
2a. *C* _X_ G1 or 2b. *C* G2 _X_
3a. G1 _X_ G2 or 3b. G1 *C* _X_
4a. G1 _X_ *C* or 4b. G1 G2 _X_
5a. G2 _X_ G1 or 5b. G2 *C* _X_
6a. G2 _X_ *C* or 6b. G2 G1 _X_

Again, each of these 12 outcomes is equally likely. If I randomly eliminate door two, then I have two winning combinations if I stay with door one and two winning combinations if I switch to door 3. My odds are still 1:3 in either case. If I randomly eliminate door three, then I have two winning combinations if I stay with door one and two winning combinations if I switch to door two. Again, my odds have not improved. This is because there is are cases where the car is eliminated when a door is opened (3a, 4b, 5a, and 6b). So why do my odds increase when we restrict opening a door to goats only? The same twelve options happen with the same frequency as without the restriction, but the possible outcomes where a car is eliminated are now swapped:

1a. *C* _X_ G2 or 1b. *C* G1 _X_
2a. *C* _X_ G1 or 2b. *C* G2 _X_
3a. G1 *C* _X_ or 3b. G1 *C* _X_
4a. G1 _X_ *C* or 4b. G1 _X_ *C*
5a. G2 *C* _X_ or 5b. G2 *C* _X_
6a. G2 _X_ *C* or 6b. G2 _X_ *C*

Now it becomes clear (at least to me). For the 12 possible outcomes after I pick a door and a goat has been revealed, sticking with door #1 has only four ways to win (out of 12). However, limiting open doors to goats only means that in _every_ instance where a goat is under door one, switching will get me the car. My odds of winning by sticking with door 1 are 4:12 (1:3). My odds of winning by switching are 8:12 (2/3).

Edited for formatting

[Dashi]

This is the first time I've ever seen the Monty Hall problem explained on a larger scale eg 1/100 vs 99/100. When you inflate to that scale, you get a gut feeling that the odds are in your favor to switch because the chance you picked the car on your first try is 1/100 which are extremely low odds. Extremely effective way to present the problem, and one I intend to use in future. Thank you!

[SamueITan]

"The math thing isn't the problem, you just need to bone."

[numberphile]

Wow, it can be frustrating reading these comments... Am glad this helped some people and sorry others remain perplexed.

Everyone who understands the solution seems to explain it in different ways, and I enjoy reading them here in the comments… Thank you.

The absolute key here is to ALWAYS remember the following... Monty KNEW what was behind the doors and his opening of a "zonk" was not a random act… There was NO CHANCE the car could be revealed at that moment. He was REQUIRED to reveal a zonk from one the two unchosen doors!

In fact if your door 1 had a dirty zonk (a 66% chance when you first chose it), you just ENTIRELY FORCED MONTY'S HAND and made him reveal the car's location because he had no other door to open but 2 (leaving that shiny new car at door 3).

If your door 1 did have the car (33% chance), you DID NOT FORCE MONTY'S HAND and he could reveal either 2 or 3. Not helpful… But that is only a 33% chance, so better to go with the previous and more likely scenario don't you think?

:)

[ryhs]

for anyone confused still
think of it like this,

the only reason its not a 50/50 is because he lets you choose a box before he reveals one,
since you choose a box, just because he now reveals one doesnt actually change the fact you had a 33% chance to land on the box you are on.

now for the complicated part.

You had a 66% chance to be wrong, and a 33% chance to be right, he now halfs your chances of being wrong by removing one of the incorrect answers, theres a 33% chance we are right still, but now there are only 2 answers, meaning the other answer must have a 66% chance of housing the right answer.

"in order to lose by switching, you must have chosen right at the beginning, so you will lose 1/3 of the time by switching, so you should switch" - someone in the comments

[TheGlobalAtheism]

I'll try and explain this:
If you pick a Goat/Zonk thing first (which is 2/3) when you swap you will always get the car - because he would show you the other Zonk.
If you pick the car first (which is 1/3) when you swap you will get the Zonk.
So if you pick Zonk A: he will show you B and you will swap to the car
If you pick Zonk B: he will show you A and you will swap to the car
If you pick the car then you will swap and you will pick either Zonk A or B
So if you 'try' and find the Zonk you are twice as likely to win!

[chrischampagne]

Another way to see it is to look at all the possibilities. At first there are three choices (Door 1, Door, 2, Door 3) and then you get two choices (Switch or Stay.) Therefore, there are six possible scenarios:

Car --> Stay = Car
Goat1 --> Stay = Goat
Goat2 --> Stay = Goat
Car --> Switch = Goat
Goat1 --> Switch = Car
Goat 2 --> Switch = Car

As you can see, always choosing to switch makes the probability of any of the first three outcomes 0% and results in winning a car 2 out of 3 times.

[jaromanda]

I've always thought it was easier to think of it this way ... lets say the car is behind door 1.

If you pick and stay. 1 = win, 2 = lose, 3 = lose

Pick and swap,
1 = lose (because you'll swap to 2 or 3, whichever is not opened),
2 = win (because 3 will be opened, so you can only swap to 1),
3 = win (because 2 will be opened and you can only swap to 1)

Who needs "concentrating probabilities" if you have logic :D

[axor]

I think the reason this problem seems counterintuitive is that the host first opens a door and only then asks you to switch, giving you the wrong impression to be chosing between two closed door. In reality opening the door doesn't change anything because when the host proposes you to switch they are asking you to chose two doors out of three.
You can also imagine that the host first asks you to switch and only after you agree or not they open a door (of the two that were offered in the exchange), this way it appears clear that they are asking you to chose two doors out of three