The Monty Hall Problem: Switch Doors or Not?

#shorts This video presents a problem related to the famous Monty Hall Problem. You may also like this longer explanation that considers more the 3 doors.

[nicholas_kujawa]

The math checks out. I wrote a python script to simulate 1000 games where the first pick is used and another 1000 games where the switches door is used. Keeping the first pick wins ~33.3% of the games. Switching doors wins ~66.7% of the games.

[JizzWrld]

I've tried to simplify this down as much as I can:
You initially choose a door at a random 1/3 chance. Let's say, Door 1. Essentially, this means that there is a 2/3 chance that the car will be in Door 2 or 3. Since there is a 2/3 chance it is behind either 2 or 3, you can be fairly certain that their is also exactly one goat also behind one of those 2 doors you did not choose (assuming you chose a door with a goat- 2/3 chance you did). The host will now open a door you did not choose, in this case, either 2 or 3. The crucial part, is that the host knows where the car is, and will only ever pick the door that has a goat. Let's say he opens up Door 2. Now you have an option between Door 1 and Door 3. You knew this entire time, however, that 2/3rds of the time, there will be one car, and one goat behind either Door 2 or Door 3. So since you now know that Door 2 is the goat, that means the 2/3rd probability of a car goes straight into Door 3! Remember, you picked Door 1 out of conplete randomness, but with the newfound knowledge that either of the other two doors likely harbor a car and a goat 66.7% of the time, you've filtered the one of the two doors with the goat, and now the door you're left with is most likely the car!

Here's are 3 example games where we switch, with all 3 possibilities of the car placement.
We will choose Door 1 as our first option for each example.
In scenario 1, let's say the car is behind Door 1, and we choose Door 1. This is the 33.3% of the time you picked the correct door on the first try. The host reveals any of the other doors having a goat (it doesn't matter, because they both have goats), you switch, and you lose .
In scenario 2, let's say the car is behind Door
2. Again, Door 1 is our option, which is a goat (door 2 has the prize). The host is forced to reveal door 3, since he cannot reveal the prize door (Door 2), nor the door you initially chose (Door 1). Since the only options are 1 or 2, you switch to door 2, and you win.
In scenario 3, the prize is in door 3, and once again, we choose door 1 for our initial option.
This time, The host is forced to reveal door 2, since he can't reveal the prize (door 3) nor the door you chose (door 1). Now you have the option between door 1 and 3. You switch to 3, and you win.

See, 2 out of 3 scenarios, you win the car. And notice that in those two scenarios, the doors you didn't choose had exactly one goat and one car, meaning that it is likely that after that one goat is revealed, the car is going to be behind the other door you didn't choose.
Hope this helps!

[AlluckyTV]

The host will never eliminate your door therefore the odds that’s your door is empty is higher than the door remaining

[TheGijow]

How dare you detective Dias, I am YOUR SUPERIOR OFFICER!

Bone...
BONE!?!?

[stevehaire]

You should always change your choice for the best odds.

There are only three scenarios (a, b, c).
There are only three positions for the prize (W), and six positions for no prize (L).

a) W/L/L
b) L/W/L
c) L/L/W

It doesn't matter which door is chosen, so long as we use the same door for each test (as these are scenarios, not different games).
We will therefore initially choose door 1 every time.

Scenario 1: W/L/L
One L is romoved, we then change to door 2 or 3. We lose.
(Sticking wins).

Scenario 2: L/W/L
The only remaining L is removed, we change to door 2. We win.
(Sticking loses).

Scenario 3: L/L/W
The only remaining L is removed, we change to door 3. We win.
(Sticking loses).

In conclusion; changing doors has a 2/3 chance of winning.
Sticking with the initial door has a 1/3 chance of winning.

[BrVn]

So the thing that matters here is your first pick.
If u pick the right one, there will be two wrong option for the security to tell u so picking the right one first has a chances of 1/3 which is also the chances of u getting it wrong after switching which makes picking the right one has a chances of 2/3.
Even simplier explanation:
Picking the right one first= picking the wrong one (1/3)
Picking the wrong ones= picking the right one(2/3)

[ruthb]

Best explanation I've found so far!

[JohnPaulCarmona-sh]

Everyone makes this so complicated for no reason. This is the way I finally understood it. Odds are you choose a goat at the first pick. Now you can stick with your 1/3 chance, but the odds it’s the other remaining door is 2/3. (After Monty exposes the other goat). The remaining door survived the first round elimination. Your door didn’t have to compete for survival in that as he will never open yours. That gives a 2/3s credence to that door and 1/3 to your door. Odds are you picked the damn goat first anyways and the other door has the car.

[dantedt]

A rule to remember: Sum of all probabilities should be 1. So initially each door has 1/3 chances of having the cash, since 1/3 + 1/3 + 1/3 = 1, right? So if one of the doors without the cash is opened, then we now have 1/3 + 2/3 + 0 = 1. Therefore, the door you initially didn't choose has a 2/3 chance of having the cash. So switching has a greater chance of winning the cash .

[Sarwaan]

If you think it’s 50-50, it’s not because you got rid of one of the doors that specifically is the wrong answer. If it was just a random door then yeah, it’s going to be 50-50.

[PolarEyes]

I just watched three videos on this subject, and this is the first one that made sense, instantly. Nice job!

[Whoopingcoughalig]

Everyone agreeing that it's a 50/50 when it's not lol

The hosts knows where the money is and will always open a door with no money behind

The chances of picking an empty door are 2/3
The host will then remove the other empty door of the equation.
The chances of picking an empty door are still 2/3 and so by switching it means you have 2/3 chance on switching into the door with money

[gvlacic]

This problem changes every time somebody new presents it

The assumption is on how the host is picking.

At random, or picking a goat on purpose.

The rest should be simple.

I've been stuck and stubborn on one side and then elementary math on the other.

It's not so cut and dry.

The framing of it is what leads people one way or another.

If he's picking at random its 5050. If hes picking with intent switch.

[Anonymous-utbg]

The one more thing that almost no one mentions is that isn't there's a 50-50 chance after one door is revealed? But what you don't think is that you should change door because the probability of you picking the wrong door is higher than picking the right one in the first step.

[ninadnagpure]

Best explanation on this...!!! Short and straight...

[Dautar]

This will only make sense IF he cant reveal that your door has a goat

[mmh]

Easy explanation, thank you. Yes, I totally agree.

[Cj]

Still trying to wrap my head around this! Don’t the odds change after the first door opens to 50/50?

[SawkTheFighter]

Knowing this, if i was the game show i would show him the door ONLY if he picked the car bc the player would always change. And if chose a goat, id say 'okay here u go'

[wesremy]

The reason this feels counter-intuitive is because of two things:
1. The host reveals a door with no prize money
2. The host doesnt reveal your door

Because of 1., the proportion doors with prize money becomes more favorable (1/2 instead of 1/3)
Because of 2., the chance of us initially picking prize money remains unchanged. So its still our initial 1/3 chance.

This means our chosen door has a worse chance than any other door, since our doors chance didnt chance while.others became more favorable.