The Monty Hall Problem: Switch Doors or Stay?

Subscribe for weekly math and science videos that'll explore challenging problems, common misconceptions, and intriguing concepts.

FIND US ON:

[blackpenredpen]

Nicely done! This is my favorite way to explain the problem to others!

[-TheBugLord]

So you have a 1/3 chance to pick the car door, so when he opens a door with a goat, there was a 2/3 chance the original door you picked was a goat, so switching would give you a higher chance (2/3) of getting a car

[Boca-do-rio]

I have something nice for you to think about. Lets bring the quizmaster back and 3 new doors. Today there are 2 guests, its me and it is you. You get asked to pick the first door and after you picked yours i pick mine... You choose 1 and i choose for 3. At this moment we both have 33% chance to have the right door, as you and i could stand before a bad or good door, even when you choose first... Door number 2 is opend and behind door 2, there is no new car to be found. Now tell me... Why would it be not good for me to switch and only for you?

[SvFilipov-qbio]

I understand the logic behind the explanation but I’ve always had an issue with it:
1. You choose 1 out of three (33.3% correct)
2. Host opens one door revealing a goat
3. And this is where I’m puzzled - at this point for you (and actually anyone watching from the beginning), the door you’ve picked at the beginning still has 33.3% vs. 66.6% for the other remaining door, but if another person enters the room at this point, having no knowledge of what just happened and this person is asked to pick a door, he would evaluate both remaining doors to be at an equal 50% chance to win the car.
So basically, two people at the same time and place should theoretically have different chance of winning the car solely based on their knowledge of the past. It gets even more extreme if say originally you pick one out of 100 doors and the host then opens 98 revealing goats. At this point you should have a 99% chance of winning the car if you switch doors, but the “new guy” would still be at 50%? Not sure I totally agree, or am I getting this all wrong?

[joaohonradodasilva]

How do we come up with that result at 2:11, the two thirds? I can't figure it out

[imagiFantasy]

I think you are comparing the chances, but i still don't understand your perspective. If the host revealed one of the door that has a goat behind, your chances of choosing the right door will be 50-50, sure its better than choosing between 1/3 doors but at the end we won't know the right answer

[GabriTell]

This is actually an EASY problem, I solved it in my mind in 5 minutes with no difficulty 🤷‍♂

[arjuncrm]

Now, say if you were presented with 3 doors with one of it already opened in front of you that has goat. Now if u r asked to pick one door of the 2 closed ones, is your chances of getting a car still 2/3???

[dawyer]

Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a door with sheep, leaving only two pairs of doors, one sheep and one car for the player to choose. If the player predicts the door of the car is correct, what will the host do? of couse Want the player to change their choice. The host asks the player to calculate the odds with misleading the player already have once choice before. player swapping his choice, and saying 2/3 chance will win.

On the contrary, the player predicted the door with the sheep is correct. What will the host do to make the player still choose the door with the sheep? you can tell.

if you are the player, do you think have any advantage from changing your decision?



Dawyer's door problem, calculate the chance of the host winning.

[TristanSimondsen]

I completely agree; it is not a 50% to get the car from the 3 doors; it is a 33.3% to get the car from the 3 doors. The 50% is in reference to stay or switch to get the car, as in you have 1/3 probability to get the car and 1/3 probability to get the goat.

33.3% or 1/3 is the probability of the car behind one of the 3 doors. The remaining 66.6% is then in reference to the probability of the car behind the OTHER TWO doors, not the goat. Because when a car is behind a door, it can only occur 1/3 of the time, not 100% of the time. Which means during this 1/3 of the time, the other 2 doors are goats, and ONLY during this 33.3% or 1/3 of the time.

At NO POINT in time will the 2 goat doors contain 66.6% probability:

Goat Door B & C - 33.3%
Goat Door A & B - 33.3%
Goat Door A & C - 33.3%

If 2 doors or 2 goats are 33.3%, then each only has a value of 16.67% or 1/6.

You're mixing probabilities of car and goat. 1/3 of something doesn't mean it's 2/3 of ANYTHING else.

[jaybefaulky]

*I can prove the true nature of this 'problem'!* .. they trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose, you 'reserve', to protect one door and let one goat be eliminated from the game, 1st round is now done! Now they start the second game where the REAL choice is allowed, there are no longer any 'reserved' doors and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose' instead of 'reserve/protect', which is very different than the first game's round of choices..it's TWO different games NOT one, so if you do not account for this by creating two separate math problems then you are NOT accurately representing the true nature of the game, as you have to change the odds for all 'unknowns' every time you eliminate possibilities. This logic applies to different examples of this 'problem' as well. The key is understanding that the only real game is when the final choice is made, and that everything before that is just changing the parameters, you have to make the math adjustment for the new parameters as they change, it all comes down to the *state of the parameters* when you *actually choose* and NOT when you are simply 'negotiating' the parameter changes, ..thanks and you're welcome;)

[dawyer]

Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing.
There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3.

Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"? In fact, this is just an ideology produced under the Goebbels effect.

[dienekes]

FAILURE. This is very disappointing for this channel. The ONLY thing that the idea that the game is played out this way is that the odds for round 1 are meaningless because the round will NEVER be resolved and it will be forced into round 2, where the odds remain 50/50. Nothing in round 1 changes this.

I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter garbage. What is presented here is called, in mathematical and scientific nomenclature, a "model". There are 2 ways to prove a model: direct testing and observation and applying it to similar yet slightly different scenarios and having it be consistent. Since testing isn't possible in this venue, I will prove this model is wrong by the second method.

Core Scenario: A contestant is presented with 3 doors. They get to choose 1. For simplicity, we'll say they pick door 1. Each round, the contestant get to switch to a different door or stay with the door they have. There is only 1 door that has a car, but 1 door always has a car. So, at this point, each door has a 1:3 chance of winning.

If the round is resolved, the contestant has 1:3 chance of winning because they only got to choose 1 out of the 3 doors. If anyone disagrees with this, please comment below.
The host opens one of the doors the contestant didn't choose (door 3). If the car is behind the door the host opened, the contestant loses. If the host opens a door with a goat, then we move to round 2.
The contestant is allowed to switch to door 2 or stay with door 1.
I think we are all in agreement up to this point.

According to the model presented here, the odds of the eliminated door gets shifted over to door 2. This is based solely on the fact that the contestant picked a door in round 1 (an action that played absolutely no role in establishing the odds in the first place).
According to the standard model, the odds of the eliminated door get evenly distributed between the existing doors. This is based on the number of doors remaining.

In order for a given model to be correct, the outcome must equal 100%. There is a 100% chance that the contestant will win or lose. All odds needs must equal 100%.

So, in round 2, according to the presented model, there is a 1:3 chance you win if you stay with door 1 and a 2:3 chance you win if you move to door 2. This equals a 3:3 chance, which is equivalent to 100%. We're good so far.

But what about the standard model? In round 2, with an eliminated door, since there are 2 doors, the odds of either door having the car is 1:2 (also described as 50%). Since each door has a 1:2 chance, that adds up to 2:2 or 100%.

So, both models are seemingly viable at this point.

So, let's apply these models to 2 slightly different scenarios:

Alternate Scenario 1:
In round 2, the round isn't resolved. The host opens door #2, which has a goat. What are the odds of the door the contestant has picked having the car? Let's go back to our models:
In the presented model, remember that the ONLY criteria of the odds for being correct are the doors that the contestant DIDN'T pick. So, according to this model, if the contestant stayed with door #1, there is a 1:3 chance of winning. But the 2 other doors, having been opened and showing goats, is clearly 0:3. What does that add up to? 1:3 + 0:3 = 1:3. So, if the car isn't behind door 2 or 3 and is only behind door 1 1/3 of the time, where is the car the other 2/3 of the time? The total for winning and losing here is 33%. So this model fails.
What about the standard model? Once door 2 is eliminated, the odds of each remaining door get recalculated and distributed evenly. So, if there is only 1 door, that means that the chance of that door being correct is 100%. So the standard model is superior and proven correct so far.

But let's move to alternate scenario 2 and see what happens:
Let's play the exact same game except with 2 contestants:
In the 1st round, contestant A chooses door 1. Contestant B chooses door 2.
The host opens door 3 with a goat). Where do those odds get shifted?
In the presented model, the odds can't be shifted because there is no door that hasn't been picked, so the odds are:
Door 1 -- 1:3
Door 2 -- 1:3
Door 3 -- 0:3 (shown to be a goat)
This adds up to 2:3 or 66.6%. So this model fails because it doesn't add up to 100%. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both lose. If they both lose, where is the car?
OR
The contestants both switch to the other door. According to the model, "the other door" gets the additional odds. Okay, so the odds end up like this:
Door 1 -- 2:3 (because contestant 2 didn't pick it, so it gets the odds from door 3)
Door 2 -- 2:3 (because contestant 1 didn't pick it, so it gets the odds from door 3)
Door 3 -- 0:3 (exposed and eliminated as a possibility)
So, the odds (2:3 + 2:3 + 0:3) equals 4:3, or 133%. Once again, a failure. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both win. We've established that there is only 1 car behind 1 door, so it is impossible for both contestants to win.

So, what does the standard model say? The odds get evenly distributed to the remaining doors. How do those odds look?
Door 1 -- 1:2 (one out of 2 remaining doors)
Door 2 -- 1:2 (one out of 2 remaining doors)
Door 3 -- (not in the calculation because it was eliminated by being exposed as a losing door)

1:2 + 1:2 = 2:2 or 100%. Once again, the standard model is victorious.

So, in both of these scenarios, the presented model fails miserably and the standard model stands.

Can anyone come up with an alternate scenario where the presented model works, but the standard model fails? I doubt it, but please present it if you can think of one.