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This video features Lisa Goldberg, an adjunct professor in the Department of Statistics at University of California, Berkeley.
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Let's say after he opens the first door, that contestant dies of a heart attack, and a random person from the audience is called up to finish the round. They know door the original contestant picked, but are unaware of this mathematical phenomenon. One of the doors is already open. Does this contestant now have a 2/3 – 1/3 chance, or is it 50/50?
Probability doesn’t lock in, the car is behind 1 of the 2 doors it’s 50/50 either way!!
What happens if you chose the correct door first?
Still not convinced. I cant see the logic behind this explanation. For me, opening a door simply lowers the options so from 1/3, it becomes 1/2 or 50/50 chance. That’s just my opinion though. I think Raymond Holt is correct.
The fact that there's 66.6percent probability is not wrong.
But the fact that you can always win is wrong for sure.
I must be really slow with maths/probability. I still don't get why the odds aren't 50/50 if you now have two doors to choose from.
easy, you dont think of the switching as switching, you just see it as a new oportunity to choose between two options. now your choice is more likely to bear fruit since you have a 50/ 50 chance instead of 1/3
but.. idk if im too pragmatic for this of not.. if you choose the door you originally chose a second time.. your chances of winning are the same as if u chose the other door, no?
50/50. sure, the likelyhood of winning is now higher than the 33,3% in the first round but since you switch based on a random gut feeling anyway, that doesnt really make any difference…. or am i wrong?
This pisses me off so much. I've watched a few explanation videos, and visited a site with a playable version and the win/lose history proving that it wo,rks…. But in my head it still doesn't. In my head, if you have two doors with two outcomes, it's 50/50, anything before that being irrelevant.
WRONG
After door 2 is selected, time is frozen, I incernated door 2, and repaint door 3 so it reads door 2
I induced amnesia to everyone there about the previous, long gone door 2
I unfreeze time
NOW WHAT IS THE PROBABILITY OF PICKING $$$$ BETWEEN DOORS 1 and "2"?
1/2.
best explanation ive ever seen
I found a super simple way to understand why switching is better: in short, Monty will NEVER open the door with a car in it (defeats the purpose of the game), that’s why the unequal probability happens and you SHOULD switch. So imagine case 1 where you already chose the car first, then switching is bad (1/3). In case 2 where you chose a zonk first, Monty will delete the second zonk, leaving the car available (2/3). It’s that simple.
This is a bad and wrong video.
I was having trouble wrapping my mind around this so I decided to write a script that would simulate and count the number of time switching would be right only to find that, if you switch, the only time you will not get the car is if you had picked the car initially hence the 2/3 probability when switching.
Short explanation:
The initial door pick, no matter what, has a 1/3 chance of being the car. From the other 2/3 doors, no matter what, Monty will always show a goat. Switching is just moving your choice from the 1/3 chance of your initial pick, to the 2/3 doors.
Long explanation:
This is what I stumbled with when writing the decision logic for the script. Doors are labeled A, B, C. If I initially pick door A, Monty will have to pick something from either door B or C, never from A.
If door A has the car, then doors B and C both have goats. If he shows B, I switch to C and get a goat…if he shows C, I switch to B and get a goat. Ok Monty, you got me this time.
Now, if door A has a goat, Monty is in trouble. Why? There's only one goat in B or C, and he is forced to show that door while the other one will be hiding the car. Since I initially picked a goat (although unbeknownst to myself), and Monty showed me another goat, switching means that I will be choosing the car's door.
So there are two scenarios: I initially picked the car and switch to a goat, or I initially picked a goat then switched to a car. But there are 3 initial options (doors), 2 of those options being picking the goat. This means that if I always switch, the 2/3 chance of picking a goat, is now a 2/3 chance of picking a goat door but switching my choice to a car.
As if we choose door 2 and door 3 having 66.67% probability on door 3(after goat on 2),we can also think of 66.67% chance on door 1 if we take door 1 and 2.
Am I missing something?
What if you pick the right one then switched to the wrong one?
To me it seems so simple.Initially the probability of the car being behind each door is 33.33%. Once the player chooses a door, the probability of the car being behind his door is 33.33% and the probability of the car being behind the other 2 doors is 66.67%. Once one of the other 2 doors is opened and is shown to have nothing, the probability of the unopened, unchosen door having the car goes up to 66.67%. But the probability of the car being behind the chosen door remains 33.33%. So it's always better to switch to the other door which has a 66.67% chance of being the right one Correct me if I'm wrong. I'm not trying to be too smart and have an iq of around 140. Yet I've been playing chess all my life and still my online rating never crosses 1700. I wish I could play better chess which is my passion. Cheers from India.
The best way to explain this!
The question is, which strategy, staying or switching, yields a greater probability of winning? The host knows where the car is, and he will always remove a goat before offering you the option. Therefore…
If you decide to stay:
Choose right the first time (1/3 chance) you always win.
Choose wrong the first time (2/3 chance) you always lose.
If you decide to switch:
Choose right the first time (1/3 chance) you always lose.
Choose wrong the first time (2/3 chance) you always win.
Great explanation. Came here after watching 21 film with Kevin Spacey to understand how variable probability works.
Each door has a probability 1/n to have a car behind.
That means that each door has an equal probability compared to the rest doors of the population.
When you open up doors, you actually decrease the population
For each door of the population, the probability of having something valuable behind is still equal to the probability the other doors of the population have.
It becomes 1/(n-1), 1/(n-2), ….
When you end up with two doors, each one has a probability of 1/2=50%.
Thus, you are indifferent in the choice.
Can anyone tell me if my reasoning is correct?
I still don’t agree with this. I don’t get how the probability goes from 1/3 to 2/3 (using the doors example)
To me, when the host opens the door to reveal a non-desirable option. That door is eliminated from the equation so your choice becomes a 1 in 2 so both remaining doors have a 50% chance of having the car behind them.
With that said, I would like to pose another error I find:
Once one of the three doors is eliminated, why does only one of the remaining doors become a 2/3 chance. Shouldn’t that also apply to the door you have already chosen as well?
Can anyone explain to me…what if they chose a random number to just distract you?so if you switch to a trap you can get nothing😢???this is on the many door thing
The example with the 100 makes it very clear
Very clear explanation!
Ahh an actual video of quality compared to vox
Its 50/50 if Monty opens a door at random. If Monty specifically chooses the door he KNOWS has a goat than it is 2/3 and you should in fact switch.
Won't it become 50/50 once the door opened up by Monty is eliminated? Like you'll have 2 doors, behind one there's a goat and your car behind the other. If i am wrong then i am happy to be corrected 🙂
The part that always got me is that my brain likes to think that probability is absolute for everyone always. This isn’t true because obviously Monty Hall has a 100% chance of getting the car since he knows which one it is. If you participate in the first part of the game then you have the information to know which of the 2 remaining choices is more likely, but if you just walked into the room and had no idea what was going on and just saw the 2 options you’d have a 50/50 chance.
So you can't just multiply the first probability of getting a goat (2/3) with 1/2 if you switch (cause then there are only 2 options left; getting a goat or a car) to find out there is a 2/3•1/2=1/3 probability of getting a goat when switching, thus 2/3 of getting a car.
I find it confusing when there's no one using multiplication for this case, since that's standard for calculating probabilities. Maybe this does not apply to this problem, correct me if I'm wrong, cause clearly there's some manipulation going on during the process since one goat gets revealed, but does that make it a conditional probability?
If anyone is skeptical, watch the mythbusters episode.
Both arguments are right based on their own perspectives.
33-67 argument:
Compared to your initial situation of 33.33% chance, after revealing a goat, by switching, you'll have the other 66.67% chance of success.
It's like, effectively, you were given a chance to select two doors "initially" instead of one.
50-50 argument:
Once a goat is revealed, you re-evaluate the chances and don't compare with previous scenario as the data available has now changed.
Even though effectively you were given a chance to select two doors "initially" instead of one, now with the available data, it's like you were asked to include one wrong door among the two, which doesn't increase the winning chances.
dear Numberphile – how about if in 2:09 we just switch the captions to say: p(door 1 + door 2) = 2/3, p(door 3) = 1/3, then after the reveal when p(door 2) = 0 we get the complete opposite result! Any brave soul to disprove that claim? I'm tempted to do a simulation here!
To put it simply, there are two goats and one car. The chance of picking a goat is 2/3. Monty HAS TO pick a goat (because he knows where they all are) and so if the door you pick is a goat and Monty picks the other goat, switching will give you the car. This means that if the door you initially pick is a goat, switching is a guaranteed win, and because there is a 2/3 chance of initially picking a goat, there is a higher chance of winning (Only if you switch).
It works under the premise that Monty is always going to open another door with a goat
How is it possible to have a ⅔ probability of winning AfTeR a door has been eliminated¿?
There's only two options left!
Or I just shouldn't have had whiskey for dinner.
It should be emphasized that this only happens with two conditions: The host knows where the car is, and he always reveals a zonk on his 'turn'. If neither of those conditions are met, you retain a 1/3 chance.
They do not replace what are behind the doors. Well then, do not make choice yet. Each door has equal chance of having a car. Let's assume Monty gave you a hand and removed one option. So you are left with two options. Now you are with 50/50 and choose which ever you want.
The way I like to explain this problem simply is to imagine that Monty doesn't open a door. You choose a door, and then Monty gives you a choice – do you want to stick with your door, or do you want to switch to the other two doors combined? You'd be a fool not to switch. Monty opening a door to reveal a goat doesn't change this one iota. Since he always opens a door to reveal a goat and never the car, and since you know for a fact that there is at least one goat behind one of the two doors you didn't choose, he's not revealing any new information to you when he opens a door to reveal a goat. The opening of the door is a red herring that doesn't change anything about the probabilities. Any which way, you'd be crazy not to choose two doors over one.
Personally, I would cry tears of joy if I got to keep the goat.
O yea, okay lets switch it around lets say that you have a probability of 66.7% when chosing the first and second door of getting it right and you pick the first door then he shows you the second door, and it's a goat, assuming by this logic you would consider picking the first door since you have higher chances although you disregard a door.
What im trying to pick out is that as a metric luck is interwoven in this question and the idea of you chosing it wrong the first time is an unknown so saying that you have more chances of getting it wrong at the first time is based on an expandable axium that does not fit the calculation as it's predicated on misfortune which cannot be known.
However this method of calculation is based on statistics that may fit contestant at the moment and with this calculation I did the same as student did since choice on a door is an illusion and by not opening door number 3 I repeated his step.
still doesn't make sense fully, something is missing
What happen if the door you choose is a car?
How is this possible when, in reality, by choosing not to switch you are still choosing a door after the first "zonk" it is just that you are choosing the same door as before instead of the other. Like if you were to not choose at first and then pick on of the two remaining after the "zonk" is revealed that would be a 50/50 split right? How does the original choice have any effect.
nope
everybody is here because of b99
Choosing the door with a GOAT is 66% chance. You are more likely to choose a GOAT. The Game Show Host will always open the door with the other GOAT in it. This means, that whenever you choose a GOAT (without knowing), and switch your door — you win. Which means 66% of the time when you choose a GOAT and SWITCH, you win!