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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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## AXON Fahim

23.08.2022lol… its not 2/3

both 1 and 3 door is 1/2 chance…. idiot oversmart people

## ABCD abcd

23.08.2022Whether door 1 or door 3, you may have to stuck to one option until win…. this probability logic explained in this vedio seems not convincing… other than a gamblers logic…

## Samar Kamat

23.08.2022Superb explanation

## Lozzie74

23.08.2022That was the best explanation of this that I’ve ever seen. I always accepted it was better to switch, but never fully grasped why. You explained this easily and gave me 100% clarity! To understand something is one thing, but to explain it so easily (and then use an improved example to prove it) is a whole other level. You are a master!

## Susan Kleinheksel-Horn

23.08.2022That is completely wrong. Yes when you choose between the 3 doors, you have a 1 in 3 chance. BUT, when one of the of the choices is shown to be wrong, the law of independent probabilities comes into to play. You now have a 1 in 2 chance of being right, so the odds for the second guess is 50/50, so it doesn't matter if you change or not.

## Joahhh 294

23.08.2022I still don’t really understand. If you were on the door number 1 and he opened 98 other doors revealing number 37 wouldn’t it be the same chance you get the car if you started on number 37 and he did the same thing (opening 98 doors) except not opening door number 1? (Other way around)

## Erik de Jong

23.08.2022how can it be one out of three if there is only two options

## BlackKnight 6660

23.08.2022The reason this idea is BS: there isn’t a 2/3 chance of getting the car behind the second door.

It’s only a 2/3 chance when you could have picked BOTH doors, this was never an option.

When the number in the pool is reduced from 3 to 2 the total odds become 1/2 instead of the original 1/3.

It’d only be a 2/3 chance if you were picking BOTH doors that you didn’t originally pick, this however was never an option.

## George Attieh

23.08.2022Question:

When Monty opens the door and asks the contestant if they want to switch, don't we use conditional probability? I mean, isn't it the probability that the contestant chooses right knowing that door 2 has a zonk? Which gives the other 2 doors equal probabilities of 0.5 each?

## Brooklynguy

23.08.2022Please excuse my ignorance about this.

(I think) I see this as changing the problem. Originally, you have a 1 in 3 chance of picking the correct door, or 33.333(etc.)% chance.

When you eliminate 1 door, you have changed to 1 out of 2, or 50%. Where did I go wrong?

If you really don’t mind and care to answer me, please do a reply to this, and a ‘like’ on my comment so I can be,alerted that you have answered. Of course the explanation could be long and in mathematical terms I don’t know, in that case could you steer me to a primer on this?

In either case, I’ll do more layman’s research and and see if I can figure it out.

— BG

## Erik J

22.08.2022This is the best explaination I have ever seen…. well done.

## David Croft

22.08.2022Has anyone ever combed through the Price is Right shows and shone this to be true?

## Ollie Jackson

22.08.2022this is wrong because u could just reverse that statement and say that initially there is a 1/3 chance that it's behind door number 3 and 2/3 for 1 and 2. It doesn't matter what happens before, the situation changes and if u were to run a thousand tests u would notice that it would be somewhat 50/50

## Ash Wynn

22.08.2022Can I use this in gambling or betting somehow?

## Nicholas Young

22.08.2022I see it like this. G is goat (1&2) and C is car. An arrow is change and a full stop is stop.

These are all the things that can happen:

G1 > C

G2 > C

C > G

Changing doors will win 2/3 times

G . G

G . G

C . C

Remaining put will win 1/3 times

## Omnium Mysterio

22.08.2022i don't fully understand this. if you ended up choosing three first, and then he called that there was a zonk behind door two, wouldn't that make it so that door one now has a 2/3s chance? i am very confused

## William Patrick

22.08.2022It only work on 4 doors or more if only three it still a 50 50. Why? Because when he open a zonk door its not his door but our door so it would be 1.5 / 1.5 which mean still 50 50. But when it more than 3, let's just say 4, when he open a zonk door your door will chage into 1.5 door and his door will change into 1.25 each, and when he open another zonk door your door will change into 2.125 and his door 1.875 and we'll minus your and his door by 1 each because its the nature chance of the door. So, your door will have value of 1125 vs 875 for getting zonk. It just a immediate thought and sorry for my bad English 🙂

## Wizard Flame

22.08.2022Yeah this is wrong. There is 3 variables, you remove 1 you have 2 its 50 50

## dilshadahmed vp

22.08.2022😂😂u guy's beleive in this sht

## Christian Swierc

22.08.2022Its is still a 50% chance.

## Rimuru Kemono

21.08.2022At last I understand this. Thank you for the explanation.

## Nate Pearce

21.08.2022You explained 2/3s rule in a way I understand it now thank u!!

## whashack

21.08.2022I think Monty is just a frame work to ignore history

## Adam Jeffries

21.08.2022I have watched this 3 times now. And I still for the life of me, can't understand why, when presented with the 2 doors, the chance doesn't then become 50/50 instead of 1/3 and 2/3???

Also what if you're a farmer and you already have 4 cars but need a goat, do you now have a 1/3, 2/3 or 50/50 chance of getting the goat?

## Jay28u

21.08.2022And that is why you should split your eights!

## Winston Neville

21.08.2022But if you switch, it changed the door you were at to a 2/3 probability?

## Randall Emmons

21.08.2022I think the important thing to remember here, is that the host choosing KNOWS what is behind the doors, and always chooses a goat. So in the many many trials you could run, he is either picking one of two goats, or the ONLY goat on the right side. This little extra bit of information shows you why you are always at an advantage to chnage. I think most people think of it as they door the host opened was "random" and had a goat behind it, and if you ran that trial 1000 times, and you "lost" every time the host revealed the car, THEN the data would line up with the 33% of you stay, 50% if you change. It's the act of removing the wrong answer, and NEVER the right answer that shifts the probability into the 2/3. You can say it always works that way, even if the host doesn't know, but that's simply not true. If the host doesn't know, and opened a random door, then you'd gave to include all the instances where there would have been a car in the math. Sure, you can say well just look at the instances where there was a goat, and it would show the same probability, but now you have to imagine it as this. No door is initallay chosen. Host reveals a goat and now you must pick a door. Does both doors have a 2/3rd chance? This is where the line blurs between hard math for me. People like to use the sand on the beach, but if the person along away all the extra sand has no idea where the extra piece of sand is, it doesn't matter if you pick the one in a billion, or the two in a billion.

End of the day, the host KNOWING where the car is changes the probably. Somehow.

## Az Sky

21.08.2022Makes more sense now with a 100 doors. Technically when 2 doors are left it is just 50/50 but you knew you had a 1/100 chance initially and you know the host knows not to open the door with the car.

## J J

21.08.2022This is a misleading fallacy. The odds remain 50/50. It is correct that there was a 2/3 chance of the other two doors having the car behind one of them but opening the door with the goat behind it DOES NOT transfer the 2/3 chance to that one door. This is the fallacy. Even if it could, there is then the issue that the first door has been ignored, since it can also be considered that the 2/3 chance gets transferred to that door because the comparison could have just as easily been between the first door and the one with the goat in it. Which means the odds between the two remaining doors balance out, back to a 50/50 chance anyway.

## Nie Mand

21.08.2022the solution is bull since the prize could have been behind door #1 after all. There is no logic behind that at all alas the show host made the same "mistake" every timeand everyone could have won quite literally speaking. Which is not the case. If you base a math solution on a show host doing the same thing every time and call that genius? What am I?

## Rcee

20.08.2022Why doesnt the problem change after the doors with goats are shown?

## David Davids

20.08.2022jeezuss. vos savant and others were NOT correct. after the initial door selection, the game is, effectively, RESET. and, that reset resulted in improved odds of winning; to one in Two. imo, (an improvement to) fifty-fifty does not support making a change in door selection. gooogletranslate

## Angel Alvarado

20.08.2022“BONNNNEEEEEEE!?!?!??”

## John Smith

20.08.2022Seen this problem before and still think switching is a red herring, if you have 2 doors to choose from then the odds are 50/50 switch or no switch

## Pesho

20.08.2022BEST explanation ever

## My573ry

20.08.2022Why wouldn’t the probability be equal?

What makes one door have 2/3 a chance? This is what I’m thinking:

Each door has a 1/3 chance that it’s the correct door.

I have a 1/3 chance that the door I chose is the right door. There is is a 2/3 chance that it is in another door I haven’t chosen. One of those doors is revealed to be the wrong one, so now I only have 2 doors where the car could be in. Therefore, wouldn’t it be 1/2 for both doors?

Also, why can’t the door you already chose be 2/3 instead? It doesn’t make sense that the 2/3 has to be the 3rd door. So shouldn’t it be just 1/2 and 1/2?

Also, this scenario is assuming that you always choose the 1st door first. I could be looking at all of this the wrong way, could someone please help me understand?

## Kanhaiyalal Rajput

20.08.2022I have a question:

Suppose, all the conditions in the problem(in case of three doors) are same except there are two peoples you and your friend who can choose two distinct doors. You choose door no.1 and your friend choose door no.3. Monty opens second door behind which there is no car. According to your explanation, the probability for you of the car behind third door is 2/3, while the probability for your friend of the car behind first door is also 2/3. How can it be that the probability of the car being behind first and third door both are 2/3?

Same with the case of 100 doors.

English is not my native language.

## phi1111ip

20.08.2022this is Rubbish

## Sam Wh

20.08.2022I'm not convinced yet

## KING K

20.08.2022this is quite easily the simplest explanation

## Galaxy

19.08.2022Let's turn it the other way around and see if you geniuses still think switching increases your chances. There's two doors, one goat one car. You pick door one. After you pick door one, they add a door three with a goat in it. Does switching now increase your chances? No, because it's a 50/50 between 1 and 2, always has been.

## sourav pr

19.08.2022Well i have a different perspective towards this….. probabilities are to be taken when a choice has to be made … If i make a choice then i need to find the probability for the success or failure of it, in this case, no matter which ever i choose the first time, at the second round i will be given with two options to choose so in that way I'm actually making a choice in the second round and not actually in the first round

btw..I have understood what professor explained.

## DaveHuncho

19.08.2022Wealth u(W) = ROOT W, W=10, probability 0.3 to turn to 100, 0.7 to lose everything, wants to avoid risk, how much is he willing to pay? (0,1,2,3)

## Ahmed Omer

19.08.2022I actually disagree, because, well i just picked the door, i did not open it, so when he opens the middle one, the chances actually increase to the two close doors, and me picking one of them won't make any different, i guess it's more logical that after elemenating one door the other two have equal chances

## onemediuminmotion

19.08.2022The "hidden variable"? Since

Ionly get to be on the show once, my chances are 50/50? If I got to do it over and over again, I'd always choose the other door.## Prof X

19.08.2022Dear Numberphile I wanted to let you know that you have some sad fans that need help they wait for anyone to comment on your video that says 50% then they insult them please tell them that it's not healthy nor productive and that frankly that makes them assholes

## Check pfp

19.08.2022This is ridiculous

## alissa

19.08.2022i still dont get it why is it not 50/50😭😭

## DeadByMeme

19.08.2022when i have 3 options and i picked one, then one option is revealed to be a

bad option and so now here lies the question. Is it smart to switch my option?

Does the Monty Hall Problem apply here?

Btw my options were:

-Go outside and touch more grass

-smok weed

-Do homework

## Don Wilson

19.08.2022Rubbish. What everyone is ASSUMING is that the SAME person is going to be on the show EVERY DAY but they are NOT. The contestant is only playing this game ONCE. So these repetitive odds everyone is imagining DO NOT EXIST. The prize is in a fixed location. It doesn't matter if there are 3 doors or 3 million doors. No matter how many doors you open or close the location of the prize remains constant. The contestant only has one shot at it. If you don't pick the winning door you loose. If you only play this game ONE TIME there is absolutely no reason to switch doors. Simple as that.