When he said “Hannah, pick a door” I got scared because that’s my name

If the goat's behind the door long enough, there will be poop too.

This helped me grasp the real issue here, you probably picked the wrong door in the first place so you should switch. Let's say that there are not three, but ten thousand doors. You pick one at random and Monty Hall opens not one but 9,998 of the remaining doors to reveal empty rooms. Now there are just two closed doors, your first choice and the one Monty did not already open. Now, you could look at the remaining closed doors and tell yourself that it is a fifty/fifty chance so you might as well stick with your first guess. However, there is only a 1/10,000 chance you chose correctly the first time. There is thus a 9,999/10,000 chance that the other door Monty left closed is the winner. Because you almost never pick the correct door when up against 10,000 choices, Monty opening all the other doors except one is telling you that THAT is the door with the prize behind it. Now 2/3 is not as obvious as 9,999/10,000, but the principle is the same -- the odds are you chose the wrong door which means Monty is showing you the correct one. Take it.

I think a better way of looking at it is to imagine if Monty didn't open ANY doors but instead asked you if you would be willing to give up your chosen door in exchange for being allowed to open up BOTH of the other doors. That is essentially what this problem and your choice comes down to. Him opening up the one door ahead of time is simply a distraction from that choice.

"A great analogy for the Monty Hall problem is this sack" *reaches into pants*

There's actually very few possible outcomes, so let's just list them all: 1) You pick a door with a goat. You don't switch. You get the goat. 2) You pick the other goat door. You don't switch. You get the goat. 3) You pick the money door. You don't switch. You get the money. 4) You pick the goat door. You switch. You get the money. 5) You pick the other goat door. You switch. You get the money. 6) You pick the money door. You switch. You get a goat. Those are all the possible outcomes. Of the three not-switching outcomes, only one got you the money. 1/3 chance. Of the three switching outcomes, two of them got you the money. 2/3 chance. You have a 1/3 chance of winning if you don't switch, and a 2/3 chance of winning if you do switch. So there you go.

Great explanation - the absolute key to this is that the host KNOWS where the money is and NEVER opens a door with the money. This fact is the crucial piece of the rules that makes this the case.

Easiest way to explain this is by pointing out that there's never actually a point where you are working with 50/50 odds. You might think you are, but you're either sticking with your 1/3 odds door, or Monty's 2/3 door.

Alternate title for the video: Michael writhes around on a table for 14 minutes talking about goat poop

"You'd have to take it home and take care of it, all that normal poop stuff."

The door you pick has a 1/3 chance of winning. That means the remaining 2 doors collectively have a 2/3 chance of winning. The host eliminates one of the doors meaning the remaining door you didn’t choose has that 2/3 chance of winning all to itself. So it’s better to switch

There are 2 goat doors, and the host will always open a goat door. Therefore, switching will always result in the opposite of your first choice. There is a 2/3 probability of being wrong with your first choice, and therefore switching in either of those 2 cases results in switching to the winning door. 2/3 is the chance of winning *if you switch doors.*

I already understand it, I am just here to watch Micheal lying on a table

0:33 "But maybe it does" _VSauce music starts playing_

WHEN FIRST CHOSEN, The chance of any door being right was 33%. So total = 100%. When one door is removed, the chance of the chosen door being the right choice is still 33% so the chance of the remaining door being right must be 100%-33%=66%.

Draw a tree of possible outcomes given every possible combination of choices, and it becomes clear that the mathematically correct choice is to switch.

0:21 "Kevin's recent video" Recent: More than 5 months ago

Best explanation I've heard: Imagine there were 100 doors instead of 3. 99 of them have goats and one has the money. You pick one door, and the host opens 98 doors to reveal goats, leaving only one other door closed. It should be DEAD OBVIOUS that it's the money door. The only way it won't be the money door is if you picked the money door initially, which is only a 1 in 100 chance.

I think it's cool that he did this in (apparently) one take, and ran the experiment at least twice in the video having it work out in the most pedagogically desirable outcomes both times.

The easiest way to explain it is this: Imagine that you’ve played with 1000 doors instead of 3. And after you’ve chosen the door Monty Hall opens 998 other doors to show that all of them have goats in them, leaving only your and another one door closed. Would you then switch to the other door? Yes, of course switch. Because you’ve had a 1/1000 chance of choosing the correct door in the beginning. And if you don’t switch - you’re still betting that you’ve won a 1/1000 chance by accident (or more likely - 999/1000% of losing), ignoring 998 doors that were opened. But since we know that Monty Hall cannot open the door with the Money - he HAD to leave the door with the Money closed. Door with the Money AND your first chosen door, that is the most probable scenario here. You will have 99,9% of winning money if you switch the door in that example. And if you think about it - in the original Monty Hall problem with 3 doors, Monty does the same thing: he opens ALL doors other than your and another one. Yes, you are not as guaranteed of winning as in the 1000 doors example, but you still are more probable to win if you switch doors. Fun fact: when I was a student about 8 years ago I could not believe it wasn’t a 50/50 chance. So I asked a friend of mine to determine the probability by making a lot of blind tests. I’ve put a coin in one of 3 cups and asked my friend to choose, then revealed one empty cup and asked her to always switch the cup. Repeated that 100 times. If I was correct the result would more incline to 50% of choosing the other door. Like a coin toss. But we ended up with like 62/100 winning (or 64, I don’t really remember now). Much closer to a 66% than 50%. So this little “field test” proved me wrong =) Took me a while to understand why I was wrong though))

I don't know whether I should be scared for Michael or of Michael.

"Oh hey this is a cool video to show to my fami-" "GRR BABY GRR"

Can we just appreciate that he got 11 minutes into this video in one cut and then proved how it works with the marbles

I'd thought I'd heard enough explanations of the Monty Hall problem at this point, but I'm glad I watched this one. For some reason I never thought of it as simply as when you're using the always switch strategy, then if your choice of door at the start is wrong, you win. When put that way, the 2/3 odds are dead simple to see.

The phrase that got me to understand was "2/3rds of the time you pick the wrong answer"

Michael, reaching into his pants: "And it contains a sack"

One more way to help simplify this: the host, in opening one of the incorrect doors, could also have just said, “you can have what is behind both of the doors you didn’t choose, or keep the one you have.”

I just watched this and, while I've seen explanations of the Monty Hall problem before, this is the first one that made sense to me insofar as I ended the video with understanding. Your explanation of the rules of the game and the host's obligations and the fact that they reveal additional information (somewhere around the mid-point) caused the concept to click in my mind. I really appreciate that. Thank you.

Me: *trying to watch something educational in front of parents* Michael: HeY DiNGaLnGs!

Michael: WHATS UP DINGALINGS KZfaq: *Demonetized*

This is a really good explanation but the one that convinced me was the adding more doors explanation. Lets say you gave 1000 doors with 999 goats and 1 money door. You have a 1/1000 chance of getting the money. Same as the monty hall problem when you choose the host will open 998 doors with goats and you can choose between the one you chose and the one left. With more doors it becomes increasingly more obvious how the math behind the problem works its just hard to understand the three door one because it's not that large of a difference

I first came across this in a Parade column in a newspaper over 30 years ago.I was shocked that mathematicians would write in explaining that it wouldn’t benefit you to switch.

Michael: if you pick the door with the poop, like the goat you would need to take it home, take care of it, feed it, all that poop stuff *incredible*

I thought Michael had fused with that table, now I'm disappointed.

This is the only video that has made this make sense, thank you so much! I was so focused on the probability after the door was removed, it didn't occur to me to think of the probability beforehand!

It's helpful to group the "doors" together. Instead of doors, you have 3 boxes on stage, and one has a prize. When you choose a box, they move it down into the audience. It's easy to understand that there is a 1 in 3 chance the prize is in the audience and there is a 2 in 3 chance the prize is still on stage. Opening an empty box doesn't change the odds that the prize is on stage, so of course you should switch to the remaining box on stage

Michael Stevens is a committed man. Micheal- "Let's do a whole episode laying on a table." Hannah- ".........Ok."

This is the first time the monty hall problem has really "clicked" for me. 1/3 of the time you will pick the door with the money, and when you switch you will lose the game. However, the other 2/3 of the time you will pick the door without the money. In these cases, the host will remove the other door without the money, meaning that the only door that you can switch to is the door with the money behind it. Therefore, if you follow this strategy, you will win 2/3 of the time. Thanks for this explanation!

When explaining this I like to elaborate and say let’s do it with 10 doors. One has money behind it, 10% chance of picking money, remove 8 doors with goats behind them, and you’ll win 90% of the time if you switch to the remaining door. People for some reason don’t understand the small scale of 3 doors when I explain this to them. But you did a great job explaining it 👍

8:25 for anyone who’s a skeptic, that 30 second clip is all you need. The real curiosity in this is why people who try explaining seem to ALWAYS unnecessarily over complicate it. The door that you don’t choose has a higher probability of having the prize because there is a 2/3 chance that you chose the wrong door to begin with. The host then reveals another door that is not a winner, telling you 2 things: The door that you chose only has a 1/3 chance of having the prize. That you already knew going in. And the second: the door that the host does open has a zero percent chance of having them prize. Therefore, the remaining 2/3 probability lies with the remaining door.

The fact that he's lying on a table actually makes it more interesting

This video is weird 1. "What's up dingalings" 2. Goat poop 3. Lying on a table the whole video 4. The ending

Basically it's not 50/50 because you're using host's knowledge of where the money is which gives you 33.3% more chance to win when you switch.

A goat's friendship is worth much more than a million dollars

Who misses the iconic Vsauce tune when something trippy happens. I know this ain't Vsauce.

Whenever I come across folks who struggle with the math, I always extend the scenario out and say, "There are a BILLION doors in front of you with poop behind all of them but one, which has a million dollars!. Pick a door." I then take away ALL the doors except the one they chose and another door (say No. 171). I then ask them if they want to change the door they chose. Believe it or not, some people still insist it's still "50/50" :D

I understood that problem only when I tried to build a simulation on python. So I thought how could I imitate that change of the chosen door and it came to my mind that: 1. If you chose the right door and changed your door later -- you 100% got the wrong door. 2. If you chose the poop door and changer your door later -- you 100% got the right door. AND: as there are 2\3 poop doors, your overall chances to get the right door by changing your decision are 2/3.

today we are going to discuss the monty hall problem _maybe its goat poop_

"Fewer people want poop-some still will-but the point is you're supposed to want the money."

In a sentence: "Switching doors is a bet that your original choice (only 1/3 chance of being right) was actually wrong, and that's a good bet."

This is the best, most intuitive, clear explanation I have seen of this puzzle. So, thank you for creating “yet another” video about it!

Michael is the only one that'll create an educational video lying down on a table and still get 2.7m views with 100k+ likes

2017: Vsauce is a science channel 2019: Vsauce is a meme channel. Plus, it's mostly DONG content

Your original pick has 33% chance of being right. The other 2 door each have 33% chance of being right, for a total of 66%. But the nice host eliminates one of of those 2 doors, so that last possible door contains the entire 66% chance. So you are twice as likely to win if you switch.

The reason you gave at the beginning for presenting "yet another" explanation is so true. Wise, even! And you really did help me with this. Especially when you began using the marbles-in-a-bag version. Then it became so clear--very nearly intuitive!

2017: Michaels head 2018: Michaels head and shoulders 2019: All of Michael 2020: Michael invisible

Never thought I'd see Michael lay on a table and pull a sack out of his pants on KZfaq.

Youre more likely to pick wrong than right (1/3 compared to 2/3) so you have better chances to switch to a "right" door than to stay on an unlikely right initial door.

basicly you switch the wrong answer 2/3s of time and get the correct answer vs the switching the correct answer 1/3 of the time for a incorrect answer

Michael lies on a table for 14 minutes

9:22 - Michael squirms across the table top moaning before pulling out his sack and proceeding to tell us about how he is going to pull one out.

Simply put, your probability is based on your initial pick being 2/3 (66%) likely to be wrong thus switching would net a 66% chance of being right.

If you pick wrong first and switch, you win. If you pick right first and switch, you lose. So the question becomes how often do you pick wrong first. Two out of three are wrong.

4:51 No, that's a $ on a yellow background. That's demonetization.

Why doesn't he just ask youtube for the password to vsauce?

There’s 2/3 chance the money is behind one of the two doors you didn’t pick. The host removes one. The remaining door has 2/3 chance of money all by itself.

Yeah it's simple. The first door you choose has 1/3 of having the $. The remaining 2 doors (together) have 2/3 chance of having the $. After the host eliminates one of those 2 doors, the last door now has the 2/3 probability all for itself. So switching to it would be the smart thing to do.

michael : “ you want the black marble” me : “ i want the black marble”

This video perfectly describes why Michael is just the internet's dad, trying to make educational topics interesting by trying to imitate how kids talk and do things.

Suppose I can choose the door 9 times. So 3 times I'll choose the car. Then if I switch I'll make a mistake 3 times. 6 times I'll choose the wrong door. Then the host is forced to choose the door with a goat. That means the other door has the car. So, by switching I'll choose the car 6 times. Double than the mistakes.

Another way to look at it is from the host’s perspective: He always sees one of three potential combos: poop/cash, cash/poop, or poop/poop. In all three cases one poop will be removed, leaving three possible outcomes: cash, cash, or poop - hence the two-thirds odds.

Have you tried "password" as the password for the Vsauce account?

This problem can be thought through an exaggeration: Imagine there were 100 doors instead - behind them are 1 case with a million dollars and 99 poops. If you now choose a door and Monty Hall opens 98 doors with poop, leaving your door and one other door, either of which has the money, it's obvious that you should switch since initially picking the money was so unlikely. EDIT:Wow I'm surprised this got so many likes and sparked such a rich discussion :D

If you were initially wrong(2/3rd times) and you switch, u will definitely be right If you were initially right(1/3rd of the time) and switch, you'll definitely be wrong. That is, by switching, you'll finally be right 2/3rd of the time. Not switching is just like guessing 1 of 3 options and nothing further than that, ie, you're right only 1/3 of the time.

Seems intuitive. You pick 1 of 3. Then one is eliminated, but never the correct one. 1/3 you chose correctly, an incorrect is eliminated, and an incorrect remains. 2/3 you chose incorrectly, host eliminates the only other incorrect, making the last door the correct one. Your initial choice forces the host's hand. If you were wrong, which you are 2/3 of the time, the host must tell you which door is correct by eliminating the other wrong door. Simple. Just remember that none of the actions are independent. His choice depends on yours.

The best example I've seen that makes understanding the gain in information when Monty opens a door really intuitive is this: Imagine there are 1000 doors, $1 000 000 behind one of them, goats behind the other 999 doors. You pick one of them, obviously not really expecting to pick the right one. Then Monty opens *998* of the remaining doors, guaranteeing that the money is not behind any of them. After rounding up the 998 rowdy goats, there is the door you picked and one other door _mysteriously singled out_ from the other 999. Now it feels pretty obvious to me that switching is a good idea. The 3 door problem is the same thing but the information gain is less obvious because it's not as big.

The easiest way to understand is that if you pick incorrectly, switching will automatically get you the money. And you pick incorrectly 2/3 of the time, so switching gets you the money 2/3 of the time (whenever you pick wrong)

Just think of it like this: Switch camp: Win: you first picked a goat Win: you first picked the other goat Lose: you first picked the money Stay camp: Win: you first picked the money Lose: you first picked a goat Lose: you first picked the other goat Switch: 2/3 wins Stay: 1/3 wins

my mom understood it with this explanation: you have a 2/3 chance of picking a goat at the start. the host will ALWAYS eliminate the other goat. since now you only have two doors to choose from, and your original door has a 2/3 chance of being wrong, that means the other door has only a 1/3 chance of being wrong, and therefore you should pick the other door.

Does the world need another Monty Hall video ? Michael "Well, no, but actually yes."

I agree with everything thing this video has to offer. The solution is solid, the table seems to be a good choice of support substrate, swapping goats for poop. 10/10.

Another way of thinking about it is this - whatever you choose, goat or money, if you switch, you ALWAYS change to the other option. Goat will always become money, money will always become goat. 2/3rds of time you pick a goat. So if you always swap, 2/3rds of the time you will swap to the money.

I have struggled wrapping my head around this for so long and I would argue for hours about it with anyone that would listen but you today have made the arguement that I needed to finally wrap my head around it

my favorite way of understanding this very quickly and easily is this: There are 100 doors. one has money, the rest have goats. the game is played the exact same way, but rather that revealing one goat, the host reveals 98. so here's how the game goes - you pick a door (1/100 chance you got the money), the host reveals 98 of the goat doors, leaving one left. stick or switch? well lets reasonably assume you didn't get the 1/100 chance. the door he leaves MUST be the money. you switch.

"True understanding comes from hearing as many voices as possibe" damn, watching michael justify his schizophrenia is sad.

The best way I've seen to explain why it's 2/3 and not 50/50 is just do the same thought experiment but with 100 doors. You pick one, the host opens 98 doors that aren't the cash and asks if you if you'd like to switch. The odds you picked the right one to begin with are so low, you are almost 100% guaranteed to win if you switch. With 3 doors, you still have an advantage but the fact that there are only 2 doors after he opens 1 tricks people

Best explanation ever. You actually convinced me. Wow. Very well put.

Marbles: You pick a marble. Monty then says, you can keep the marble, or switch for this sack with 2 marbles. You say switch. Monty removes one white marble from the sack. Still want the sack? Of course.

“There’s an analogy that makes this more clear” *reaches for his sack*

"Fewer people want poop. Some people still want it" Words to live by.

You had 3 choices. You picked 1. It has a 1/3 chance of being right. What anyone does to anything else doesn’t change that. If only 1 alternative remains, it has to have a 2/3 chance.

Unpicked two doors have 2/3 chance combined. The host removes one door but the total probability is still 2/3.

"Kevin's recent video about paradoxes" - Apr 24, 2018

"Fewer people want poop, some still will, but THE POINT IS..."

Yup. Every time you pick a goat, which is 2/3 of the time, the host will be forced to open the other door with the goat, meaning the door the host didn't open is the one with the money. So yeah, 2/3 of the time, the door the host didn't open has the money, so you have a better chance of winning by switching.

The crucial thing to understand is that the only "chance" step in the entire game, is the first choice of the first door. Everything else is set at the beginning and isn't remotely chance related. So, when you first choose, the chance of picking the $ door is 1/3, and the chance that $ is behind the other doors is 2/3. If you can get to the other two doors then your chances would be 2/3. Great, you switch, and if there was any $ there, you get it. You're not flipping a coin to decide whether to switch. You always switch. The "game" is, in fact, that you pick a door and win if the $ is behind one of the other two doors. (If you think of the game as being the pre-written plan that you pick a door and the host shows you which other door to not pick of the two and you switch. None of that is a probability after picking the first door, it's all known data.)

So nobody going to talk about how he called us ding-a-lings? No? Okay lol

2:24 "you gotta take the poop home take care of it feed it and all that normal poop stuff"

You have 2/3 to bring the goat at first so is more porbable to get it wrong than wright, but if you are wrong than you have a 100% probability that the Car is in the other door because the conductor will always show the other goat

You're telling me the Monty Hall problem was this easy to understand? Wow I've seen such a good explanation