Probability and the Monty Hall problem | Probability and combinatorics | Precalculus | Khan Academy

Here we have a presentation and analysis of the famous thought experiment: the "Monty Hall" problem! This is fun.

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[moltrev]

Okay, let's go through every single possible outcome of this situation (Note: if you're too lazy to read all of this, skip to the end to read a summary!):

-Door 1 has a car, door 2 has a goat, and door 3 has a goat.

-You pick door 1.

-Door 3 is revealed to have a goat.

-You switch to door 2.

-You lose.

-You stay with door 1.

-You win.

-Door 2 is revealed to have a goat.

-You switch to door 3.

-You lose.

-You stay with door 1.

-You win.

-You pick door 2.

-Door 3 is revealed to have a goat.

-You switch to door 1.

-You win.

-You stay with door 2

-You lose.

-Door 3 is revealed to have a goat.

-You switch to door 1.

-You win.

-You stay with door 2

-You lose.

-You pick door 3.

-Door 2 is revealed to have a goat.

-You switch to door 1.

-You win.

-You stay with door 3.

-You lose.

-Door 2 is revealed to have a goat.

-You switch to door 1.

-You win.

-You stay with door 3.

-You lose.

-Door 1 has a goat, door 2 has a car, and door 3 has a goat.

-You pick door 1.

-Door 3 is revealed to have a goat.

-You switch to door 2.

-You win.

-You stay with door 1.

-You lose.

-Door 3 is revealed to have a goat.

-You switch to door 2.

-You win.

-You stay with door 1.

-You lose.

-You pick door 2.

-Door 1 is revealed to have a goat.

-You switch to door 3.

-You lose.

-You stay with door 2.

-You win.

-Door 3 is revealed to have a goat.

-You switch to door 1.

-You lose.

-You stay with door 2.

-You win.

-You pick door 3.

-Door 1 is revealed to have a goat.

-You switch to door 2.

-You win.

-You stay with door 3.

-You lose.

-Door 1 is revealed to have a goat.

-You switch to door 2.

-You win.

-You stay with door 3.

-You lose.

-Door 1 has a goat, door 2 has a goat, and door 3 has a car.

-You pick door 1.

-Door 2 is revealed to have a goat.

-You switch to door 3.

-You win.

-You stay with door 1.

-You lose.

-Door 2 is revealed to have a goat.

-You switch to door 3.

-You win.

-You stay with door 1.

-You lose.

-You pick door 2.

-Door 1 is revealed to have a goat.

-You switch to door 3.

-You win.

-You stay with door 2.

-You lose.

-Door 1 is revealed to have a goat.

-You switch to door 3.

-You win.

-You stay with door 2.

-You lose.

-You pick door 3.

-Door 1 is revealed to have a goat.

-You switch to door 2.

-You lose.

-You stay with door 3.

-You win.

-Door 2 is revealed to have a goat.

-You switch to door 1.

-You lose.

-You stay with door 3.

-You win.

Now let's look at the numbers:

-Number of times you played: 36

-Number of times you won: 18

-Number of times you lost: 18

-Number of times you switched: 18

-Number of times you stayed: 18

-Number of times switching made you win: 12

-Number of times switching made you lose: 6

-Number of times staying made you win: 6

-Number of times staying made you lose: 12

I'll let you come to your own conclusions about this data...

[MrManisangsu]

Thanks for the nice video. Especially the Intuitive way that "if you initially chose wrong you should always win by switching" - which is 2/3.
After a lot of confusion I really understand this now. The key point is that Monty opening a door is not a "random" variable.

[rogerbodey]

A two line solution:

1. By swapping you always get the "opposite" of your original choice.
2. Two thirds of the time, your original choice will be a goat.

That's it.

[TraceguyRune]

I am a software programmer. I wrote a simulation with 2, 000, 000 contestants split into 2 control groups. Control group A stayed with their first choice. Control group B switched.
Control group A won the car 33% of the time.
Control group B won the car 66% of the time.

[youracherrynut]

6:59 - 7:18 made sense to me and helped me understand the problem. At the outset, you have a 1/3 of picking correctly and a 2/3 chance of picking incorrectly. Also at the outset, if you consider two doors together, you have a 2/3 chance of picking correctly. When one of the doors is eliminated, you essentially "capture" that 2/3 probability by switching doors from your initial selection. It's almost as if you got to pick two doors at the same time, but not really. More information - Monty revealing one of the doors that has a goat - improves your odds if you switch.

[briant]

I finally got it from this video, after about 2 hours of reading articles and watching videos.

It goes like this. If you stick with your original choice your odds are only one third all the way through. BUT... When you switch your odds DOUBLE because there is an EXTRA door! Monty HELPS you by telling you where the prize is not, thus increasing your odds by another one third, equaling 2/3. Lol

[MrOoooskar]

Thanks, Khan Academy, I finally understood the Monty Hall problem. You have by far the best explanation.

[TraceguyRune]

If I am married, and have 2 mistresses, and my wife finds out, should I stay with mistress #1 or pick Mistress #2?

[rohanchung]

Khan you always explain things in order for viewers to easily understand. God bless.

[ieornl]

Great explanation! I now understand the 2/3 chance of winning when switching! I mean I knew it's 2/3 but never really understood it.

[jomi]

You should have known going in, or realized very quickly (because it stands to reason), that Monty Hall would show you a goat because that's the only way the game works, he can't show you the car or it's game over.  So you have to look at it as if one door is already removed, you just don't know which one until he shows you.  Your probability of picking car or goat initially is still 1/3 and 2/3 respectively.  When he shows you a goat he's only giving you the illusion of new information, when really he's just revealing which door was the removed one.  So it doesn't change the probabilities at all.  Your choice isn't between 2 doors, it's between staying with your 1/3 odds that you picked car right away, or 2/3 odds that you picked goat and are switching to car.  Smart money says take the 2/3.

[timrussell]

FACTS:
1) Chance of a door hiding car = 1/3
2) Monty picks from 2 doors = 2/3 chance one hides the car.
3) Monty ALWAYS picks a goat...
4) ...leaving you CERTAIN to win a car in the 2 out of 3 cases Monty has one.

The *only* way you can lose by switching is if you correctly guessed the car originally (1/3 chance). Monty eliminates a goat in the other two scenarios, leaving only the car. Switching gives you a 2/3 chance of winning.

You're welcome!

[Aletheia]

The MH problem is great because it appears at first glance to be counter-intuitive and posing the problem to friends reveals just how stereotypical our decisions are when made under conditions of incomplete knowledge. This reliance on heuristics unveils the systematic biases in our decision-making yet problems like these can show us normative methods of reasoning like actually applying probability theory to problems involving uncertainty instead of "thinking with yer gut" in every situation.

[abeda]

Yes! Absolutely enjoyed every second of that! Especially the fact that that first explanation came from a student's point of view 👏🏻

[Esus]

Based on the way he is describing it Monty is showing you the goat curtain for no apparent reason. However, according to the laws of game show economics Monty cannot be giving out cars (which were always meh level but that's not important ) to every person who comes on the show. When the producers signed Monty and explained the routine they noted "Now look Monty forget about any fantasies about taking spokesmodel Carol Merrill to the Bahamas for a tryst if you let players win cars all the time. Because it's completely out of your bonus buddyboy. Monty thinks: I had better make sure they lose as much as possible. So I if you pick the goat curtain Monty's goto line is "You picked curtain number three, well lets see if you won that [Oldsmobile your grandmother always wanted]! At this point it gets all mathy which is I'd not really my thing.

[kingxerocole]

Here's another way of thinking about it:

Say that, instead of 3 doors, the game involves 99 doors. After picking one door, the host removes every other door except the one you chose and one that he chose. He then tells you that either your door or his door has the car.

Picking the host's door is a no-brainer at that point; you have a 98.9% chance of being wrong the first time. The reason the Monty Hall problem is so tough is that the difference between 1/3 and 2/3 isn't THAT big. You could still have easily picked the right door the first time. And, if you lose, knowing that the odds were technically in your favor won't make you feel better.

[AdeshAtole]

The only video on this topic which made sense to me

[satisfiction]

I agree fully. This is a 50/50 shot after the revealed curtain.

[richardbuxton]

approve of your analysis - the host opens one of your two alternative doors for you - he shows you your own goat - what a nice man.

Then he makes the switch offer.

The revealed goat is not a pointer to the car - it's just one of two goats and can always be shown. The probabilities with the doors all closed are exactly the same when all the doors are opened - the door that actually hides the car has only a 1/3 chance probability of having the car.

Two doors are better than one.

[DanielWillen]

So what you don't know, but picking random there's a 2/3 chance you randomly select a door with a goat. Then the host MUST open the only other door with a goat in it since he can't open the one with the car, you switch and you win. If you're REALLY sure the car is in door X, then you choose that door in stick with it, but you don't know so the chances are 1/3 if you don't switch.