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

On August 10, 2019 by Raul Dinwiddie

Let’s now tackle a

classic thought experiment in probability, called

the Monte Hall problem. And it’s called the

Monty Hall problem because Monty Hall was the game

show host in Let’s Make a Deal, where they would set up

a situation very similar to the Monte Hall problem

that we’re about to say. So let’s say that

on the show, you’re presented with three curtains. So you’re the contestant, this

little chef-looking character right over there. You’re presented with

three curtains– curtain number one, curtain number

two, and curtain number three. And you’re told that behind

one of these three curtains, there’s a fabulous prize,

something that you really want, a car, or a vacation, or some

type of large amount of cash. And then behind the

other two, and we don’t know which they

are, there is something that you do not want. A new pet goat or an

ostrich or something like that, or a

beach ball, something that is not as good

as the cash prize. And so your goal is to try

to find the cash prize. And they say guess

which one, or which one would you like to select? And so let’s say that you

select door number one, or curtain number one. Then the Monte

Hall and Let’s Make a Deal crew will make it a

little bit more interesting. They want to show you

whether or not you won. They’ll then show you one

of the other two doors, or one of the

other two curtains. And they’ll show you one

of the other two curtains that does not have the prize. And no matter

which one you pick, there’ll always be at

least one other curtain that does not have the prize. There might be two,

if you picked right. But there will always be

at least one other curtain that does not have the prize. And then they will

show it to you. And so let’s say that they

show you curtain number three. And so curtain number

three has the goat. And then they’ll

ask you, do you want to switch to curtain number two? And the question here is,

does it make a difference? Are you better off holding

fast, sticking to your guns, staying with the original guess? Are you better off switching

to whatever curtain is left? Or does it not matter? It’s just random

probability, and it’s not going to make a difference

whether you switch or not. So that is the brain teaser. Pause the video now. I encourage you

to think about it. In the next video, we will

start to analyze the solution a little bit deeper, whether

it makes any difference at all. So now I’ve assumed

that you’ve unpaused it. You’ve thought deeply about it. Perhaps you have

an opinion on it. Now let’s work it

through a little bit. And at any point,

I encourage you to pause it and kind of

extrapolate beyond what I’ve already talked about. So let’s think

about the game show from the show’s point of view. So the show knows

where there’s the goat and where there isn’t the goat. So let’s door number

one, door number two, and door number three. So let’s say that our

prize is right over here. So our prize is the car. Our prize is the car, and

that we have a goat over here, and over here we

also have– maybe we have two goats in

this situation. So what are we going

to do as the game show? Remember, the contestants

don’t know this. We know this. So the contestant picks door

number one right over here. Then we can’t lift

door number two because there’s

a car back there. We’re going to lift

door number three, and we’re going to

expose this goat. In which case, it

probably would be good for the person to switch. If the person picks door number

two, then we as the game show can show either door number

one or door number three, and then it actually

does not make sense for the person to switch. If they picked

door number three, then we have to

show door number one because we can’t

pick door number two. And in that case, it

actually makes a lot of sense for the

person to switch. Now, with that out of

the way, let’s think about the probabilities

given the two strategies. So if you don’t

switch, or another way to think about this strategy is

you always stick to your guns. You always stick to

your first guess. Well, in that situation, what

is your probability of winning? Well, there’s three doors. The prize is equally likely

to be behind any one of them. So there’s three possibilities. One has the outcome

that you desire. The probability of winning will

be 1/3 if you don’t switch. Likewise, your

probability of losing, well, there’s two ways

that you can lose out of three possibilities. It’s going to be 2/3. And these are the

only possibilities, and these add up to

one right over here. So don’t switch, 1/3

chance of winning. Now let’s think about

the switching situation. So let’s say always–

when you always switch. Let’s think about how

this might play out. What is your

probability of winning? And before we even

think about that, think about how you would

win if you always switch. So if you pick wrong

the first time, they’re going to show you this. And so you should always switch. So if you pick door

number one, they’re going to show you

door number three. You should switch. If you picked wrong

door number three, they’re going to show

you door number one. You should switch. So if you picked wrong and

switch, you will always win. Let me write this down. And this insight actually came

from one of the middle school students in the summer camp

that Khan Academy was running. It’s actually a fabulous

way to think about this. So if you pick

wrong, so step one, so initial pick is wrong,

so you pick one of the two wrong doors, and then step

two, you always switch, you will land on the car. Because if you picked

one of the wrong doors, they’re going to have to

show the other wrong door. And so if you

switch, you’re going to end up on the right answer. So what is the probability of

winning if you always switch? Well, it’s going to be the

probability that you initially picked wrong. Well, what’s the probability

that you initially picked wrong? Well, there’s two

out of the three ways to initially pick wrong. So you actually have a

2/3 chance of winning. There’s a 2/3 chance you’re

going to pick wrong and then switch into the right one. Likewise, what’s

your probability of losing, given that you’re

always going to switch? Well, the way that you would

lose is you pick right, you picked correctly. In step two, they’re

going to show one of the two empty

or non-prize doors. And then step

three, you’re going to switch into the

other empty door. But either way, you’re

definitely going to switch. So the only way to lose, if

you’re always going to switch, is to pick the right

the first time. Well, what’s the

probability of you picking right the first time? Well, that is only 1/3. So you see here, it’s

sometimes counter-intuitive, but hopefully this makes

sense as to why it isn’t. You have a 1/3 chance of winning

if you stick to your guns, and a 2/3 chance of winning

if you always switch. Another way to

think about it is, when you first make

your initial pick, there’s a 1/3 chance

that it’s there, and there’s a 2/3

chance that it’s in one of the other two doors. And they’re going to

empty out one of them. So when you switch,

you essentially are capturing that

2/3 probability. And we see that right there. So hopefully you enjoyed that.

When switching you still only have 1/2 chance of winning ,saying 2/3 in incorrect.

i knew this from mythbusters

but 2/3 of the choices is wrong, and when it's wrong, it's in the other door.

I chose to ignore your facts and go with mu gut on this and i'm hopping you,ll see the error of your way,s and apologize to queen Sheba because the voices in my head cant come to the phone.

The explanation indeed only makes sense in this specific scenario where one door being eliminated is part of your strategy. If you would repeat this scenario often enough, you would find that you win 66% of the time by switching. Under any other circumstances it would basically boil down to 50%.

Fascinating.

Mythbuster explained and did an experiment on it too .. !

LOOK for YouTube user NIANSENX and play his "The Monty Hall Problem" several years later and with over a million views he used Goats and a Car too. Actually this is better done than Kahn's but I still give Kahn much respect for what he is doing. I only found it strange he would pick something so widely distributed already.

he didnt stay with the first choice

This problem is flawed in the assumption that I would not want a goat.

I wonder how this concept would have worked on "Deal or no Deal" all with briefcases with various amounts of cash prizes in them ranging from $1 to $1million. I think there were 23 or so briefcases. You picked one at the start, and then eliminated the others but you were always given a chance to swap out your chosen case with others. According to this, my guess is that it would have been wise to always switch there too.

What if I want the Goat ? ðŸ™‚

a year ago I had a hard time explaining the solution of this problem to one of my probability "Professors" ðŸ™‚

If I want the ostrich how does my strategy change? I'm thinking of starting a farm.

Thank you so much, I finally understand this ðŸ˜€

I admit that you start out with a 1in 3 chance them Monty shows you a door which gives you a 1 in 2 chance ,it was alway,s 1 in 2 ,not 2 in 3 ,but im not to smart and never met Monty so it matter,s very little to me .

It becomes really clear if you have 100 doors and the game-master opens 98 of them for you and you're allowed to pick another one each time. The likelyhood that there is a car behind the other door is much higher.

Yup your right!

I finaly got it ,your right ,thank you !

Did you really just come in here, watch the video, not understand it and then try prove khan wrong?

Did the host always offer a switch?

That is really interesting, didn't expect the correct answer at all (I didn't think it would make a difference). I'm really enjoying these more informal lessons, they're fun and neat and the uber upper level calculus ones are still over my head x)

This was on MythBusters

What if you want to win the goat?

You're right (like you need me to confirm that!). I looked into it a bit more last night, and found a site that emulated the scenario. I performed 100 passes with each strategy (always stay, always switch). 100 tries is a small sample, but it was enough to show me the probablities at work.It's just counter-intuitive to my thought process.

I know, what my odds are, but thanks for clarifying :)! But your last sentence is correct….

This problem is flawed in that you are calculating based on three doors. The probability of the show revealing one of the goats is 100%. So you are only choosing between two doors. So applying probability to the choice will be 50%. So switching is not good our bad. Probability of winning the game as a whole is 66.6% because they are going to reveal a door the is incorrect but that has nothing to do with switching

Wrong. And here's why: if we generalize this to 100 doors, and the host opens 98 doors, while KNOWING which one has the goats behind them, and being unwilling to open the door with the prize behind it, then it would be advantageous to switch. Switching is much more likely to get you the correct one, because your original guess had a lower probability than the UPDATED guess with the new information.

I find it hilarious that people here have the audacity to tell Sal Khan that he's wrong.

The problem is famous for a large number of people not being able to understand it.

this is BS i'm sure it's 50 50 either always switch or no switch

Nope. Read my response below, also look it up on wikipedia and read the section on Bayes' Theorem if you don't believe me. It's a fact that switching is always better. I've also written a simulation in Python if you still don't believe me: codepad.org / 2FCv9dQv (remove the spaces)

you are always gonna see a false door no matter what you chose and then the whole game changes to 50 50 chance of correct door

No, it doesn't. Since the host knows which door is the correct one, and he will never open the door that has the prize behind it, he effectively eliminates one choice. Now with that choice eliminated, the probability that the other door is the correct one is higher than your original probability of 1/3. You can even try this with a deck of cards if you don't believe me. Get 3 cards, one an ace, and the other ones 2s, and then get another person to play the game.

You would have a 95% chance of winning if you switched with 23 briefcases.

you were not really given the low priced cases by Howie, you were given a chance to open cases at random throughout the game, but once you opened it, the cash in that case was eliminated. So, there was a chance of eliminating the million throughout the game; however, if you had two cases, one with the million, then switch, because there is a 22/23 chance the other case is the million. The case you chose has only a 1/23 chance.

Probability is about the unknown. Using three doors or 100 does not matter if in the end you only have to choose between 2 doors. Because probability is concerned with unknowns. Kahn is either trolling you or forgot that is not how this thought exercise work.

enlightenment at last, this one has bugged me for a while, thanks!!

ðŸ™‚

I finally understand this MH problem completely. Thanks!

If you want a goat, simply never switch. The reasoning holds true ðŸ™‚

If you get the goat, good.

If you get the car, sell it. Buy goats.

ðŸ™‚

I know you should switch the solution provided confuses people. Because I people think in the positive. You are calculating a negative outcome to confirm your decision. You switch because you 2/3 likely picked the wrong one.

When they show you a wrong box and they ask if you want to switch, they basicly change the game into: 'Here you have two boxes, one of the boxes contains a car while the other contains a goat… pick one'. In that case it doesn't matter what you pick. Or am I misrepresenting the situation now?

It's all about what you did in the beginning. The odds of picking a goat in the beginning was 66%, so most often you'll have a goat. When the other goat is revealed, the more likely case is that the car is behind the other door. Basically, when it's down to two options, it's more likely that your door is a goat door.

I dont see why switching changes your odds. You dont know if you picked right or picked wrong. they are always going to eliminate one of the goats. So it seems at that point its a 50/50 choice and if you dont know if you picked right or wrong it doesnt matter if you switch. Id like to see if there have been experiments on this to see if there is an advantage either way.

There has been millions of computer simulated trials showing its 33/66

WRONG IT'S ACTUALLY 1/5

I don't see what all the arguments are about, this problem is basic. If you choose a door with a goat in it to begin with (that's a 2/3 chance you do) the host has to open a door with the other goat, leaving only the car if you switch. Hence if you switch there is 2/3 that you win. If you stick to your door from start, no matter what, it's only 1/3 of a chance.

but you don't know if your initial door has a goat or a car. (i'm assuming all the curtains aren't revealed til the end of the game.)

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.

you are wrong.

Since there are 2 goats and 1 car, you are 2 out of 3 times more likely to pick the goat on the very first try. So that means whatever curtain you pick you will more likely to have a goat so you should switch your answer when he reveals the other goat.

I love you.

https://xkcd.com/1282/Â

I'm going to be rich next year because of this.

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. Â 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. Â Your probability of picking car or goat initially is still 1/3 and 2/3 respectively. Â 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. Â So it doesn't change the probabilities at all. Â 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. Â Smart money says take the 2/3.

best video

This is ridiculously over complicated.Â

If the originally chosen door is a 1/3 chance of the car – and you reject it – then any other option must be a 2/3 chance.

Monty's revelation of a goat does not help the contestant in any way – Monty shows a goat where a goat was bound to be.

Interesting that a problem of quantum mechanics can be explained by a version of this brain teaser.

I made a video about it, you can watch it, click my avatar icon.

this is true it is a 2/3 chance switch

@oggo loggoÂ I 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.

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

onlyway 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!

The reason people are so confused is because they keep thinking 50% or 50-50 chance. Guess what? That's wrong and makes no sense. You begin with a 2/3 chance of losing and if you stay you still have 2/3 chance of losing. Simple as that. For example, you have 3 doors, door 1, 2, and 3. Let's say that the car is on door 2. Now if you choose door 1 or 3 initially and did NOT switch, you lose. However, if the host reveals a goat (and he always picks a goat) if you picked door 1 or 3 initially and switch, you win! You have a 2/3 chance of winning of you ALWAYS switch. If you stay, your odds remain the same 1/3 of winning. So stop thinking 50% chance because that makes no sense. Just remember switching when the goat is reveled always gives you a better odds of winning.

ha ha wow! ðŸ™‚

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…

i think switching works more when more doors are involved

Another way you can think about it is this:

There are 3 doors to choose from. 2 doors have a goat and one has a car. The host asks you to choose a door. You have a 1/3 chance of picking the car and a 2/3 chance of picking a goat. Now, the host, without opening the door that you chose, opens another door. Of course, the door that he opens will have a goat behind it. The host asks if you want to switch your door with the one that is not open yet. Do you switch or not? Let's looks at the probabilities.

If you do not switch, you are staying with the original probability before the host opens the door of a goat. The probability of winning is still 1/3 and the probability of losing is still 2/3. You have more of a chance of closing that you do of winning.

If you do switch, think of it this way:

When the host opens the door (which has a goat), that eliminates that door from one you can choose. He asks you if you want to switch or not. This takes you back to the beginning, except instead of having to pick between 3 doors, you only have to pick between 2 doors: either the one you chose or the one you didn't in the first round of choosing. The third door has been eliminated because you now know what is behind it (the goat). So instead of having a 1/3 chance of winning, you now have a 1/2 chance of winning. If you do not switch, you are sticking with your original choice, where you had a 1/3 chance of winning and a 2/3 chance of losing. If you do switch, you have narrowed your chances down to 1/2. You can either keep the one you picked or switch. You should always switch because if you do, you have a 1/2 chance of winning instead of your original 1/3 chance of winning.

Dude in between monty and the chef is super creepy.

willddddddddd thats so crazy thanks!

Why do I care about the third door with the goat he shows us if he shows us the goat every single time, doesn't it take 3 out of the equation, and narrows it to 2 every time?

There are mathematically identical versions of this puzzle that you can play with drunken patrons of a bar to rook them of money:http://dennishodgson.blogspot.com/2016/09/open-box.html

thank you

finally I understood the solution of the Monty Hall problem

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.

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.

Thank you!

But… steel is heavier than feathers….

Class assignment was about this problem, we discussed the statistics, and everyone STILL chose to stay. What is it with this mentality? I think my class mates hate me.

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

I understand this answer but why can't it be 1/2 if you ignore the fact that you chose anything before the first goat was revealed?

So basically, you have a higher chance of landing on a goat (2/3) than actually landing on the car (1/3) meaning that it would be a better idea of switching than staying on the same door because it would raise you odds if you switched theoretically.

If you don't switch, your odds of winning the car is 1/3.

If you switch, your odds of winning the car is 2/3 because your odds of picking a goat from the start was 2/3.

but in the end tho it doesn't really matter right? because then it comes down to either we switch or don't…?

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.

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.

in 1971sept episode of this show monte hall himself says it's 50/50 chance after he opens door

1st:Â Sal Khan is a legend. The world owes you, your vision and the people who have helped you put your platform together a great deal. May you have many more years of sustained effort and resources toward your project.

Now to the problem: How about 2 doors?

Then if there were 2 doors with 1 goat, I choose a door he shows me a goat, I switch every time and my odds of winning are 1/2? Yes, 1/2. Yes, the logic works here, but its really quite nonsensical to switch every-time. If he shows me the goat, I should make the logical choice and win the car every time.

Let's extrapolate this problem into a larger numbers of doors:

Then if there were 4 doors with 1 car and 3 goats, I pick 1 door, he shows me 1 goat, then my odds of winning when switching are 3/4?

…

Then if there were 100 doors with 1 car and 99 goats, I pick 1 door, he shows me 1 goat, then my odds of winning when switching is 99/100?

NO!!! The host is not going to show me 98 goats. The "explanation" only holds in 1 case of 2 goats and 1 car. Its a trick. Once he shows you one door, the probability changes, it becomes a different situation. This is my belief.

So are the classical explanations to this problem illogical junk or perhaps is this a true paradox, were 2/3rds and 1/2 are both correct answers?

Does anyone have a verifiable computer simulation with code that actually represents the problem and the choices being made with results? Mind you, writing code that just uses the classical logic in dispute is just "begging the question." You can't use a word you are trying to define to define itself.

Or You Could just put your ear to the door and see if you could hear a goat.

Or you could just bone

Dude why would I not want a pet ostrich

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

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.

They never taught this in undergrad econ or high school math. Interesting thought-experiment.

If you pick wrong and switch, you will always win because picking wrong means you picked a door that has a goat behind it and after Monty reveals the other door that has a goat behind it, switching means you switch to the only door that's left aka the one with the car

Please help me solve :

Consider a game-show called deal or no deal", where you have to choose between

three suitcases labelled (A, B, C), one of which contains Rs. 1 crore and the others are empty. On this show

contestants pick a suitcase (say suitcase A) and keep it temporarily without opening it. The host (who already

knows what's in the suitcases) then opens up one of the other two cases (say suitcase B) which is empty. The

contestant is now offered the option to keep the suitcase (A) or switch it with the remaining unopened suitcase

(C). Suppose that you are on this show, would you switch or keep your suitcase to get the prize? First make a

guess.

Now let us analyze this using likelihoods. Let Hi denote the hypothesis that the prize is in suitcase i where

i = A;B;C.

a) What are the a-priori probabilities of Hi before you pick a case?

b) Calculate the likelihood of each hypothesis Hi given that the host opened suitcase B after you picked suitcase

A.

c) Compute the a-posteriori probabilities of each Hi after the host has opened case B and shown that it was

empty.

d) To get the prize should you (i) keep A, (ii) switch to C, or (iii) does it not make a difference? Justify.

so I believe that you should switch for the reasons you said above but there is a common belief that once the first door is eliminated you have been technically been given the choice of one of 2 doors, one of which has the grand prize so there you have a 50-50 chance and the choice does not matter

There is a glaring problem with any explanation of the Monty Hall problem. On the big deal of the day there are two players, both of whom chose a door. If neither pick the big deal, Monty opens one of the contestants door, usually the one that's a zonk. I watched some videos, and after opening the first door, the contestant is not given a chance to switch. If they picked the big deal, then the door no one chose is opened first, and the big deal is revealed last. If the contestant didn't pick the big deal, the contestants door is revealed, followed by the big deal no one chose. The Monty Hall problem is based on one player, not two, and the contestants on the show aren't given the option to switch doors. I have yet to find a video of Let's Make a deal that's a true reflection of the actual problem as stated.

The problem should be renamed. As Monty Hall never optioned to switch doors on his game show. He only optioned to trade your choice for money.

I see this as two separate equations. The 1st is 1 in 3 chance. The second is 50/50 chance.