3 doors - One contains a brand new car, the other two contain a goat. Choose.

B.A._Baracus

Was watching Brooklyn Nine Nine (really good show I think) and they talked about what's known as "The Monty Hall Problem" (It's been mentioned in movies such as 21 too) It has since become a good example of probability sparking debate of is the choice "50/50" or "2/3" (you'll understand as you read on)

So you are on a game show and have three doors to select from. 2 of the doors will have a goat (a dummy prize) and the other will have a brand new car. But of course you don't know what's behind each.
After you select one of the doors the host (Monty Hall, american tv host) would open up another door to reveal another goat. Monty knows what's behind each door. He then says "Do you want to swap your door?" - here is the dilmemma, do you switch or stay? is it 50/50 (or 50%) at this point or is it 66% (of winning the car) since you made your choice before Monty revealed one of the doors.


^ just an visual example

They say you should switch every time. AS you now have a 66% chance of winning. But what would you do? switch or stay?
As for me, maths says switch. But I only get to play this game once. If I played it 100 times I would switch everytime. But that's the dilemma for me Play the probability game or trust my gut.

vandriver

I want a goat.

blinding

With Oscar Pistorius I would be extremely worried for the Goats and the Car .

blinding

Have I time to put on my Giddy Goat Perfume .

vandriver

vandriver wrote: »

I want a goat.

Same here. I have a car. I don't have a goat.

But what to do to nail down that goat...

The_Valeyard

Ive played Goat Simulator



I know what they can do.


Give me the Goat.

Avatar MIA

I've heard of this before.

And I still don't understand the logic behind switching, and why it goes from 50/50 to specifically 66% (not 65 or 67, but 66%) chance of success, are we to assume the presenter is disposed to us winning the car?!?

Princess Consuela Bananahammock

It's not a 66% chance since Monty opened the second door. It;s a 50/50 and, like deal or no deal, a complete waste of a good hour's entertainment.

Fr_Dougal

Gimme the car.

Are Am Eye

The back of the goat at door 3 is behind door 2.
It looks to me like those two goats can move around back there.
Is there partitions?
The whole thing is rigged I think. You wouldn't even fit a car.

WoollyRedHat

Have the proper procedures been followed and have the goats consented to appearing in the show?

Avatar MIA

Avatar MIA wrote: »

are we to assume the presenter is disposed to us winning the car?!?

Essentially yes. When the host opens one of the other doors - after your decision - he has to open a non-winning door.

So whether you pick the winning door or not - the host _must_ open a losing door.

That is the crux of why you should always switch - and why switching gives you a 66.66666666666666666666666666666667% chance of success.

Ficheall

Avatar MIA wrote: »

I've heard of this before.

And I still don't understand the logic behind switching, and why it goes from 50/50 to specifically 66% (not 65 or 67, but 66%) chance of success, are we to assume the presenter is disposed to us winning the car?!?

66% ~= 2/3.

Avatar MIA

Are Am Eye wrote: »

The back of the goat at door 3 is behind door 2.
It looks to me like those two goats can move around back there.
Is there partitions?
The whole thing is rigged I think. You wouldn't even fit a car.

Fr_Dougal wrote: »

Gimme the car.

Here you go...



The car is free, but you need to send me €500 for postage...

Outlaw Pete

We discussed it a few years back in the mathematicsame forum too:

http://www.boards.ie/vbulletin/showthread.php?p=70063175

Tigger

Deleted User wrote: »

Essentially yes. When the host opens one of the other doors - after your decision - he has to open a non-winning door.

So whether you pick the winning door or not - the host _must_ open a losing door.

That is the crux of why you should always switch - and why switching gives you a 66.66666666666666666666666666666667% chance of success.

i dont think thats it

McMurphy

vandriver wrote: »

I want a goat.

Just don't get a Kevin.

https://www.youtube.com/watch?v=yuwprXAaSv0

dmc17

I suppose on the first go you have ~66% chance of picking a goat, so chances are you've got a goat. This is why you'd switch on the second go unless you fancy curried goat for the next week?

Cortina_MK_IV

I'm against this animal testing.

Avatar MIA

dmc17 wrote: »

I suppose on the first go you have ~66% chance of picking a goat, so chances are you've got a goat. This is why you'd switch on the second go unless you fancy curried goat for the next week?

^^^ doesn't clarify - I feel like I should be wearing a dunces hat.

I can't watch the above 21 clip. I'm in company - I'd rude enough to be on a PC, but not sufficient rude enough to put on the sound. Will look at it later.

Tigger

Tigger wrote: »

i dont think thats it

It really is yes.

The host has to remove all the wrong doors except one basically. The one being either the one you picked - or the one remaining that you did not pick.

Picture it with 1000 doors. You pick one. The host has to remove 998 other wrong doors leaving your one - and one other.

This is why you switch

Capt'n Midnight

It's not that difficult.


Pick the door that doesn't smell of goat.

Ficheall

Here:

mikeymouse

Avatar MIA wrote: »

I've heard of this before.

And I still don't understand the logic behind switching, and why it goes from 50/50 to specifically 66% (not 65 or 67, but 66%) chance of success, are we to assume the presenter is disposed to us winning the car?!?

It's 66.66...., or two turds if you like!!
I remember in the pre Google era a discussion about this went on for ages on some forum.

harry Bailey esq

It depends on what you need. Wanna get somewhere you need to be without relying on public transport? It's got to be the car. If your eldest daughter happens to be getting married to the son of a wealthy merchant however,the two goats will be more than adequate,and will increase your status with the in-laws no end.
Personally I'd take the car and sell it to buy a herd of goats,set up a fake marriage network and get some driving lessons from the profits.

shaunr68

Ficheall wrote: »

Here:

Surely here case 3 has been eliminated because we have just been shown door 3 which contains a goat.

Therefore at this point the option to switch consists of two possible rewards, one a goat and the other a car.

Ficheall

It's showing you all of the initial possible layouts. Your selection is door 1 (because it doesn't make any difference), then you can see what happens if you switch or don't switch after being shown a goat from behind door 2 or 3. Since all initial layouts are equally likely, switching is better 2/3 of the time.

shaunr68

Ficheall wrote: »

It's showing you all of the initial possible layouts. Your selection is door 1 (because it doesn't make any difference), then you can see what happens if you switch or don't switch after being shown a goat from behind door 2 or 3. Since all initial layouts are equally likely, switching is better 2/3 of the time.

I think I understand

So let's say, again because it makes no difference, that the host shows you door 3 and that contains a goat.

We now know that case 3 cannot be true, and therefore at this point there are two possible outcomes with 2 possible rewards

Of course this works the same way if he shows you door 2 which discounts case 2

jamesbere

I don't want either. I want the door, would come in very handy

shaunr68

In not very mathematical terms, I'd say that the teaser only works on the premise that there are three possible outcomes and the probability is being calculated based on the situation at the beginning of the game show. However at the point the contestant is asked to choose, one of those outcomes has now been discounted and this of course changes the odds.

corner of hells

jamesbere wrote: »

I don't want either. I want the door, would come in very handy

I'd like a door too , my shed door is bollixed.

PARlance

As my grandfather once said. Stick with your goat.

Wonderful man, never had any kids.

Ficheall

shaunr68 wrote: »

I think I understand

So let's say, again because it makes no difference, that the host shows you door 3 and that contains a goat.

We now know that case 3 cannot be true, and therefore at this point there are two possible outcomes with 2 possible rewards

Of course this works the same way if he shows you door 2 which discounts case 2

You're not looking at it in a convenient way.

Pretend the doors on the show aren't numbered yet.
You pick a door - call THAT door 1.
Label the other doors 2 and 3.
Then look at the table again.
The three cases listed show the three possible scenarios in which you now find yourself. In each scenario, pretend Monty opens one of doors 2 or 3 to show you a goat. Then consider what you ought to do, bearing the table in mind.

Hannibal_Smith

I just got totally distracted by the very cute visual example

Burial.

Switch every time. The only way you lose is if you picked the car initially.

I love that movie and seeing that maths problem rattled my brain as a young fella and ultimately made me take the math of poker seriously and made me lots and lots of money down the road.

Your Face

'Let's take a look at what you could have won'

Avatar MIA

Burial. wrote: »

Switch every time. The only way you lose is if you picked the car initially.

Which you may have, which was a 331/3% chance at the time. The other two choices being 66 2/3 %.

So, as it stands the car is mostly likely in the subset you've not chosen.

The logic I've seen so far even though one of those choices is proven not to contain the car (when the presenter opens one of the boxes) is that this subset is still 66 2/3% likely to contain the car.

THIS is the bit I don't get .

Burial.

shaunr68 wrote: »

I think I understand

So let's say, again because it makes no difference, that the host shows you door 3 and that contains a goat.

We now know that case 3 cannot be true, and therefore at this point there are two possible outcomes with 2 possible rewards

Of course this works the same way if he shows you door 2 which discounts case 2

You pick door 1 with a goat, he shows you the other goat behind door 3, switch to 2 you win. You pick door 2 with the car, he shows you goat behind 3, you switch to 1 you lose. You pick door 3 with the goat, he shows you door 1 with the other goat, you switch to door 2 you win. You win 2 out of 3 times when you always switch, thus 66.6%.

Initially you have a 33.3% chance, pretty straight forward. Host shows you one of the goats. To consider it now 50/50 to get the car is discounting the information he's given you. If you pick the car initially the host has a free choice of choosing which goat to show you. However if you pick a goat (66.6% chance initially) he's no choice but to show you the other goat. He has no choice 66% of the time, he can pick a door at random 33%. You're effectively trying to find out if you gave him a choice to choose or not. 66% of the time he has no choice.

shaunr68

Ficheall wrote: »

You're not looking at it in a convenient way.

Pretend the doors on the show aren't numbered yet.
You pick a door - call THAT door 1.
Label the other doors 2 and 3.
Then look at the table again.
The three cases listed show the three possible scenarios in which you now find yourself. In each scenario, pretend Monty opens one of doors 2 or 3 to show you a goat. Then consider what you ought to do, bearing the table in mind.

Thanks, I still can't see it but it's been a long day, I can't get past my conclusion that this teaser doesn't take into account that probability will change based on new information. Fascinating thread, though where's the "Makes no difference" option?

shaunr68

shaunr68 wrote: »

However at the point the contestant is asked to choose, one of those outcomes has now been discounted and this of course changes the odds.

It doesnt though, that is the whole trick - and the reason it messes with peoples heads so bad.

shaunr68 wrote: »

Thanks, I still can't see it but it's been a long day, I can't get past my conclusion that this teaser doesn't take into account that probability will change based on new information.

Ok here is another way to phrase the explanation that usually gets people to see the "trick" for me in the end.

Forget for a moment the door numbers. Just imagine you pick a door. Call your door "over here" and all the other doors "over there".

Now the chances you picked the right door are 1/3. That means that the chances the right door are in the _other_ doors "over there" is 2/3.

Now it does not matter if the host takes away a wrong door - it changes nothing. The chance still is 2/3 that your original choice was wrong and that therefore the right door is more likely not the one you picked and is in the group "over there".

Imagine the same thing with 100 doors. You pick one. There is a 1/100 chance your door is right. There is a 99/100 chance the right door is "over there".

Now he can take away 98 wrong doors from "over there" just leaving one. But the chances the right door is "over there" is still 99/100. Solely because the chances your original choice was right is 1/100.

Avatar MIA

Deleted User wrote: »

Now he can take away 98 wrong doors from "over there" just leaving one. But the chances the right door is "over there" is still 99/100. Solely because the chances your original choice was right is 1/100.

Yes, I get it now. And that's why it's key that the presenter knows which door holds the car, and which one doesn't.

Pheeeeeeeeeeew :pac:

Pherekydes

What type of car?

Are the goats brand new also?

Switch is the answer. You can write a simple simulator in your preferred language. If you switch you win 2/3 of the time. If you don't switch you win 1/3 of the time.

Spudmonkey

The Monty hall problem is easier to understand if you increase the doors.

If you had a hundred doors and you made your choice. Then 98 were opened to reveal the goats, you are almost always better off switching.

If you are now down to two doors, one with a goat and one with a car. Of course the likelihood of you picking the correct door in the first place was 1/100. Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

Avatar MIA

Spudmonkey wrote: »

The Monty hall problem is easier to understand if you increase the doors.

If you had a hundred doors and you made your choice. Then 98 were opened to reveal the goats, you are almost always better off switching.

If you are now down to two doors, one with a goat and one with a car. Of course the likelihood of you picking the correct door in the first place was 1/100. Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

Sometimes when you know something too well it can be hard to tell it correctly.

To improve your explanation above, I'd humbly add, that when they are showing you the 98 empty/goat hiding doors they know that the car/prize isn't behind them. So, when they open the 98 doors there's zero chance they'll reveal the car/prize.

shaunr68

Spudmonkey wrote: »

Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once all the others (whether one or 998) had all been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Anyway, I will ponder this tomorrow and maybe put it to my two expert mathematician pygmy goats, Ted and Dougal! Await their expert analysis! Anybody want to swap them for a car?

Avatar MIA

shaunr68 wrote: »

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once all the others (whether one or 998) had all been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Anyway, I will ponder this tomorrow and maybe put it to my two expert mathematician pygmy goats, Ted and Dougal! Await their expert analysis! Anybody want to swap them for a car?

When you figure it out, you'll kick yourself. I KID you not

Outlaw Pete

shaunr68 wrote: »

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once one of the others has been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Well, the choice would absolutely be a 'one in two' chance if the host was not aware of what was behind each of the doors and where it only accidentally becomes apparent that there is a Goat behind Door No.3. Under those circumstances, there would be no advantage of switching at all.

You see, when the host purposefully reveals that one of the doors has a goat behind it, they really aren't showing you anything you don't already know yourself and so that action therefore doesn't alter the mathematical odds in any way shape or form. They might as well have showed you nothing and just offered you the chance to swap Door No.1 for Door No.2 and Door No.3 combined.

Which is why I have always maintained that the Monty Hall dilemma is really more of a trick question than an actual mathematical problem.

Omackeral

Me mate Anto got caught out with something similar before and they made him bring it home.

somefeen

This doesnt make any sense.

Nothing changes if you switch. You played a game with three doors and lost and now you are playing a game with two doors. The probability of the car being behind the door you choose is the same as the other remaining door.

I'm no stats genius, but all this seems to do is show the limitations of probability calculations.

Digital Solitude

somefeen wrote: »

This doesnt make any sense.

Nothing changes if you switch. You played a game with three doors and lost and now you are playing a game with two doors. The probability of the car being behind the door you choose is the same as the other remaining door.

I'm no stats genius, but all this seems to do is show the limitations of probability calculations.

Nothing changes if you switch, but its more likely that the door you originally chose was a goat than car.

I think this explanation makes sense.

There's a 1/3 chance you choose the car first time, and a 2/3 chance its in one of the other 2 doors. After one of the other doors is opened, there's a 2/3 chance the other door has the car. As there's still a 1/3 chance you chose correctly.

Its not a maths problem at all, just a brain teaser imo.

greencap

Digital Solitude wrote: »

Nothing changes if you switch, but its more likely that the door you originally chose was a goat than car.

I think this explanation makes sense.

There's a 1/3 chance you choose the car first time, and a 2/3 chance its in one of the other 2 doors. After one of the other doors is opened, there's a 2/3 chance the other door has the car. As there's still a 1/3 chance you chose correctly.

Its not a maths problem at all, just a brain teaser imo.

its weird though, because how does reality decide youre still playing the original game.

for example what if you were on the stage, having chosen the first door and gotten a goat, and then you said to Monty 'hang on I'll be back in 5 minutes' and then you go have a drink and take a sht.

Then you come back and theres a stage, a host, 3 doors (one which happens to have a goat standing in it).

Now your job is to choose a door. Door 1 or door 2.
Aren't you a different person now, different from the you of 5 mins ago.
Or do your prior decisions follow you forever?

Ok .... now if Monty says and does nothing, and you just walk up and choose a door, then thats a 50/50 choice surely, right?

But if while you're walking up to the doors Monty says 'would you like to switch' then the maths goes all funny.

How does reality keep track of your odds.

Like if I'm playing and get the first goat and monty says 'stick or switch' I'm now facing mathematical laws cause ->I<- chose a certain door.

But say my mate bob is in the audience and I'm standing there faced with a goat and 2 doors and I wave at bob and say 'hey bob pick a door' then that HAS to be a 50/50 for bob, thats the very definition of a 50/50.


dat iz mental innit.