# Boards

## The Monty Hall Problem

I like this.

Suppose you are on a game show in which you can win a car, and you are given the choice of three doors - behind one door is a car, the others, nothing.

You pick a door, say no.1. The gameshow host, knowing what is behind the doors, then shows you one of the other doors (say no.3) which has nothing behind it.

The gameshow host then asks you if you would like to change your choice to door number 2. Is it to your advantage to switch?

## no

## *QI buzzer*

## pants :(

explain - with long winded references to animal genitals, plants with rude names or why alan davies is a an idiot

## ok...

Let's say what is behind the doors - number 1 has nothing, 2 has the car, 3 has nothing. Let us see what happens when you stick in these scenarios:

SCENARIO ONE:

You choose door number one. The host is the forced to show the only other empty door, which is 3. You stick. YOU LOSE.

SCENARIO TWO:

You choose door number 2. The host then opens either of the other doors. You stick. YOU WIN.

SCENARIO THREE:

You choose door number 3. The host is forced to open door 1. YOu stick. YOU LOSE.

So, if you stick, you have 1/3 chance of winning.

Now, what happens if you twist in all those scenarios?

SCENARIO ONE:

Choose door 1. Host is forced to open the only other empty door, 3. You switch. WIN

SCENARIO TWO:

Choose door 2. Host opens either of the other doors. you switch to the one he didn't open. LOSE.

SCENARIO THREE:

Choose door 3. Host is forced to open the only other empty door, 1. You switch to 2. WIN.

Chance of winning by switching = 2/3

-----

I got this wrong first two. Amazingly, so did over 1000 people with Maths PhDs when this was originally published. Also it was on horizon tonight, which had Alan Davies on it, he got it wrong too :)

## too*

## i dont get it. at all.

## ^

SOHCAHTOA

## Yes

## Yes

## Yes it is

:)

## ...

I don't see how it would matter.

## explanation

http://en.wikipedia.org/wiki/Monty_Hall_problem

## ...

Ah.

...but in this version you don't know that the host knows there's nothing behind the door he himself has picked. So (seeing as you have no new information), the odds haven't changed, right? I ain't no math-doer...

## No, he opens the door to show that there's nothing inside it.

Therefore you know that has nothing behind it.

## ...

Ah of course.

In that case... I don't know. Yes?

## this might be a bit clearer if you ain't read it already

http://drownedinsound.com/community/boards/social/4172551#r4477573

## Or just

it's more likely that you picked a bad door than a good one, so it's more likely that switching will be in your favour.

## ^

## Yeah, but that

doesn't really explain it fully.

## Does really

## Sure.

But not really.

## Go on

## Well, because it's also

about the host knowing what's behind the doors. If he doesn't know, then I think it doesn't matter whether you switch or not, chances are the same.

http://en.wikipedia.org/wiki/Monty_Hall_problem#Why_the_probability_is_not_1.2F2

## You're right

but the reason you should change - given that the host knows what is behind the doors and deliberately leaves the winning door closed, if you didn't pick it - is still that you probably picked a bad door initially.

## If the host doesn't know, how can he show you and empty door?

## He could open a door at random

and happen to open an empty one.

## But then the game show is fucked

cos he could open the one with the car behind! Therefore giving you the choice of losing and, um, losing.

## But then he couldn't offer the swap if he unwittingly revealed the car, could he?

## Nope

but you could still ask the question - if he happened to open an empty door, should you swap? And the answer is, it wouldn't make any difference.

## It surely would make a difference?

He's just ruled out 1/2 of the empty doors leaving you with a greater chance of a car if you swap.

## As I understand it,

the fact that he doesn't know removes from the consideration the two scenarios in which he unwittingly reveals the car (since these don't fit the description of the scenario given in the original problem).

This eliminates two scenarios in which switching leads to a win, thereby reducing the probabiliy that switching is advantageous.

## FUCK! *an

## Haha I was waiting for this to come up on DiS,

I spent the whole of Christmas arguing with family members about it!

It's the whole 50/50 problem that throws people off, offer them the swap and they think 'it's either this one or that one, doesn't matter what I do I've got the same chance of losing as winning.'

The simplest way we found of explaining why wou should always swap is the fact that at the start, you're twice as likely to choose nothing instead of a car, therefore you should always swap.

Or the 1000 doors problem that I think is mentioned on wikipedia.

## But surely...

...after the first door has been opened then you're dealing with a different scenario.

Yes, if you switch, then statistically you have a better chance of winning a than you had AT THE START because one door has been eliminated and the chances of being WRONG are reduced. However, after this first door has been eliminated you're simply dealing with a different problem, which cannot be analysed using the same maths as before. In the original scenario you have a choice between 3 doors. After the first door has been opened then you have a straight choice between two doors. Therefore, doesn't the fact that the first door is empty become completely irrelevant?

The "problem" has changed! Whilst you can write some numbers to mathematically prove that based on the ORIGINAL scenario you have a greater chance of winning if you switch, after the first door has been opened you're dealing with a different problem which requires different maths... aren't you? Surely it just boils down to to 4 possible outcomes

STICK:

1 - You keep your original choice, the car is in there you WIN

2 - You your original choice, the car is not in there you LOSE

1/2 chance

CHANGE

1 - You change your original choice, the car is in there, you WIN

2 - You change your original choice, the car is not in there, you LOSE

1/2 chance

Both of these new outcomes have a 1/2 chance, don't they? Whilst switching may make sense based on your original starting point of THREE doors - you now only have TWO so... your chances of winning are 50/50.

I mean I ain't no mathematician but can someone tell me how what I've written above is incorrect? It's gonna bug me otherwise!

## nah man, there are three possible outcomes, as there are three different doors to choose

You are therefore likely to pick an empty door first 2 out of 3 times. In both these situations, the host must show you the only other empty door. Thus by swapping here you win (2/3)

The only outcome where you won't win by swapping is if you choose the car originally, which you have a 1/3 chance of doing.

## But...

...after the host has opened one door, and shown you there's nothing in it, then there are only 2 doors to choose from. After that first door opens, the problem changes because you have a different CHOICE to what you had at the start.

Or have I got that spectacularly wrong?

## Yes, but the original choice you make is done with a 1/3 chance of getting the car

and a 2/3 chance of rock all. If you stick with your choice, you therefore don't change your choice and thus still have 1/3 chance. If you twist, by logic, it your chances increase to 2/3.

## Read Verbal's explanation above

Your original choice had a 1/3 chance of winning and 2/3 chance of losing. If you stick you still have a 2/3 chance of losing. As you made the choice before the door was revealed.

If you switch you turn the odds around to take advantage of the fact that 1/2 doors will jow win you a car.

## Yeah...

...but you've been given ANOTHER choice.

My understanding is that the host offers you a choice at the start when there are 3 doors. You choose a door. You have a choice of 1 door from 3 to win a car. Cool.

The host then reveals an empty door. This door is then eliminated.

The host then gives you another choice. This time you can choose between 1 of 2 doors. The car is in one of them. You have a 1 door in 2 chance of winning a car.

I understand how the maths works, but what I don't understand is how you can apply the original logic and original probability from the original choice made when there was 3 doors, when now there's only 2. The scenario shifts too drastically throughout the problem to make this useful.

The reality of the scenario that the contestant ultimately finds themselves in is that they have to choose between 2 doors to win a car. Their odds at the start when they had to choose between 3 doors to win a car is irrelevant, isn't it?

I'm confused.

## When it's down to 2 doors,

you're not simply choosing 'one or the other'.

You're choosing to stay with a door that already has a 1/3 chance of being a car, or switch to a door that already has a 2/3 chance of being a car.

## Hmm...

...that makes more sense but in my head you are still choosing between 2 doors - regardless of what odds they had of containing a car at the start of the game. The situation is now different so those odds become irrelevant. To me anyway...

## If you want to put money on it,

I could set up a little game for us.

## I don't think I can explain it further sorry!

I think it's one of those things that you either 'get' or don't. Lots of my family were exactly the same, they could accept that you should swap but couldn't get round the 50/50 part of it either.

Hope that doesn't sound patronising!

## Not at all!

I've been tinkering with one of those applets and it proves you lot right. It DOES work! I don't know how, but it does so... I'll be quiet now.

Thanks all for your help in opening up a part of my brain which don't work as good as it should though!

## That is the logical way of thinking about it,

but you can't just ignore the original scenario because that's when you chose your door. The fact that one has been eliminated from your choice makes no difference to the odds.

The way that changed the way I thought about it was imagining 100 doors. You choose 1 and the host proceeds to show you 98 empty doors (basically all but your door and one other, just like when there's 3 doors) you'd be crazy not to change. It's the same with 3 doors, just with smaller odds of success.

That probably doesn't explaing it very well, sorry. To put it simply, if your twice as likely to have chosen wrongly, you should always sway given the choice.

## *swap. Swaying would do you no good whatsoever and you'd probably get kicked off the show.

## Aha!

The thing I will never understand about this is the fact that a door being eliminated doesn't change your odds. Especially when you're asked to pick again! THIS DOES NOT MAKE SENSE! The 2 problems to me are different.

I understand the maths behind it, totally. Ultimately though - the maths doesn't help the contestant does it?

Right - new thing. If this problem was carried out as an experiment. 2,000 people took part in the game show. The same situation happened. 1,000 of them stuck, and 1,000 of them changed.

Are you saying that if this experiment took place then 333.3 of those who stuck would win a car, and 666.6 of those who changed would win a car? This again doesn't make sense to me. The maths is fine! I just don't see how it can practically help someone win a car!

## See my above post,

It's not simply a choice between two doors, they are 'loaded' with odds behind them already.

As with all probability, that should statistically happen. There are numerous simulators out there that'll carry it out for you, have a look:

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

## Or this one, you can make it carry 100 at a time and change the number of doors:

http://www.shodor.org/interactivate/activities/GeneralizedMontyHall/

## You can do the sums, that is not the same as understanding the maths.

We are saying roughly 333 of those who stuck would win and 667 of those who switched would win.

If you take the example abovce with 100 doors. there are two reasons why your door will be in the last two. 1) You picked thw 100/1 chance of the correct door. 2) The host has KNOWINGLY removed every door that was not a winner, leaving your choice and one other door.

You chose at 100/1, if you stick witht he original door your odds have not changed, as the host has not been choosing at random. Had he been chosing at random, by the end you would have a 1/2, but in reality this situation would only come up very rarely as the chat show host would almost certainly pick out the winning door himslef before it got to the final two.

## We're assuming that the host does know where the car is though right?

Whether or not you've got 3 doors or 100 doors, the host will reveal every empty door except yours and 1 other (in the case of 3 doors he's only revealing 1 empty door but it's the same in theory as 100 doors. Just with greater odds of success.)

## Yes, the host knows.

The fact that the odds change is dependent on the host knowing the winning door, or else the final odds on two doors would be 1/2, but this situation would not be reached often as the host would sometimes open the winning door.

I was trying to explain a situation where the odds would be 50:50 to be compared against this problem.

## Ah I see, makes sense now :)

## they did this on Horizon last week with Alan Davies and some maths guy

basically, the guy who switched got a more than 2/3 haaul of 'cars'

I can't find the vid online, but as in all instances of ideological conflict, Hollywood may have the answer

http://www.youtube.com/watch?v=cXqDIFUB7YU

(psst! I can't listen to this at work, so I've no idea if the sound works.)

## Two boys problem:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

## 1/3?

I'm not sure I understand the question.

## 2/3

## essentially rephrased question of this:

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

## ...

100%. I'm absolutely certain of this. Never been more sure of anything in my life.

## answer is 1/3 btw

## YEAH!

## there are four possible outcomes for any set of two children

girl, girl

boy, girl

girl, boy

boy, boy

-----------------

now if you know that one of the kids is a boy, and want to find out what the chances are that the other one in that set will be , simply delete the sets which have no boys in them - in this case, the girl, girl one. You are left with three sets. Only one of these sets has two boys in them. Therefore answer = 1/3

## ...

reminds me of genetics last year :S

## i would like to bump this to disagree

your assessment is only correct if you take the genders of the two dogs as NOT being independent of one another. If you take aside one dog, which you know is male, then you are left with a second dog which has a 50-50 chance of being male, completely regardless of the other dog's gender.

So I would argue that the answer = 1/2. But I might be getting this horribly wrong.

## My argument against the 1/3 answer is that boy, girl is the same as girl, boy.

So really you're counting one possibility twice.

## but it should be counted twice

because it's twice as likely as either girl/girl or boy/boy.

## Yeah.

Maybe if you phrase the question as "a baby is born in USA and a baby is born in the UK, at least one of them is a boy, what are the chances of the other one being a boy" then that makes it clearer that BG and GB are distinct possibilities.

## But still what we are essentially saying is:

If you have a child then it's only a 1/3 chance the next child you'll have is a different sex.

I'm not sure if that could ever be seen as logically true. Though evidence might prove it is, which is just mind blowing.

## Well if you've already had a child then that eliminates one of the scenarios

e.g. if you've had a boy then that eliminates GG and GB, leaving BG and BB. So it would be 1/2.

## Ignore me,

I didn't properly read what you posted.

## Yes.

The issue I have is:

- are we seeing an honest statistic that simply goes against what you'd expect?

- Or are we seeing maths used incorrectly but in a way that seems correct?

- Or maybe we're just missing something.

## I think

it's a problem with our intuition and how we estimate probabilities intuitively. Like, we're used to things being well-defined and so when we're told that one of the children is a boy, we separate off that child and put in in a box marked 'boy' and consider the other child in isolation.

## Yeah but, surely they ARE in isolation?

The puppies thing is interesting because all the puppies are born at the same time whereas the children are born years apart.

I guess I answered the question there: you can't just apply statistics like that. In the case of human, single birth children the chance is always 1/2 because the 'order' is locked and Boy Girl is not the same as Girl Boy.

With a litter of puppies there isn't really an ordering. Though I guess there is actually in terms of when a puppy is born.

So without knowing which puppy we talk about, older or younger the chance is 1/3 but in reality chance is actually 1/2 because the one we know about IS older or younger even though we don't know it.

## Yes they are in isolation in reality.

But in the context of the problem they're not because we're denied access to the kids(!). They're in a kind of superposition to us.

But the 1/3 chance is a real probability. As an analogy, if I tossed 2 coins 100 times and disregarded all the HH outcomes and offered you a bet, would you bet on there being more outcomes where both coins were tails or where the coins came up different?

## Yours is a very bad analogy.

If you tossed two coins, revealed the face to be heads on one and asked what are the odds that the other one was heads.

Both heads and tails have 1/2 chance.

Same withh the children/puppies.

## Well yes.

I don't think the problem involving the puppies is a good one for the reason that it ISN'T the analogue to a real situation, as we've just established.

## I didn't say anything about revealing one of the faces.

That was the whole point.

## I see what you are saying now.

What threw me off was the whole debate about whhat is essentially maths a 13 year old could do. Reading it a week late rather than in real time made it slightly confusing as to who was agreeing/disagreeing with what.

As you point out, where the puppies/babies problem falls down is in observing the sex of one. If it were left at "neither were girls, what are the chances that they are both boys" it would be 1/3. In looking at one dog, it removes it from any calculation of probability on the sex of the other dog.

## Yeah.

Most of the difficulty with it is the language used in the problem and the assumptions it causes to naturally make.

## ah yes but you don't know which dog is male

if we split the four possibilities by category, say oldest and youngest...

ELDER YOUNGER

girl girl

girl boy

boy girl

boy boy

So, you hav been told that at least one of the puppies is a boy, but you cannot determine whether it is the eldest or youngest. So you have to take into account the three possibilities with at least one boy in hence 1/3. Well that's how I see it anywho.

## i do see what you're getting at now

it's all about dependent and independent probablities I guess. But yeah you're pretty much right.

## I had all but managed to erase the memory of the last years doing Quantitative Methods modules

THEN YOU POST THIS! (Its what we used to do pretty regularly last year when we were doing probability)

So yeah, thanks for that.

In all seriousness, its actually quite interesting when I don't have a ballbag of a lecturer ruining it, and making it incomprehensible (in a desparate measure, me and a friend only understood it a week before the exam by watching solutions videos on youtube - THAT'S how bad the lecturer was!)