nav[aria-label="Primary Navigation"] { padding: 0; & ul { list-style: none; width: 100%; display: flex; flex-direction: row; justify-content: start; align-items: start; gap: 30px; padding: 0; & li { margin: 0; } & ul li { list-style: none; } } }

Child Item

Probability Question.

rediguana

I'm reading "Butterfly Economics" and this one has me stumped:

"A Game Show contestant has won the right to open one of three doors, A, B, or C. Behind two doors are useless duds, and behind one is E10,000. He wins whatever is behind the door he chooses. Amidst the mounting excitement of the audience, Door A is chosen. The host heightens the tension by opening Door C (the host knows what is behind each door) to reveal a dud. The contestant is given the chance to alter the decision, to choose B rather than A... What should he do?

Find more posts tagged with

Quick Links

Comments

tdow

this was explained to me by a statistician once

iirc you should alway change your choice. The idea is when you choose door A there is a probability of 0.3333 of winning the E10000 but after you're told that there is nothing behind door C , there is a probability of 0.5 that it is behind door B

kjt

tdow wrote:

this was explained to me by a statistician once

iirc you should alway change your choice. The idea is when you choose door A there is a probability of 0.3333 of winning the E10000 but after you're told that there is nothing behind door C , there is a probability of 0.5 that it is behind door B

There is also a 0.5 probability of it being door A so why would you change the door....?

rediguana

Correct and right. The book says:

"In practice, most people refuse to change their decision. But switching to Door B is always the best strategy to follow, for the simple fact that it doubles your chance of winning the E10,000. There is a 1/3 chance of the money being behind Door A, and a 2/3 chance of it being behind B and C together. If you know that there is a zero chance of it being behind C - because the host has opened it for you - then there is a 2/3 chance of it being behind B"

rediguana

kjt wrote:

There is also a 0.5 probability of it being door A so why would you change the door....?

THIS is my issue with it!

I was thinking I must be missing something though. The book also says "Many people find it hard to understand the reasoning, even when it is explained to them".

tdow

the book's explaination make a lot more sense than mine especially since my total probality didn't equal 1 :rolleyes:

rediguana

tdow wrote:

the book's explaination make a lot more sense than mine especially since my total probality didn't equal 1 :rolleyes:

Sorry so, I mean "Incorrect and Wrong"

Well, at least you won the money. Let's hope that if you ever get onto "Blankety Blank" or whatever, you don't have to show your rough work.

Seriously though, what's the story? Surely, at the stage of Door C having been eliminated, Door A now has a BETTER chance than one-in-three of being the right one?

tdow

its easier to think about it in terms of the door you picked

probability of door A being the correct door = 0.3333

probability of door A being the wrong door = 0.6666

these probabilities never change

rediguana

I don't really understand, even when it's explained to me! I should stick to the Food & Wine Forum

Can't probabilities change? Surely, when new information comes to light (eg. Door C is a dud), they can?

I'm making an assumption that the eminent Paul Ormerod, and the book's editors (not to mention the Maths Boardsters), know their Maths better than me. So I must be wrong. But still, I can't get my head around it...

I'll check back later and see if I've become any brainier.

Any further elucidation welcome.

Pherekydes

This is the Monte Hall problem. You can also try here. Marilyn Vos Savant was the brainy woman who took on the Maths Professors and humiliated them. That's how counter-intuitive it is! There is a Monte Hall simulator here where you can run the problem and see the results for yourself.

DeVore

Draw out the decision tree, branching at each choice at each point. You will find that always switching is the right choice.

The key point is well hidden... the host kindly ALWAYS removes a dud. If he randomly chose a door (and you lost the prize if he picked the money) then that would be different. However he ALWAYS excludes a dud.

So, the times you choose the door with the money at the start (1 time in 3) he will exclude a dud and changing will lose you the cash. The times you DONT (ie you choose a door with a dud at the start... that is: 2 times in 3), the host helpfully excludes the other remaining dud and switching will GUARUNTEE you the money. So, always switching is the better option.

DeV.

DeVore

ps: this stumped me for a few days too and my gut reaction was "it doesnt matter" but when I worked it out on paper it became clear.

DeV.

denachoman

I think the concept of switching being right is easier to grasp if you make the numbers involved bigger.

For example if there are 100 doors and you choose one at random everyone will hopefully agree that you have a 1% chance of picking the correct door. If after you make your choice the host then shows you 98 of the remaining 99 doors are duds it is fairly intuitive that you should switch from your original choice to the other remaining door. Your original choice only had a 1% chance of being right, this hasn't changed because the host has opened some dud doors. By switching you have a 99% chance of winning the money.

Pherekydes

Another probability problem was the Sally Clark case. It shows that probability can play a very serious and tragic part in life when the principles are not fully understood.

Edited to add: the previous post shows how our Stats lecturer explained it to us. It became crystal clear after she used that example.

rediguana

denachoman wrote:

I think the concept of switching being right is easier to grasp if you make the numbers involved bigger.

For example if there are 100 doors and you choose one at random everyone will hopefully agree that you have a 1% chance of picking the correct door. If after you make your choice the host then shows you 98 of the remaining 99 doors are duds it is fairly intuitive that you should switch from your original choice to the other remaining door. Your original choice only had a 1% chance of being right, this hasn't changed because the host has opened some dud doors. By switching you have a 99% chance of winning the money.

Eureka.

BlackIguana

denachoman wrote:

I think the concept of switching being right is easier to grasp if you make the numbers involved bigger.

For example if there are 100 doors and you choose one at random everyone will hopefully agree that you have a 1% chance of picking the correct door. If after you make your choice the host then shows you 98 of the remaining 99 doors are duds it is fairly intuitive that you should switch from your original choice to the other remaining door. Your original choice only had a 1% chance of being right, this hasn't changed because the host has opened some dud doors. By switching you have a 99% chance of winning the money.

This is a good way of explaining it - just a slight correction...

You should have said "By switching you have a 50% chance of winning the money".

The rest of your logic is 100% correct. By switching you have changed your chances from 1% to 50%.

Cheers.

Rediguana - this is a classical stats question. It's about as interesting as stats gets!

Michael Collins

BlackIguana wrote:

...You should have said "By switching you have a 50% chance of winning the money".

The rest of your logic is 100% correct. By switching you have changed your chances from 1% to 50%...

No, he's right. You have a 99% chance of winning if you switch. There's only two boxes left at the end,

P(box1) + P(box2) = 1 (since the prize is definately behind one of them)

but we know that P(box1) (the original box) is 1%, so the other box has to be 99%.

aequinoctium

switch

i've heard this one many times before...including in the show Numb3rs

BlackIguana

Michael Collins wrote:

No, he's right. You have a 99% chance of winning if you switch. There's only two boxes left at the end,

P(box1) + P(box2) = 1 (since the prize is definately behind one of them)

but we know that P(box1) (the original box) is 1%, so the other box has to be 99%.

Apologies denachoman - I stand corrected.

cheers.

humbert

DeVore wrote:

So, the times you choose the door with the money at the start (1 time in 3) he will exclude a dud and changing will lose you the cash. The times you DONT (ie you choose a door with a dud at the start... that is: 2 times in 3), the host helpfully excludes the other remaining dud and switching will GUARANTEE you the money. So, always switching is the better option.

DeV.

That's by far the simplest explanation of that problem I've heard.

elivsvonchiaing

humbert wrote:

That's by far the simplest explanation of that problem I've heard.

I'm in awe here -- everyone cool if I put this in my thesis? :rolleyes:

Colonel Sanders

intuitively it obviously sounds wrong but when I thought of it the thing that swung it for me was the fact the host KNOWS which door the car is behind. Therefore he must eliminate a door with a goat behind it.

I assume there is possibly a bayesian argument using prior and posterior probabilities?

rediguana

Colonel Sanders wrote:

intuitively it obviously sounds wrong but when I thought of it the thing that swung it for me was the fact the host KNOWS which door the car is behind. Therefore he must eliminate a door with a goat behind it.

I assume there is possibly a bayesian argument using pior and posterior probabilities?

Many helpful posts and much soul-searching later, I actually DO get it. But what the hell is a Bayesian argument? I have no maths background, by the way!

Pherekydes

There are two basic approaches to probability, frequentist and Bayesian.

Have a read and then come back and we'll have a lengthy discussion.

SGKM

I got asked this problem in a job interview about a month ago and I knew that you always switch but I couldn't explain why. I tried to logic it out but it was tough under pressure.

Buffer

Colonel Sanders wrote:

I assume there is possibly a bayesian argument using prior and posterior probabilities?

Wikipedia has a pretty good presentation of the prior and posterior probabilities, including the effect of changing the prior information (for example, if we know that Monty Hall likes to open one door more than the others, so selects it more often).

babytooth

is this jsut not Bayes Theroem.

Actions impart information etc etc...all coming from Game theory.

interesting subject, massive in finance.

Colonel Sanders

just thinking about this today. I don't doubt the answer as I have seen it derived using conditional probability, just wanna get the understanding clear. Is my logic correct or anywhere near correct?

assume contestant picks door 1. p =1/3 (p=probability that car is behind door 1)

=> q = 2/3 (q=probability car is not behind door 1)

we know that there is at least one goat behind doors 2 & 3 (1 if the car isn't behind door 1 and 2 if the car is behind door 1). We know that monty will open one door and this door will be a door containing a goat. therefore by opening the door he tells us nothing that we already don't know. The probability that the car is behind either 2 or 3 (eventho we know whats behind one of these two doors) is still q.

therefore as he has shed no further light on the problem the posterior probability that the car is behind door 1 remains 1/3.

anyone?

Michael Collins

Colonel Sanders wrote:

...therefore as he has shed no further light on the problem the posterior probability that the car is behind door 1 remains 1/3.

anyone?

Yep that sounds right to me, but since the car IS behind one of the remaining doors, the probability that it's behind the unopen one is now 2/3. So, you should go for that.

Buffer

Colonel Sanders wrote:

Is my logic correct or anywhere near correct?
...
We know that monty will open one door and this door will be a door containing a goat. therefore by opening the door he tells us nothing that we already don't know.

No, this is the standard mistake. People assume that there is no new information to be derived from him opening the door, which is why they assume there's no reason to change their minds, but this is wrong.

As pointed out before, MH can only open a door with no prize. He knows where the prize is, he's not opening a door at random.

I previously omitted the link to the wikipedia article on this: http://en.wikipedia.org/wiki/Monty_Hall_problem.

Buffer.

Colonel Sanders

Buffer wrote:

No, this is the standard mistake. People assume that there is no new information to be derived from him opening the door, which is why they assume there's no reason to change their minds, but this is wrong.

I said in my explaination that we know he has to open a door with a goat behind. Thats what swings it and ensures that the probability isn't a half.

"We know that monty will open one door and this door will be a door containing a goat. therefore by opening the door he tells us nothing that we already don't know."