3 Doors...1 car = 1/3 ??? I thought.

[Pinch Paxton]

Its the last round of a game show. You have to choose between three doors labeled One, Two, and Three. Behind each door is a prize. Now two of the doors contain goats while one of the doors contains a brand new car.

 

Your asked to choose a door number. Lets say you choose 1. The Host then reveals a door other than the one that you choose, lets say that it is door number 3, and it has a goat behind it. You are then asked if you would like to change your choice to number 2 or stick with number 1.

 

Now the riddle is, does this make a difference? Should you change your answer or stick with it?

 

 

Have you all heard of this wierd puzzle? I have a book about it, but the answer is surely misleading.

[Skye]

I'll take door number 1. Goats are fun to ride.

[blike]

well if theres a goat behind one that he revealed, and its not the one you chose, then you have a 50/50 change that yours is a car. Your chances will not increase if you change doors.

[Pinch Paxton]

Yes that sounds perfectly correct, but mathematically your odds are supposed to improve if you change your mind. I shall post the maths later, but first I want to see if anyone can work it out. Supposedly, if you make a computer program that keeps changing its mind, you do actaully win more often. This completely confuses me.

[Pinch Paxton]

It's because if you picked one of the wrong ones, they eliminate the other wrong one, meaning if you change your answer, it will be the right one. And since there's twice the chance of picking the wrong (2 doors are bad, 1 is good), that means there's twice the chance of it being right if you switch. Hence the 2/3 probability or 66% of getting the car.

 

Look at it like this, consider door 3 is the winning door.

 

You pick 1. They eliminate 2. You switch to 3. You win.

You pick 2. They eliminate 1. You switch to 3. You win.

You pick 3. They eliminate 1 or 2. You switch to 2 or 1. You lose.

 

If you stayed:

 

You pick 1. They eliminate 2. You stay on 1. You lose

You pick 2. They eliminate 1. You stay on 2. You lose

You pick 3. They eliminate 1 or 2. You stay on 3. You win.

 

 

 

But even though the maths says that you have an advantage, my computer program below shows no increase of wins

 

 

 

 

Rem Change Choice

 

Randomize Timer()

 

For n = 1 to 100000

 

Car = rnd(2)+1

choice = rnd(2)+1

Goat = car+1

if Goat = 4 then Goat = 1

change = goat + 1

if change = 4 then change = 1

if change = choice then change = change +1

if change = 4 then change = 1

if change = Car then inc Changewin

next n

 

Rem Stick to Choice

 

For n = 1 to 100000

 

Car = rnd(2)+1

choice = rnd(2)+1

 

if choice = Car then inc Choicewin

next n

 

Print "Change and win = "; Changewin

 

Print "Stick to First choice = "; Choicewin

 

Suspend for key

[Cap'n Refsmmat]

Erm... your math shows it equal.

[Pinch Paxton]

I think it is equal, but it apparently isn't. Read this link...

 

 

http://mathworld.wolfram.com/MontyHallProblem.html

[Crash]

crikey!

[Skye]

Ok, that makes more sense, I thought the door opening bit was random.

[Sayonara]

Can't I just have the goat?

[iglak]

that link doesn't explain anything though. all it says is "switch because i said so". it's math is wrong.

It said that you first pick with a 1/3 chance. that's correct.

then it said that if you stay with that choice, it's still a 1/3 chance. that's incorrect because one choice has been eliminated and you are making a new choice (switch or stay) with a 50:50 chance.

It also said that switching increases the chances up to 2/3. that's very incorrect. this is the same new choice with a 50:50 chance. in fact, wich only one good prize, there is no possible way to have chances greater than 50:50 (assuming whole numbers)

 

your math is correct, but it would be better represented like this:

Look at it like this, consider door 3 is the winning door.

 

You pick 1. They eliminate 2. You stay on 1. You lose.

You pick 1. They eliminate 2. You switch to 3. You win.

You pick 2. They eliminate 1. You stay on 2. You lose.

You pick 2. They eliminate 1. You switch to 3. You win.

You pick 3. They eliminate 1. You switch to 2. You lose.

You pick 3. They eliminate 1. You stay on 3. You win.

You pick 3. They eliminate 2. You switch to 1. You lose.

You pick 3. They eliminate 2. You stay on 3. You win.

 

this totals to a 50:50 chance

[Pinch Paxton]

Yeah, I agree, and my computer program agrees, it's 50/50 on the second choice, and you can't change that no matter how you get there. The confusing part to me is that this probabilty idea has sold many books, and is often quoted as a certain fact that your chances improve by swapping choices.

[BrainMan]

iglak-

 

You are artifically increasing the odds of winning by staying because you list four cases of door 3, when the cases are not actually distinct- there are only two distinct cases at door three. What it should look like is this:

 

You pick 1. They eliminate 2. You stay on 1. You lose.

You pick 1. They eliminate 2. You switch to 3. You win.

You pick 2. They eliminate 1. You stay on 2. You lose.

You pick 2. They eliminate 1. You switch to 3. You win.

You pick 3. They eliminate 1 or 2. You switch to 1 or 2. You lose.

You pick 3. They eliminate 1 or 2. You stay on 3. You win.

 

 

The number of times you win by staying is 1/3. The number of times you win by switching is 2/3.

[Pinch Paxton]

I think that the maths error is actually at the very beginning. Your first choice is not 1/3 of picking any object. It's a certainty! It doesn't matter what you pick on your first choice, because you get the option to swap choices. So I think I have found the error.

 

 

No I've changed my mind again. I can't decide.

 

Pincho.

[Pinch Paxton]

Here are 3 * 1Million runs of my program....

 

Change = 333532 = 334360 = 333511 ......3 runs

 

Stick = 333404 = 333496 = 332950.......3 runs.

 

So That's a total of 6 million competitions, and swapping was slightly better overall, but not significantly better.

 

Pincho.

[BrainMan]

You were using the wrong equations in your program.

[Pinch Paxton]

Maybe, but I can't find anything wrong with it. I have checked it several times.

 

Car is stored in a box. = car

You choose a box. = choice

man tells you a goat box = goat (solved by not car)

 

change = not goat, not choice.

 

I can't see any errors in that. So I think my program is OK.

 

Pincho.

[Pinch Paxton]

Ahh, I have found the error in my program. The host of the show was taking the goat from any of the two doors, but that was allowing him to take the goat out of your chosen door. I have stopped him from doing that now. This gives me new figures...

 

Swap = 666699

stick = 334077

 

You should always swap doors!!!

 

Wow amazing!

 

Pincho.

[Pinch Paxton]

The new computer program.

 

 

 

 

Rem Change Choice

Amount = 1000000

 

Randomize Timer()

 

For n = 1 to Amount

 

Car = rnd(2)+1

choice = rnd(2)+1

Goat = car+1

if Goat = 4 then Goat = 1

if goat = choice then goat = goat + 1

if Goat = 4 then Goat = 1

change = goat + 1

if change = 4 then change = 1

if change = choice then change = change +1

if change = 4 then change = 1

if change = Car then inc Changewin

next n

 

Rem Stick to Choice

 

For n = 1 to Amount

 

Car = rnd(2)+1

choice = rnd(2)+1

 

if choice = Car then inc Choicewin

next n

 

Print "Change and win = "; Changewin

 

Print "Stick to First choice = "; Choicewin

 

Suspend for key

[iglak]

for those of us that aren't very programming laguage literate, could you explain each line and tell us what programming language you are using?

 

<edit> ah, i see.

by that program, change only loses if both choice and car are the same (1/3 chance of that)

and stick to choice only wins if choice and car are the same (1/3 chances, again)

[iglak]

BrainMan said in post # :

You pick 3. They eliminate 1 or 2. You switch to 1 or 2. You lose.

You pick 3. They eliminate 1 or 2. You stay on 3. You win.

i'm having trouble understanding why these are not 2 seperate cases each, especially the top one.

what i was thinking is that you (whoever wrote it first, whatever) are artificially increasing the odd of winning, instead of me artificially decreasing the odds of winning.

 

P.S. either way, it's still 50:50 chances

the way you listed it shows 3 wins and 3 losses.

they way i posted it shows 4 wins and 4 losses

[iglak]

ah... if you pick a loss and change, it BECOMES a win... that's 2/3

if you pick a win and change, that's a loss, the other 1/3

but if you pick a loss and stay, that's 2/3

and if you pick a win and stay, that's the other 1/3

 

<edit> so now the question is: is it a seperate choice with 50:50 odds, or is it just the end result of the first choice?

 

lets see...

write a computer program that choses between switch or stay AFTER the goat was revealed.

 

like:

 

Amount = 1000000

Randomize Timer()

For n = 1 to Amount
Car = rnd(2)+1
choice = rnd(2)+1
Goat = car+1
if Goat == 4 then Goat = 1
if goat == choice then goat = goat + 1
if Goat == 4 then Goat = 1
swich_stay = rnd(1) + 1 

if switch_stay == 1 then
- if choice != car then inc changewin 
endif

if switch_stay == 2 then
- if choice == car then inc choicewin
endif
next n

Print "Change and win = "; Changewin
Print "Stick to First choice = "; Choicewin
suspend for key

 

P.S. hehe... 3 posts in a row with no double posts and no non-relevent posts...

[Pinch Paxton]

Yes you have figured it out. It's surprising how clear it becomes when the computer program works.

[iglak]

yeah, okay, so it IS better to switch than to stay...

i was tired when i wrote that, but now it makes even more sense.

^^;

[wolfson]

This problem confused the hell out of me when i met it at university last yr! the point is that the probabilities are not equal to 1/3 because the situation is not symmetric - if i said there are 10 red balls, ten green balls and ten blue balls in a bag, what's the probability that i'll pull a blue ball out you'd say 1/3, right? and you'd have got this by symmetry.

 

but if i then told you that the red balls are stuck to the sides you'd have to "revise" your estimate!

 

the monty hall problem is basically the same thing; the reason it's confusing is because the actual physical situation does not change at all in the monty hall problem, but the probabilities do.

 

it's quite a lot like the EPR paradox in quantum physics. have you heard of that?