Puzzle - which door?

[Nox]

the only way i can get my head around this at this time of day is to look at it in a much easier way (im still only on my first cup of coffee)

choose door 1, 1/3 chance of winning

look at doors 2&3 as a single entity, choosing this 'pair' of doors gives you a 2/3 chance of winning. As the wrong one is removed you are effectively getting two guesses for the price of one.

i think...

Nox

[Paul Adams]

It took me ages to even accept it was right, there are sites all over with mathemtical proof of it, but I'd most likely not understand it - the one I "got" was this, though:

Imagine there are 100 doors, not 3.
You still pick 1, but of the remainder 98 are opened with nothing behind them - now you are left with 2 doors as before.
Your initial chance of success was 1% - what is the chance of the other door now having the prize? 99%

Still convinced it's a 50/50 chance of your door being the right one?


I love things like this that just seem to defy logic, and unless you've come across it before, or are really smart, you'd most likely say it makes no difference.
Great for pub arguments

[Blub2k]

possibility vs probability is a head melter, cant be dealing with it at all. Fair play to the people who get their heads around it is all I can say.

[Iain]

Answer to the twins thing

They are part of a set of triplets, quadruplets, quintuplets, etc

[Rythmic]

Imagine there are 100 doors, not 3.
You still pick 1, but of the remainder 98 are opened with nothing behind them - now you are left with 2 doors as before.
Your initial chance of success was 1% - what is the chance of the other door now having the prize? 99%

Still convinced it's a 50/50 chance of your door being the right one?

Yup I'm still convinced - because in the opening of the 98 doors you had a 98% chance of one of them having the prize behind it. By making the assumption that you can open 98 doors - you're changing the odds.

I'm amazed that this kind of nonsense gets perpetuated (I personnally studied a bit of probability in Maths at Universtity). But then again even Einstein couldn't except it (look up his quote "god does not play dice (with the universe)".

And most people can't accept that the numbers 1, 2, 3, 4, 5 and 6 have the same chance of winning as any other set of numbers....

[Devilbod]

Right I've gone away and had a read about this because it's doing my nut in and it's two different branches of maths Probability verses Statisics, so both answers are correct.
So Apologies to AZ .

Either way I can understand the probability side of it, it's like when you hear a statictic that something happens to 1 in 10 people yet if you got 10 people at random chances are none will have that problem.
To be honest I still er on the side of the statictics purely because if you take the probability you could go buy a lottery ticket and then change your mind about the numbers a million times and then win.....

Little Java Applet with the door and statistics.
http://www.shodor.org/interactivate/activities/monty3/

[Paul Adams]

Yup I'm still convinced - because in the opening of the 98 doors you had a 98% chance of one of them having the prize behind it. By making the assumption that you can open 98 doors - you're changing the odds.

I'm amazed that this kind of nonsense gets perpetuated (I personnally studied a bit of probability in Maths at Universtity). But then again even Einstein couldn't except it (look up his quote "god does not play dice (with the universe)".

And most people can't accept that the numbers 1, 2, 3, 4, 5 and 6 have the same chance of winning as any other set of numbers....

This is not pure probability - it relies on knowledge that the door with the prize is not one of the 98 (in this example) opened.

The gameshow host knows which door has the prize behind it, and will never open that door, hence it is not a possible outcome.


Another way to look at this:
Initially there is a probability of your uneducated choice being correct of 1 in 3, let's call your choice set "A" and the remaining "B" (which contains 2 items).
The probability of A is 1/3, and the probability of B (or, if you prefer "not A") is 2/3.
B is reduced to a set of 1 item, but its probability cannot change as the reduction occurred under knowledge that the "correct" result cannot be revealed.
Your choice was A (1/3), you change your choice to "not A" (B) and it it (2/3).


This is not as simple as making uninformed choices, rolling a die or flipping a coin N times.

[Paul Adams]

Ah, thanks Devilbod, I coulnd't find a website with a decent emualator on it

Games stayed: 30
Games stayed and won: 10
Experimental probability to win: 0.33333334
Games switched: 30
Games switched and won: 21
Experimental probability to win: 0.7

[Raz316]

First choice.
1/3 - right door
2/3 - wrong door
More chance of being wrong on your first go. That means, it is more likely that you will need to swap after a bad door is revealed to get the correct door.

To put it to the extreme...
there are 1000 doors, you pick one, 998 incorrect doors are opened, would you swap your choice?

E> oops didnt see third page :x

[Devilbod]

That about what I got lol but did it to 50.

[Big RICHARD]

Right

Three doors, one prize so to make this easy you pick door 1 that gives three possible outcomes:

prize is behind door 1. You win.
prize is behind door 2. You lose.
prize is behind door 3. You lose.
chance of winning 1 in 3 = 33%

Reguardless of the choice the orignal chance of winning is one in 3.

And this bit's for AZ your maths is wrong your point (3.) is infact two seperate possible loosing situations if you switch and two possible chances to win if you stick.
I'll explain buy breaking this down into all possible outcomes:
So lets say your told to pick a door and and then one of the wrong doors will be opened and you'll have the chance to change.

You pick door 1.

Door 2 is opened, you stick to door 1, the prize is behind door 1. You win.
Door 2 is opened, you stick to door 1, the prize is behind door 3. You lose.
Door 3 is opened, you stick to door 1, the prize is behind door 1. you win.
Door 3 is opened, you stick to door 1, the prize is behind door 2. You lose.

Door 2 is opened, you switch to door 3, the prize is behind door 1. You lose.
Door 2 is opened, you switch to door 3, the prize is behind door 3. You win.
Door 3 is opened, you switch to door 2, the prize is behind door 2. You win.
Door 3 is opened, you switch to door 2, the prize is behind door 1. You lose.

So out of 4 possible outcomes with sticking you can win twice 2 in 4 chance of winning which is the same a 1 in 2 chance = 50%
Swtiching 4 outcomes result in the 2 win, 2 in 4 which again is the same as 1 in 2 = 50%.

Defy my logic I dare you. Muhahahahaha

You didn't use your method on all possible outcomes. What if you picked door 2 or door 3 to begin with? Your reasoning is flawed because you were not exhaustive of all possibilities....

[Rythmic]

This is not pure probability - it relies on knowledge that the door with the prize is not one of the 98 (in this example) opened.

The gameshow host knows which door has the prize behind it, and will never open that door, hence it is not a possible outcome.

Thats exactly what I'm saying. Because the host will only ever open a non winning door, your initial choice has no bearing on the actual chance of winning.

Another way to look at this:
Initially there is a probability of your uneducated choice being correct of 1 in 3, let's call your choice set "A" and the remaining "B" (which contains 2 items).
The probability of A is 1/3, and the probability of B (or, if you prefer "not A") is 2/3.
B is reduced to a set of 1 item, but its probability cannot change as the reduction occurred under knowledge that the "correct" result cannot be revealed.

Why can B's probability not change? The contestant will always reach the second stage, of one door being correct, and one being not. It's simple 50/50

This is not as simple as making uninformed choices, rolling a die or flipping a coin N times.

It is. The final chance bears no relation to what was picked last time.

[fondie]

I was trying to show that if you take all the possible outcomes provides a different answer and more accurate, with your example:

you choose box 1 and stick to box 1. you win!
you choose box 1 and change to box 2. you lose!
you choose box 1 and change to box 3. you lose!

you choose box 2 and change to box 1. you win!
you choose box 2 and stick to box 2. you lose!
you choose box 2 and change to box 3. you lose!

you choose box 3 and change to box 1. you win!
you choose box 3 and change to box 2. you lose!
you choose box 3 and stick to box 3. you lose!

Sticking gives you 1 in 3 chance.33%
Switching gives you 2 in 6 chance.33%
Overall gives you 3 in 9 chance. 33%
It's all the same.

In the door example the odds are 50% because a door is removed.

you choose box 2 and change to box 3. you lose!
you choose box 3 and change to box 2. you lose!

these outcomes are not possible, because if the prize is in box 1 and you choose box 2, the host MUST open box 3, therefore you can't choose it, and similarily if you choose box 3, he MUST open box 2. you're only option when you switch in these cases, is box 1, herein lies your advantage.

[Rave]

This is why I hate statistics and probability- and the people who can understand it all. It's enough to provoke feelings of irrational violence, as Vaul has already discovered.

Rich :¬)

[Big RICHARD]

Rave mate, it's ok to be a little slower than the rest of us. We'll speak slowly for you. I-s t-h-i-s s-l-o-w e-n-o-u-g-h?

[Rave]

It doesn't matter if you explain statistics to me in words of one syllable or in the most technical language possible- I'll still want to nut you.

Rich :¬)