The Monty Hall Dilemma........goats And Caddys!

[ToraToraTora]

http://www.bbc.co.uk/news/magazine-24045598
I have come across this before and like most initially I always thought that switching or not made no difference but once the penny dropped I saw it. I have had many a difficult time trying to explain why you should switch to people and my explanations have been somewhat clumsy but this explanation is as neat as I have come across:
"In the game you will either stick or switch. If you stick with your first choice, you will end up with the Caddy if and only if you initially picked the door concealing the car. If you switch, you will win that beautiful automobile if and only if you initially picked one of the two doors with goats behind them."
So do you get it? Never got it? don't accept it? ......

[mibn2cweus]

In summation;

Of the three equally likely doors, the Caddy is twice as likely to be behind one of the two (out of three) doors not originally chosen.

Monty reveals which of these two (out of three) doors it's not behind without changing the (two out of three) probability that the Caddy is (twice as likely to be) behind the one remaining door.

[New Judge]

A bit late (I've been away). BUT, a very simple way to understand this is thus:

You can only lose by changing if you chose correctly in the first place. Since the chances of choosing the winner from the outset is one in three the chance of losing by changing is one in three, so the chance of winning by changing is two in three.

[pixie373]

Strangely, i hadn't seen this thread, but was talking to OH about it earlier. Derren Brown has explained the maths several times, but although i understand the maths, i still don't get the logic. Surely, it is where it is?!

[pixie373]

Also, just having read the link, i don't agree with part of it. If you have two unopened doors, it is a "new game" and the chances of opening the "right" one WILL be 50/50. Surely?!

[factor-fiction]

It is indeed where it is, pixie, but initially only the host knows where it is- the contestant doesn't know. But if you play the game enough times your first choice has a one in three chance on average of being the right one; sticking doesn't change that probability. Following the elimination of one door (which the host knows never holds the top prize) you are left with a remaining door which must be right two thirds of the time.
There is a simulator. Try it and see.
http://www.grand-illusions.com/simulator/montysim.htm

[pixie373]

I just don't see why it's still 2/3rds (whether the host knows where it is, or not- does that make a difference?). One has been ruled out, so the probability of it being in each of the other two, is 50%. No?

[pixie373]

Ok. I tried the simulation, factor, twelve times. I won the car 33% of the times, without changing any choices. Which is what i would have expected. Am i missing something?

[factor-fiction]

The probability of the first being right hasn't changed - it is still one in three.
The host deliberately chose one knowing it had no car. So the remaining unopened door has a two in three chance

[factor-fiction]

Now try it again but switch- you should win around 67% of the time if you do enough trials

[LazyGun]

Pixie - I have run that same simulation 100 times, and changed my initial choice every time. Result? 75/100 I would have won the car, compared to your 33%.

[LazyGun]

@ff excellent link/simulation, by the way :)

[pixie373]

Hmmm. I have now done it twelve times and changed my choice every time. Won 55%. Bamboozled! Is it fixed? Lol

[pixie373]

But why doesn't the remaining choice of two unopened doors have a 50/50 probability? It's in one and not the other. That's what i can't get my head around.

[LazyGun]

@Pixie - No, its mathemagic :)

It just illustrates how difficult humans find probability. Often our intuitive feel is wrong.

You are right incidentally, when you say that on your first choice you have a 33$ chance of winning. What changes is that the host of the programme gives you some very valuable information - they show you that one of the remaining doors has no prize and gives you a chance to reselect. Now you have a 1/2 chance of winning, rather than a 1/3 so your best chance, proven by the maths, can seem counter-intuitive, but changing your first choice now becomes the correct strategy :)

[pixie373]

I do understand that, lg, just not why changing your mind should increase your chances of winning. When Derren Brown explained it, i understood it perfectly all the way through. Then got to the end and thought - what??! :-)

[LazyGun]

@ Pixie - I can give you another webpage which goes through the problem in some detail. Once read I think it does become clear.

Only 1/3 times is your initial guess going to be the right one. That's the best chance you have at that point. This also means that you are going to be wrong 2 out of 3 times with your initial choice.

Have a read, if you fancy it, see it that page makes it any clearer, rather than me trying to restate it here ;)

http://montyhallproblem.com/

(And apologies if this link has already been posted; I did not read through all posts exhaustively first)

[pixie373]

Thank you, LG, i have read that link very carefully. It is exactly the way i saw derren brown describe it. I'm clearly very dense though, because when he said

//behind any two doors considered together, the probability of there being a prize is 2/3.//
(written alone in bold). I still see that as a 50/50 probability, no matter how many doors you started off with, this is what you know now. I fully accept I'm just too stupid to understand it and if i am ever on Deal or No Deal, i will swap:-)

But i do find it frustrating that i just can't accept that two closed doors ever have a probability of 2/3. I did get an A in A-level Maths, but you wouldn't think so!

[New Judge]

I thought A-Level Maths was becoming a trifle easier, pixie! But you don’t have to worry about maths. This is simply a matter of logic.

An understanding of this is easier to gain if you follow the idea suggested by ellipsis earlier: expand the game to 10,000 doors. You choose one (and so have a 1 in 10,000 chance of choosing the winner). The host then reveals another 9,998 losers. You now have the choice of sticking with your original choice (which, remember, had a one in 10,000 chance of being the winner) or changing to the one remaining door. Do you still think the odds you now have are 50:50? Of course not. The host has provided you with 9,998 losers from the original choice of 10,000. If you originally chose correctly (remember, 1 chance in 10,000) you would lose if you switched. If you did not originally choose correctly (9,999 chances in 10,000) you would win if you switched. The game’s original odds do not change by the opening of the losing doors - your chances of choosing the winner originally are still 1 in 10,000. And the chances of the winner being behind one of the other doors is still 9,999 in 10,000. All that has happened is that you have subsequently been provided with information that all the other doors bar one are losers. The odds of either of those doors being the winner is not 50:50. The odds of your original choice being the winner is 1 in 10,000. The odds of the winner being ANY ONE of the others is 9,999 in 10,000. So effectively by switching you are taking those odds, except that 9.998 of your choices have been thoughfully eliminated by Monty.

Is that any easier?

[pixie373]

Ermmm... Thank you for the explanation, new judge. I still see that if there are 1000 doors, 998 are opened an shown to be empty, the remaining two are 50/50. Obviously, i must be wrong, but I'm convinced still that the odds must change, the fewer choices that are left. It doesn't matter if you start with a million doors, when you are down to two and the prize is behind one, you are equally likely to get it right or wrong. It shouldn't matter how many are "ruled out". I am happy to accept the explanation by all those more knowledgeable than me. I just wish i could actually understand why.

[mibn2cweus]

LG, this has been explained a number of times for anyone who has read the thread.

If there are three doors to chose from, the odds are obviously twice as good that the Caddy waits behind one of the two doors not initially chosen. Having then been shown which of these two doors it's not behind in no way alters the 2/3 likelihood of the Caddy being behind the other door.