Why?
The obvious answer is 50/50 … I asked my friends and they agreed too.
Brian, you’re wrong. It’s a matter of conditional probability. I’m not 100% clear on the exact explanation myself, but it’s not that simple. It’s explained by Bayesian statistics.
And Kairos - that’s what I thought too. But a simple probability tree gives a different result: there are 3 possible choices for the first choice, and adding in the second choice makes a total of 6 possible outcomes (two possible branches for each first choice). Of those, you’ve got a 1 in 6 chance of being right if you stay with your first choice (that is, of the six outcomes, only one is being right both times).
But, even that explanation isn’t right, because it fails to take into account information provided by the host by way of which door he opens. The actual mathematical proof is based on conditional probablities and, like I said, Bayesian theory, which is where my understanding gets a bit fuzzy.
I think to put it simply, Bayesian theory is based on the concept that probability is subjective.
Therefore to use Brian’s deck of cards example, to the person sitting at the table watching Brian discard all the other cards save one, the odds are overwhelmingly in favour for that person that the Ace of Spades is the card in front of Brian. However, if they both leave the room and a new person then walks into the room with no knowledge of what has been happening and is told ‘one of these two face-down cards is the Ace of Spades. Which one?’ from his point of view the probabilities are 50/50. Head hurts now 
I wrote a little web-script to test this out about a year ago:
yule.org/brainteaser/monty.html
This allows you to run the experiment hundreds/thousands of times, so you can see for yourself how often either ‘sticking’ or ‘switching’ lets you win…
Until the door is opened, there is no way to know if it contains a goat or a car, in fact it contains both a car and a goat at the same time, until the moment the door is opened. This is known as Schroedinger’s goat.
[quote]Brian, you’re wrong. It’s a matter of conditional probability. I’m not 100% clear on the exact explanation myself, but it’s not that simple. It’s explained by Bayesian statistics.
[/quote]
Bullocks. It’s very simple. Forget my last post. Go back to the one before. There’s no way that it is wrong. Show me where that is wrong. That’s simple statistics and logics. Don’t give me any more of this Bayesian bollocks.
Brian
I think you are all crazy. With your first choice you either pick the car or you don’t. The host then choses one of the remaining doors with a goat. That leaves two doors. Your chances are 50/50.
Sorry, what was the question again?
Now that I know that one particular door contains a goat my odds of being right are increased from 1:3 to 1:2, so my chance of being right has now increased.
However I have a 50/50 shot at choosing the right door. It’s either the one I already chose, or it’s the other one. I’m being asked to make a straight choice, this one or that one. The fact that I have already made a choice is irrelevant because we never opened the door, so I didn’t really make a choice. The fact that there used to be three doors, and the odds used to be different, is also irrelevant. I now have a straight 50/50 shot.
Speaking of shots, consider Russian roulette.
You try once and have a 1 in 6 chance of blowing your brains out. You try again and you still have a 1 in 6 chance, and that doesn’t change however many times you spin the barrel and pull the trigger.
Each time you try the odds are 1 in 6. But the odds against a 1 in 6 probability not occuring out of 100 attempts are not 1 in 6.
The game show is a similar situation. You have a 1 in 3 chance the first time, and if you were to go to another set of doors (or spin the barrel again) you would have a 1 in 3 chance again. But you don’t go to a second set of doors, you just take away one of your three chances. So now you have a 1 in 2 chance of being right, but the odds are against you being wrong twice.
I need a drink.
Look, screw computer simulations. Can we do this for real at Alleycat’s on Friday? I’ll bring my car, although I can’t promise to be able to get it down the stairs. Does anyone have a couple of goats?
No that’s clearly wrong. Look at what I said about the cards a few posts back to see why.
If you change your choice, you have a 2/3 chance of winning. The reason is very simple. There is a 1/3 chance that the door you first picked was right. Only 1/3. If you were wrong (and there’s a 2/3 chance of that), you will automatically win, by changing your choice. Thus there’s a 2/3 chance of winning if you always change your choice.
Brian
[quote=“stragbasher”]Sorry, what was the question again?
Now that I know that one particular door contains a goat my odds of being right are increased from 1:3 to 1:2, so my chance of being right has now increased.[/quote]
ACK! NO! Your chances of being right if you stick with the door you chose first are exactly the same as they ever were. 1 in 3. Why would it have increased? Nothing about your door has changed. It’s the same door.
However, if you switch doors, your chances of being right are 2 in 3.
Let’s try this another way. I show you three doors. This time, you can pick TWO of them. If the car is behind either of them, you win it. Good odds, right? So, you have a 2 out of 3 chance of winning. Now, I open one of your two chosen doors to reveal a goat, which you knew would happen, since there’s only one car. Your chance of winning is STILL the same as it was at the beginning, right? 2 in 3. It’s just now you know that one door has probability zero, so the remaining door carries the full 2 in 3 weight.
Therefore the other (single) door has the probability of 1 in 3 of having the car. If you’d chosen it and stuck with it, you would therefore be stuck with that 1 in 3 chance. By switching you are effectively gaining control of two doors (one known, one unknown) instead of one.
Your odds are only 50/50 if you don’t get to pick until AFTER the goat is revealed.
Look, if there were 1,000 doors and I let you pick 999 of them, and then opened 998 of the ones you’d chosen to show goats (which promptly formed a huge mob and attacked the studio audience who were dying of boredom anyway), there would only be two doors left, but wouldn’t you still be pretty confident that you’d won the car?
That is, assuming it is the car that you want after all… 
I figured I’d have to bust this link out. Brian is right with his explinations (that’s the way I like to explain it too). It’s really this simple. If there were 100 doors and you picked one door the chance of being right is 1/100. After 98 doors are opened there are 2 doors left and one of them has the prize (visualize please). Now do you really think you got it right the first time (1/100 or 1% chance of being right) or would it be better to switch to the other door (which has a 99/100 or 99% chance of being right)? The same princible applies to 3 doors, it’s just harder to see/imagine.
If that didn’t do it for you, on to the link.
[CLICK HERE]
I like the way Dr. Math thinks.
PS: I knew this was gonna get booted out of the open forum, but it’s all good.
I think it’s a good idea to do this at Allet Cats if there are still nay sayers. All we would ne is 10 cards.
Miltownkid, I personally blame you for starting this annoying thread and I hold no responsibility for posting the following question:
Three chaps go into a restaurant. They buy a meal for
I reckon Science Daily explains it better, but I’m a sucker for mathematical proofs:
I almost had it when Brian explained it but then dassgirl came along and confused me again.
[quote=“Dangermouse”]The same three chaps are being tested by a wise man.
“I have three white caps and two black caps”, he says, “You will close your eyes. I will put a hat onto each of your heads, and hide the others. You will not be able to see your own hat. You have to deduce what colour hat you are wearing.”
So he does, and the chaps open their eyes, and there is a pause, and then one of the chaps says…[/quote]
“Give me back my wallet, bitch!”
DM, I think you have the puzzle wrong.
If it’s the same one that I am thinking of, the wise man says he has three white hats and two black hats, but without telling anyone he stashes the black ones and pops a white hat on each of the three men. Then he asks them to work out whether they are wearing a black or a white hat.
How does the brainy man deduce that he’s wearing a white hat?
[quote]A Czar invites the three smartest philosophers to his palace and showes them five hats (three white and two black). He seats them in a circle and tells them to close their eyes. When they close their eyes he places the three white hats on their heads (hiding the remaining two hats under his throne), he tells them that those are three out of the five hats (that he previously showed to them), but he doesn’t tell them which ones (which color). The philosophers open their eyes. Each of them sees the other two and the two white hats they wear, but doesn’t see the color of his own hat. They are told not to talk to each other. Then the czar asks them if any of them knows the color of the hat he wears. After a couple of minutes of thoughts, one of the philosophers says: " - I wear a WHITE hat."
Question: How did he know?[/quote]
It’s a fairly tough one (for me!). Even when I cheated and read the answer it took me a while to get it. 
Chaon has an even tougher puzzle involving trunks, padlocks and keys which I still refuse to believe is actually possible so I’ll save that 'til later.
[quote=“miltownkid”]If that didn’t do it for you, on to the link.
[CLICK HERE]
I like the way Dr. Math thinks. [/quote]
Hey! Those were my goats! Fine. Dr Math can clean up the mess 
Dangermouse’s are fairly standard.
3*9 (what was paid) = 25 + 2 (cost of meal + tip)
A man who sees two black hats will know his must be white, since there are only two black hats. But I like Tetsuo’s answer better
Edited: Oh yeah, I remember Spack’s version, which went a step further - something to do with IF any man could see two black hats, he would know his was white. However, since no-one has identified themselves quickly, no-one can see two black hats, and therefore his own is white. Something like that anyway, but I can’t remember the number or tyep of hats involved!
The other two are each wearing a white hat. How does the man know that he is also wearing a white hat (or is not wearing a black hat)?
The other two are each wearing a white hat. How does the man know that he is also wearing a white hat (or is not wearing a black hat)?[/quote]
Yes, that’s true of your puzzle - however Dangermouse didn’t specify that in his version so I only thought of the most obvious answer (I typed my response before reading your edited reply, if that makes any sense)! I’m sure that’s the one he meant, though, now I read it.