@Groove: If Monty Hall showed you a random goat out of the three doors (possibly your own door), then yes, it would be a 50/50 split (I think). However, because he is always showing you the “loser” in the bigger group, it is better to choose the bigger group (switch).
RE: Monty Hall Problem
Think of it this way. You have a choice of one door or two doors(which has been collapsed to one by elimination). Wouldn’t you pick two doors over one if you could?
Sorry if somebody already said this, I did not read every comment.
It’s only logical to suspect the game show host has more information than contestants about what is behind the doors, at least at some point, or that the game is rigged in some other way, but the question does not give enough information about these things, so there are at least a couple of different possibilities to consider.
First, either contestants are always given a second choice, or they are not. A situation where the game show host is not obligated to give contestants a second choice is not likely since for maximum entertainment purposes, and thus the game show’s popularity, so a second choice would likely always be offered. So it’s more logical to assume that you’ll be allowed to change your mind and given a second choice.
Second, if the game show is rigged because what’s behind all the doors, or even just the two remaining doors, can be switched by the game show operators after the game is started, then it makes no difference if you switch or not, since the odds are solely what the operators choose them to be, which arguably is for more winners than only one third of the contestants to keep the game more popular than if only one third, or less, of the contestants win.
Although it’s highly unlikely the game is rigged in this way since that would not be ethical, and in any case it could not be expected to remain a secret, but with the limited information given you cannot be absolutely sure, so it’s better to switch for your second choice because of the following reasoning.
If you picked the door with a goat, the other doors have a car and a goat. You can certainly reason that the host then can’t open the car door because that would ruin the game for the audience and thus the show’s popularity. The probability that you first picked the car door is one out of three, so there is a two-thirds chance that you first picked the one of the two goat doors. Thus, without knowing the history of the show, you can reasonably assume that it’s at least rigged to the extent that the host knows what is behind the two remaining doors, so he will open the goat door. That means the other door now has a two-thirds chance of having the car, and your initial door selection has only a one-third chance.
So considering everything what don’t know and the new information the host gives by selecting a goat door, you have a much better chance if you switch.
Holy crap I finally understand it and here’s now facepalm I hope it makes sense.
Take a pizza divide it into 3 pieces. Make 2 plain and 1 pepperoni. You want the pepperoni. Choose 2 of the 3 pieces randomly. Chances are that you picked the pepperoni (2 of 3 slices 66%). (Or 33% chance you didn’t). But you want the pepperoni, even if it’s only one slice. You trade the 2 slices 1 slice w/pep. If you did this over and over, 66% of the time = keep and 33% = trade? Simple math, no?
So that means 66% of the time you wouldn’t trade but would have 2 slices(or 2 doors if you look at it that way). Pretend those slices are instead doors. Let’s looks at Monty Hall problem from Monty’s point of view and in reverse (sort of). We know where the car(pepperoni) is, so that no longer a mystery.
Pretend the 1 pizza slice was the 1 door pick by the Monty Hall contestant. We already said before that the 1 pizza slice(door) could only be the pepperoni(car) 33% of the time and is only traded 33% of the time, right?
So if you were the contestant and knew your first choice of doors was the car (or pepperoni) only 33% of the time. Wouldn’t you trade your 1 slice(door) for the other 2 slices(2 doors)? Since we know the 2 of 3 slices(2 doors) will contain the pepperoni(car) 66% of the time? Him removing a door is a kin to you dropping a piece on the ground when exchanging the 1 piece for the 2. Even though the 1 piece(door) is not there in the very end doesn’t remove the fact that the pepperoni(car) was one of those 2 pieces 66% of the time.
Change the rules: 3 doors - 1 car 2 donkeys, you can have 2 doors or just 1 door. If you choose 2 doors (aka switch your choice after the first pick) Monty is willing to take one donkey off your hands. You pick the 2 doors even though one’s a donkey for sure.
Damn that’s a lot easier to think about than to try and put into words.
just pick the wrong door and he’ll pick the right door for you. with 3 doors,t here are two wrong ones, and with 1000 doors, there are 999 wrong ones. in any case, you have over 50% chance of having picked a wrong door.
You are amazing. That just made so much sense unlike the textbooks i’ve read with all these formulas.
Another way of explaining -
Let there be 3 doors - A,B,C. You pick A.
Let goat=0, car=1
All Possibilities:
A B C
0 0 1 - B is revealed to be a goat. C left.
0 1 0 - C is revealed to be a goat. B left.
1 0 0 - Any one is revealed. Other left.
After revealing goat,
A Other gate
0 1
0 1
1 0
A’s winning chance = 1/3
Other gate’s winning chance = 2/3
The other gate S (switched) is actually an OR function.
S = B or C (ie, best of B and C)
Since the other gate S is “BEST OF TWO”, it is a better option.
Hope you don’t mind my comparision with binary functions :D:D
Valid - very well done. I admit I get a weird pleasure from seeing the logical knots the non-switchers tie themselves into. One correction, in comment 32 you said to Frank that Monty is revealing new information with the pick. But the reason the switch probability is 2/3 is precisely because Monty has not revealed any new relevant information by showing you a goat. He can always show you a goat. So your original probability of guessing correctly of 1/3 is unchanged by the revelation of a goat behind one of the remaining doors.
I have not read all of the above, so if this is already stated then … sorry you have to read it again.
This doesn’t even have to be a question of math or probability.
Given three cards: one Ace and two 9’s
You pick one of the cards but you are not allowed to look at it.
Some bozo tells you that at least one of the other cards is a 9 (if you accept this like it means something special then you are a bozo as well … because every time at least one of the other cards is a 9).
You place the single card face down and next to it you place the other two card face down.
Now the bozo tells you that if you pick the “stack of cards” that has the ace in it you win a car.
Which of the “stacks” do you pick up … the one with a 1/3 chance of winning or the other pile that has two cards in it?
It has been twenty years since I thought of this scenario … thanks Kalid, for the opportunity to keep thinking about stuff.
@Trevor: Hah! It’s funny, I think everyone starts as a non-switcher until you slowly start to realize… =). Good point, no information was actually revealed since one of them had to be a goat.
@Don: Thanks for the comment, glad the site’s giving you something to chew on :). Actually I find learning most enjoyable when you come back to it, vs. having to memorize some explanation about why the Monty Hall problem works (vs. understanding why it works).
I like the analogy – the key is seeing you’re given the choice between 1 card (your original guess) or the best of two other cards. Monty just does you a favor by throwing one away early. Nice insight.
Actually Monty does you a confusing (to most people) disservice by throwing one away early. If he left the unpicked two doors closed and said you could open them simultaneously, and if you see a car … you win. Almost all would then trade for the two doors.
Gary’s articulation from the top of the comments helped me to make the connection and understand what the problem was getting at: “The odds of the selected door having the prize hasn’t changed at all. You just know more a lot about the doors you didn’t pick.”
Oh and creating a program to practically prove it is very good too and simple. Luv this 
Simple problem, big explanation xD.
Think visually not with arithmetic and it makes sense instantly.
This is a problem I have always been stumped by (and I do understand probability very well) because of the apparent conflict between the model and the reality.
At the purported start of the game, I have no door and so no chance of having the car. If I choose 1 of the 3 doors I have a 1/3 chance of choosing correctly (~33%). Due to the nature of the game, the actual game doesn’t start until Monty reveals the door with the goat behind it and asks if I want to switch.
At this point - the true start of the game - I have a 1/2 chance of having the car (50%).
If I switch, I have a 1/2 chance of choosing the correct door (50%) - but I already have a 50% chance of having the correct door.
Since probabilities are multiplicative:
Stick with original choice: 0.50
Switch: 0.50 x 0.50 = 0.25
So mathematically - if I switch it reduces my likelihood to end up with the car from 0.5 to 0.25. However, it never works out that way in the simulations.
I guess I need to spend the time to program it the way I conceptualize it and see what the results are.