No, your intuition is misleading you: it's not a choice between two doors. If the probabilities don't convince you, you can easily simulate this by using three cards and considering one is the car and the other two goats. Pick one at random, then pick the remaining "goat" as if you were Monty; lastly, switch your initial choice as a player. Repeat the experiment 10 times and count how many times you win; remember you must always switch. You will win more often than lose! I tried this, that's how I know this will convince you.
Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.
The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
So when they don’t switch each one has probability 1/3 of getting the car? Where does the car go if no one gets it?
And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing?
Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again.
Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
> So when they don’t switch each one has probability 1/3 of getting the car?
Yes.
> Where does the car go if no one gets it?
I don't understand your question. What does this have to do with probabilities?
> And if they switch both have probability 2/3 of getting the car?
Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously.
They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's.
> Is this some form of car sharing?
That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch.
> Maybe you should try to play with three cards and one ace.
There are only 3 doors, so you should only use 3 cards.
> Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
Player #1 had a 1/3 probability of picking the right door. Therefore, he has a 2/3 probability of winning if he switches.
Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches.
In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).
No. Because the experiments are completely independent (i.e. player #1 & player #2 don't interact at all; their choices don't interfere), the events are effectively "choosing door A and switching to door B" and "choosing door A' and switching to door B'". You have doors A, B, A' & B'. This of course assumes all doors are revealed simultaenously; i.e. no player gets to see the result of the other's choice beforehand.
You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent.
Hang on, I think you've missed something. They are not completely independent, because if they pick different doors, that forces the host to pick the last door. Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math. Having two players with the ability to choose different doors (but still 3 doors and the host must choose the third) makes it a different game with different dynamics. It doesn't work the same because it's not the same game.
Or, because they are two independent games, the host can flip different goats for different people. You would have to add an element of "no screen peaking" where the contestants cant see which door opened for the other person.
So you have Player A who chose door 1. The host flips the goat in door 3.
You have Player B who chooses door 2. For them the host flips open the goat in door 1.
Player A should switch, and wins the car. Player B should switch, but had the car and loses it. The right move in both their cases was to switch.
Really, its no different than playing two separate games. Unless, like you said, you somehow change the rules and limit someones choices, in which case Im not sure why its a part of the conversation, other than being introduced through confusion.
If the players pick different doors, the host will open the third door. That prevents him from being sure to get a goat, in the same way that opening a door at random in the single-player case prevents him from being sure to get a goat.
Well, you're right: in fact, in some cases Monty cannot open any door! That is, if players A & B both chose goats, then the remaining door is the car, and Monty cannot open it without ruining the game.
For this reason, both players should be playing "blind", without seeing the other's choices or the door Monty opens for the other player. Anything else doesn't make sense for the game.
> Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math.
As it does removing the host's ability to choose a non-winning door in the variant where "the host had just picked at random from the two doors, and it happened to show a goat". That doesn't work the same either because it's not the same game.
These are two independent games, so for every player:
a) The probability of winning if staying: 1/3
b) The probability of winning if switching: 2/3
c) The probability of the car being behind door the player chose is the same as a): 1/3
d) The probability of the car being behind the door Monty didn't open is 2/3
e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat!
Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
There is one single game. It's my version and I make the rules!
Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door."
Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it."
In that scenario, what are the probabilities?
I agree it's different from the original problem, that's the point!
Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door."
Scenario0: "The player picked one door, the host opened another door and there was a goat behind it."
But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem.
Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door."
Scenario1: "The player picked one door, the host opened another door and there was a goat behind it."
The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different.
However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors.
They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat.
Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door."
Scenario1': "The player picked one door, then opened another door and there was a goat behind it."
Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.
The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)