Q: How does the probability of winning the car change in the Monty Hall problem?

[VRT]

Hi everyone,

2 goats and the car are from . The monkey is for you to remove from my back – it’s killing me!

The problem also was in “21” movie.

The setup is you are on a game show, there are 3 closed doors with 2 goats and a car behind the doors, you do not know what is behind which door, the host does. You sure want to get the car, not a goat. So it begins

1) you pick a door;
2) the host opens 1 of other 2 doors where a got is;
3) host gives you a choice to reconsider and pick the other closed door

Q: is the probability to get the car any different between the door you picked on step 1 and the other closed door?

A: door from step #1 has 1/3 probability to win, the other closed door has 2/3 probability to win (so, you always should changed your original choice to increase your chances).

I do understand why it is – the door from step #1 has 1/3 probability to win from the very beginning and the fact that another door is opened on step #2 doesn’t change this fact (from here you have 2/3 for the last door, because it’s the only door left at this moment).

The monkey on my back screaming that after step #3 you have a brand new fresh choice to make between just 2 doors, so probability must be ½ for each door. Please kill this damn monkey – it didn’t let me sleep last night. I’ve written a simple simulation that randomly places the car, makes my random choice, pretends the host opens a door with a goat and does or doesn’t change original choice. I ran it up to 100000 games for it always keeps the original pick and the same number of games where it always changes the original choice. The simulation confirms that 1st door wins in 1/3 of games and the last door wins in 2/3 of games (after 15-20 games the picture starts to show up, after 100 it’s already roughly about 1/3 and 2/3 and by 100000 games it’s even closer to 1/3 and 2/3). But the monkey doesn’t shut up! Help me!

Thanks.

 

[Office_Shredder]

To get a better level of intuition, let's step up the game. You pick a card randomly from a deck. Your objective is to have the ace of spades. After you pick the card, I search through the deck and find 50 cards which are not the ace of spades, and show them to you. You can either keep your original card, or you can take that last remaining card in the deck. What do you think the odds are that your card is the ace of spades? It is far more likely that the ace of spades was in the deck, and my procedure was simply a way of asking you "do you think you picked the ace of spades at the start, or was it one of the cards in the deck?"

Back to the original game, when the host reveals one door to you, he is asking you "Do you think that you won the game on your first pick, or was it one of the other two doors?" Clearly 2/3rds of the time it is one of the other doors.

To bring some math into it: the probability that you win given that you don't change is clearly 1/3... if you don't switch doors, the fact that the host revealed a goat behind a different one isn't going to change anything. Meanwhile, the probability that you win given that you switch doors is going to be:
P(Win after switching) = P(Win after switching|your original door was the winning door)*P(your original door was the winning door) + P(Win after switching|your original door was a losing door)*P(your original door was a losing door).

P(Win after switching|your original door was the winning door) = 0
P(your original door was the winning door) = 1/3 for the same reason as before,
P(Win after switching|your original door was a losing door) = 1, since if your original door was a losing door, then the only door left to pick when you switch is the winning door,
P(your original door was a losing door) = 2/3.

So doing all the math says that the probability that you win if you switch is 2/3

 

[economicsnerd]

To supplement Office_Shredder's great explanation...

Consider the following game:
- First, you pick a door from {1,2,3}.
- Then you pick from {A,B}, where A means "Get the car if and only if your initial choice was correct", and B means "Get the car if and only if your initial choice was incorrect".
How would you play this game? Would you choose A or B?

What Office_Shredder is pointing out is that this game is really equivalent to the game you described. That is, picking Stay in the old game is equivalent to picking A in the new game, while picking Switch in the old game is equivalent to picking B in the new game.

 

[VRT]

Thank you for replies,

The whole picture is understood (and even confirmed). What is missing is why the part of the game after host opens a door is not an independand game? How it's different from the game where you wouldn't have to make 1st pick (1st pick was totally random anyway)? Just 3 doors, host opens a goat door - you just woke up and have to choose between 2 doors - both must have 1/2 probability. The same with the deck of cards. What's the difference (and I know there is a difference)?

 

[Xaotique]

Lets start from the beginning ..

You pick a door at random.
There's a 2/3 chance of choosing the wrong door and a 1/3 chance of choosing the correct door.

If you chose one of the incorrect doors, there is left a correct and incorrect door.
They would flip the incorrect one.

Changing means you go to the only remaining door, the correct one.

So, choosing the incorrect door at the start means you win. (2/3 chance you win)

If you choose the correct door to start, they will flip an incorrect one and you'll move to the remaining incorrect one.

There's only a 1/3 chance of that happening though. Therefore, there's a 2/3 chance of winning if you switch. If you don't switch, you stick to the basic 1/3 chance of winning.

 

[D H]

What is missing is why the part of the game after host opens a door is not an independand game?

Because you've already picked a door. Whether there's a car behind either of the two remaining doors depends on whether you happened to have picked the right door in the first place.

The question as specified assumes a knowledgeable yet dumb Monty Hall. Suppose Monty Hall is instead devious: He only shows you a door if you happened to have picked the right door in the first place. This Monty isn't quite devious enough. There is still information of which you can take advantage. Even more devious: Monty flips a coin. Heads, he shows another door, tails he doesn't. Now your choice to switch or stay is irrelevant. Your odds are 1/3 whether you switch or stay.

That Monty needs to know where the car is another factor. He has to know which door *not* to open. Suppose Monty asks Carol to open a door, and voila! It's a car. "Thanks! I'll switch to that door!"Ultimately, this is a question of how to take advantage of partial information in an uncertain world. A whole lot of very useful mathematics has been created to address this general question. It's the basis for how we put men on the Moon in the 1960s, for example.

 

[economicsnerd]

Even more devious: Monty flips a coin. Heads, he shows another door, tails he doesn't. Now your choice to switch or stay is irrelevant. Your odds are 1/3 whether you switch or stay.

Really?

If I'm interpreting correctly, you're saying:
- You choose a door.
- He flips a coin.
- If tails, you get no other information.
- If heads, he opens a door with no car behind it.
If that's the game you had in mind, you still face the usual trade-off conditional on heads. That is, if it comes up heads you switch.

 

[D H]

You're right. I'll have to rethink that.

 

[Office_Shredder]

I tried calculating - if he has probability p₁ of opening a door if you have a winning door, and probability p₂ of opening a door if you have a losing door, then for P(you have a winning door | no door opens) = 1/3 (i.e. you gain no information), you must have p₁ = p₂, and for P(you have a winning door | a door opens) = 1/2 (i.e. again you gain no information) you must have p₁ = 2 p₂, i.e. the only way he can give you no information is by never opening a door (assuming he has a mixed strategy contingent only on the state of the game after you pick a door).

If you don't know what his p₁ and p₂ are though (e.g. he picks them randomly before the game begins and doesn't tell you what they are) then you might not know which direction you are gaining information in, so it might be possible for him to do something tricky still.

 

[economicsnerd]

If you don't know what his p₁ and p₂ are though (e.g. he picks them randomly before the game begins and doesn't tell you what they are) then you might not know which direction you are gaining information in, so it might be possible for him to do something tricky still.

As long as you have a belief over what p₁ and p₂ are and maximize P(car) with respect to that, your previous argument works: just replace p₁ and p₂ with their expectations.