Why are the two probabilities in the Monty Hall problem dependent on each other?

Question

Like most people, one of the first things I did after ringing in the new year was get into a discussion about the Monty Hall problem. Past discussions typically amounted to the other person saying, "It's obviously 2/3rds" over and over again, in various ways. This conversation, though, helped my understand why I find the Monty Hall problem difficult to understand.

I've looked at the de facto post on the topic: The Monty Hall problem and it doesn't quite get at the issue I'm having understanding the problem, so I'll try my best to lay it out below.

The probability of initially picking the correct door is 1/3. So you pick door 1 and Monty opens door 2 and shows you a goat. I have the same issue as the person asking the question in the link above: why is the new probability not 1/2? What clicked the other night is that the probabilities are linked, though I'm not sure why. For example, let's say I flip a quarter twice and both times it comes up heads. If I ask, "What is the probability that the next quarter flip is heads?", is the answer 1/8 or 1/2? If you interpret the question as "What is the probability the quarter comes up heads three times in a row?" you'd say 1/8, but if you interpreted it as, "What is the probability a quarter flip comes up heads?", you'd say 1/2.

Or, say that instead of it just being one player, there were two players.

1. Player 2 sits in a sound/light proof room
2. Player 1 chooses door 1
3. Monty opens door 2 and reveals a goat
4. Player 1 goes into the sound/light proof room and Player 2 comes out

It's obvious that Player 2 has a 1/2 chance of guessing the door with the car behind it, so why would Player 1 have a different probability?

Answer 1

Probability depends on one's knowledge. If I roll a die behind a screen so that you can't see the outcome, what's the probability it came up 3? But I can look down and see a 3 so the probability is different for me. For you, it's 1/6. For me it's 1/1.

In your scenerio, Player 1 has knowledge that Player 2 does not. It's not that the probabilities are related, but that the different players have different knowledge. Probability isn't a characteristic of the situation, it's a characteristic of the inside of your head.

Answer 2

Player $2$ doesn't know anything about the situation, so yes, Player $2$ has a $50\%$ chance of guessing right. But I suspect that the reason you think this is so is incorrect.

Let me modify your setup. On your $4$th step, let's say Player $1$ said that he guessed door $1$ first to Player $2$ while walking out. In this case, Player $2$ also has a $\frac23$ chance of winning, since the player can simply follow the same strategy as player $1$.

The reason why the probability is $\frac23$ is simply that for the Monty Hall problem to work, the game host has to know which doors have goats, so that the host can always reveal one, regardless of which door is chosen. It has nothing to do with the player, but rather that choice.

Answer 3

The actual discrepancy between switching and not switching is because of a rule that isn't always stated in spelling out the rules of the game: Monty will always show a goat, even if you guessed wrong and thus one of the doors that you didn't open has the car. Thus there are two scenarios after your initial selection:

• You guessed right, in this case Monty picks one of the other two doors (both of which are losing doors) at random. If you switch you lose, if you don't switch you win. This scenario has probability 1/3.
• You guessed wrong, in this case Monty picks the remaining losing door, leaving the winning door there for you to switch to. If you switch you win, if you don't switch you lose. This scenario has probability 2/3.

The symmetric version of the game is slightly different: in it, if you guess wrong, Monty has a 50% chance to show the car in which case you just immediately lose. In this case there are three scenarios instead:

• the first one above (still with probability 1/3)
• the second one above (now with probability only 1/3)
• the immediate loss scenario (with probability 1/3).

Compared to this game, in the actual game, all the probability that was associated to the immediate loss scenario gets rolled into the second scenario, none of it into the first one, which creates the asymmetry in the game.

A way to trick your intuition into understanding what is going on is to imagine a version of Monty Hall with a million doors. You pick a door, Monty opens 999,998 doors that all have goats. Is it a 1 in a million chance that your original choice was right, or a 1 in 2 chance?

Answer 4

One thing I found helped was to imagine that there are one thousand doors and a monty will always show you $998$ goats.

Now imagine your person 2 in a sound proof room. He is led out and he sees that door #138 and door number #569 are closed but all the other ones are open with goats.

Monty asks player number #2 what he thinks happened.

Player #2 says: "Well, since the rules of the game say you never open the door player #1 guessed, then player 1 must have either guessed door #138 or #569. And the rules are that you don't show the car, so the car is behind door #138 or #569.

"So I figure one of four things could have happened:

"1) Player 1 guessed door #138 but the car was behind door #569.

"2) Player 1 guessed door #569 but the car was behind door #138.

"3) Player 1 guessed door #138 and by great luck guess right and you randomly chose door #569 to show but that requires player one guessing right which could only happen $1$ is $1000$ times.

"4) Player 1 guessed door #569 and guessed correct and you randomly chose door #138 to show.

So Monty asks. "So what door do you pick?"

Player 2 says: "I don't really know. Situation 1 and 2 are equally likely so if player 1 guessed wrong which she almost certainly did, I don't know which door she didn't pick. And Situation 3 and 4 are both unlikely but again, I know nothing. I figure I have no idea... I'm going pick Door number ..."

At this moment Player 1 blurts out "I picked door #569!"

Now.... my question for you is: Did Player 1's blurting out her door change anything? And which door should Player 2 pick?

........

Now if we did the same thing with only three doors. Cut, paste and replace numbers:

.... $\require{cancel}$

Now imagine your person 2 in a sound proof room. He is led out and he sees that door $\cancel{\#138}$ $\color{blue}{\#1}$ and door number $\cancel{\#569}$ $\color{blue}{\#3}$ are closed but all the other one$\cancel{s}$ are open with goats.

Monty asks player number #2 what he thinks happened.

Player #2 says: "Well, since the rules of the game say you never open the door player #1 guessed, then player 1 must have either guessed door $\cancel{\#138}$ $\color{blue}{\#1}$ or $\cancel{\#569}$ $\color{blue}{\#3}$. And the rules are that you don't show the car, so the car is behind door $\cancel{\#138}$ $\color{blue}{\#1}$ or $\cancel{\#569}$ $\color{blue}{\#3}$.

"So I figure one of four things could have happened:

"1) Player 1 guessed door $\cancel{\#138}$ $\color{blue}{\#1}$ but the car was behind door $\cancel{\#569}$ $\color{blue}{\#3}$.

"2) Player 1 guessed door $\cancel{\#569}$ $\color{blue}{\#3}$ but the car was behind door $\cancel{\#138}$ $\color{blue}{\#1}$.

"3) Player 1 guessed door $\cancel{\#138}$ $\color{blue}{\#1}$ and by $\cancel{\text{great}}\color{blue}{\text{slight}}$ luck guess right and you randomly chose door $\cancel{\#569}$ $\color{blue}{\#3}$ to show but that requires player one guessing right which could only happen $1$ is $\cancel{1000}3$ times.

"4) Player 1 guessed door $\cancel{\#569}$ $\color{blue}{\#3}$ and guessed correct and you randomly chose door $\cancel{\#138}$ $\color{blue}{\#1}$ to show.

So Monty asks. "So what door do you pick?"

Player 2 says: "I don't really know. Situation 1 and 2 are equally likely so if player 1 guessed wrong which she almost certainly did, I don't know which door she didn't pick. And Situation 3 and 4 are both $\cancel{\text{un}}\color{blue}{\text{less}}$ likely but again, I know nothing. I figure I have no idea... I'm going pick Door number ..."

At this moment Player 1 blurts out "I picked door $\cancel{\#569}$ $\color{blue}{\#3}$!"

Answer 5

I'd like to expand on your coinflip example a little.

For example, let's say I flip a quarter twice and both times it comes up heads. If I ask, "What is the probability that the next quarter flip is heads?", is the answer 1/8 or 1/2? If you interpret the question as "What is the probability the quarter comes up heads three times in a row?" you'd say 1/8, but if you interpreted it as, "What is the probability a quarter flip comes up heads?", you'd say 1/2.

We can look at this another way, which may be instructive here - it's a similar kind of "what question is being asked?" angle to the one you considered, but brings in the idea that what you know informs the calculation.

I flip two coins behind a screen. I then tell you I'm about to flip a third coin, and ask you for the probability that all three flips were heads.

You don't know anything about the situation, so it's $(1/2)^3$ - the 1/8 you identified.

I flip two coins behind a screen. I then tell you I'm about to flip a third coin, show you both previously flipped coins are heads, and then tell you I'm about to flip a third coin and ask for the probability that all three are heads.

Even though I'm asking for the probability of flipping three heads in a row, you'd probably say 1/2.

And if I showed you only one coin, you'd have $1*(1/2)*(1/2)$.

I could even do these at the same time - if I have three people watching, and show the first no coins, the second one, and the third two (all heads), person 1 would say it's a 1/8 chance, 2 would say 1/4, and 3 1/2. They'd all be right about the calculation they were making.

What we're actually doing here is calculating conditional probabilities - P(Coin 1 yields Heads and coin 2 yields Heads and coin 3 yields Heads) given either no condition, "coin 1 yields Heads," or "coin 1 yields Heads and coin 2 yields Heads."

We can expand this out (I promise this is allowed!) to P(Coin 1 yields Heads given Coin 1 yields Heads) * P(Coin 2 yields Heads given Coin 2 yields Heads) * P(Coin 3 yields Heads) - fairly clearly those first two are 1, but the crucial point here is that we're not just taking P(Coin 3 yields Heads) - those other two variables still exist, they've just been shown to yield a particular result, so we're multiplying by 1.

The exact logical structure in Monty Hall is a little more tangled. The two crucial points are that the events aren't independent and that we're not looking at "door 1 wins given door 2 loses" but a more complex "door 1 wins given the player chose door 1, then the host followed the rules to show door 3 loses."

In your locked room example, player 1 is calculating the probability that door 1 wins given the player chose door 1, then the host followed the rules to show door 3 loses; player 2 doesn't know which door player 1 picked, so they can only calculate P(door 1 wins given door 3 loses). These are different conditions, so they give different results - in fact, exactly the 1/3 and 1/2 we're familiar with as the two ways of looking at this problem.

The calculations to show that are complicated enough that I think they're more likely to confuse than clarify; the real point here is that we need to state exactly what we know, and superficially trivial simplifications like saying "door 3 loses" instead of "the player chose door 1, then the host followed the rules to show door 3 loses" make a significant impact on the system.

Answer 6

A second answer to deal with the coin tossing issue:

The coin flip problem is that there are 8 possible outcomes equally likely. $(\color{red}{TTT; TTH; THT; THH; HTT; HTH;}\color{blue}{HHT;}\color{green}{ HHH}$

But by seeing the first two flips you know that $6$ of them are now impossible ($(\color{red}{TTT; TTH; THT; THH; HTT; HTH;}$). That leaves only $2$ two possible ($\color{blue}{HHT;}\color{green}{ HHH}$) both equally likely. But if you didn't see the flip you don't know they are impossible so you think all are eight were equally likely.

Let's add an irrelevant step to the Mony Hall problem that changes nothing. After the player picks a door Monty flips a coin. He only looks at it if the player picked the door with a car and both remaining doors are goats. If it's heads he shows the first door with a goat. If it's tails he shows the second door with a goat. However if the player chose a goat he flips the coin but shows the remaining goat either way.

Here are the possible outcomes:

Car in door 1: Player picks 1: coin is Heads: Monty shows door 2.

Car in door 1: Player picks 1: coin is Tails: Monty shows door 3.

Car in door 1: Player picks 2: coin is Heads: Monty shows door 3.

Car in door 1: Player picks 2: coin is Tails: Monty shows door 3.

Car in door 1: Player picks 3: coin is Heads: Monty shows door 2.

Car in door 1: Player picks 3: coin is Tails: Monty shows door 2.

Car in door 2: Player picks 1: coin is Heads: Monty shows door 3.

Car in door 2: Player picks 1: coin is Tails: Monty shows door 3.

Car in door 2: Player picks 2: coin is Heads: Monty shows door 1.

Car in door 2: Player picks 2: coin is Tails: Monty shows door 3.

Car in door 2: Player picks 3: coin is Heads: Monty shows door 1.

Car in door 2: Player picks 3: coin is Tails: Monty shows door 1.

Car in door 3: Player picks 1: coin is Heads: Monty shows door 2.

Car in door 3: Player picks 1: coin is Tails: Monty shows door 2.

Car in door 3: Player picks 2: coin is Heads: Monty shows door 1.

Car in door 3: Player picks 2: coin is Tails: Monty shows door 1.

Car in door 3: Player picks 3: coin is Heads: Monty shows door 1.

Car in door 3: Player picks 3: coin is Tails: Monty shows door 2.

..........

Now lets suppose that Player 1 picks door $2$ and Monty shows door $1$.

Now that rules out $15$ of the $18$ outcomes as impossible. $\require{cancel}$. (They just didn't happen).

$\cancel{\text{Car in door 1: Player picks 1: coin is Heads: Monty shows door 2}}$.

$\cancel{\text{Car in door 1: Player picks 1: coin is Tails: Monty shows door 3}}$.

$\cancel{\text{Car in door 1: Player picks 2: coin is Heads: Monty shows door 3.}}$

$\cancel{\text{Car in door 1: Player picks 2: coin is Tails: Monty shows door 3.}}$

$\cancel{\text{Car in door 1: Player picks 3: coin is Heads: Monty shows door 2.}}$

$\cancel{\text{Car in door 1: Player picks 3: coin is Tails: Monty shows door 2.}}$

$\cancel{\text{Car in door 2: Player picks 1: coin is Heads: Monty shows door 3.}}$

$\cancel{\text{Car in door 2: Player picks 1: coin is Tails: Monty shows door 3.}}$

Car in door 2: Player picks 2: coin is Heads: Monty shows door 1.

$\cancel{\text{Car in door 2: Player picks 2: coin is Tails: Monty shows door 3.}}$

$\cancel{\text{Car in door 2: Player picks 3: coin is Heads: Monty shows door 1.}}$

$\cancel{\text{Car in door 2: Player picks 3: coin is Tails: Monty shows door 1.}}$

$\cancel{\text{Car in door 3: Player picks 1: coin is Heads: Monty shows door 2.}}$

$\cancel{\text{Car in door 3: Player picks 1: coin is Tails: Monty shows door 2.}}$

Car in door 3: Player picks 2: coin is Heads: Monty shows door 1.

Car in door 3: Player picks 2: coin is Tails: Monty shows door 1.

$\cancel{\text{Car in door 3: Player picks 3: coin is Heads: Monty shows door 1.}}$

$\cancel{\text{Car in door 3: Player picks 3: coin is Tails: Monty shows door 2.}}$

Of the three remaining $2$ of them have the car in door $3$ and only $1$ has it in door $2$. Player should switch.

.......

Now let's march out player 2. All player 2 sees is that Door $1$ is open. She doesn't know what player one guessed. The possible things that could have occured as for as she knows are:

$\cancel{\text{Car in door 1: Player picks 1: coin is Heads: Monty shows door 2}}$.

$\cancel{\text{Car in door 1: Player picks 1: coin is Tails: Monty shows door 3}}$.

$\cancel{\text{Car in door 1: Player picks 2: coin is Heads: Monty shows door 3.}}$

$\cancel{\text{Car in door 1: Player picks 2: coin is Tails: Monty shows door 3.}}$

$\cancel{\text{Car in door 1: Player picks 3: coin is Heads: Monty shows door 2.}}$

$\cancel{\text{Car in door 1: Player picks 3: coin is Tails: Monty shows door 2.}}$

$\cancel{\text{Car in door 2: Player picks 1: coin is Heads: Monty shows door 3.}}$

$\cancel{\text{Car in door 2: Player picks 1: coin is Tails: Monty shows door 3.}}$

Car in door 2: Player picks 2: coin is Heads: Monty shows door 1.

$\cancel{\text{Car in door 2: Player picks 2: coin is Tails: Monty shows door 3.}}$

Car in door 2: Player picks 3: coin is Heads: Monty shows door 1. (This didn't happen but player 2 doesn't know that.)

Car in door 2: Player picks 3: coin is Tails: Monty shows door 1. (This didn't happen but player 2 doesn't know that.)

$\cancel{\text{Car in door 3: Player picks 1: coin is Heads: Monty shows door 2.}}$

$\cancel{\text{Car in door 3: Player picks 1: coin is Tails: Monty shows door 2.}}$

Car in door 3: Player picks 2: coin is Heads: Monty shows door 1.

Car in door 3: Player picks 2: coin is Tails: Monty shows door 1.

Car in door 3: Player picks 3: coin is Heads: Monty shows door 1. (This didn't happen but player 2 doesn't know that.)

$\cancel{\text{Car in door 3: Player picks 3: coin is Tails: Monty shows door 2.}}$

As far as she knows there are six possible options as so for as she knows maybe player 1 could have picked door three.

Of these six options: $3$ of them have the car in door $2$. (And in two of those player $1$ guessed wrong). And $3$ of them have the car in door $3$. (And in two of those player $1$ guessed wrong).

Player $2$ doesn't know which to pick but she does know that whatever player one picked he should switch.

Answer 7

An easy way to obtain probabilities is to take advantage of "law of large numbers," according to which the more times you repeat an experiment, the proportion of successes will tend to approach the expected value. For example, if you flip a coin, you expect to get each result $1/2$ times. Maybe you flip it $4$ times and get a proportion very different to $1/2$, but if you flip it $1000$ times, each result can move away from $500$ and the proportion will still be close to $1/2$. It would be very difficult to get $100$ times "heads" and $900$ times "tails", for example.

So, imagine you play the Monty Hall game $900$ times. The car should appear in each door in about $300$ games, and suppose you always pick $Door1$.

$1)$ In $300$ games the car appears in $Door1$ (yours). In this case, the host is free to reveal each of the other doors, because both have goats. We don't know if he has preferences for one over the other, so we are not sure about which door he will reveal once you have selected the correct door. We can only suppose he reveals each with $1/2$ probability.

__ $1.1)$ In $150$ of them he reveals $Door2$.

__ $1.2)$ In $150$ of them he reveals $Door3$.

$2)$ In $300$ games the car appears in $Door2$. The host can only reveal $Door3$, because he can never discard yours and neither which has the car.

$3)$ In $300$ games the car appears in $Door3$. The host can only reveal $Door2$.

Let's suppose, for example, that $Door2$ is revealed. You could only be in case $1.1)$ or in case $3)$, which is a subset of $450$ games. You win by staying in $150$ of them $($case $1.1)$ and by switching in $300$ of them $($case $3)$. So, you win by switching twice the times as by staying.

As you see, when the door was revealed not only the case in which it has the car was discarded; also half of the case in which the car is in your door was discarded, because in the other half the host would have revealed the other door.

So, the confusion is because when people see that the probabilities of have gotten the correct door remain in $1/3$, they think that we are not modifying the sample space, like we were counting the original cases. In fact, the sample space is modified, only that both the cases in which you could have failed and the cases in which you could have been successful were reduced by half, and to reduce both by half results in their respective proportions being maintained.

Seeing it another way: you picked $Door1$. The probabilities of the car being in your door $($number $1$ in this case$)$ and later the host revealing $Door2$ are $1/3 * 1/2 = 1/6$. The probabilities of the car being in $Door3$ and later the host revealing $Door2$ are $1/3 * 1 = 1/3$. You can make the same reasoning with all the other cases.