Am I wrong about the Monty Hall Problem?

Question

First off, I don't think its 50/50 always. That's not what this is about.

I know that simulations show that if Monty randomly picks a goat, it IS 50/50. I always thought I understood why this was, but recently I've been looking at discussion boards and the explanation doesn't seem to match mine.

I always thought that if Monty chose a goat randomly, the probability was actually more complicated. It ends up being 50/50, but there are more possible scenarios.

However, I've been seeing lots of explanations that go something like this: "If Monty purposefully reveals a goat that means that he's given you more information, but if he's chosen randomly he tells you nothing." It has me wondering if maybe I've overthought it. Is there actually a more simple explanation for why a randomly revealed goat gives you 50/50?

**By "randomly chooses a goat" I mean that either Monty has no idea what is behind each door or that if he does know he ignores this knowledge in favor of a random choice (ie. coin flip determines which door he opens). Monty only has a choice between the two doors that you have not chosen. Sorry I should have been clearer.

**** Okay let me put it this way. Everyone says that if you choose at random you get no new info. What I don't understand is why this is true. Why does the random reveal not also give you more info? It seems to me that it does..

Answer 1

Let $P_i$ be: you pick door $i$

Let $C_i$ be: door $i$ has the car

Let $E_i$ be: Monty shows a goat behind door $i$

Let’s go with: you pick door $1$ and Monty shows a goat behind door $2$. Now what?

Well, let's assume Monty always makes a random choice when opening one of the two remaining doors (maybe because Monty doesn't know where the prize is, or because he always just flips a coin to determine which one to open, or ....). That is, after you picking door $1$, he could just as well have opened up door $3$, and indeed in this scenario it is possible that Monty sometimes does end up opening a door that has the car.

Thus, we have:

$$P(E2|C1,P1) = P(E2|C3,P1) = \frac{1}{2}$$

Notice that of course we do have that

$$P(E2|C2,P1) = 0$$

since the event of Monty opening up door 2 as having a goat is impossible if it has a car!

OK, so this gives us:

$$P(E2|P1) = P(E2|P1,C1)\cdot P(C1) + P(E2|P1,C2)\cdot P(C2) + P(E2|P1,C3)\cdot P(C3) =$$

$$ \frac{1}{2}\cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + \frac{1}{2}\cdot \frac{1}{3} = \frac{1}{3}$$

And therefore:

$$P(C3 | E2 , P1) = \frac{P(E2|C3 , P1) \cdot P(C3|P1)}{P(E2|P1)}=$$

$$\frac{\frac{1}{2} * \frac{1}{3}}{ \frac{1}{3}} = \frac{1}{2}$$

Now, compare this with the situation that Monty does not always pick randomly one of the remaining doors, but will always choose the one that has the goat if one of the remaining two has the car (of course, if both remaining doors have a goat, we'll assume that Monty does pick randomly one of those two).

OK, so now we still have:

$$P(E2|C1,P1) = \frac{1}{2}$$

and of course we also still have:

$$P(E2|C2,P1) = 0$$

but we get a different value for:

$$P(E2|C3,P1)=1$$

For now Monty is sure to open door $2$ when you pick $1$ and the car is behind $3$.

So this changes the value of:

$$P(E2|P1) = P(E2|P1,C1)\cdot P(C1) + P(E2|P1,C2)\cdot P(C2) + P(E2|P1,C3)\cdot P(C3) =$$

$$ \frac{1}{2}\cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} = \frac{1}{2}$$

And therefore of:

$$P(C3 | E2 , P1) = \frac{P(E2|C3 , P1) \cdot P(C3|P1)}{P(E2|P1)}=$$

$$\frac{\frac{1}{2} * \frac{1}{3}}{ \frac{1}{2}} = \frac{1}{3}$$

So, there you go: Monty's intent makes all the difference: if the opening of the door with the goat was the result of him always just randomly picking one of the remaining doors, then switching does not matter, but if he will make sure never to show the car, then switching does matter!

Answer 2

I don't know what is meant by the "no new information" argument. However, here is an argument based on symmetry.

Let $A_i$ denote the event that the car is behind door $i$, and let $O$ denote the collection of events we've observed, including Monty opening door 3 and revealing a goat. If neither you nor Monty know where the car is, the probability of your collective actions can't depend on where the car is. Also in both cases $A_1$ and $A_2$, door 3 will reveal a goat. Thus $$ P(O|A_1)=P(O|A_2). $$ Suppose that the car placement is uniformly random, so $P(A_1)=P(A_2)$. Now applying Bayes' theorem, $$ P(A_1|O)=P(A_2|O). $$ More generally, this argument shows that no matter how elaborate the game is, the closed doors are all equally likely to have the car provided:

• The car placement is uniformly random,
• No participant in the game knows its location,
• Only goats have been revealed.

To illustrate with an example, suppose we arrange the probabilities in a table with a column for each car position, and a row for each possible sequence of observations. After we make some observations, to determine the probability of the car being behind each door, we should find the appropriate row in the table and compute the relative probabilities in that row (that is, divide each probability in the row by the sum of the row).

To make it clear that it doesn't matter how you and Monty play, suppose we have a biased coin with head probability $3/5$. First you flip and pick door 1 on heads and door 2 on tails. Then Monty flips and selects the leftmost free door on heads. The initial probabilities are $$\begin{array}{c|ccc} \text{Observations}&\text{Car behind }1&\text{Car behind }2&\text{Car behind }3\\ \hline &1/3&1/3&1/3 \end{array}$$ After the first flip: $$\begin{array}{c|ccc} \text{Observations}&\text{Car behind }1&\text{Car behind }2&\text{Car behind }3\\ \hline H\,(\text{You pick }1)&3/15&3/15&3/15\\ T\,(\text{You pick }2)&2/15&2/15&2/15 \end{array}$$ After the second flip: $$\begin{array}{c|ccc} \text{Observations}&1&2&3\\ \hline HH\,(\text{You pick }1\text{, Monty picks }2)&9/75&9/75&9/75\\ HT\,(\text{You pick }1\text{, Monty picks }3)&6/75&6/75&6/75\\ TH\,(\text{You pick }2\text{, Monty picks }1)&6/75&6/75&6/75\\ TT\,(\text{You pick }2\text{, Monty picks }3)&4/75&4/75&4/75 \end{array}$$ In each row, all three entries are equal; they must be because the process so far has been independent from where the car actually is. Finally we find out what's behind Monty's door: $$\begin{array}{c|ccc} \text{Observations}&1&2&3\\ \hline HH;\,\text{Pick 1, open 2, reveal goat}&9/75&0&9/75\\ HH;\,\text{Pick 1, open 2, reveal car}&0&9/75&0\\ HT;\,\text{Pick 1, open 3, reveal goat}&6/75&6/75&0\\ HT;\,\text{Pick 1, open 3, reveal car}&0&0&6/75\\ TH;\,\text{Pick 2, open 1, reveal goat}&0&6/75&6/75\\ TH;\,\text{Pick 2, open 1, reveal car}&6/75&0&0\\ TT;\,\text{Pick 2, open 3, reveal goat}&4/75&4/75&0\\ TT;\,\text{Pick 2, open 3, reveal car}&0&0&4/75 \end{array}$$ Now we do get some information; if we find a goat behind Monty's door, the car definitely isn't there. However if a goat is revealed, the probabilities behind the other two doors are still equal; they are the same as in the previous table. Thus the car is equally likely to be behind either closed door.

Answer 3

Imagine you have picked door $1$. If Monty picks the door at random, there are the following possibilities:

$$\begin{array}{c|c|c|c}\text{Car behind}&\text{Monty picks}&\text{You win}&\text{Probability}\\\hline 1&2&Yes&1/6\\1&3&Yes&1/6\\2&2&No&1/6\\2&3&No&1/6\\3&2&No&1/6\\3&3&No&1/6\end{array}$$

So, in the rows $1,2,4,5$ where Monty has revealed a goat (probability $4/6$), rows $1,2$ (probability: $1/3$) give you the victory if you don't switch (conditional probability: $\frac{1/3}{2/3}=\frac{1}{2}$).

If Monty picks the door with a goat on purpose, then there are the following possibilities:

$$\begin{array}{c|c|c|c}\text{Car behind}&\text{Monty picks}&\text{You win}&\text{Probability}\\\hline 1&2&Yes&1/6\\1&3&Yes&1/6\\2&3&No&1/3\\3&2&No&1/3\end{array}$$

and the probability that you win with no switching (rows $1,2$) is $1/3$.

Answer 4

The other answers have already presented versions of the "more complicated" explanation. It is, in my opinion, the better way to approach the problem: it does a better job avoiding the possibility of intuition leading you astray. Nevertheless, I'm posting a defense of the "no new information" argument because I think it is sound if presented carefully.

As you noted, you can't just say there is no new information -- obviously you get new information when the door opens. Beforehand there was a possibility that the car was behind the opened door, now there is not.

To make the argument sound, we have to present some version of the following. Suppose the door opened was door #3. Originally, the car is equally likely to be behind door #1 and behind door #2. Now two things happen: door #3 is chosen by Monty, and door #3 is opened. Since the random process by which door #3 is chosen does not depend on the location of the car, the choosing of door #3 does not give any new information about the location of the car. The opening of door #3 (and revelation of a goat) doesn't give any new information about the relative probability of the car being behind door #1 vs. door #2. So overall you get no new information about the relative probability, and they remain equally likely -- 50/50.

This is in contrast with classic Monty Hall, where the opening doesn't give any new information about the relative probability, but the choosing does, since the choosing process is different depending on whether the car is behind the door the contestant originally picked (in which case the choice is random between the two other doors) vs. whether the car was behind the other door (in which case the choice is determined).

Is this "simpler" than the other explanations posted? It involves less drawing out charts, but more careful thought and avoiding pitfalls. So it's up to you to decide.

Answer 5

Your primary question seems to be: is there a simple explanation of why the probability you have chosen the car is $\frac12$ if Monty chooses randomly and reveals a goat?

What counts as a simple explanation is partly a personal thing, but here is an argument - or perhaps we should call it a thought experiment - which for me captures the intuitive reason it is $\frac12$:

Suppose your friend holds out a standard deck of $52$ cards, face down. You choose one at random and take it but don't reveal it. Your friend then accidentally drops one of the remaining 51 cards. It falls to the floor and you see that it is a club. What is the probability that you now hold a spade?

Well, the $51$ cards that have not been revealed are all equally likely to be yours, since there is no reason that any of them should have different likelihoods of being yours than any other. Therefore since $13$ of those are spades the probability that your hold a spade is $\frac{13}{51}$.

The Monty Hall problem where Monty chooses at random and reveals a goat is exactly analogous. Out of the remaining two doors, one is a car and so the probability you have chosen a car is $\frac12$.

The question naturally arises: if Monty deliberately chooses a goat, how does this change things?

In my cards analogy, this corresponds to the situation where your friend deliberately chooses a club and drops it on the floor. In this case, the $51$ unrevealed cards are not all equally likely to be yours; the $12$ unrevealed clubs are slightly more likely to be yours than the others, since their having escaped the deliberate club-dropping constitutes some (weak) evidence that they were "hiding" in your hand!

If we want to know the probability that you now hold a spade in this case, we can argue as follows: before the club was dropped, the probability that you held a spade was $\frac{13}{52}$ = $\frac14$. Now the revealing of the club gives you no new information about probability of your having originally chosen a spade, since we knew all along that a club would be revealed no matter what. Therefore the probability you hold a spade is still $\frac14$.

Analogously in the Monty Hall problem, the probability you originally chose the car is $\frac13$, and if Monty deliberately reveals a goat that gives you no new information about the probability that you originally chose a car (since we knew all along that a goat would be revealed whatever), and that probability remains $\frac13$.

I speculate, since there was come confusion about it above, that the "no new information" argument actually refers to this, i.e. it is when Monty chooses a goat deliberately that we get "no new information", not when he does so at random.

Answer 6

I think this question may be most easily answered by working backwards. And also for convenience let’s say we’re dealing with car vs. empty, instead of car vs. goat.

When you’re faced with a choice between two visually identical doors, you’re inclined to think that because it’s one or the other, it’s a 50/50 proposition.

The comedian Gallagher once joked that a weatherman saying there’s a 50% chance of rain was just saying it will or it won’t rain. But of course ‘it will or it won’t’ encompasses a greater range of probabilities. An either-or description may refer to many ratios: 50:50, 60:40, 1:99, etc.

So we start by recognizing that this may not be an 50:50 situation.

If I told you that previously the car was known to have been behind one of three doors instead of two, but one of them was eliminated when a child wandered up and opened it, revealing no car to be present, that information would correctly do nothing to make either remaining door more attractive to you.

If, however, you learn that prior to the child opening the door, the contents of that room had been added to the one to the left of it, you’d realize that the room to the left was doubly likely to conceal the car.

It would have 2/3 of the contents of the three rooms; whereas, the other remaining choice would contain 1/3.

In our problem, the host produces this effect by deliberately showing you that B is empty if A holds the car, and showing you A is empty if B holds the car.

Answer 7

However, I've been seeing lots of explanations that go something like this: "If Monty purposefully reveals a goat that means that he's given you more information, but if he's chosen randomly he tells you nothing." It has me wondering if maybe I've overthought it. Is there actually a more simple explanation for why a randomly revealed goat gives you 50/50?

That is backwards.

• By cheating, Monty denies information.   Monty the Cheater was certain to have revealed a goat, so revealing a goat tells you nothing more than that.   There was $1/3$ probabilty that you picked the car, and that remains so after revealing one of the goats.

• On the other hand, Monty the fair had $1/3$ probability of revealing a car, and so revealing the goat tells you the condition of the world.   You are among the possible worlds where Monty didn't reveal the car: Bayes' Rule revises the (conditional)probability for you picking a car to $(1/3)\div(2/3)$.   That is $1/2$.

Answer 8

I don't believe that if he shows the goat by random then it's 50/50. It shouldn't matter whether he knows or not since you're the one making the decision based on the information you have. He is revealing new information regardless.

There is initially a 1/3 chance that the door you selected is the correct one and therefore a 2/3 chance that one of the other doors is correct. After finding out that one of the other doors is incorrect, there is still a 2/3 chance that one of the other doors is correct, except now there's only one door.