# Am I wrong about the Monty Hall Problem?

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..

• What do you find unconvincing about the "no new information" argument as you've quoted it? – Y. Forman Feb 6 '18 at 21:10
• I suggest being more explicit about what you mean by "randomly picks a goat", since the exact strategy used by Monty affects the answer. – stewbasic Feb 6 '18 at 21:11
• Simpler than the 'no new information' argument? That's hard to do .... – Bram28 Feb 6 '18 at 21:11
• Suppose you have chosen a door. Then you walk to some other door and in complete ignorance you open it. Behind that door you find a goat. Now do you think that your chance on winning a car will change if you choose for another door on base of your discovery? The answer is: no. The discovery does change something, though. Your original chance on winning was $\frac13$. But after the discovery your chance is $\frac12$. – drhab Feb 6 '18 at 21:29
• You begin your question with "First, I don't think it is 50/50 always" . No, of course not. In the original problem, with the original Monty Hall TV program, Monty does know behind which doors there is a goat and behind which there is a nice present. That's all the difference of the world and thus you will always have $\;2/3\;$ to win if you change your first choice. Now, assuming Monty does not know behind which doors there's a good and he just opens by chance any door, makes the odd 50/50 (it's just the smae to remain with your original door than switching)... – DonAntonio Feb 6 '18 at 21:47

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.

• btw if you prefer tables to probability notation, you can interpret an event $E$ as a set of table cells, $P(E)$ as the sum of probabilities in those cells, and $P(E|F)=P(E\cap F)/P(F)$ as the fraction within $F$ of probability that also lies in $E$. – stewbasic Feb 6 '18 at 22:07
• I have the feeling that this is the answer I'm looking for and yet I am completely unable to understand it due to lack of probability training. (100% my fault 0% yours) – Nina Feb 6 '18 at 22:56
• @Nina I added an example with probability tables, hope it helps. – stewbasic Feb 7 '18 at 0:11
• thank you!! def more clear what you're doing now! only question - in this situation, you never consider what happens if the contestant originally chooses door 3, why doesn't that have to be considered? i understand that with the parameters you have laid out, it will never happen, but is it fair to lay out parameters where this is the case? doesn't that take away from the random variable? thanks!! – Nina Feb 7 '18 at 1:01
• @Nina I picked an example where you never pick door 3, just to keep the tables small. Hopefully it's clear that the conclusion (two nonzero entries are equal in all goat rows) will hold no matter what method you and Monty use to make the choices (as long as they are independent from the actual car location). – stewbasic Feb 7 '18 at 1:13

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 $1/3$.

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.

• I'm sorry I'm not quite getting the card analogy. It seems much more complicated than Monty Hall – Nina Feb 6 '18 at 22:54
• I'm confused about the situation that corresponds to the classic Monty Hall Problem. Your second paragraph seems to contradict the first. The second paragraph seems right to me, but I don't understand why the first is necessary to reach that conclusion. In the first you claim that a deliberate club revelation means it's more likely that you picked a club. If your friend will always reveal a club, you've learned nothing about the card in your hand. You still have a 1/4 chance of having chosen a club, as you explain in paragraph #2. – Nina Feb 8 '18 at 2:20
• In the random drop scenario, the random revelation of a card does tell you something: you're more likely to have originally picked a non-club. Additionally, this scenario is not directly analogous to the random choice variation of Monty Hall. It's much more complicated. – Nina Feb 8 '18 at 2:30
• I assume you're referring to the paragraphs beginning "In my cards analogy..." and "If we want to know..." You're right, the first paragraph is not necessary for the second. They are independent arguments serving different purposes. The first is an explanation of why the earlier argument leading to a probability of 13/51 for the random-drop scenario no longer works in the deliberate-club-drop scenario, since the 51 remaining cards are no-longer "equivalent" in a sense. The second finds the probability in the deliberate-club-drop scenario. – alcana Feb 8 '18 at 18:47
• I agree that in the random-drop scenario, the revelation of a card tells you something. I disagree that my scenario is not analogous to Monty Hall: if you imagine my scenario but where a deck of cards consists of 2 goats and 1 car, instead of 13 clubs, 13 diamonds, 13 hearts and 13 spades, then I think it replicates Monty Hall. It is of course more complicated in terms of having bigger numbers, but somehow that helps me to think about it more clearly, by getting away from small numbers like 1 and 2... perhaps that's a peculiarity of me! – alcana Feb 8 '18 at 18:52

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!

• Full disclosure: I'm not great at reading probability notation. (I always use tables.) I think, however, that this is the explanation I have been using in the past. I'm wondering if there is a way to explain that's easier than the original problem, because the reasoning I always see is that "two outcomes, hence 50/50" works when the door is opened randomly. It is 50/50, but its not because there are only two outcomes..I think? Thank you so much for this answer!! – Nina Feb 6 '18 at 21:57
• @Nina I agree ... "two outcomes hence 50/50" is of course bad reasoning! – Bram28 Feb 6 '18 at 21:58

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.

• TL;DR: The "no new information" argument works if it means Monty's choice of door gave no new information, in contrast to classic Monty Hall. The opening of the door gives information but doesn't affect the equal likelihood. – Y. Forman Feb 6 '18 at 22:43
• The opening of door #3 to reveal a goat does depend on the location of the car. It's more likely to happen if you've originally chosen the car. Sorry if I'm missing something – Nina Feb 6 '18 at 22:48
• @Nina It is true that door #3 revealing a goat is more likely to happen if the car is not behind door #3 (100% chance) vs. if the car is behind door #3 (0% chance). But the fact that there is a goat behind door #3 does not give any new information as to whether the car is behind door #1 vs. door #2, so they remain equally likely. – Y. Forman Feb 6 '18 at 22:54
• @Nina The important point here is to focus on the relative probability of door #1 vs. door #2. We get new information about door #3, and therefore about the relative probabilities of #1 vs. #3 and #2 vs. #3 (and #1 vs. #2-3 combined, etc.), but we get no new information about the relative probability #1 vs. #2. – Y. Forman Feb 6 '18 at 22:55
• The problem relies on a guess prior to the revelation of the goat. If you don't guess, then door #3 gives you no info. But it's not the Monty Hall problem if you don't guess. (also - i can add "the revelation of the goat" to the things i never thought I'd say list lol) – Nina Feb 6 '18 at 23:01

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.

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 $$1/2$$.