# If the door in Monty Hall is opened by a person without knowledge of where the prize is, and also does not reveal prize, do the odds change?

Here is the quiz variant.

After the candidate chooses one door for the first time, and before the master (who knows which door hides the prize) intentionally opens another one to reveal no prize, an uninformed person appears and opens a door, which happens not to be the one chosen by the candidate, and, which, also, happens to reveal no prize.

Does this affect the quiz strategy (in terms of probabilities) for the candidate? What are the probabilities for the candidate's remaining second choices now?

This is the "ignorant Monty" variant of the problem and the probability is

50/50 whether you switch or keep your original choice.

Proof:

Without loss of generality, assume the candidate chooses door #1 and then the uninformed person appears and opens door #2, which happens to reveal no prize.

The probability that door #2 would've revealed no prize, given that the prize was behind door #2, is of course $$0$$. The probability that door #2 would've revealed no prize, given that the prize was behind door #1, is $$1$$, and similarly the probability that door #2 would've revealed no prize given that the prize was behind door #3 is also $$1$$. Thus the probability that the prize is behind door #1, given that door #2 was accidentally opened and revealed no prize, is $$1/2$$.

• Thanks for reaction ... I personally believe it is exactly so as well, but, I have not tried to formally prove this yet ... perhaps you have a reference. Have a nice day! Commented Jun 28, 2022 at 23:52
• Wikipedia is a good starting point. This is a well-known variant of the classic problem. Commented Jun 29, 2022 at 0:05
• By the way, this change in probability also happens when Monthy is unaware which door the candidate opened, he just chooses a non-winning door and happens to not open the one chosen by the candidate. Commented Jul 1, 2022 at 16:09
• @justhalf If the person knows what's behind the doors we lose generality by fixing their door choice to door #2 since depending on where the prize it there is no longer an equal chance of opening doors 2 and 3. Commented Jun 28, 2023 at 14:31
• Whoa, nice. Thanks @GoblinGuide. Commented Jun 28, 2023 at 16:33

I mean the obvious answer with regards to whether you should revise your strategy, that apparently no one dared to produce is:

No

No you don't need to revise your strategy because. Either you stumbled into a masked version of the original Monty Hall Problem in which case it would yield an advantage to switch or you'd have the same chance for both doors, in which case you'd also switch as that doesn't do any harm given a 50:50 chance. So no reason to change the strategy.

Now with regards to the probability:

I'd argue that for this particular case the odds are even.

which is an answer already given, but the explanation for that is a little sketchy at best.

So allow me to try something different. Essentially you have 2 outcomes for your initial guess, car or goat. Car has a probability of 1/3 and goat a probability of 2/3.

Now if you correctly guessed the location of the car, the accidental Monty is left with 2 option to reveal something, either he opens your door which also happens to be where the car is, which he has a 1/3 chance of doing or he has a 2 in 3 chance of picking one of the two goats. So the chance of you having picked the car and a goat being revealed by a falling Monty is 1/3 for the car times 2/3 for the goat or 2/9.

The other option is that you initially picked a goat. Which has a higher probability of 2/3 compared to picking the car at 1/3. Now it's again Monty's turn to fall. Though now he has 3 different options. Either he reveals the car, or he reveals the door that the player has chosen (which is different as the player has chosen a goat in the initial round) or Monty happens to reveal the 2nd goat in the game. So now the chance to reveal the 2nd goat is just 1/3. So the combined chances of picking a goat initially and revealing one by a falling Monty is again 2/9 just that now the initial probability is 2/3 and the falling probability is 1/3 (so the opposite as before).

So while the original Monty had you with a chance of 2/3 of initially picking a goat and only made a cosmetic change to that in the reveal, thus making it preferential to swap, in this case the odds of the goat reveal happening are equal for both an initial car pick and an initial goat pick. So there's nothing stopping you adhering to the standard strategy, but you also don't have to stick with it.

Btw the other 5 out of 9 options are 3 times directly revealing the car and 2 times revealing the the contestants goat (obviously in 1 of the 3 car reveals, the car is also the contestants chosen door).

So all in all:

It's probably best to stick with the strategy of switching as for a random reveal that accidentally ends up creating a Monty Hall Problem setup, it is either a 50:50 chance in which case it wouldn't matter anyway or OP/Monty is so hard wired on creating that particular effect that it actually means you're playing the original MHP all over again in which case switching is even beneficial over staying.

Edit: Also as OP seems to have been fixated on the increasing N > 3 idea:

Let's go through the same idea. Suppose at first the contestant picks the car (1 in 4 chance). Now Monty's reveal leaves him with a 3/4 chance not to hit he car in the first round and a 2/3 chance not do to it in the second reveal. So a combined 1/4*3/4*2/3 = 6/48 or 1/8 ~ 12.5% chance of achieving this feat.

Now the reverse case. The contestant picks an initial goat (3 in 4 chance). Now Monty is left with a 2/4 option to dodge the two bullets of the car and the contestants choice and a 1/3 option to do that again. So now it's 3/4*2/4*1/3 which again is the same 6/48 just differently organized.

Now does that increase the understanding of the problem? I don't think so. What it undoubtedly did though is to make this event less likely to occur. So N=3 had a combined probability of Monty avoiding the pitfalls and revealing a goat was 4/9 while for N=4 that already dropped to 12/48 or 1/4 So from nearly 1/2 to 1/4. So if that trend continues the higher you grow your N the more unlikely this instance will become and the more likely it will appear to the contestant that they're fooling him on the claim of purely random reveals and that they are just running the original MHP because that delays the reveal and amps the tensions...

Unlike with the original problem where increasing the N really drives home the point because your chances of picking the car initially are 1/N while your chance of picking a goat are N-1/N. So the more N grows the more certainly your initial pick is a goat while the car is in the other pile. Now if you know that Monty only eliminates fake options you know, that doesn't change anything as you know there are N-2 fake options in the pile of size N-1 so eliminating all of them does nothing to change the probabilities which are still 1/N for your initial guess and N-1/N for the other, so a swap becomes more and more beneficial with growing N.

• The original MHP strategy can be kept indeed but the odds did of course change for MFP. As you, and @SQLnoob, indicate and is easy to prove: for any $N$ MFP odds are $50:50$ so one might stick to original MHP strategy (where odds relate as $1/N:(N-1)/N$). OP question was about odds, not about strategy. I agree OP question may be considered ambiguous. I should only have asked for probabilities (and not use the word odds which is a poor mathematical term) and I should not have mentioned the concept strategy. Thanks. Commented Apr 22 at 20:27
• I will perhaps try to change title to mention probabilities in stead of odds. And again, yes switching to other door is in both cases a best strategy. There are only two strategies (stick to first choice or switch to other remaining door). In MHP switch is the only best strategy, in MFP both strategies are equally good. Thanks again for your answer which I very much appreciate. Commented Apr 22 at 20:55

The problem with this scenario is that it totally ignores the case where the uninformed person opens the door and reveals the big prize, something that will happen ⅓ of the time.

Then, having magically discarded a third of the possibilities, we're left with the two remaining doors, each of which has equal probability of having the big prize.

The real question is, how does the game work when the reveal is the big prize. Does the player simply lose immediately?

• When the reveal is the big prize (say car) I guess some new rule is needed to decide if player looses or not. But that was not the case. Reveal was the small prize (say goat). So the quiz can proceed without new rules. Commented Jun 29, 2022 at 21:17
• @FirstNameLastName, in which case, as I've answered: “we're left with the two remaining doors, each of which has equal probability of having the big prize”. Commented Jun 29, 2022 at 21:23
• @FirstNameLastName, yes, having a high reputation can sometimes be a problem. On another site, I used to suggest changes, and often they were accepted, sometimes not. But I can no longer do that; now, if I submit a change it immediately happens. So I no longer suggest changes unless I'm sure they are appropriate (e.g. trivial spelling mistakes). That's the unfortunate result. ¶ When I've mentioned this, the response is that the OP can always revert the change, so don't worry about it. I'm not comfortable with that. But be aware that you should revert the change if you don't agree with it. Commented Jun 30, 2022 at 0:56
• @FirstNameLastName, … many of us have high reputations because we've been around for a long time. We might have more experience and knowledge of how things work, but it doesn't necessarily mean that our views are any more valid than yours. Commented Jun 30, 2022 at 0:59
• "The problem with this scenario is that it totally ignores the case where the uninformed person opens the door and reveals the big prize, something that will happen ⅓ of the time." - That's a true statement, but I don't see how it's a problem. By analogy, if someone poses the puzzle "you roll two dice, their sum is 7, and one of them is 3; what's the other one?" then we don't object that the puzzle totally ignores the case where neither die is 3, or where the sum isn't 7. Commented Jun 26, 2023 at 5:07

The current structure of the question and accepted answer is misleading. This is because it conflates two distinct items: 1) the knowledge of the person who reveals a door (whether host or interloper), and 2) the rules of the game from the outset.

In the situation described in the question, the answer is that the player's strategy should not change at all, and the probability of winning the car by switching the door remains 2/3. Why is this?

This is because in the situation described, there is no distinction from that of the classic question. The conditions are...

• There are two goats (losses) and one car (win)
• The player chose effectively at random between the three
• A single unchosen door has been opened to reveal a goat

This yields the conclusion that there is a 1/3 chance that the player's original choice was the car. The chance that the remaining door contains the car is thus 2/3. It is easier to illustrate this by expressing it thusly; "all the doors are opened until there remain only two doors; one door which the player chose, and one door which is the opposite of the door that the player chose." So long as the original choice was a 1/3 chance of a win, the second choice will have a 2/3 chance of a win by swapping.

The fact that the person revealing the door was unaware of whether it contained a goat or a car is immaterial. What is important is the truth that the door which was revealed did not contain a car, with probability 1. The very fact that it has happened, as has been stipulated by the question, means that it has probability 1. "But", some may object, "the person opening the door didn't know! They could have revealed the car!" Yet the question does not countenace this possibility. It tells us, explicitly, that the revealer reveals a goat. Only adding, as if it were trivia, that "they did not know it to be a goat before openining it".

As is such, the contestant will find that they are simply in a single instance of the very same monty-hall problem that has been posed before: there is 1 prize, they must choose from 3 doors, then 1 of the remaining doors is opened to reveal a loss. Is it a good idea for them to switch doors?

In the single case, the answer is "yes" or "no" based upon whether or not they have already chosen the prize. We can only speak about probabilities in the abstract case. And the described situation is an instance of the standard question's abstract case. This is because it precludes the possibility of the revealer accidentally revealing a car. The revealer's knowledge is therefore completely apropos nothing.

If generalised, then the way that this question is posed would thus be as below...

A contestant has a choice of 3 doors. Behind 1 door is a car (a win), and behind the other 2 are goats (losses). After the contestant chooses a door, but before they see what is behind it, a person will open one of the remaining two doors to reveal a goat (although the revealer didn't know it was a goat). After a door which does not contain a car is revealed, what is the probability that the contestant can win the car by trading the door that they chose for the remaining closed door?

The only change being that in bold. It seems nonsensical, because in the real world it is. Somehow, by divine fate or obscure physical laws or even a superpower, this completely uninformed person simply cannot open a door that hides a car. That's stipulated.

So why the answers of 50%? Because they are answering a (slightly) different question. The question which they are answering should be posed as follows:

A contestant has a choice of 3 doors. Behind 1 door is a car (a win), and behind the other 2 are goats (losses). After the contestant chooses a door, but before they see what is behind it, a person will open one of the remaining two doors at random and reveal what is behind it. In the case that this door reveals the car, the game will be restarted. After a door which does not contain a car is revealed, what is the probability that the contestant can win the car by trading the door that they chose for the remaining closed door?

The answer to this question is 50%. The bolded segment is the critical section missing from the question, and discussions of the answer. At least, it must be, because this is the only way to conclude that the player has only a 50% chance of winning the car.

---Edit---

To expand upon why 50% is the answer to the second question, please see below:

Fundamentally, the dominant strategy for the monty hall problem allows the contestant to win when they do not choose the car on their first pick. This is because by choosing the opposite of their first choice, they invert their original 1/3 chance of choosing the car to be a 2/3 of trading for the car.

But the new rules mean that, in the case that the contestant successfully fails (lol) to choose the car on their first selection, there is now a 50% chance that the uninformed revealer will accidentally reveal the car, and force the game to restart. This steals away half of the 2/3 chance of winning that a prepared contestant would begin with. But since the stolen half of the 2/3 is not turned into a loss, but simply removed from circulation, as if it never happened, it means that a single attempt at the game now has the following outcomes...

• I. 1/3 chance of the player choosing a goat, the revealer revealing a goat, and the player swapping for a car. (WIN)
• II. 1/3 chance of the player choosing the car, the revealer revealing a goat, and the player swapping for a goat. (LOSS)
• III. 1/3 chance of the player choosing the goat, the revealer revealing the car, and the game restarting (GAME CONTINUES)

So the only game-ending results are the I. and II.. Their probabilities are equal, so between the two of them, it's a 50-50 shot. The player can feel free to swap, or not. Doesn't matter anymore. (Of course, it might take a few tries to get to the game end: there's a 1 in 3 chance that the game will just keep going every time).

----EDIT #2----

The critical comments are making the n = 1 error. Statistics can say nothing practically useful about the individual case. Proving that your strategy provides a 2/3 chance of winning a car will not win you 2/3 of a car, if you happen to choose a goat. And it will not take away 1/3 of your car, in the cases where you win. In the single case you will either win or lose. 1, or 0, regardless of how statistically sound your logic is. If you try to force statistics into a corner by saying "here is a single case of some larger unspecified monty hall process, which door should the person choose?" then the unsatisfying answer that statistics will give is "they should choose the door with the car behind it. Only I certainly don't know which door that is."

What statistics does tell us, is about the expected outcomes over arbitrarily large repetitions of a process.
If you choose the dominant strategy on the Mony Hall problem one time, you will either have a car or you will have a goat. Full stop.
But if you play the Monty Hall problem over and over, and repeat the dominant strategy, then over a sufficiently large number of trials, you will find that you have won a car 2 out of every 3 times you play, more or less.

So far so good. But repeating a process requires knowing the specifics of the entire process. And this is the ambiguity in the question, that makes both it and the accepted answer misleading.

As it is written, the question limits the selection to only cases in which a goat is revealed. There is no possibility that the uninformed revealer will reveal a car. How is this so? We do not know. But we know that the question selects only for legal reveals. When we structure the probability tree for this result, it is clear: there is no change from the standard Monty Hall case. All the probabilities are exactly the same, and the only question that remains is "how in the world does this uninformed revealer manage to never reveal the car?"

Here are some aids: some R code, and the resulting graphs, and some excel figures, that illustrate this.

library(tidyverse)
library(scales)

# Choose our number of trials. Let's do 100,000!
trial_count <- 100000

# Prepare a dataframe for the results
results <- tibble(trial_number = 1:trial_count, final_unrevealed_door = NA_character_, win_or_loss = NA_character_)

# There are the following 3 choices, overall
choices <- c("car", "goat1", "goat2")

# For each trial...
for (i in 1:trial_count){
# Contestant chooses one of the doors at random
contestant_first_choice <- sample(choices, 1)

# The remaining choices are those != contestant's choice
remaining_choices <- choices[choices != contestant_first_choice]

# the revealer will choose one of the remaining doors, at random
revealer_choice <- sample(remaining_choices, 1)

# But the revealer is not allowed to choose the car! We are selecting only for the situations in which the uninformed revealer *does not reveal the car*. You tell me how this is possible.
while(revealer_choice == "car"){
revealer_choice <- sample(remaining_choices, 1)
}

# the absolute final prize, that the contestant can choose for, is self evidently the one that the revealer did not choose.
final_unrevealed_door <- remaining_choices[remaining_choices != revealer_choice]

# If the final unrevealed door is a car, then the player will win by swapping
win_or_loss <- case_when(
final_unrevealed_door == "car" ~ "Win",
T ~ "Loss" # Otherwise, the player will lose.
)

# We will record which prize is behind the final unrevealed door
results[i,] <- list(trial_number = i, final_unrevealed_door = final_unrevealed_door, win_or_loss = win_or_loss)

if(i %% 1000 == 0){print(paste0("Finished trial ", i, "."))}
}

# Finally, let's plot our results
results %>%
ggplot(aes(x = win_or_loss, fill = final_unrevealed_door)) +
geom_bar() +
labs(
title = "What if the revealer is uninformed, but reveals a goat anyways?",
subtitle = paste0(comma(trial_count), " trials. Horizontal dashed lines are at 1/3 and 2/3 of the total number of trials."),
fill = "Prize behind final door",
y = "Number of trials",
x = "Result if contestant swaps doors after the reveal"
) +
scale_y_continuous(labels = comma) +
geom_hline(yintercept = trial_count / 3, linetype = "twodash") +
geom_hline(yintercept = trial_count * 2 / 3, linetype = "twodash")


library(tidyverse)
library(scales)

# Choose our number of trials. Let's do 100,000!
trial_count <- 100000

# Prepare a dataframe for the results
results <- tibble(trial_number = 1:trial_count, number_of_restarts = NA_integer_, final_unrevealed_door = NA_character_, win_or_loss = NA_character_)

# There are the following 3 choices, overall
choices <- c("car", "goat1", "goat2")

# Initialize counter variables
i <- 1
number_of_restarts <- 0

# For each trial...
while (i <= trial_count){
# Contestant chooses one of the doors at random
contestant_first_choice <- sample(choices, 1)

# The remaining choices are those != contestant's choice
remaining_choices <- choices[choices != contestant_first_choice]

# the revealer will choose one of the remaining doors, at random
revealer_choice <- sample(remaining_choices, 1)

# If the revealer reveals a car, this starts the game over!
if(revealer_choice == "car") {
number_of_restarts <- number_of_restarts + 1
next
}

# the absolute final prize, that the contestant can choose for, is self evidently the one that the revealer did not choose.
final_unrevealed_door <- remaining_choices[remaining_choices != revealer_choice]

# If the final unrevealed door is a car, then the player will win by swapping
win_or_loss <- case_when(
final_unrevealed_door == "car" ~ "Win",
T ~ "Loss" # Otherwise, the player will lose.
)

# We will record which prize is behind the final unrevealed door
results[i,] <- list(trial_number = i, number_of_restarts = number_of_restarts, final_unrevealed_door = final_unrevealed_door, win_or_loss = win_or_loss)

# Reinitialize the restart counter
number_of_restarts <- 0
# increment i
i <- i + 1

if(i %% 1000 == 0){print(paste0("Finished trial ", i, "."))}
}

library(colorspace)
# Finally, let's plot our results
results %>%
group_by(win_or_loss, number_of_restarts) %>%
summarise(count = n()) %>%
ggplot(aes(x = win_or_loss, y = count, fill = number_of_restarts)) +
geom_col() +
labs(
title = "What if the revealer is uninformed, and revealing the car \nstarts the game over?",
subtitle = paste0(comma(trial_count), " trials. Horizontal dashed line at 1/2 of the total number of trials."),
fill = "Number of times game \nrestarted by a revealed car",
y = "Number of trials",
x = "Result if contestant swaps doors after the reveal"
) +
scale_y_continuous(labels = comma) +
geom_hline(yintercept = trial_count / 2, linetype = "twodash") +
scale_fill_continuous_sequential(palette = "Dark Mint")


Finally, the probability tree for the situation described in the question: the uninformed revealer reveals (and only will, and only can) a goat.

• @FirstNameLastName I've just drafted this for you. It should make the probabilities easier to visualize and understand. i.imgur.com/yVza05E.png Commented Jun 28, 2023 at 13:56
• I don't see a difference between the situation in the first quote block and the ultimate game in the situation in the second quote block. You might object and say that in the second situation the person could still have opened a prize door but we know that this situation eventually terminates with them opening a no-prize which makes it the same as the first situation. Commented Jun 28, 2023 at 14:41
• "how in the world does this uninformed revealer manage to never reveal the car?" He doesn't. But we're only interested in the 4/6 cases where he does manage. Commented Jun 29, 2023 at 2:01
• @user99478 the difference between playing a single MHP and a single MFP is that MHP host knows where car is and can, and on purpose will, always open a door revealing a goat. i.e. with probability 1. But MFP intruder is uninformed and opens opens a door revealing a goat with different probability. So it is a matter of different conditional probabilities, even that singe time. Information can affect probabilities. Commented Jun 29, 2023 at 12:59
• I think there is a mistake in the simulation, at the line labeled "But the revealer is not allowed to choose the car:" if they choose the car you have to throw out the whole sample instead of picking again. Commented Jun 30, 2023 at 0:13