The Monty Hall Problem

Forum Games
1

I want to offer you a game, a mini-game, if you will. It’s actually just a one-player game, that you will play by yourself, but everyone can play it in parallel. Just follow the instructions below…

Welcome to Let’s Make a Deal! I’m your host, Monty Hall!

Today, you’ll be playing a game with three doors, where behind one of them, I’ve hidden a brand new car! And it can be yours, if you just open the correct door! But if you open a wrong door, all that you will find is a goat.

Ok, here’s how it works. First, you will select a door, but you won’t open it just yet. Instead, I will first show you that one of the other doors has a goat behind. Then, you must choose whether you want to keep your original selection or switch to the other remaining door that has not yet been opened.

Alright, got it? Let’s play. Which door do you want to select?

Door 1

You have selected Door 1.

Monty Hall opens Door 2 to reveal a goat.

Do you want to switch your selection to Door 3 or keep Door 1?

Switch to Door 3

You have switched to Door 3 and open it to reveal a goat
:goat:

Keep Door 1

You have kept Door 1 and open it to reveal a Brand New Car!
:red_car:




Ignore these options below…

Door 2



You have selected Door 2.

Monty Hall opens Door 3 to reveal a goat.

Do you want to switch your selection to Door 1 or keep Door 2?

Switch to Door 1

You have switched to Door 1 and open it to reveal a Brand New Car!
:red_car:

Keep Door 2

You have kept Door 2 and open it to reveal a goat
:goat:




Ignore these options below…

Door 3



You have selected Door 3.

Monty Hall opens Door 2 to reveal a goat.

Do you want to switch your selection to Door 1 or keep Door 3?

Switch to Door 1

You have switched to Door 1 and open it to reveal a Brand New Car!
:red_car:

Keep Door 3

You have kept Door 3 and open it to reveal a goat
:goat:







After playing, please answer these polls

How did your game go?
  • Kept original door and found the car :red_car:
  • Switched doors and found the car :red_car:
  • Kept original door and found the goat :goat:
  • Switched doors and found the goat :goat:

0 voters

Which door did you select at first?
  • 1
  • 2
  • 3

0 voters

:goat:

6 Likes
2

If you want to compare, I was introduced to the problem in high school, so 10-11th year of school for reference. In math class, naturally. And at the time the answer really didn’t make sense to any of us in the class. It just didn’t make any sense. Of course, when you write it out it’s evident, but even after that the lingering feeling of suspicion didn’t disappear. On intuition level it didn’t click even after an elementary proof.

Fortunately, now I’m better. Though I don’t know how much knowing the answer contributes to it.

I think part of the magic is the number of doors. You pick one of three so 33/33/33. Then one door is opened and one stays closed so it’s kinda 50/50. And then you switch or not so 50/50. Like that, it seems as if everything is symmetrical, no asymmetry is introduced. Maybe that’s why it feels like there’s no difference between switching and not switching.

On the other hand, let’s take, for example, the version of the game from Zero Time Dilemma. The number of doors is 10, and 8 wrong doors are opened. Immediately you notice asymmetry so intuitively it’s much easier to imagine that one option is better than the other.

3 Likes
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Yep I did advanced math in both highschool and university and strongly opposed the accepted answer for this for a long long time; well beyond university and into adult life.

I had to invent your Zero Time Dilemma for my self (though the numbers I chose were 100 and 98) before my brain could finally accept it for the 3 and 1 version.

2 Likes
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I couldn’t even accept the 9998/10000 version the first time I was introduced to this question.

Say, if there are 1000 doors, only one of which there is a car behind. You have chosen door 1. You get a chance to peek into door 2-999 and find there is no car behind them. It is a pure accident, nobody expected you were able to peek. Then in this scenario switching wouldn’t make any difference.

After thinking about it for a long time I understood that there is a difference between “behind door 2 is a goat” and “they tell you behind door 2 is a goat”.

3 Likes
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On one hand, I’m thinking “Why can’t you apply the same logic here?” i.e. “The probability that your first choice (door 1) is the car is 1/1000 and thus the probability that the other door is the car is 999/1000 and you should switch”?
I understand that, if we calculate the conditional probability, that is, the probability that door 1 is the car given that doors 2 - 999 are goats, then it is 1/2. So what am I missing? How is it possible that these two arguments are contradicting yet both logical? Is the problem that we first need to define which probability space we are assuming, and depending on this both could be true (but not in the same probability space)?

2 Likes
6

Wrong, switching does make a difference: in the case where Monty chooses the doors, switching changes your chance from 1/1000 to 999/1000. In case you peek or where the doors are opened accidentally, switching changes your chance from 1/1000 to 1/2.

As @S_Alexander pointed out below (and I myself pointed out as well, further in this post, ironically), the chance doesn’t change when switching. However, because of the other doors being opened the chance does become 1/2 regardless of switching or not.

The way I like to explain it is with a diagram. Suppose the car is behind door 1.

image
image1802×1030 137 KB

At first you choose at random, with 1/3 probability for each door, assuming you picked randomly. Next, Monty Hall will never open a door with a car, thus the scenario where we pick door 2 or 3, and door 1 is opened, is not reality, but also has 0 chance to happen.
If we picked the door with the car first, which happens in 1/3 of the cases, switching will give us a goat. If we picked the door with the goat first, which happens 2/3 of the time, then in 100% of the cases Monty will show us the other goat, and switching will give us the car.

Pick 1, reveal 2 has 1/6 chance, and switching is bad
Pick 1, reveal 3 has 1/6 chance, and switching is bad
Pick 2, reveal 1 has 0 chance, and switching is bad
Pick 2, reveal 3 has 1/3 chance, and switching is good
Pick 3, reveal 1 has 0 chance, and switching is bad
Pick 3, reveal 2 has 1/3 chance, and switching is good

So in conclusion there is 2/3 chance that switching is good.

If, however, Monty Hall trips and accidentally opens a door that I have not chosen then there is an equal probability between the two doors he can accidentally open:

image
image1738×1026 140 KB

It just happens to be revealing a goat, so we cannot be in the universe where a door with a car is opened. But, importantly, since Monty opens the door accidentally, this non-reality has a 1/3 chance of happening.

Here, when I pick the car, with 1/3 probability, switching will give me a goat. But if I pick a goat first, which happens 2/3 of the time, then in half of those cases Monty accidentally opens the car door, which we have to include in our calculation (even though it happens not to be reality).

Pick 1, reveal 2 has 1/6 chance, and switching is bad
Pick 1, reveal 3 has 1/6 chance, and switching is bad
Pick 2, reveal 1 has 1/6 chance, and switching is bad ← not reality
Pick 2, reveal 3 has 1/6 chance, and switching is good
Pick 3, reveal 1 has 1/6 chance, and switching is bad ← not reality
Pick 3, reveal 2 has 1/6 chance, and switching is good

So, in half of the cases corresponding to reality (which each have equal chance) switching is good. Hence there is 1/2 chance that switching leads to the car.

Or, worded differently, there’s 1/3 chance a car is revealed, there’s 1/3 chance a goat is revealed and the switch is good, and there’s 1/3 chance the goat is revealed and the switch is bad. Hence under assumption that a goat is revealed, there’s 1/2 chance that the switch is good.

3 Likes
7

I’d like to propose a probability space where the result is not 50-50 in this scenario.

Provided that the placement of the car and the initial choice of door is independent and uniformely distributed, we have 9 cases each with probability 1/9. I’d like to denote them by (x, y), where the first coordinate stands for the placement of the car, and the second is the initial choice of a door.

Next a door is chosen at random, again independent to the placement of the car and the initial choice, and the door opens. We gain two bits of information: 1.) a door is opened that is not the door we initially chose and 2.) behind the door is a goat.

We’ll denote the new cases by (x, y, z) where x and y are as before and z denotes the number of the opened door. However we don’t consider cases where our door opens or the door with the car opens, as obviously those things didn’t happen. For each case we calculate the probability that door z was opened given the two bits of information from above. In the 6 cases where we selected a door that is a goat, this probability is 1, and thus the probability of the whole case is 1 * 1/9. In the cases where we selected the door with the car, the conditional probability is 1/2 and thus both branches will have probability 1/18. So I get the following 12 cases:

(x, y, z) - P(x, y, z)

(1, 1, 2) - 1/18
(1, 1, 3) - 1/18
(1, 2, 3) - 1/9
(1, 3, 2) - 1/9
(2, 1, 3) - 1/9
(2, 2, 1) - 1/18
(2, 2, 3) - 1/18
(2, 3, 1) - 1/9
(3, 1, 2) - 1/9
(3, 2, 1) - 1/9
(3, 3, 1) - 1/18
(3, 3, 2) - 1/18

Given this discrete probability space, the strategy of changing door has a chance of 2/3 to get the car.

2 Likes
8

It is important to consider those cases as well, though. That’s the whole distinction between Monty Hall knowing where the car is, or opening the doors accidentally. To discard scenarios that did not happen, they either need to have 0 chance to happen, or they need to be independent of the other events (including the doors you could choose at the end) to not affect the calculation.

With Monty Hall having knowledge, the chance of opening a door with a car is 0, thus this can be discarded (or not, but adding 0 to something has no effect, of course).
If we include the chance that it’s raining, this is independent of the calculation, thus this can be discarded (or not, but the probabilities of everything under assumption it rains, and of everything under assumption it doesn’t rain are equal, so adding these together will not change the result)
Monty Hall accidentally opening a door is not independent of the whole situation: in 1/3 of his accidents he “ruins” the game by revealing the car, after which the player’s options are clearly influenced (switching or not: the player can’t choose the car anymore). Hence in this case, the non-reality cannot be discarded.

Also, if we assume the initial choice of door is uniformly distributed, and the placement of the car is as well, we need to only factor in one of these two variables (since they’re independent, so (1, 1) will have the same probability as (2, 2), for example).

2 Likes
9

I need to make a confession. I’d like to think that I know a thing or two about probability theory, but it’s only ever made sense to me when a probability space is already given. In theory that is the first thing you begin with, and basically gives all needed information. In practice however you’re asked to calculate probabilities without a clear given probability space. How am I supposed to do that? Who am I to decide which probability space is suitable to describe the experiment? This has always bugged me.

I understand that, supposedly, the correct probability space to consider in this case is given by the following: With the notation above, there are 27 cases denoted by (x, y, z), each with probability 1/27, taking into account cases where our door is opened or the door with a car is opened. Then we can construct the probability space conditioned on the two bits of information we have, that is, a different door than ours is opened, and behind it is a goat. Out of the 27 cases, 12 fulfill this criteria and so the probability of this event is 12/27. So the probabilities conditioned on this event are (1 / 27) / (12 / 27) = 1 / 12. This gives us the same discrete probability space from above, except with an uniform distribution. And in this probability space, the strategy of changing door has a chance of 1/2 to succeed.

So how to decide which probability space to assume? This is something that has always confused me.

2 Likes
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I think that’s why we start assuming any probability space, usually something that seems intuitively plausible, but I don’t think it really matters that much, and then have to rely on Bayes’ theorem to update our assumption. Through repeated application (= statistics), we will eventually converge towards the actual probability space.

But probability and statistics are incredibly non-intuitive. I’d say it’s about as weird as general relativity to me.

As far as the Monty Hall Problem is concerned, we are given the probability space, though. Either by the fact that Monty Hall has knowledge about which door he opens, or that he does not (in which case we use the standard “we don’t have any more specifics given, thus we default to uniform distribution”).

We could also reformulate the question with something like “Monty Hall always chooses a door with a goat, but prefers the right-most door to not have to walk so far” or “Monty Hall has a 1/10 chance to have forgotten where the car is”, etc.

3 Likes
11

I’ve known for a long time that one is “supposed” to switch doors.

That answer has never made sense to me, so I chose to stand on my own reasoning.

To me, the much more interesting result is that I’ve so far been the only person who decided to stick with their original door. This means that everyone else either:

  1. Switched doors because they genuinely understand (or “understand”) the Monty Hall problem.

  2. Switched doors solely out of whimsy.

  3. Switched doors because they had learnt that they “should” do so, or because they were worried that someone would think them ignorant for not doing so.

  4. Switched doors because they had observed that other players had switched doors, acting out of mimicry or peer pressure.

My interest is in how many voters fall into each group.

I would much rather be incorrect as a result of what I believe to be correct reasoning than correct as a result of imitation or faith.

4 Likes
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If you are interested in that, I wonder why you didn’t set up a poll for it?

That’s most atypical around here! :smiley:

4 Likes
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I admit I had completely forgotten the problem and did not read at all carefully, I thought the revealed door containing the goat was the one chosen. So I was correct for “correct” reasoning based on incorrect information :joy: (in my defense I was on my phone waiting in line for coffee not my brightest moment)

2 Likes
14

I was reading your post for quite a while trying to understand what you wrote. Are you saying that under the assumption of “peeking” the probability of winning still different for switching and not switching? And for not switching it’s 1 /1000, for switching it’s 1/2 and thus it’s better to switch anyway?

But that doesn’t make any sense. If a car is behind the other door (to which we can switch) with 1/2 probability then it’s behind the original chosen door with 1-1/2=1/2, because there’s nowhere else for it to be.

And you wrote exactly that later.

It’s 1/3 switching is bad, 1/3 switching is good, and other 1/3 never happened. So 1/3 vs 1/3, it doesn’t make any difference to switch or not. Which is what terrific said. But you wrote as if you disagree. Which is why I want to clarify whether you merely worded it weirdly.

2 Likes
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Hm, yes, you’re correct. You don’t need to switch to update your chance, because the accidental opening of the doors already changed that chance from 1/1000 to 1/2.

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Hidden poll about bugcat’s question above.

In the poll above you:

  • switched doors because you genuinely understand the Monty Hall problem.
  • switched doors solely out of whimsy.
  • switched doors because you had learnt that you “should” do so, or because yoou were worried that someone would think you ignorant for not doing so.
  • switched doors because you had observed that other players had switched doors, acting out of mimicry or peer pressure.

0 voters

It’s anonymous, to make the voters more honest. :slight_smile:

1 Like
17

How about “I switched doors because, even though I don’t genuinely understand the Mounty Hall problem, I am convinced that switching doors does not lower my chances.”?

Because that’s my reasoning :wink:

1 Like
18

That sounds like whimsy to me.

1 Like
19

It’s in between genuinely understanding and whimsy: you genuinely understand it doesn’t lower your chances, and then decide to switch based on whimsy :stuck_out_tongue:

I think it’s more towards genuinely understanding, though.

1 Like
20

So there is ignorant whimsy and confident whimsy at play, then.