Understanding the Monty Hall Problem

There’s one thing that never gets brought up by people when explaining the odds: the door reveal is a red herring.

People explain that because it’s a choice of three doors the odds of picking the right one is 1/3 therefore swapping gives you 2/3 odds. Odds presented by rigidly examining the game over the course of many static trial runs. Whenever people try to explain these odds, they will use a macro scale of the game. Playing it 50 times, 100 times, 300 times, a 1000 times. Throw in more doors. Run through every possible scenario. And the math of those odds are based on that rigid viewing of various possibilities. And that is all, technically, statistically correct.

But if you’re actually playing this game as it is presented in the monty hall problem, then you’re playing it once. Just, once. And when you are playing it, no matter what door you choose, a door will be revealed. That revealed door is always going to be A) not your door and B) a wrong choice. Which means you know, before you even pick any door, that the probability of the door that’s going to be revealed being the correct choice is 0%. And the game is rigged to where the 0% door is -never- going to be your choice, because they get it out of the way before your final choice. That means no matter what door you choose, you know beforehand you will always be left with 1 winning door and 1 losing door, regardless of the initial choice.

I’m sure people are ready to bum-rush my post and defend the 33%/67% odds with a smug sense of superiority. Don’t bother; I ‘get’ it. I understand the statistics. I understand the logic within the explanation. But playing to the odds is just a great way to make a wrong decision with confidence.

889
"But playing to the odds is just a great way to make a wrong decision"
No it isn’t, that’s a clearly ridiculous statement. If you have a 66.7% chance of winning and a 33.3% of losing, and assuming you want to win, then playing to the odds is definitely making the right decision.

890 Way to quote snipe. And to completely miss the meaning too. Good job.

Thank you.
What did you mean then by “But playing to the odds is just a great way to make a wrong decision with confidence”?

The right decision is the one that wins you the car. The odds it takes for you to get it aren’t as important as, you know, actually winning the prize. So if you lose, regardless of your odds, you still made the wrong decision. As you just so showed, you would confidently make that wrong decision 33.3% of the time.

895
There’s nothing fallacious about my statement. Nor was I attempting to smudge the data. The comments throughout this thread prove it to be quite accurate with a general attitude that you should always switch because it’s better odds. They all seem quite confident, comfortable, ready and willing to make that kind of decision based solely on their odds. A decision could still very well net them a loss, and they seem quick to call others fools if they, say, had a gut feeling that they shouldn’t switch despite the odds not being in their favor. Even if they won. This attitude is clear because it’s always been discussed by people just looking at the data. I still would think it would matter to the individual playing.

If you were really advising that many imaginary people, why not refine your advice to tell them to pick the door that last won then switch since the odds of it being there multiple times in a row is statistically less. Or the door that already won the most for that day so far. Or the door that’s won the most in general up to that point of time. Go for broke.

If your statement about making “a wrong decision with confidence” is not fallacious, everyone who switched would do so with confidence. But no one relinquishes their initial pick easily. That is because they feat the considerable, 33%, probability that they will be giving up a winning position.

The recommendation to always switch is simply evidence-based advice that provides the better outcome on average, not in every single instance. It is like other evidence-advice such as:

  • Don’t smoke (even though there are a few octogenarian smokers who do not succumb to smoking-related illnesses);.
  • Wear a seat belt (even though some people may have had better collision outcomes without a seat belt fastened);
  • Look both ways (even though small children have run across roadways without being hit by cars).

No one is saying switch and you will win. Yet you appear to object to the switching recommendation merely because it may fail to yield a better result than gut feeling in one or more particular instances. In fact, switching will do worse than gut feel around 33% of the time.

But it is simply a case of which odds you prefer because looking backwards at results of two cohort s of sufficient sample size, the cohort that always switched would do around twice as well as the cohort that never switched, even though individual results may vary.

As for “going for broke” by providing even more advice, you assume each contestant knows which door concealed the car last time and/or that there is some rhyme or reason to how that door is picked. If there is no rhyme or reason, the chances of each door concealing the car are equal and not influenced by the previous game.

As the game is reset before it is played each time, none of the trends to which you refer are likely to be instructive; the car can be placed behind the same door multiple times in a row. At the start of each game, each door has an equal chance of concealing the prize, irrespective of how many times it has yielded a prize in the past.

Whatever the case, what advice would you provide that group of 100 eager “Let’s make a Deal” contestants? Will you tell them to Always go with their gut; Always switch; or Always stay?

They are their options and, if adopted en masse, each can be expected to cost the show’s producers different amounts in prizes.

A frequentist would say to an individual: “I cannot advise you with confidence because you are only playing once.”

A Bayesian might say: “I cannot guarantee any particular result, but if you played the game 100 times you would win around twice as often if you always switched. Therefore, the probability of winning any one game will increase accordingly if you switch”

893 Anonymous: "The right decision is the one that wins you the car."
And the odds of making it are …?

I presume you are quoting yourself. Wouldn’t it be helpful to put a name, any name, to your comments?

But “odds” are what a bookie offers based on probability, which in terms of picking the car on the first guess reflects the fact that a contestant should do this 33% of the time.

Even if the game is only being played once, the probability is still useful as it indicates that a contestant has a reasonable chance of picking the car on the first guess, making sticking a reasonable strategy for those who prefer instinct to hard numbers. However, in a game of 4 or more doors, one would be taking a much greater risk following only their instinct.

I seem to think that this is not as complicated as one would believe. Let’s look at it this way; if I initially chose Door 1 and was then given the option of swapping my choice of Door 1 for both Doors 2 and 3, I would obviously choose to swap my 1 door for both the other doors as my chances with 2 doors would be doubled. When choosing the option of 2 doors, I know that at least one of them will contain a goat. With either Door 2 or Door 3 having being revealed as a goat and I am then allowed to switch, I am in effect still being offered the 2 for 1 switch, so my odds must still be double those of my initial pick.

The issue with MHP is not so much that it is complicated as that it is counter-intuitive: Instincts betray most people by telling them that, if there are two doors from which to choose and only one prize, each door has a 50% probability of concealing the prize.

As I have noted before to no response, in the simulation, if the computer opens the right-most of the two doors there is a 100% chance that switching will win as the simulator always prefers to open the left-most door when it can and only opens the right-most door when the car is to the left.

This means that people who like to play against the odds and follow their gut-feel on occasion can elect to do so only on games where the computer opens the left-most door. Doing so can result in some results significantly higher than 67%, although not generally in the long run.

I use to have a lot of trouble trying to explain the solution until I started showing people that by switching you always get to pick two doors. If I want two doors, let’s say doors 1 and 2, then I would select door 3 and then open the door that doesn’t get open. People seem to understand this pretty quickly.

  1. "As I have noted before to no response, in the simulation, if the computer opens the right-most of the two doors there is a 100% chance that switching will win …"
    Yes, there is a shortcoming in the simulation if as you say the computer prefers the leftmost door. The MHP contestant has the information in the MHP question to go on, and no more, which gives them no indication of the host’s preference or whether he has any.

905/6 The simulation does demonstrate 2/3 by switching, however it’s a case of ‘right answer, wrong working’.

The simulation violates the assumption that the host has no ulterior motive, and thus opens a random goat when there are two, though does not necessarily violate the condition that the contestant is unaware of any such motive. The instructions partly replicate that condition by suggesting play-and-hold fifty times and then play-and-switch twenty times (which prevents the contestant selectively switching), then to vary the strategy over a few dozen more (which, if they spot the pattern, does not). As stated, the simulation reveal should really have two goats randomised not leftmost.

However, as pointed out, in the simulation as it is switching still achieves 2/3, never more, as the opportunity for 100% (rightmost goat) occurs 1/3 of the time and is balanced by the greater frequency of the 50% chance (leftmost goat) 2/3 of the time.

The way I explain to friends is this:

Let’s make it a iron-clad rule that I switch, after Monty opened another door to reveal a Goat. And of course, the initial choice gives me a 1/3 chance of picking the Car, and a 2/3 chance of picking a Goat.

This means if I picked the Car at first, I will invariably lose it, as the switched choice will invariably be a Goat. And the opposite is true: if I picked a Goat at first, I will invariably switch to the winning door with the Car behind it. In other words, if I picked right at the beginning I lose, and vice versa.

Don’t forget: the odds of the initial choice remains the same, but would would just reverse your outcome. So, at first I had a 1/3 chance of picking the Car but I am bound to lose it, but at the same time, I had a 2/3 chance of picking a Goat which means I will get the Car.

If I stay with the initial choice, Monty opening another door would not change anything: I would still be stuck with an unfavorable odds of 1/3 of picking the car.

Your Door simulator needs to be reworked. Right now, I have 1/3 chance of a guaranteed door, and a 2/3 chance of a 50/50 choice. This is because the program places a goat tag on the first goat door that isn’t selected by the player. So, If I always select Door 1 as mine, If Door 2 is goat, then 50/50 chance between Doors 1 and 3. If 3 is goat though, it means that door 2 is guaranteed to be the car, or it would be labelled goat. This is nothing against the concept itself, and the probabilities still work (1/3 + 0.5*2/3 = 2/3) with this code, but it could still be improved to be a more realistic depiction of the concept.

While I do not believe you have, with respect, explained the point as plainly as might be possible (“tags”?), you’re on to something Pat, as I have noted several times above, most recently at 902.

To be clear, the location of goats is determined when all doors are closed.

The anomaly I have observed as I describe it is that

(a) if you pick the correct door on first guess;

(b) the “Host” always opens the left-most of the two remaining doors.

This means that whenever the Host opens the right-most of the two remaining door, a switch will be successful 100% of the time.

This is caused by the Host’s undiagnosed OCD IRT the order in which doors should be opened.

The treatment is some random-number-generating code that varies the door to be selected when the contestant has picked the car on first selection, n’est ce pas?

Mr. 50/50 takes the stage…

Monty… “Congratulations Mr. 50/50, you are our game’s finalist and I will give you the opportunity to win our grand prize of the day. See here, there are 3 doors, 2 of which are hiding a goat, and 1 door hiding a brand new car. Now all you have to do is guess which door you think the car is behind. After your guess I will show you one of the doors you didn’t choose that hides a goat. And as a bonus you will have the chance to switch your guess to the other closed door at that time.”

Mr. 50/50…“Ok Monty, I’m going to choose door 2.”

Monty…“Mr. 50/50 chooses door 2, let me show him a door that he didn’t pick that has a goat behind it.”

Mr. 50/50…“It’s ok Monty, you don’t have to.”

Monty…“Huh?”

Mr. 50/50…“Monty, if you open door 3 to show a goat, then the car is either behind my guess or door 1, right?”

Monty…“Yes, that sounds reasonable.”

Mr. 50/50…“And if you opened door 1 to show a goat, the car is either behind my guess or door 3, right?”

Monty…“Right again Mr. 50/50, I follow you so far.”

Mr. 50/50…“So no matter which door you open to show a goat my guess is only up against one other door.”

Monty…“So exactly where is this going?”

Mr. 50/50…“It means that no matter which door you open, I will have a 50/50 chance against the other. So you don’t have to open EITHER door, I will just stay with my guess!!! Open door 2 for me Monty.”

Monty…“Wow, I never thought of that, and all this time I was opening doors with goats behind them for nothing.”

Mr. 50/50…"Open door 2 will you!!!

Monty…“I’m so sorry Mr. 50/50. I was going to show you that door 3 has a goat behind it, and since you decided to stay against the one other closed door you are going home with a goat. Too bad because the car was behind door 1. It must have been just plain bad luck. Maybe you can be a contestant again in the near future because with those odds you are giving yourself you are bound to win!!!”

Yeah, exactly. It is because in simple terms you have only two choices: either you win or you lose. There is no third option that is why it is hard to understand probability and all this mathematics when it is easier to think that way.
Also, all people that cry Heureka and say, now I understand why it is better to switch perhaps think it is a ‘winning’ strategy although it is not a ‘winning’ (100%) strategy but only a ‘more winning’ (66%) strategy.
One thing that comes to mind is Russian roulette. With one bullet in the revolver you have “much better” chance to stay alive and you’re still scared to shit the same way if you had 5 bullets because you “either die or survive” which seems 50/50.