I Please Explain (actually explain) The Monty Hall Problem

[Timbre]

TL;DR Summary

Without presuming anything, can anyone explain why the general consensus is that there are two questions?

Hello,

I'm joining this forum to ask two questions which have nagged me for some time. They both are presumed obvious, yet don't make sense to me. Nobody will explain their positions, which is...uh...aka science. I also have a thread for the other question.

But this one involves probability, known as the Monty Hall Problem. Please see any number of YouTube videos on this for an explanation, I'll leave it to them to explain it.

I question the predicate of all those who answer this problem. I only see one question, and one slight-of-hand suggesting there is a question when there really is not one. It appears to me that all who "explain" this get way out ahead of themselves, and are in the weeds where there is no need, and they're then speculating on (argue) stats theories which aren't based in reality, so have no answer, only argument (is it 1/2 or 1/3 probability).

Others apparently consider there to be two questions. My question is why one would consider there two questions. Please explain why the first question is a question (of any meaning/why would it factor in any statistical sense). If you're going to dictate that this is obvious, please refrain from comment. It's not obvious and it appears to me that the whole field, the many YouTube channels, everyone, is incorrect on this. I'm not trying to be a contrarian for no purpose, I genuinely see it this way and not one person who asserts that there are two questions has been able to explain, why.

The first "question", of which door to choose, is immaterial. The simplest reason that their response doesn't factor, is that the game will continue in effectively the same way no matter what they respond. There is no answer to the first question, no new information added to the paradigm. That is, no matter what, the game will continue to the second round of "question" with 2 options; one a winner and one a loser.

For the first question, the contestant could say Door 1, 2, or 3, or respond "the sun", or "blue flowers", or not answer at all, and it'd all effectively be the same. It's not a material question, it's speculation, and the result of that is never revealed to any consequence. It has no real bearing on the rest of the game. It's game show dazzle, nonsense, filling airtime, drawing attention. There is the suggestion that this factors, but it does not, it is pure suggestion.

I can illustrate if one would like:


Let's say the contestant chooses 1 of the doors without the prize, a losing door. Let's say the prize is door 2, and they choose 1 or 3. The game host removes one of the losing doors no matter what. The guest chooses 1, he removes 3. The guest chooses 3, he removes 1. Simple. Because we always know the host will remove a losing door.
The game continues, the guest "chooses" a second time, and the probability is 1/2 they'll choose right. What bearing did the first speculation, have?

Or let's say they choose door 2 with the prize. If the guest chooses 2, the host removes either 1 or 3, it doesn't matter. That's an arbitrary choice of the host. The contestant will not be told they've won, there is no new information given, the game will continue to the second round of "choice" no matter what, and, again, it'll be 1/2 probability they'll choose right.

There is never a 1/3 choice. We know the game will continue past that first round of question, and one losing door will be eliminated, and then the guest will have a 1/2 choice at the end. No matter what. This is how it works.

Will someone please explain why they think the first "choice" is in any way material?

Thank You, all. :)

 

[Dale]

It doesn’t matter if the first question is a question or not. What matters is that one of the other doors is opened. They could sing campfire songs, or have a belching contest, or whatever.

What matters is that the contestant picks a door and the host reveals one of the other doors. That action provides information.

 

[Timbre]

Yes, exactly. Providing information.

I realized that I could have written that there is only 1 question/answer combination. The first question/response provides no new information, and therefore does not factor. (Editing now)

 

[Dale]

The first question/response provides no new information

It does provide information. To see the information that it provides, consider this variant:

There are 100 doors, only one has the prize. You pick one, and then Monty opens 98 of the others, all without the prize. Do you still think no information was provided?

 

[Timbre]

Yes, I think no relevant information was provided. What's the phrase; plural of anecdote is not data?

If the game is prescribed to be played as you say, with the reveal coming down to 2 options, and the host will always remove losing doors, then it only matters what the final question/answer is. A 1/2 choice.

 

[Dale]

(Editing now)

Please don’t do that. It makes things confusing for subsequent readers unless it is done very carefully

 

[jedishrfu]

The way i like to explain it is you have ten doors behind nine are goats. Only one door has a car.

You choose a door and the MC chooses a door. The MC knows whats behind each door. The MC opens his door to reveal a goat.

Question: The MC asks if you want to change.

You analyze the situation and conclude YES lets pick a new door.

Why? Because you know that its likely your first choice was a goat so given the opportunity its good to change your mind.

—-

Reduce it down to three doors. Your first choice is likely a goat since two doors have goat and only has the car. The MC knows whats behind the door he opens, its a goat.

Its basically the same situation as the ten doors. You most likely chose a goat so by changing: Yay you win the car, and federal taxes on your winnings.

Whats distracting here is the MC opened a door.

There is a chance that you were lucky on your first choice and had actually chosen the car. In this case which occurs 1/3 of the time and switching the door means you lose the car.

Since you cannot know what you selected you must use the 2/3 of the time your first door selection is a goat. Consequently, in switching you’ll win the car 2/3 of the time and lose the car 1/3 of the time.

 

[Dale]

A 1/2 choice.

There are two options, but the outcome is not equiprobable. The coin is not fair. Why do you assume that it is?

 

[Timbre]

The way i like to explain it is you have ten doors behind nine are goats. Only one door has a car.

You choose a door and the MC chooses a door. The MC knows whats behind each door. The MC opens his door to reveal a goat.

Question: The MC asks if you want to change.

You analyze the situation and conclude YES lets pick a new door.

Why? Because you know that its likely your first choice was a goat so given the opportunity its good to change your mind.

—-

Reduce it down to three doors. Your first choice is likely a goat since two doors have goat and only has the car. The MC knows whats behind the door he opens, its a goat.

Its basically the same situation as the ten doors. You most likely chose a goat so by changing: Yay you win the car, and federal taxes on your winnings.

Whats distracting here is the MC opened a door.

There is a chance that you were lucky on your first choice and had actually chosen the car. In this case which occurs 1/3 of the time and switching the door means you lose the car.

Since you cannot know what you selected you must use the 2/3 of the time your first door selection is a goat. Consequently, in switching you’ll win the car 2/3 of the time and lose the car 1/3 of the time.

To stay focused: Could someone answer why the first question is relevant?

I've now seen my concern has been expressed previously on this forum but not posing it the way that I am. Why is the first "question", relevant at all. I do not accept the presumption that it is relevant. Again, the contestant guest could say anything as a response to the first question and the game would continue as it always would.
If the prize is door 2, and the guest chooses 2, the host removes option 3. If the guest chooses 1, the host removes option 3. Either way it's the same path to the second question which is 1/2.

 

[Timbre]

I'm challenging the notion that there are two questions in the Monty Hall Problem.

I place the onus on those suggesting that there are 2 relevant questions, to explain why there are 2. Why does the first one factor at all, what relevance does it have, what bearing does it have on the game play, what new or relevant information does it provide that changes the trajectory of the game?

It's fluff. It has no bearing on the game. It is suggested that the contestant may win the prize after the first round, therefore it matters, but the contestant winning on the first round is never an option. It's akin to that game that'd ask for an answer, then ask again "final answer?". The first time they respond is immaterial. Final answer is all that matters.

 

[Dale]

I'm challenging the notion that there are two questions in the Monty Hall Problem.

It isn’t a relevant challenge. It doesn’t matter if there is a first question or not. What matters is that information is provided.

 

[Dale]

It's fluff.

Indeed, it is fluff. You are missing the whole point of the problem because you are focused on which part of the fluff is fluffier.

The point of the problem is to determine what strategy is best: switch or stay. You are focused on the questions and answers, but the problem is about the strategy.

If we do the 100 door version and, without speaking at all, you point at a door and then I open 98 other doors. Do you really believe that the switch strategy will work just as often as the stay strategy?

 

[FactChecker]

Yes, I think no relevant information was provided.

I disagree. In @Dale 's 100-door example, our selected door initially had a 1/100 chance of being the correct door. The other 99/100 chances were in the other doors. The host is not allowed to open your door or the winning door. He opens 98 other doors. That gives you a lot of information -- there are 98 doors that you know are losers. Furthermore, you know that his selection of those 98 doors did not miss the prize out of blind luck. So the 99/100 chance is now associated with the one remaining door that you did not select.

 

[PeroK]

Yes, I think no relevant information was provided. What's the phrase; plural of anecdote is not data?

If the game is prescribed to be played as you say, with the reveal coming down to 2 options, and the host will always remove losing doors, then it only matters what the final question/answer is. A 1/2 choice.

Two points:

1) You can write a computer simulation of both games. And count how often the stick or switch decision wins. In one sense, you don't need any explanation beyond the numbers.

2) Probablity is ultimately about calculation. You could claim that the probability that a die comes up 6 is 50-50. No one can disprove that, other than by repeating the experiment many times and finding that the number 6 does not come up half the time. If you find by calculation that the number 6 comes up one time in 6, then you have to accept that is the probablity.

 

[PeroK]

The two versions of Monty Hall, in terms of calculation look like:

Monty Hall (proper game) has three equally likely scenarios:

Contestant picks correct door; Monty opens wrong door; stick wins
Contestant picks first wrong door; Monty opens wrong door; switch wins
Contestant picks second wrong door; Monty opens wrong door; switch wins

Conclusion: switch wins 2/3

Monty Fall (alternative game) has six equally likely events:

Contestant picks correct door; Monty opens first wrong door; stick wins
Contestant picks correct door; Monty opens second wrong door; stick wins
Contestant picks first wrong door; Monty opens wrong door; switch wins
Contestant picks first wrong door; Monty opens correct door; Monty wins!
Contestant picks second wrong door; Monty opens wrong door; switch wins
Contestant picks second wrong door; Monty opens correct door; Monty wins!

Conclusion: stick wins 1/3, switch wins 1/3, Monty wins 1/3

 

[Ibix]

TL;DR Summary: Without presuming anything, can anyone explain why the general consensus is that there are two questions?

The simplest reason that their response doesn't factor, is that the game will continue in effectively the same way no matter what they respond.

This is not true.

If the contestant picks a loser at the first stage then Monty can only open one door because there is only one loser remaining. But if the contestant picks a winner, Monty must choose one of two doors to open. The contestant knowing this asymmetry in Monty's behaviour gives them some weak deductive power.

That said, the simplest way to understand the Monty Hall problem (IMO) is to notice that you have a 1/3 chance of picking a winner at the first stage. "Stick with your initial choice" is therefore only correct 1/3 of the time.

An even simpler option, but more time consuming, is to play the game. You can do it solitaire. Get three playing cards, a picture and two aces. Shuffle and deal them face down side by side. You start as the contestant: pick a card, and turn it sideways to mark your choice. Now you switch to being Monty: flip all the cards and pick an ace that the contestant didn't choose (if there's more than one unchosen ace, you may choose which one any way you like) and put your finger on it to indicate you'd have opened it. Now record if the contestant would have won if they'd changed to the remaining card or not. About 40 repeats should be enough to show clearly that switching is a better strategy, unless you have bad luck (with 40 repeats there is a chance - less than one in twenty - that you will not show an advantage to switching simply due to the luck of the draw). If 300ish repeats doesn't show a clear advantage to switching, buy a lottery ticket immediately.

 

[Dale]

1) You can write a computer simulation of both games. And count how often the stick or switch decision wins. In one sense, you don't need any explanation beyond the numbers.

This is a very useful tool in general called “Monte Carlo simulation”. I use it a lot.

 

[DaveC426913]

1) You can write a computer simulation of both games. And count how often the stick or switch decision wins. In one sense, you don't need any explanation beyond the numbers.

I wrote this sim. https://www.davesbrain.ca/science/monty-hall.html

Staying wins one third of the time.
Switching wins two-thirds of the time.
Empirical proof.

It also has the side effect of proving that Monty revealing the second door definitely does provide new information.

 

[Timbre]

I would still like to stay focused and ask those promoting the first question is relevant, to explain that.


But this is a very simple explanation, a proof, of why they're incorrect:

I've seen this nowhere on the internet,so please read...

Question (to be answered below): In the second round of choice, what knowledge does the contestant have about his chosen door, that he does NOT have about the second door he could switch his choice to? None, he has no distinguishing knowledge, no differential in the knowledge between the two remaining doors.

The proponents of this problem purport the following, and I agree:

Guest is presented with 3 doors, he has a 1/3 probability, and makes his choice.

One option/door is removed. He still retains his choice door, and knowledge that it's 1/3 probability.

That information from the first round, that his door is 1/3 chance, does transfer to the second round where it is a 1 in 2 choice.

Agreed.

Here is where they're incorrect:

They suggest that he has 1/3probability that his first chosen door is right. Agreed.

And they imply that he has NO such knowledge of the other remaining door. Disagree.

Fact:
He has knowledge of his first chosen door, that it's 1/3 the winning door..

And he has knowledge of the other remaining door that it's also 1/3 the winning door.


So he has no useful knowledge to help him in the second choice. Because he has knowledge that his chosen door is 1/3, and he also has the same knowledge of the other optional door has a 1/3 chance from the previous round of choices. So this knowledge, that both doors previously had 1/3 chance, is immaterial, useless, not really information. Now he knows they both have a 1/2 chance, and that's all the probability ever is.

The proponents state the fact he has knowledge of his chosen door after the first round (1/3), but they imply that he has no similar knowledge of the second remaining door. That implication is false. The only relevant knowledge would be if he knew something about his door that he DIDN'T know about the other door. Then he could use that knowledge. But he has no such knowledge of the DIFFERENCE between the two doors.


This is why they're incorrect. It's a½ choice the second round. Period.

 

[Ibix]

It's a½ choice the second round. Period.

Except you can play the game and prove yourself wrong on this. I suggest you do so.

 

[PeroK]

This is why they're incorrect. It's a½ choice the second round. Period.

You're not calculating probabilities here. Probabilities can be tested by experiment. If we played this game repeatedly, then I could win a lot of money off you - with you betting 50-50 on your original choice of door.

Probability relates to relative frequency (with all due respect to @Dale). What you are calling a "probability" is not a probability in this sense. It's a number that does not represent relative frequency. If you bet using your number as a probability, then you would lose more often than you'd win.

 

[Dale]

In the second round of choice, what knowledge does the contestant have about his chosen door, that he does NOT have about the second door he could switch his choice to? None, he has no distinguishing knowledge, no differential in the knowledge between the two remaining doors.

This is false.

They know that the probability of the originally chosen door being a winner is 1/N. They know that N-2 other doors have probability of 0 (this is the new information). So, since probability must sum to 1 they know that the probability of the other door is 1-1/N.

He has knowledge of his first chosen door, that it's 1/3 the winning door..

And he has knowledge of the other remaining door that it's also 1/3 the winning door.

If this were true then the total probaiblity would be 1/N + 1/N = 2/N, which is not 1 for 2<N. Total probabilities must sum to 1.

The proponents state the fact he has knowledge of his chosen door after the first round (1/3), but they imply that he has no similar knowledge of the second remaining door. That implication is false. The only relevant knowledge would be if he knew something about his door that he DIDN'T know about the other door. Then he could use that knowledge. But he has no such knowledge of the DIFFERENCE between the two doors.

They do. The knowledge that the other N-2 doors have probability 0 of being the prize door allows them to know that the probability of the remaining door is 1-1/N and not 1/N.

 

[PeroK]

For example, someone in my class at school insisted that if Liverpool Football Club played our local village team, then the result would be 50-50. He could not accept that if two teams played a game of football, the outcome could be anything other than 50-50. That wasn't a probability. That was just a statement that (excluding a draw) there were only two possibilities.

Having two possible outcomes is not the same as having two equally likely outcomes.

 

[Timbre]

Having two possible outcomes is not the same as having two equally likely outcomes.

Yes, I understand this. The difference between two possible and likely outcomes. Still, I see no difference in the two outcomes because there is no information about one that is not true about the other. Both options were 1/3 in the first round of questions.

 

[Timbre]

You're not calculating probabilities here. Probabilities can be tested by experiment. If we played this game repeatedly, then I could win a lot of money off you - with you betting 50-50 on your original choice of door.

Probability relates to relative frequency (with all due respect to @Dale). What you are calling a "probability" is not a probability in this sense. It's a number that does not represent relative frequency. If you bet using your number as a probability, then you would lose more often than you'd win.

If you can expand on the difference between my use of probability and the true definition, I'm interested in understanding what you're expressing. I thought that's what I was explaining, relative frequency. In the second round of choice, the knowledge of relative frequency for both doors is identical. They both had a 1/3 chance in the first round of choice. So there is no relative difference between the two.

 

[Dale]

there is no information about one that is not true about the other.

But there is. You keep saying this, but saying it doesn't make it true. This isn't politics.

Both options were 1/3 in the first round of questions.

And then you received additional information that you didn't have then.

 

[Dale]

Probability relates to relative frequency (with all due respect to @Dale).

I see what you did there . But it made me realize that I don't think that I have ever seen a Bayesian analysis of the Monty Hall problem. So, it is now a personal challenge to myself to try such an analysis (without looking for one in the literature, I am sure there is one but I just haven't seen it).

 

[DaveC426913]

Here are the results of the 200 game sim.

You are welcome to run analysis on the randomly generated numbers to assure yourself they're not fixed.

You can clearly see the wins converging on .67.

The fact that the player does win much more after switching than he wins without switching shows that the player is demonstrably using his gained knowledge to his advantage. This is irrefutable.

So, it's demonstrable that your logic and conclusions are flawed, the only question remaining for you to figure out is where the flaw is.

Click here for 100% image.

 

[PeroK]

If you can expand on the difference between my use of probability and the true definition, I'm interested in understanding what you're expressing. I thought that's what I was explaining, relative frequency. In the second round of choice, the knowledge of relative frequency for both doors is identical. They both had a 1/3 chance in the first round of choice. So there is no relative difference between the two.

Relative frequency means the proportion of times an outcome occurs - in an experiment or simulation.

If the player chooses a door and sticks with it, then they win the car 1 time in 3. That's a relative frequency, hence a probability of 1/3.

The possible outcomes and their relative frequency are given in post #15.

 

[Dale]

So, it is now a personal challenge to myself to try such an analysis

It was far easier than I had thought. Bayes theorem is $$P(H|E)=\frac{P(E|H) \ P(H)}{P(E)}$$ There are $N$ doors. The contestant chooses door $C$, but does not open it. Then Monty chooses $N-2$ and opens them all, leaving only door $R$ remaining that is neither chosen by the contestant nor opened by Monty. Let $H_C$ be the hypothesis that the prize is in the door chosen by the contestant. Let $H_R$ be the hypothesis that the prize is in the remaining door. Let $E_R$ be the evidence of opening all of the other doors besides doors $R$ and $C$.

The prior probabilities, before seeing evidence $E_R$, are $$P(H_C)=P(H_R)=\frac{1}{N}$$

The conditional probabilities are $$P(E_R|H_C)=\frac{1}{N-1}$$$$P(E_R|H_R)=1$$

The marginal likelihood is $$P(E_R)=P(E_R|H_C) \ P(H_C) + P(E_R|H_R) \ P(H_R) = \frac{1}{N-1} \frac{1}{N} + 1 \frac{1}{N} =\frac{1}{N-1}$$

So substituting into Bayes theorem we get by direct calculation $$P(H_C|E_R)=\frac{P(E_R|H_C) \ P(H_C)}{P(E_R)}=\frac{(1/(N-1))(1/N)}{1/(N-1)}=\frac{1}{N}$$$$P(H_R|E_R)=\frac{P(E_R|H_R) \ P(H_R)}{P(E_R)}=\frac{(1)(1/N)}{1/(N-1)}=\frac{N-1}{N}$$

 

[BWV]

This has been beat to death, but thought I would chime in that it actually does not matter whether the host knows what door the car is behind, if you pick door 1 and the host opens 2 or 3 without knowing where the goat is, there is a 1/3 chance that the host reveals the car and a 2/3 chance he reveals the goat. From your standpoint, if he does not show you the car, the problem is the same (if he does open the door with the car, then you definitely should switch ;) - there is a 2/3 chance you win by switching. The host's knowledge does not change the problem after you have been shown a door with a goat, it just prevents the awkward situation of him revealing the car.

The problem can be reduced to one decision you either bet that the car is behind one door or either of the remaining doors - so of course it makes sense to bet that the car is in the larger subset

 

[Ibix]

The host's knowledge does not change the problem after you have been shown a door with a goat, it just prevents the awkward situation of him revealing the car.

Note that your modified game is only the same as the normal one if this "awkward situation" is dealt with somehow - perhaps re-playing the game. So it still depends on the host's knowledge of where the prize is - it just changes when the host learns that fact.

 

[BWV]

Note that your modified game is only the same as the normal one if this "awkward situation" is dealt with somehow - perhaps re-playing the game.

Agree with this

So it still depends on the host's knowledge of where the prize is - it just changes when the host learns that fact.

If the host and the player have the same info at the same time, then the game cant depend on the host's knowledge

 

[PeroK]

I remember someone posted the following analysis for the Monty Hall game with 100 doors.

Let's say the contestant picks door 16 and the car is behind door 59. Monty goes down the doors in order. Opens doors 1-15, misses out door 16. Then opens doors 17-58, coughs discreetly while missing out door 59, before opening doors 60-100.

Now, @Timbre might be happy to stick with door 16, but the rest of us would have the strong suspicion that the car is behind door 59.

 

[jedishrfu]

Your friend or internet source is using flawed logic. It's the same logic people use and they get it wrong. It's why probability is often counter-intuitive to people.

Suppose we modify the game and bring a second contestant up who is free to choose one of the two closed doors. Then that second contestant has a 1/2 chance of getting it right. Now if he knows which door the first contestant chose, thought and said "hey I'm picking the other door." Then he would be using prior knowledge to inform his choice.

As has been mentioned earlier if the first contestant does nothing then chance of winning is 1/3 of the time.

This has been beat to death, but thought I would chime in that it actually does not matter whether the host knows what door the car is behind, if you pick door 1 and the host opens 2 or 3 without knowing where the goat is, there is a 1/3 chance that the host reveals the car and a 2/3 chance he reveals the goat.

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

 

[jedishrfu]

Thinking further, we could always just send you a goat. Though I don't know that I could do that since I have neither a goat nor a mailman willing to deliver it.

---

In Texas, they have an agg exemption if you raise a tribe of 3-5 goats on five acres or more. And you get property tax break. You do however have make effort make money off of the goats by selling them, or selling their milk for goat cheese.

 

[PeroK]

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

He would soon run out of space in his garage.

 

[Ibix]

If the host and the player have the same info at the same time, then the game cant depend on the host's knowledge

Well, your game is longer than the standard version because it sometimes involves replays, and whether you have a replay depends on revealing where the prize is. So in that case it depends on somebody knowing the location of the prize and that it is not known to the contestant when they actually choose.

A game that is statistically identical to Monty Hall is: contestant chooses a door, then may choose to open that door or open both the others. In that case switching is the obvious strategy, but there's no counter-intuitive behaviour.

 

[Ibix]

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

I believe that in the actual game show Monty had options other than "open a door and offer a switch", and was a skilled cold reader who would typically take the choice that he reckoned would maneuver the contestant away from the prize. So he offered you a swap if he thought you'd take it and lose, or if he thought you'd stick and lose, and he didn't think one of his other options was better.

The first article I read about commented that the experience of watching the game show may have messed with people's intuition about the puzzle when it was first stated.

 

[Timbre]

This has been beat to death, but thought I would chime in that it actually does not matter whether the host knows what door the car is behind, if you pick door 1 and the host opens 2 or 3 without knowing where the goat is, there is a 1/3 chance that the host reveals the car and a 2/3 chance he reveals the goat. From your standpoint, if he does not show you the car, the problem is the same (if he does open the door with the car, then you definitely should switch ;) - there is a 2/3 chance you win by switching. The host's knowledge does not change the problem after you have been shown a door with a goat, it just prevents the awkward situation of him revealing the car.

The problem can be reduced to one decision you either bet that the car is behind one door or either of the remaining doors - so of course it makes sense to bet that the car is in the larger subset

To my knowledge this is not how the game is played. On the first round of questions, the host will never reveal the car. That's why the first round is not an actual question/answer that lends any new information to the game.

 

[Ibix]

That's why the first round is not an actual question/answer that lends any new information to the game

Why do you keep ignoring us when we point out this isn't true?

 

[berkeman]

[Note -- the OP is on a 10-day vacation from PF.]