Monty Hall problem (again)

[phyti]

phyti:

You continue to make the same error, over and over.

For the player, at the time of the initial choice, there are only three possible choices, not four. The host has no say in the player's initial choice and no power to influence the game at this stage.

There are only three doors. The player is required to choose one door from the three available.

If the player initially chooses door 1, then in your list of scenarios, that player will then play one of two scenarios - the ones you numbered s=1 and s=2 in your second table in post #76. This is where the host can influence the game, but only to extent of determining whether s=1 or s=2 plays out. And the host's choice for the player to play s=1 or s=2 cannot affect the player's chances of winning at this stage of the game.

Each of scenarios s=1 and s=2 is played only half as often as s=3 and s=4 in the same table. This is the mistake you keep making in your analysis.

If the player is playing scenario s=1 or s=2, then after the host opens his door, the player then knows there are two doors unopened. The player only had a 1/3 chance of choosing the door with the car originally. So, at this stage in the game, the player has a 1/3 chance of winning the car if he chooses to "stick" with the original door. If he switches to the remaining unopened door, he must win 2/3 of the time.

Using the 2nd (Savant) table:
When A chooses door 1, host opens door 2, leaving the 1st prf column for the 2nd choice.
When B chooses door 1, host opens door 3, leaving the 2nd prf column for the 2nd choice.
The f for both prf columns is 1/3. The player only gets 3 of 4 possible sessions, and why the probability is 1/3.
It is independent of the initial 1st door choice. I.E., the 1st choice is redundant (changes nothing).
Try it with car in door 1. Host opens door per the rules, player chooses a door.

 

[phyti]

James R#75;
Additional response.

I don't have a clue what you mean by "biased" or "fair", in this context.

bias
1. preference: an unfair preference for or dislike of something
2. statistics distortion of results: the distortion of a set of statistical results
The lottery is a fair game of chance, where everyone who buys a ticket has the same probability of winning. The winner is decided by a random drawing within the system.

The player's chance of winning by switching is 2/3, as has been amply demonstrated, if the player is playing the game as described in the rules above.

If you have a choice of 2 things, there can only be 2 outcomes. That does not depend on what activity. It's a rule of logic. Dinner, Italian or Chinese. Tie, blue or green. To your office, stairs or elevator. People make these simple decisions every day around the world.
The 1/3, 2/3 ratios are a consequence of session frequency manipulation, (game rigging) shown repeatedly.

Savant reply to Whitaker,
"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance."
She is using the misleading visual aid of 1/3 car in each door. When door 3 is opened, she erroneously adds the 1/3 from the ghost door to door 2, an error in basic probability. Since probability is determined from a history of all possible events,
if the history is reduced from 3 to 2, you calculate a new probability as 1/2.
Possible choices = number of closed doors.

No. They don't have control over which of your "sessions" is played. Which "session" is played is determined by the player's choices and where the car happens to be at the start of the game.

You win here because I didn't express myself correctly. The distribution of prizes is the same for all doors. For the purpose of simplicity, the same setup is offered for each game with the car in door 1. Each player applies their own independent choices.

Who is Selvin?

He was the statistics student who proposed the original hypothetical variation of the game show in 1975 using 3 boxes, 1 with keys to a new car and 2 empty.
He used Monty Hall as his host, forming the association 'MH game show'.
He made the same conclusion as Marilyn Savant. When he sent his paper to 'The American Statistician', he got some negative reviews. Considering the readership of the magazine, that would be significant. He sent a 2nd letter using conditional probability.
My interpretation was negative, and did not find any reference to the 2nd response.

What is the difference between a "necessary restriction" and an "unnecessary restriction", in the game we have been discussing?

The host can only open 1 door per session.
case 1.
If the player selects door1 with the car, the host has 2 choices of goat doors.
Logic allows the host to open door 2 in 1 session and door 3 in a different session.
case 2.
If the player selects door 2 or door 3, the host opens door 3 or door 2, in separate sessions. That is a total of 4 sessions. Since the choices are random, for a large number of sessions, each would occur with a frequency of 1/4. No one has altered any factor of the game. Both Selvin and Savant thought case 1 was 1 session because there are 3 doors. Instead of playing it in 2 sessions, they played it as 1 with the host alternating opening door 2 and door 3. The stay prize was the car, but the session was played half as often as the case 2 sessions. The player wins a car 1/3 of the sessions, which favors a switch. If they had accepted the 4 session method, the win car ratio would be 1/2 and a fair game for all players. Their frequency of play created the bias that allowed a strategy to 'beat the system'. Remove the bias and you have a fair game of chance.

The Monty Hall game is not merely a game of chance. Are you claiming that it is?

The player could not predict where the car was, the game was literally a guessing game. The game did not require any special knowledge or mental processing.
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The game rules for Selvin and Whitaker were the same, and Savant understood them.

One objection is misinformation in an era of fake news and internet scams.
A greater objection is the notion of the common person not capable of making simple decisions. They do it every day world wide. The use of 'counterintuitive' cannot correct an error in thinking.

 

[Pinball1970]

If you have a choice of 2 things, there can only be 2 outcomes.

For the sweet love of Jesus...

There are NOT two things, there are THREE things to choose from. Two of the things to choose from are loosing doors, that is 2/3 or 67%. One of the things is a win, 1/3 or 33%.

The three doors are part of the game till the end, the open door sits there with it's revealed 33%.

If you do not follow the logic that you are twice as likely to have picked a losing door (two losing doors verses one winning door) then just write out the options.

You only have to do this for one door since all the doors are the same.

You pick door one which is the car, you swap you lose, that is 33% chance
You pick door one which is a goat you swap you win as car is door 2, thats 33% chance
You pick door one and swap you win as car is door 3. That's 33% chance.

Add up the swaps. It is TWICE as likely you will win if you swap.

 

[billvon]

The player could not predict where the car was, the game was literally a guessing game. The game did not require any special knowledge or mental processing.

But after Monty opens a door, the player now has special knowledge that he did not have originally.

 

[Sarkus]

But after Monty opens a door, the player now has special knowledge that he did not have originally.

Not really. At least not in the way I would explain the puzzle. To wit: you pick a door knowing you have 1/3 chance of picking correctly. You know there is 2/3 chance of picking incorrectly, and you also know that "at least one" of the doors you didn't pick is empty (or a goat, or whatever denotes a losing result). When Monty opens an empty door, you already knew at least one of those doors was empty, and Monty is just showing that. So, really, no new important information. You had 1/3 chance of winning. No new important information, so you still have 1/3 chance of winning by keeping the door you picked.
The additional information you do gain by Monty opening an empty door is that that specific door was empty. But that is not information that changes your odds of winning if you stick (1/3), not of the odds of swapping (2/3).

Imagine he didn't open a door, but gave you the choice to swap. Everyone should agree that swapping doubles your chances of winning from 1/3 to 2/3. So assume you swapped and now have 2 doors. You know that at least one of those must be empty. So opening one that is empty is not really new information, other than confirming a specific empty door.

While I don't doubt that you know the correct answer (swapping doubles your chances), as soon as I hear people trying to explain it through gaining "special knowledge" or an important change in what they know, this, to me, seems to be an argument that favours the incorrect 1/2 solution. It is incorrect, and your odds of winning if you don't swap stay at 1/3 even after Monty's reveal, precisely because you haven't gained any "special knowledge".

That's the way I look at it: no important new knowledge therefore no change in odds.
Simples.

 

[billvon]

Imagine he didn't open a door, but gave you the choice to swap. Everyone should agree that swapping doubles your chances of winning from 1/3 to 2/3.

As long as you know which one is empty. Monty can open the door or just tell you. But without that information, you do not improve your chances.

 

[Sarkus]

As long as you know which one is empty. Monty can open the door or just tell you. But without that information, you do not improve your chances.

The option is to keep the door you originally selected, or effectively to swap to both the other two. The fact that he has opened one of those other two is a red herring, designed to make you think that it's now a 1/2 chance for each unopened door. But you're in effect swapping from the original door to both the other two - one of which you knew up front was empty, whether you knew or not that a specific door is empty.

So, you make a choice..1/3 chance of selecting correctly. Do you then swap to the other two, one of which you already know is empty, or not? When viewed like that, there is no pertinent new information when Monty opens a door and reveals it to be empty.
Knowing which specific door of those two doors is empty gives you no information that affects probability.

 

[James R]

So, you make a choice..1/3 chance of selecting correctly. Do you then swap to the other two, one of which you already know is empty, or not? When viewed like that, there is no pertinent new information when Monty opens a door and reveals it to be empty.
Knowing which specific door of those two doors is empty gives you no information that affects probability.

That phrase "one of which you already know is empty" is concealing the complexity of the situation. It could mean one of two things:

1. You know that a specific door is "empty" (does not have the car); or
2. You know that at least one of those two doors is "empty", but you still don't know that any specific door is "empty".

With case 1, you have more information about where the car might be than with case 2.

In the actual game, of course, the situation you're talking about at this stage of the game (before Monty opens a door - or if he doesn't open any doors until you've made your final choice) is actually case 2 and not case 1.
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So, again....

You make an initial choice. At that point, you have a 1/3 chance of selecting the winning door.

If Monty did not open one of the other doors and show you that it is empty, then upon been given the choice "stick or switch?", you would have no new information and could therefore not improve your chances of winning by swapping doors.

In the actual game, though, Monty does open one of the other doors.

Before Monty opens that door, the two unopened doors together have a 2/3 chance of hiding the car (it will be behind one of the two unopened doors 2/3 of the time. You know that at least one of the those two doors is "empty", but you don't know which one.

After Monty opens one of those two doors, that 2/3 chance of hiding the car is no longer shared between two doors. Now, it's associated with just the single unopened door that you didn't originally choose. This is because you now know that a specific door is empty. Moreover, you are now unable to "switch" to that specific "empty" door, whereas before the door was opened that option was still "live" for you. This is why your chances of winning have just improved. In fact, the chances that the unopened door that you did not originally choose now hides the car have doubled.

So if you choose to "switch" doors, there will be a 2/3 chance that you win the car. If you stick to your original door, you'll have a 1/3 chance of winning the car.

 

[billvon]

That phrase "one of which you already know is empty" is concealing the complexity of the situation. It could mean one of two things:

1. You know that a specific door is "empty" (does not have the car); or
2. You know that at least one of those two doors is "empty", but you still don't know that any specific door is "empty".

With case 1, you have more information about where the car might be than with case 2.

Exactly. You need that additional information to increase your odds to 2/3.

 

[Sarkus]

That phrase "one of which you already know is empty" is concealing the complexity of the situation. It could mean one of two things:

1. You know that a specific door is "empty" (does not have the car); or
2. You know that at least one of those two doors is "empty", but you still don't know that any specific door is "empty".

With case 1, you have more information about where the car might be than with case 2.

In the actual game, of course, the situation you're talking about at this stage of the game (before Monty opens a door - or if he doesn't open any doors until you've made your final choice) is actually case 2 and not case 1.
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You have more information in 1, but it is not pertinent to improving your choices compared to 2.

So, again....

You make an initial choice. At that point, you have a 1/3 chance of selecting the winning door.

If Monty did not open one of the other doors and show you that it is empty, then upon been given the choice "stick or switch?", you would have no new information and could therefore not improve your chances of winning by swapping doors.

Monty isn't asking you to pick a specific door from 2 that are closed, though. In that situation, yes, the chance of switching to either of the closed doors would be still be 1/3.
The game, though, is that you have picked (1/3) and are effectively given a choice to switch to both of the other doors (2/3), at least one of which is empty. You don't know which. You don't need to know which. You know one door will be opened and one will not. You don't care which. That information about a specific door being empty or not is not relevant to the choice. You are, in effect, being asked to swap from thinking you are correct (1/3) to thinking you are incorrect (2/3). Monty opening an empty door doesn't alter that.

In the actual game, though, Monty does open one of the other doors.

Before Monty opens that door, the two unopened doors together have a 2/3 chance of hiding the car (it will be behind one of the two unopened doors 2/3 of the time. You know that at least one of the those two doors is "empty", but you don't know which one.

After Monty opens one of those two doors, that 2/3 chance of hiding the car is no longer shared between two doors. Now, it's associated with just the single unopened door that you didn't originally choose. This is because you now know that a specific door is empty. Moreover, you are now unable to "switch" to that specific "empty" door, whereas before the door was opened that option was still "live" for you. This is why your chances of winning have just improved. In fact, the chances that the unopened door that you did not originally choose now hides the car have doubled.

So if you choose to "switch" doors, there will be a 2/3 chance that you win the car. If you stick to your original door, you'll have a 1/3 chance of winning the car.

The issue here is not the result - we all agree on that - but on whether the additional information you get from Monty opening a specific door to reveal it is empty is adding anything to your knowledge that affects the chances. Billvon has said yes, and I am saying no, because you're effectively not swapping to a particular door, but to both of the other doors - albeit with Monty having opened one that is empty.

Let's say there are 10 doors - you pick one initially - 1/10 chance of winning.
Monty now asks - before opening any of the other doors - if you would like to swap to ALL the other 9.
Would you?
I would.
Now, instead of him saying "would you like to swap to all the other 9" he says "I am going to open 8 doors that are empty - would you like to swap to the one I leave closed?" - you still don't know which doors Monty will open. You don't yet have that information. But would you swap? I would. And I don't care which door Monty doesn't open. I don't need that information.

The specific door Monty opens isn't what increases the odds. It is recognising that you are swapping from 1 door to ALL the others. The opening of the doors is just a red-herring and showmanship. You are simply swapping from 1/3 chance of being right, to 1/3 chance of being wrong. The rest is window-dressing.

 

[Sarkus]

Exactly. You need that additional information to increase your odds to 2/3.

No, you don't.
I have 100 boxes, one of which contains the prize. You pick one - a 1/100 chance of winning.
I tell you that I am going to reveal 98 empty boxes. You don't know which box will be left. Without knowing which box will be left, will you swap to it? But you don't know which specific box it is? You have don't have that additional information. And, per you, without that knowledge your odds aren't increasing, right?

Yes, you would make the swap, even without the additional information. Because the game is not really that you are swapping to a single other box, but that you are swapping to ALL other boxes, with the host simply pre-opening many empty ones.

It is the mechanics of the game that give you the information, not the revelation of a specific box/door as empty.
And from the mechanics of the game you can recognise that you're not swapping from one door to one door - even if that is how it optically appears - but from one door to both of the others. You are swapping from 1/3 chance of being right to 1/3 chance of being wrong.
The opening of empty door/boxes is showmanship, but doesn't give you any pertinent information: if I give you the choice to swap to ALL 99 boxes you didn't originally pick, do you really need to know which box has the prize in it to have increased your odds? The fact that the host opens 98 empty ones is just to change the appearance of what you're being asked to swap to (one box), not the reality (all 99, at least 98 of which are empty).

Ultimately we're not arguing about the result, only in what information you think is required to arrive at the result. I think the mechanics aren't about swapping to 1 specific box, but to all other boxes, with the rest being showmanship to make you think it's swapping from 1 to 1. That's the difference here, I think.

 

[phyti]

For the sweet love of Jesus...

There are NOT two things, there are THREE things to choose from. Two of the things to choose from are loosing doors, that is 2/3 or 67%. One of the things is a win, 1/3 or 33%.

The three doors are part of the game till the end, the open door sits there with it's revealed 33%.

The open door with no car shows the cartoon analogy of 1/3 each door was false.
The player cannot select an open door!
The door opened by the host is removed from play along with its 1//3 contribution to the probability.
NOW you are playing a 2-door session.

 

[phyti]

Here are 3 ways to determine the number of a possible sessions.

There are 3 distinct prizes, car, bike, acorn.
Game rules. 1. host cannot choose player choice, nor door containing the car.

Left, all possible player-host choices of doors = 3*2 = 6.
Game rules eliminate (x) 2 leaving 4 possible sessions.

Center, a truth table showing possible host door choice given a player door choice.
Possible = 1, not possible =0.
Game rules eliminate the diagonal elements (1,1) (2,2) (3,3) and column d1,
leaving 4 possible sessions.

Right, the probability tree with 4 possible paths composed of player door choice followed by host door choice. The 4 outcomes each have an occurrence frequency of 1/4.
The occurrence frequencies of choosing the doors are not equal, since door 1 allows 2 host choices compared to 1 host choice for doors 2 & 3.
All 3 methods produce the same result of 4 possible sessions.


Craig Whitaker suggested a Monty Hall style game with a 2nd choice for the player allowing them to change their door selection. He asked Marilyn Savant if there was a winning strategy if the player switched their selection after the host opened a losing door.
In the case of s1, player selecting the door with the car, the host has 2 choices,
open door 2 and open door 3. Since Savant has assumed 1 session per door, she has to explain how to get 2 different results from 1 session. She adds a new game rule allowing the host to open door 2 half the time and open door 3 the other half.
This results in a win car ratio of 1/3 for the stay column p in the 2nd player choice. The ratio for the switch column r would be 2/3, enabling a winning strategy. This the same conclusion by Steve Selvin in his 1975 paper.
Her response was correct for Selvin's game but not for Whitaker's game that had no such rule with a reduced frequency of play. His game would have a win car ratio of 1/2 as shown in the probability tree above. If he had known of the Selvin game, there would have been no reason to ask his question.
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With the car in door 1, after the host opens door 2, there are NOW 2 possible states of the session. Event X is car in door 1 or door 3, event Y is car in door 2.
P(X) = 1, P(Y) = 0
Probability of car location = 1/2 + 1/2 = 1 for the set of 2 doors.
Probability of winning a car =1/2 since the car can only be in door 1 or door 3.

 

[Pinball1970]

The open door with no car shows the cartoon analogy of 1/3 each door was false.
The player cannot select an open door!
The door opened by the host is removed from play along with its 1//3 contribution to the probability.
NOW you are playing a 2-door session.

I will assume you are posting in good faith and not trolling.

Do you accept that at the start of the game you have a 1/3 chance of winning?

Let's take this one step at a time.

 

[Pinball1970]

There are 3 distinct prizes, car, bike, acorn.

No, there are no bikes or acorns, just answer post #94.

 

[Sarkus]

He's hooked on the notion that if there are 4 possible variations that these must occur equally. Until he can disabuse himself of that idea, and recognise the truth that two of his options happen only half as often as the other two, as others have repeatedly pointed out to him, he won't progress.

 

[Pinball1970]

He's hooked on the notion that if there are 4 possible variations that these must occur equally. Until he can disabuse himself of that idea, and recognise the truth that two of his options happen only half as often as the other two, as others have repeatedly pointed out to him, he won't progress.

This my last attempt.

 

[Randwolf]

I've never understood this puzzle, or, more specifically, I've never understood why people have such difficulty with it. It seems intuitive to me, to the point that I always feared that I must be "missing something."

I understand the "formal," statistical method of arriving at the answer, but to me that it boils down to a very simple choice - would you like to pick TWO doors, or ONE door? Which will give better odds?

If the question was posed that way from the beginning, I think most people would prefer two out of three over one of three chances, no?

Sarkus approaches the problem in a similar manner - disregard for a moment whether any of the doors are open or closed, or whether there is any "special" information in play after the host reveals one of the choices. In the end, you can either stick with your original choice (a single door) or switch to "owning" two doors. I'll take two doors over one every time.

Just because one of the doors happens to be open when I pick two out of three makes no difference, at least to me. What am I missing?

 

[Tiassa]

What am I missing?

It has always seemed like a question in which both answers are right according to their presuppositions; the question becomes which set of presuppositions is accurate.

Like the mistake people make about odds and probability: What are the odds of throwing seventeen heads in a row? But at the point that you've already thrown sixteen, the odds of throwing seventeen in a row are irrelevant; the next toss still has a probability of one in two.

Or, what does your betting ticket say?

To the other, it's math; there should be a straightforward answer. What I find fascinating about watching this thread argue about presuppositions is that it looks no different than politics.

And that actually seems to explain a lot.

 

[phyti]

I will assume you are posting in good faith and not trolling.

Do you accept that at the start of the game you have a 1/3 chance of winning?

Let's take this one step at a time.

No, since there i no prize for the 1st selection.
That happens for the 2nd selection.