Monty Hall problem (again)

[phyti]

Sarkus;
"If a person comes in from the blue and sees the two doors left closed and knows a car is behind one of them, then for thatperson the choice would be 50:50 as to which door. Because that would be an independent event to what has gone before."

If a person outside the studio was offered to come inside and fill in for the player, they would be aware of the empty door. The substitute has no less knowledge of the car location than the player and can only make a random choice like the player. How or when the door was opened is irrelevant to the current session.
The player did not gain any constructive information that would provide any advantage. The player used their 1st choice of 1 of 3 closed doors. The host opening a goat door eliminated that door. In the world of reality, the distribution of prizes changes from 1 car and 2 goats to 1 car and 1 goat. I find it amazing that no one seems capable of comprehending the fact there are not 3 ways to choose 1 of 2 things. If you think there is, what's the 3rd way?

The fair game.


A session is labeled by the initial of the player.
Player is p, host is h, remaining closed door is r.
There are 2 sessions (ways to play) if the door is a car door.
There are 3 distinct prizes in general, thus 2 different goats.
There is 1 session if the door is not a car door.
Each player begins with 3 doors.
Since only 3 of the 4 sessions can be assigned to 3 doors,
there are 2 sets of sessions S1 and S2.

Probability is the ratio of (#successful events)/(#possible events), e/s.
If S1 is used, e/s = 1/3, but B is excluded.
If S2 is used, e/s = 1/3, but A is excluded.
All possible events s = 4.
This should explain the difference why win car ratio is determined by number of choices, not number of doors.

The car door is offered twice for each round of 4 sessions. Savant and Selvin, were asking the host to open the goat doors half as often when the player chose the car door compared to when the player chose the goat doors. The fair game eliminates that bias.

The player has made their choice in column p. The host cannot alter that choice, so their choices are independent. If the player accepts the option to switch, their 2nd choice is column r which is determined by column p. I.e., is it c, g or g, c.
The p vs r column shows no advantage in switching.
That was already known, since a game of random choices is not predictable, just like the lottery. There is no algorithm or method to form a basis for a strategy. (Selvin as a statistics student should have known that.)

"Notice in your post #189 your "corrected" frequency diagram has door 1 picked 50% of the time."

"Do you accept that, in a fair game, each door has equal probability of being picked by the player? If so, that initial probability MUST be 1/3 for each door. Not the 1/2 for door 1 and 1/4 for the other 2."

As explained in 'the fair game', it depends on what you consider as all possible outcomes.

"For some reason you seem resistant to learning."

I enjoy learning, more so in the beginning with Special Relativity at Physforum. A fast moving clock running slower seemed strange but interesting.
Lost interest, partly due to the collective attitudes of some forums, and less interest in science in general. I don't have to know how far Alpha Centauri is, since I don't plan to go there anytime soon.

 

[phyti]

I know, it's a comparison. Let's see if you can understand it.

If person A removed one card from a shuffled deck, and person B took the other 51 cards - who is more likely to have the Ace of Hearts in their hand?

Gave the answer earlier. Are you paying attention?
For A, P=1/52. For B, P=1/51.

 

[Pinball1970]

For A, P=1/52. For B, P=1/51.

Took all the other cards, all 51.

 

[pzkpfw]

That's a different game.

No.

I'm pretty sure the deck of cards in this thread started at post #139, and for many posts now we've all been trying to get you to understand up to step 5.2 in the list there. It seems like you are deliberately being evasive by changing to a "different game". I understand why James has said "trolling" more than once.

Maybe we need something simpler than cards? Here are some very simple questions of probability using marbles:

---

Setting:
1. Marbles. One red, and nine blue. OOOOOOOOOO
2. Identical shape and weight, no way to tell them apart when in a closed opaque bag.
3. There are two of these bags.

---

Scenario A:
The red marble, and one blue marble, are each secretly placed in separate bags:

Bag 1	Bag 2
?	?
One Marble	One Marble

Question A1: What are the odds that the red marble is in bag 1?
Question A2: What are the odds that the red marble is in bag 2?

---

Scenario B:
The red marble is randomly mixed with the nine blue marbles.
One random marble is placed in bag 1 and the other nine are placed in bag 2.

Bag 1	Bag 2
?	?????????
One Marble	Nine Marbles

Question B1: What are the odds that the red marble is in bag 1?
Question B2: What are the odds that the red marble is in bag 2?

---

What are your answers to A1, A2, B1, and B2; phyti?
Just four numbers needed.

Please be honest and answer as asked, not changing the scenarios to something else.
(I am very happy to answer for different scenarios/questions in turn.)

 

[Pinball1970]

Marbles. One red, and nine blue

Phyti lost his at some point so he may have trouble relating to that game.

 

[Sarkus]

Sarkus;
...
If a person outside the studio was offered to come inside and fill in for the player, they would be aware of the empty door. The substitute has no less knowledge of the car location than the player and can only make a random choice like the player. How or when the door was opened is irrelevant to the current session.

And therein lies your failure.
Scenario 1: person B enters the situation with no prior knowledge. They see 2 doors and know a car is behind one of them. The chance of picking correctly is 50:50.
Scenario 2: person B enters the situation and knows that the door in front of them was initially one of three. That in itself is different knowledge to scenario 1. They know that the door in front of them has 1/3 chance of being correct, irrespective of what else happens.
So, no, you are wrong. Your failure to understand that cripples you.

The player did not gain any constructive information that would provide any advantage. The player used their 1st choice of 1 of 3 closed doors. The host opening a goat door eliminated that door. In the world of reality, the distribution of prizes changes from 1 car and 2 goats to 1 car and 1 goat. I find it amazing that no one seems capable of comprehending the fact there are not 3 ways to choose 1 of 2 things. If you think there is, what's the 3rd way?

This, in a nutshell, is where you fall down: you see the number of different outcomes as determining the probability rather than, well, the maths.
If you have a pack of 52 cards with all of them blank but only one with an X, you have 2 outcomes of picking a card: a blank or an "X". But the frequency of the X being picked is not 50:50.

The rest of your efforts to explain yourself just repeat your same mistakes, and can therefore be ignored. You are fixating on possible outcomes and assuming they are all equally likely, rather than working through the logic to arrive at their probability. Your failure is repetitive, and it is beyond boring.

All I can suggest at this stage is that you go back to college. Go and talk to someone who can lead you through your wilderness face-to-face. Because here you are simply coming across as a crank who has no capability of recognising their failures in the face of logic, rationality, and, heck, maths.

 

[pzkpfw]

... If you have a pack of 52 cards with all of them blank but only one with an X, you have 2 outcomes of picking a card: a blank or an "X". But the frequency of the X being picked is not 50:50. ...

This stuff from phyti reminds me of actual posts I've seen where people try to "mathematically prove" that God exists.
Beginning with: "God exists or doesn't so we'll start at 50% ..." ! (Sigh)

 

[billvon]

Gave the answer earlier. Are you paying attention?
For A, P=1/52. For B, P=1/51.

Nope. If person A took 1 card, and person B took the other 51, then the odds that person A has that card is 1/52. The odds that person B has that card is 51/52.

The fact that you do not understand this explains why you're not getting anything else about the question.

 

[James R]

That's a different game.

It's funny you should say that, because what you just provided, in the process of stone-walling on the Monty Hall problem again, is a completely "different game" from the actual Monty Hall problem:

The probability tree

View attachment 7693
A session is a sequence of actions by the player and host in playing the MH game defined as: player 1st choice, host choice, player 2nd choice. All numbers within cells are door ID's. Prize distribution is (c, g, g).

Left, the tree based on the Selvin and Savant interpretation with the assumption of number of sessions equals number of doors, which contradicts the tree having 4 different outcomes. There are 4 sessions of player chooses a door, host opens a door, noted as (1,2), (1,3), (2,3), and (3,2). Their solution to get 4 different results from 3 doors was to play session 1 half the time with host opening door 2 and half the time with host opening door 3. That forms a bias with half as many car wins as goat wins. That allows the strategy of switching to beat the system. The probability is formed in reverse order, left to right.

Right, the method of forming probability values based on the frequency of occurrence of all possible sessions, right to left.
The tree structures are identical, only the frequencies differ.

On the left of your diagram is the problem as described - the one you keep ignoring. The analysis of that one - by Selvin, vos Savant and all of us here apart from you - is entirely correct, of course.

Presumably, you understand that the analysis is correct for the actual Monty Hall game. So now you're playing silly buggers, trying to change the game to one of your own invention.

In post #195, Sarkus explained you why your altered game doesn't remotely match the actual Monty Hall game.

For emphasis: in the Monty Hall game, as described, the first step is that the player chooses one of the three doors. On average across many plays of the game, we expect players to choose doors 1,2 or 3 with equal probability. One-third of the time, the player will choose door 1. One-third of the time, the player will choose door 2. One-third of the time, the player will choose door 3.

Obtusely, you want to set up a different game of your own invention - one in which players are somehow forced to choose door 1 twice as often as they choose door 2 and twice as often as they choose door 3. I don't know whether you've actually bothered to think about how you would propose to run your game to achieve that outcome. I don't think you actually care.

For a fair game with no deception and offering the same opportunity to all players, the game show producers must offer the sessions with equal frequencies on average.

No. The fair game is where, at the start of the game, the host says "pick a door - any door" and the player has a free choice to pick door 1,2 or 3, on a whim.

Your game somehow has to force the player to "choose" door 1 twice as often. One way that could be done would be to have the host force the player to swap from door 2 to door 1 in exactly half of the games in which the player indicated that he wanted door 2 - and similarly for where the player wants door 3. But this isn't "fair", by any means. In fact, this pseudo-choice you want to give the player reduces their chances of being able to win the car (using a sensible strategy) from 2/3 in the real Monty Hall game to 1/2 in your rigged game.

Will you now admit that vos Savant - not to mention all the rest of us in this thread - has the analysis of the actual Monty Hall game exactly right, while you have been continuously wrong about it for over a year?

Or are you going to continue to dodge and weave and play the fool?

 

[James R]

I don't have to know how far Alpha Centauri is, since I don't plan to go there anytime soon.

What happened to you, man? Did MAGA convince you to join in the general denial of reality or something?

Basically, here, you're saying that you don't trust the generations of competent astronomers who have determined that Alpha Centauri is about 4 light years from Earth.

What is it with you? Have you decided that you're not going to believe anything unless you feel competent to work all steps of the problem yourself?

Do you believe that you, personally, are the only expert who can be trusted to be accurate with any kind of knowledge?

You're going to be a very lonely old man if you refuse to trust anybody but yourself. You'll also remain ignorant about many things. The problem with your ignorance of the Monty Hall problem is only the tip of a very big iceberg. Once you stop listening when people try to teach you things, you're going to keep making the same mistakes over and over (unless you get lucky and somehow stumble across the correct information somehow).

 

[phyti]

pzkpfw;

A
P(R in bag 1)=1/2
P(R in bag 2)=1/2

B
There are 9 possible variations of removing R from bag 2.
RBBB...
BRBB...
BBRB...
Only one will become a reality, so P(removing R from bag 2)=1/9.
You can only do that once!
You are counting possible/potential outcomes as if they are certainties.
Probability is predicting the removal of a red ball from bag 2.

A1 1/2
A2 1/2
B1 1/10
B2 1/9

focus!

 

[phyti]

Sarkus;

'"That in itself is different knowledge to scenario 1. They know that the door in front of them has 1/3 chance of being correct, irrespective of what else happens."

It's different but not helpful. Both know initially there were 3 doors and now there are 2. The 1/3 for the car door WAS correct for a set of 3 doors, but the host opening a goat door shows it was wrong. There was no 1/3 car behind that door. Now you consult the history of choosing between 2 doors, or 2 of anything including a coin toss. You can't use the 1/3 NOW since there are not 3 doors.

"This, in a nutshell, is where you fall down: you see the nuber of different outcomes as determining the probability rather than, well, the maths. "

For basic probability expressed as a ratio from 0 to 1, that IS the math.
A typical definition: P = (# ways the desired outcome can occur)/(total # of outcomes)
Check the internet, there are many articles on basic probability.

 

[phyti]

James#209;

"Presumably, you understand that the analysis is correct for the actual Monty Hall game. So now you're playing silly buggers, trying to change the game to one of your own invention."

The latest table compares the fair game with the Savant/Selvin game.

"Your game somehow has to force the player to "choose" door 1 twice as often."

False. "each player begins with 3 doors." Their choices are random and independent of other players. The host is bound by the rules. They can only open goat doors.

"In fact, this pseudo-choice you want to give the player reduces their chances of being able to win the car (using a sensible strategy) from 2/3 in the real Monty Hall game to 1/2 in your rigged game."

For a game of chance involving random choices, there is no strategy. That IS the problem. The car door is offered twice for each round of 4 sessions. [Savant and Selvin, were asking the host to open the goat doors half as often when the player chose the car door compared to when the player chose the goat doors.] That is the bias that favors the player who switches, and violates FCC regulations.

"Will you now admit that vos Savant - not to mention all the rest of us in this thread - has the analysis of the actual Monty Hall game exactly right, while you have been continuously wrong about it for over a year?"

From Selvin's 2nd letter to 'The American Statistician'.

"[Monty Hall wrote and expressed that he was not "a student of statistics problems" but "the big hole in your argument is that once the first box is seen to be emprty, the contestant cannot exchange his box."]"

It was always a hypothetical version of a Monty Hall game.

 

[phyti]

James#210;

"Basically, here, you're saying that you don't trust the generations of competent astronomers who have determined that Alpha Centauri is about 4 light years from Earth."

What I'm saying is, that fact is not important in my life and many other lives. Most of the approx. 8B people are busy surviving. It is a fact for special interest groups, NASA, Spacex, etc.

"What is it with you? Have you decided that you're not going to believe anything unless you feel competent to work all steps of the problem yourself?"

That would be foolish, knowing we depend on other people to do things when we don't have time or ability to do them ourselves. I had cataract surgery a few years ago which restored my vision to normal. I benefited from the skill of the surgeon and the engineering of the equipment.

"Do you believe that you, personally, are the only expert who can be trusted to be accurate with any kind of knowledge?"

I only attempt to solve problems that are within my abilities. The case of a simple game show is a problem in logic only requiring counting. All the examples on the internet are just people repeating the Savant response, using the term "counterintuitive" for emphasis. None have shown any original analysis of the details in the form of fact checking. I analyzed the Savant/Selvin interpretation as one of the average people who the Savant followers claim are not capable of solving the problem. No one wants to admit Savant was wrong. If they do, then they are wrong. Every one wants to be a winner.

"Once you stop listening when people try to teach you things,"

You can't teach anyone unless you know the subject.

You are not doing well with your mental analysis.

 

[pzkpfw]

pzkpfw;

A
P(R in bag 1)=1/2
P(R in bag 2)=1/2

B
There are 9 possible variations of removing R from bag 2.
RBBB...
BRBB...
BBRB...
Only one will become a reality, so P(removing R from bag 2)=1/9.
You can only do that once!
You are counting possible/potential outcomes as if they are certainties.
Probability is predicting the removal of a red ball from bag 2.

A1 1/2
A2 1/2
B1 1/10
B2 1/9

focus!

Yes, focus.

You didn't answer the question. It was not about removing R from bag 2. The question was which bag is the red marble most likely in.
e.g. say you were offered the choice of which entire bag to take. Which bag would you select if you wanted the red marble?

Question B1: What are the odds that the red marble is in bag 1?
Question B2: What are the odds that the red marble is in bag 2?

Try again.

 

[Sarkus]

It's different but not helpful. Both know initially there were 3 doors and now there are 2. The 1/3 for the car door WAS correct for a set of 3 doors, but the host opening a goat door shows it was wrong. There was no 1/3 car behind that door. Now you consult the history of choosing between 2 doors, or 2 of anything including a coin toss. You can't use the 1/3 NOW since there are not 3 doors.

If it is an independent choice, there would be no knowledge of the 3rd door for the person who comes in off the street. For the choice to be 50/50 they would only see 2 doors, and would not know anything about the third, or how it came to be 2 doors. This is all information the original player has, but the person coming in off the street would not.
If the person coming in off the street does have all that information, then the choice to swap or not is NOT independent of the player's original choice. At least not if they're understanding probability correctly.

For basic probability expressed as a ratio from 0 to 1, that IS the math.
A typical definition: P = (# ways the desired outcome can occur)/(total # of outcomes)
Check the internet, there are many articles on basic probability.

This isn't "basic probability", though. This is "conditional probability" - and you should check the internet for correct explanations to the Monty Hall problem, because you're struggling with this problem. That's not the issue: many people struggle with it, because they can't grasp the same thing you can't. The issue is that you're not even open to the idea that you might be wrong.

Look, here's a site that explains it: https://betterexplained.com/articles/understanding-the-monty-hall-problem/?utm_source=chatgpt.com
It even has a game you can play (with instructions).

There are undoubtedly other sites, other explanations. I suggest you go away and fill your boots with everyone else trying to explain it to you, because you're using up (if you haven't already) all the goodwill this site has to offer you.

 

[James R]

This isn't "basic probability", though. This is "conditional probability"

Exactly.

At the point where the player is given the choice "stay or switch?", the relevant probability question is not this:

"What is the probability that the car is behind one of two indistinguishable doors?"

The relevant question is this:

"What is the probability that the car is behind the door that the player would be 'switching' to, given that the there were three indistinguishable doors originally and the player originally chose one door at random and then the host opened one of the doors and that the host has known all along which door hides the car and that the host was not allowed to open the one with the car?"

phyti:

The latest table compares the fair game with the Savant/Selvin game.

Your latest table is as wrong as all your previous wrong analyses.

You're literally not analysing the game we've been talking about all along, and you're pretending you don't know that, like some kind of clown.

"Your game somehow has to force the player to "choose" door 1 twice as often."

False.

True. I'm not going to explain why to you again. It's a waste of my time. I'm done feeding the troll.

For a game of chance involving random choices, there is no strategy.

"Stay or switch" is not a random choice in the Monty Hall game, since 'switch' gives the player 2/3 probability of a win, while 'stay' gives a 1/3 probability of a win. If it was random, it would be 1/2, either way.

Savant and Selvin, were asking the host to open the goat doors half as often when the player chose the car door compared to when the player chose the goat doors.

Wrong. The host opens a goat door in every play of the game.

From Selvin's 2nd letter to 'The American Statistician'.

"[Monty Hall wrote and expressed that he was not "a student of statistics problems" but "the big hole in your argument is that once the first box is seen to be emprty, the contestant cannot exchange his box."]"

It was always a hypothetical version of a Monty Hall game.

It's a little late in the proceedings for you to be pretending you don't actually know which version of the game has been the subject of all of the foregoing analysis in this thread and in the thread last year in which you similarly played the fool.

I only attempt to solve problems that are within my abilities.

Then you ought to give up on the Monty Hall problem. It is quite clearly beyond your capacities.

You can't teach anyone unless you know the subject.

You can't teach anybody unless they are open to being taught.

You are not doing well with your mental analysis.

I have more than adequately demonstrated that you're unequipped to comment on such things, when it comes to the topic of probability. You have a woefully poor grasp of the subject.

Perhaps it's more that the character you are playing - the clown - is incapable. But at this point, you have made yourself indistinguishable from the clown. If you've been playing dumb, well, congratulations, I guess. You've played your part so well that you've now convinced me that you're actually dumb.

 

[Sergej Materov]

The Monty Hall problem is a classic illustration of precisely the "flow" error associated (merely an analogy) with my proven no-go theorem (The Embedded- Observer No-Go Theorem). When the host opens a door, he performs an informative action: he knows where the prize is, and his choice of door is correlated with that knowledge. A player who argues "there are two doors left -it's 50/50" commits the exact error described in Lemma B: they treat their knowledge system as evolving autonomously, ignoring the fact that a real, non-ignorable interaction (the host's action) has occurred between their subsystem (their chosen door) and the rest of the system ("R" -the true location of the prize plus the host's knowledge). A correct Bayesian update involves precisely accounting for the fact that this connection was active (non-trivial/non-product) at the moment of the host's action. Ignoring it is equivalent to assuming u(Φ)=1 (channel unitarity) when, in reality, it is less than 1.

 

[Pinball1970]

The Monty Hall problem is a classic illustration of precisely the "flow" error associated with my proven no-go theorem (The Embedded- Observer No-Go Theorem). When the host opens a door, he performs an informative action: he knows where the prize is, and his choice of door is correlated with that knowledge. A player who argues "there are two doors left -it's 50/50" commits the exact error described in Lemma B: they treat their knowledge system as evolving autonomously, ignoring the fact that a real, non-ignorable interaction (the host's action) has occurred between their subsystem (their chosen door) and the rest of the system ("R" -the true location of the prize plus the host's knowledge). A correct Bayesian update involves precisely accounting for the fact that this connection was active (non-trivial/non-product) at the moment of the host's action. Ignoring it is equivalent to assuming u(Φ)=1 (channel unitarity) when, in reality, it is less than 1.

I was going to say that myself, kind of.

 

[phyti]

pzkpfw;

You didn't answer the question. It was not about removing R from bag 2. The question was which bag is the red marble most likely in.
e.g. say you were offered the choice of which entire bag to take. Which bag would you select if you wanted the red marble?

Question B1: What are the odds that the red marble is in bag 1?
Question B2: What are the odds that the red marble is in bag 2?

Then I misinterpreted your question.

You have randomly chosen 1 marble from the original set for bag 1, which is 1/10.

There is 1 possible withdrawal for a red marble from the original set.

There are 9 possible withdrawals for a blue marble from the original set.

The red marble would be left in the original set 9 of 10 withdrawals.

The P(red in bag 1)=1/10.

The P(red in bag 2)=9/10.

I would pick bag 2. Is that the result you are expecting?