High School The Monty Hall paradox/conundrum

[Dale]

With only 1 row per car, you don't need a probability column for this problem

Personally, I disagree with that approach for constructing these tables. I personally think that you should always include one or more probability columns. Even if every row is equal probability it is good to write that down explicitly and make sure that it is reasonable to assume that each is equal probability. Again, I do understand that it is a personal preference, but it never hurts to provide a probability column and in cases like this it is very helpful, so I think you should always do it.

 

[gmax137]

I think you should always do it

Do you mean, like in the next Monty Hall Paradox thread?

Actually that's not fair, this is the first MHP thread where I feel I have really got it straight in my head. So thanks to everyone!

 

[DaveC426913]

Personally, I disagree with that approach for constructing these tables. I personally think that you should always include one or more probability columns. Even if every row is equal probability it is good to write that down explicitly and make sure that it is reasonable to assume that each is equal probability. Again, I do understand that it is a personal preference, but it never hurts to provide a probability column and in cases like this it is very helpful, so I think you should always do it.

If the table shows results - i.e. after an (ideal) set of games are played - then probabilities of what might happen - are moot.

 

[DaveC426913]

Those more complicated tables can be examined for a characteristic repetition: when the car is behind the door initially chosen, they list once for each of the 2 doors that Monty is allowed by the rules to possibly open.

You are still looking at tables of what might happen. Why not look at tables of what did happen (in a set of ideal games)? Then the whole issue of probabilities just goes away.

 

[Dale]

If the table shows results - i.e. after an (ideal) set of games are played - then probabilities of what might happen - are moot.

Only if all results are equally likely. Unless you are talking about Monte Carlo simulations or an actual experiment where you might have thousands of lines in your table. And many will be the same.

 

[DaveC426913]

Only if all results are equally likely. Unless you are talking about Monte Carlo simulations or an actual experiment where you might have thousands of lines in your table. And many will be the same.

The point is to crate a table of ideal results.

For example:
A table of rolling 2 6-sided dice will have exactly 36 rows.
Every row has the same probability.

 

[Dale]

The point is to crate a table of ideal results.

I have no idea what you mean by “ideal” results.

Not all results have equal probability. Does a table of ideal results include duplicates of results with higher probabilities? If not, how does a table of ideal results indicate the results with higher or lower probability?

 

[lavinia]

I have a lot of sympathy for other posters who are sincerely and humbly confused and don't understand the resolution. I have very little sympathy for posters like the OP who come in telling everyone else that the well-known standard resolution is wrong and that everyone else has committed a fallacy to get the wrong answer. The harshness is probably not a reaction to the bad intuition but to the arrogance.

Some people aggressively challenge things as a way of learning - even if they are challenging well understood facts. I do not see this as arrogance.

 

[Dale]

Some people aggressively challenge things as a way of learning

Frankly, that is even worse than arrogance. An aggressive challenge, by nature of being aggressive, provokes defensive replies. It makes the learning environment antagonistic and adversarial instead of cooperative and constructive. Aggression provokes hostility and aggression by a student is stupidly provoking hostility against a superior opponent.

Don’t you think it is better to cultivate strong allies rather than strong enemies? Someone who foolishly chooses this as a learning strategy will surely learn little and will have earned all of the unkind responses from the people who would have otherwise taught them.

 

[mfb]

The point is to crate a table of ideal results.

Here is an extremely simple game: You lose with probability 1/pi, you win with probability 1-1/pi. There is no way to create a table of entries with equal probabilities that accurately reflects your chance to win.

 

[Dale]

Here is an extremely simple game: You lose with probability 1/pi, you win with probability 1-1/pi. There is no way to create a table of entries with equal probabilities that accurately reflects your chance to win.

Also, a Monte Carlo simulation or a real experiment will never accurately reflect that either. You could certainly construct a confidence interval that included $1/\pi$ but you would never get that exactly.

 

[DaveC426913]

I have no idea what you mean by “ideal” results.

Not all results have equal probability. Does a table of ideal results include duplicates of results with higher probabilities? If not, how does a table of ideal results indicate the results with higher or lower probability?

I gave you an example: a table of rolls of 2 6-sided dice.
An ideal results table will have every combination of rolls without duplicates:

D1​	D2​	Sum​
1​	1​	2​
1​	2​	3​
1​	3​	4​
1​	4​	5​
1​	5​	6​
1​	6​	7​
2​	1​	3​
2​	2​	4​
2​	3​	5​
2​	4​	6​
2​	5​	7​
2​	6​	8​
3​	1​	4​
3​	2​	5​
3​	3​	6​
3​	4​	7​
3​	5​	8​
3​	6​	9​
4​	1​	5​
4​	2​	6​
4​	3​	7​
4​	4​	8​
4​	5​	9​
4​	6​	10​
5​	1​	6​
5​	2​	7​
5​	3​	8​
5​	4​	9​
5​	5​	10​
5​	6​	11​
6​	1​	7​
6​	2​	8​
6​	3​	9​
6​	4​	10​
6​	5​	11​
6​	6​	12​

The case of MHP, we should be able to do the same.

 

[DaveC426913]

Here is an extremely simple game: You lose with probability 1/pi, you win with probability 1-1/pi. There is no way to create a table of entries with equal probabilities that accurately reflects your chance to win.

This is a good counter-example, because it definitely shows that you get what I'm striving for.

But it's hardly fair: it's using irrational numbers. In a game using only integers, I'm not sure that problem would arise.

You're sort of saying "In a game of musical chairs, with fractional chairs, no one can win". Well, sure - but the game isn't played with fractional chairs.

 

[Dale]

An ideal results table will have every combination of rolls without duplicates

In general not every combination of rolls will have the same probability. So your “ideal” results table in general should include a column on probability. The probability is not inherently encoded in an ideal results table.

 

[DaveC426913]

In general not every combination of rolls will have the same probability.

Yes they will. One in 36. I listed them in post 103.

I think what you mean is every sum will not have the same probability.

So your “ideal” results table in general should include a column on probability.

Every row has 1 in 36 chance. That's implicit.

 

[Ibix]

Yes it will. One in 36.

Not if the dice are biased.

 

[DaveC426913]

OK, this is the example I've been trying to get my head around that may blow my own idea out of the water.

A penny is tossed.
There are exactly three possibilities: heads tails, and edge.
That cannot be represented with an equal-probability-per-row table.

 

[DaveC426913]

Not if the dice are biased.

Red herring.

 

[lavinia]

Frankly, that is even worse than arrogance. An aggressive challenge, by nature of being aggressive, provokes defensive replies. It makes the learning environment antagonistic and adversarial instead of cooperative and constructive. Aggression provokes hostility and aggression by a student is stupidly provoking hostility against a superior opponent.

Don’t you think it is better to cultivate strong allies rather than strong enemies? Someone who foolishly chooses this as a learning strategy will surely learn little and will have earned all of the unkind responses from the people who would have otherwise taught them.

Provocation can be pedagogically productive. Cultivating allies or enemies is irrelevant to pedagogy. No one earns an unkind response in a learning situation. The only thing that matters is desire to understand. Bruised egos do not belong. The teacher is not superior to the student. Part of leaning is having the confidence and audacity to challenge. The teacher should welcome it.

 

[Ibix]

Red herring.

Why? That different outcomes need not be equiprobable is the general problem with your approach. You cannot, for example, analyse a variant of Monty Hall where the car is twice as likely to be behind door 1 than 2 or 3.

 

[Dale]

Yes they will. One in 36.

Not in general. That is only true for a pair of fair dice. Not all dice are fair. Also, not all games are dice or reducible to equivalent dice games. Many games of chance have very non-uniform probabilities. An ideal table, as you describe, would not be able to capture such games.

In the specific case of the Monte Python game the ideal table (one line for every result with no duplicates) will have an equal number of lines representing wins for switch as stick precisely as shown in the OP.

 

[Dale]

Red herring.

It is not a red herring. That is exactly the case for the Monty Hall problem. There are four possible outcomes, with unequal probabilities. Two are a win for stick (P=1/6) and two are a win for switch (P=1/3)

 

[Dale]

Cultivating allies or enemies is irrelevant to pedagogy

This is completely and demonstrably wrong. The social interaction between teacher and student and between students is very relevant to the effectiveness of teaching. I am actually surprised that is not obvious to you.

No one earns an unkind response in a learning situation

Any form of aggression earns an unkind response, especially in a learning situation where it is so egregiously inappropriate and unproductive for the goal of learning. It is also quite a double standard to demand that a teacher respond with kindness when a student is being aggressive.

Part of leaning is having the confidence and audacity to challenge.

Do you have any peer reviewed scientific study to back up this claim. I am certain that “having the confidence and audacity to challenge” is not necessary for learning, and I strongly doubt that it is beneficial to learning.

 

[sysprog]

OK, this is the example I've been trying to get my head around that may blow my own idea out of the water.

A penny is tossed.
There are exactly three possibilities: heads tails, and edge.
That cannot be represented with an equal-probability-per-row table.

If the coin lands on its edge, you just disregard that toss, and toss again.

Not if the dice are biased.

Red herring.

It is not a red herring. That is exactly the case for the Monty Hall problem. There are four possible outcomes, with unequal probabilities. Two are a win for stick (P=1/6) and two are a win for switch (P=1/3)

The case of a biased die is not parallel to or otherwise reasonably analogous to that of a twice-represented car location in the Monty Hall problem.

According to the game conditions as presented, there is no bias in the door selected, and no discernible bias in the door opening procedure.

The use of more than 1 row for the car location, when the car is behind the initially chosen door, reflects the obfuscatory nature of the door-opening part of the procedure; it doesn't reflect 2 different actual final outcome possibilities each with a probability of 1/6; those 2 'possibilities' are in fact 2 aliases for the same 1 possibility. At the end of the game, 3 doors are opened, and which 1 of the 3 doors concealed the car is revealed, and there are only 3 possible car location outcomes. Which door was opened before the second choice was offered is of no consequence concerning those 3 possible actually significantly different outcomes.

Regardless of which door is opened, the opening of a non-chosen door after the first choice is made and before the second choice is offered, collects the pair of 1/3-each probabilities for the 2 doors not selected into a single 2/3 probability for the remaining unopened not-selected door; when the originally chosen door conceals the car, Monty's selection of which non-car door to open doesn't split the 1 possibility for the originally chosen door into 2 possibilities of 1/6 probability each; there are only 3 possibilities for the car location, not 4 and not 6.

 

[DaveC426913]

If the coin lands on its edge, you just disregard that toss, and toss again.

No, I'm considering it as a valid outcome in a hypothetical game. "If it lands on its edge, you win a car!" or some such.

I'm simply pointing out what Dale was after - to prove (to myself) that fixed integer starting options can lead to outcomes that can't be represented as an equal-probability-per-row table.

 

[DaveC426913]

Not in general. That is only true for a pair of fair dice. Not all dice are fair.

Then you might as well include spurious winds and inclement weather in the table!

No, both the Monty Hall problem and dice-rolling are fair, have simple rules, everybody knows them, and there's no hidden machinations. Loaded dice are totally inappropriate as an analogy.

It's a red herring inasmuch as - while it may technically be true - it only obfuscates, rather than clarifies, the solution.

 

[mfb]

I'm simply pointing out what Dale was after - to prove (to myself) that fixed integer starting options can lead to outcomes that can't be represented as an equal-probability-per-row table.

That was true for my game already. Make the coin very thick and you can get a 1/pi chance that it lands on the edge.

Even in games where all outcomes have an integer fraction as probability it can be impossible to make such a table. Throw a die. 1,2,3 you win, 4,5 you lose, 6 roll again. The game finishes in finite time with probability 1 - but there is an infinite number of cases.

 

[Dale]

it doesn't reflect 2 different actual final outcome possibilities each with a probability of 1/6; those 2 'possibilities' are in fact 2 aliases for the same 1 possibility.

Goats are not electrons. They are distinguishable. The two different goats are two distinguishable outcomes and can reasonably be counted as separate rows in a table, just like different faces of a dice are put in separate rows even if for the purposes of a given game they are equivalent.

The point is that there is no single procedure or rule for writing tables that will have the following properties for all games: no lines are repeated, each line has equal probability. Therefore, you should augment the table with one or more columns for probability. That way you can choose the rules for writing your rows as you like and always come out with a correct analysis.

Without the probability column, no rule for writing the rows will give correct probabilities for all games. With the probability column, any rule for writing the rows will give correct probabilities for all games.

The problem in the OP is not that he wrote a table with incorrect rows, but that he assumed that the rows were equal probability.

 

[lavinia]

I understand that there are other threads on this, (e.g. https://www.physicsforums.com/threads/monty-hall-vs-monty-fall.661985/ which gives a thorough account) but they support the proposition that a 'swap' scenario results in a 2/3 win probability rather than the 'intuitive' 50:50 assumption.

I want to discuss the 50:50 conclusion, which I think is correct and the 2/3 is a fallacy which I will explain.

To summarise the paradox, if you are presented with 3 doors in a competition and there is a car behind one (that you want to win) and a goat behind each of the other two (the booby prize), you pick one door and the chance of you getting the door with the car behind it is obviously 1/3. The host opens one of the other doors to reveal a goat (they know which ones have goats) and asks 'do you want to swap your choice of door'? A cursory inspection of that moment suggests the chance of you now getting the car is 50:50, but a slightly deeper inspection suggests a 2/3 chance if you go with a 'swap' strategy.

One might argue that your chance of picking the door with the car was 1/3, so it therefore remains 1/3. Therefore, all other options (i.e. swapping from that door) must then be 2/3, so it is better to swap, if you want to win.

I have to disagree for the following observation (which I will then back up with a reason the 2/3 is a fallacy); let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

This makes absolutely no sense, and the fallacy exists here:- once the host opens one door to a goat, the 1/3 probability that you picked the car in the first 'round' then changes. It is no longer a 1/3 chance that you picked the car, it is now a 1/2 chance. It just is, it is clearly no longer the same probability as before. This is nothing less than Bayesian probability in which the confidence in a given observation changes according to prior observations. In this case, the later observation that there is a goat behind one door increases the probability of your original choice to 50:50. It does not remain the same. This is nothing less than an axiom of science; if you derive a hypothesis that has a very low probability (e.g. your prize car is behind the first door) you then test it, and by making observations that do not disprove that original hypothesis, then the probability of that hypothesis improves.

This is precisely the Monty Hall scenario: The probability your choice of door has a car behind it increases once you get a further non-contradictory observation.

OK, so that deals with the fallacy in words, but this doesn't un-stitch the description of events that lead to a 2/3 'count' of outcomes, if laid out sequentially in some structured matrix. So, where is the fallacy in that?

It is as follows; the 'misdirection' is the focus on the options for what happens when the game player picks a goat, so they could swap and win the car? Right? Well, yes, but what's missing is that there are TWO options that the host can follow if YOU pick the car correctly in the first place! TWO outcomes - they can either pick the 'leftmost' goat or the 'rightmost' goat. These are TWO options, not one, and this is where the 'mathematical' fallacy exists.

See, like this;

Door 1​	Door 2​	Door 3​	​	I pick​	Host Opens​	I stick​	I swap​
car​	goat​	goat​	​	1​	2​	win​	lose​
​	​	​	​	1​	3​	win​	lose​
​	​	​	​	2​	3​	lose​	win​
​	​	​	​	3​	2​	lose​	win​

(It doesn't matter what the actual combination is behind the doors, the same would be the case for each combination.)

The point is that if you DO pick the car, then the host has TWO options. One outcome they pick one goat, the other they pick the other one. In a typical explanation of this, this is simply put down as 'the host picks a goat', as if that was one outcome. It isn't, it is two outcomes that look like one. The host can ONLY pick one other option if I pick a goat, but he has TWO options if I pick the car.

There are 4 options for any given combination behind the doors. Two are 'wins' and two are 'loses', whether you fix to one strategy or the other.

It is a 50:50 chance once the host opens a goat door. The question is whether he has opened 'the other' goat door or 'one of' the goat doors. These things are not equal.

This closes the circle between the 'obvious' fact that any 'new' contestant who comes into the competition just as the host opens a goat door is, clearly and obviously, presented with a 50:50 chance, yet the maths didn't say this. The reason is the 'two choice' option the host had is 'hidden' within the definition of the question.

I would welcome a confirmation or rebuttal on this, I have been mulling this over for a couple of weeks and I simply could not reconcile the 'new contestant' scenario with the apparent numbers from the mathematical description. But once you realize the host is actually making one of two choices (in effect, they are pre-selecting two of the contestant's options and reducing them to one) then you realize it is a 50:50 outcome after all. At least, I think this is the case, because the alternative strays so far from intuition it begs us to look for the fallacy in the maths, and I believe this is it.

@cmb

There is a 2/3 chance that the car is in one of the other two doors. The problem is you don't know which one. But the MC always shows you the door that it isn't so 2/3 of the time it is in the second one.

On the other hand if the game started out with the MC choosing a door that had a goat, then choosing one of the other two would be 50/50. So what is the difference?

I found it instructive to make a spread sheet that simulates many repetitions of the game.

 

[sysprog]

No, I'm considering it as a valid outcome in a hypothetical game. "If it lands on its edge, you win a car!" or some such.

I'm simply pointing out what Dale was after - to prove (to myself) that fixed integer starting options can lead to outcomes that can't be represented as an equal-probability-per-row table.

Unlike the standard probability distribution for a coin toss, which is used as a model for a 50:50 chance, the probability of a coin landing on its edge is inestimable and extremely small. Your reply to @Dale in response to his saying that not all dice are fair, "Then you might as well include spurious winds and inclement weather in the table!", is equally applicable to your acceptance of an edge landing for a coin toss as a possibility that should be reflected in a table of coin toss outcomes.

A proper table for a single coin toss has 2 cells, one for heads and one for tails, unless there is some other outcome condition that pivotally matters.

Using 2 table rows of 1 cell each, 1 row for left-handed tosses that result in heads, and 1 for right-handed tosses that result in heads, with only 1 table row for a toss, with either hand, that results in tails, when the bet is won or lost depending only on whether the actual outcome is heads or tails, is unnecessarily informative, and could invite the mis-supposition that there are 3 possibilities, and therefore 1/3 chance for each.

I think that saying that there are 3 possibilities, 2 with 1/4 chance each, which should then be summed to 2/4, and 1 possibility that is 1/2, for the 2 'outcomes' that have heads in common, along with the 1 tails outcome, is not the best remedy for the potential of some persons to be misled in the matter, and that a better remedy would be ignoring the superfluous information of which hand was used to perform the toss.

I think that such a scenario of using and explaining 3 table rows for a 2-sided coin toss is functionally similar to, albeit somewhat simpler than, the using and explaining of 4 table rows for 3 car locations scenario that was presented at the inception of this thread.

Including repetitions of the same outcome on the grounds of there being or not being something irrelevant that took place in conjunction with the pivotal event, e.g. music was or wasn't playing during the event, is a different error from including a remote possibility that the conventional procedure as normally stated does not include as a possibility.

Let's please look further at the example of a table of outcomes for a throw of 2 dice -- a pair of two fair cubic dice is thrown -- no edge or vertex landing -- proper 1-face-up landings only, 36 possible pair outcomes, 11 possible sums -- some of the sums will have higher probability than others -- each cell on the outcome table represents a single ordered $\mathtt {(An, Bn)}$ or $\mathtt {(Bn, An)}$ pair:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline\ \ \ \ \ &\ \ \mathtt {A1}\ \ &\ \ \ \mathtt {A2}\ \ &\ \ \ \mathtt {A3}\ \ &\ \ \ \mathtt {A4}\ \ &\ \ \ \mathtt {A5}\ \ &\ \ \mathtt {A6}\ \ \\
\hline\ \ \ \mathtt {B1}\ \ &\ \ \mathtt 2\ \ &\ \ \mathtt 3\ &\ \ \mathtt 4\ &\ \ \mathtt 5\ &\ \ \mathtt 6\ &\ \ \mathtt 7\ \ \\
\hline\ \ \ \mathtt {B2}\ \ &\ \ \mathtt 3\ \ &\ \ \mathtt 4\ &\ \ \mathtt 5\ &\ \ \mathtt 6\ &\ \ \mathtt 7\ &\ \ \mathtt 8\ \ \\
\hline\ \ \ \mathtt {B3}\ \ &\ \ \mathtt 4\ \ &\ \ \mathtt 5\ &\ \ \mathtt 6\ &\ \ \mathtt 7\ &\ \ \mathtt 8\ &\ \ \mathtt 9\ \ \\
\hline\ \ \ \mathtt {B4}\ \ &\ \ \mathtt 5\ \ &\ \ \mathtt 6\ &\ \ \mathtt 7\ &\ \ \mathtt 8\ &\ \ \mathtt 9\ &\ \ \mathtt {10}\ \ \\
\hline\ \ \ \mathtt {B5}\ \ &\ \ \mathtt 6\ \ &\ \ \mathtt 7\ &\ \ \mathtt 8\ &\ \ \mathtt 9\ &\ \ \mathtt {10}\ &\ \ \mathtt {11}\ \ \\
\hline\ \ \ \mathtt {B6}\ \ &\ \ \mathtt 7\ \ &\ \ \mathtt 8\ &\ \ \mathtt 9\ &\ \ \mathtt {10}\ &\ \ \mathtt {11}\ &\ \ \mathtt {12}\ \ \\
\hline
\end{array}
The table shows one cell per ordered pair outcome. When each of the 2 dice shows the same number there is only 1 cell for that pair, whereas when the 2 numbers are different, $\mathtt {(An, Bn)}$ is a different outcome from $\mathtt {(Bn, An)}$, just as on a planar $(x,y)$ graph, when $x=y$, then only the same point is defined as when $y=x$, not 2 distinct points, as when $x \neq y$. The cells contain the sums of the values on the 2 dice.

The number of possibilities for each die is 6, and the number of possible outcomes for the 2 dice is the number of possibilities per die times the number of dice: 6(6)=36. The chance for each outcome is therefore 1/36, but that's not the total chance for any 1 outcome sum; the number of cells for each possible sum is not the same as the number of cells for each other possible sum.

The point is that there is no single procedure or rule for writing tables that will have the following properties for all games: no lines are repeated, each line has equal probability. Therefore, you should augment the table with one or more columns for probability. That way you can choose the rules for writing your rows as you like and always come out with a correct analysis.

Without the probability column, no rule for writing the rows will give correct probabilities for all games. With the probability column, any rule for writing the rows will give correct probabilities for all games.

If the purpose of constructing the table is to learn the total per-throw probability for each of the sums, then it doesn't make sense that the table should have to contain in advance an explicit reference to the probability for each sum.

Instead, we can use the table to help us to sum the number of possibilities for each possible sum, 2-12 inclusive, and thereby determine the total probability per throw for each possible sum.

It can be seen at once that the sums in the cells along each of the $\mathtt {(An \to Bn)}$ diagonals, i.e., those diagonals such that $\mathtt {(A1 \to B1 \dots A6 \to B6)}$, are the same, and that no equal sums appear elsewhere on the table. That makes the following table, of total probabilities per throw for each sum easy to construct; the number of instances of each sum is the number of instances of it in its diagonal, i.e., the total number of cells in the $\mathtt {An \to Bn}$ diagonal in which exclusively that sum exclusively appears:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline \mathtt {sum~of~dice} & \mathtt {2} & \mathtt {3} & \mathtt {4} & \mathtt {5} & \mathtt {6} & \mathtt {7} & \mathtt {8} & \mathtt {9} & \mathtt {10} & \mathtt {11} & \mathtt {12}\\
\hline \mathtt {num~of~cells} & \mathtt {1} & \mathtt {2} & \mathtt {3} & \mathtt {4} & \mathtt {5} & \mathtt {6} & \mathtt {5} & \mathtt {4} & \mathtt {3} & \mathtt {2} & \mathtt {1}\\
\hline \mathtt {num/36 (dec)} & \mathtt {.02 \overline 7} & \mathtt {.0 \overline 5} & \mathtt {.08 \overline 3} & \mathtt {. \overline 1} & \mathtt {.13 \overline 8} & \mathtt {.1 \overline 6} & \mathtt {.13 \overline 8} & \mathtt {. \overline 1} & \mathtt {.08 \overline 3} & \mathtt {.0 \overline 5} & \mathtt {.02 \overline 7}\\
\hline
\end{array}
The tables above, and the table that you (@DaveC426913) posted in post #103, do not over-represent any of the possible actual outcomes, and in not so over-representing, they do not thereby omit any actual outcomes.

Goats are not electrons. They are distinguishable. The two different goats are two distinguishable outcomes and can reasonably be counted as separate rows in a table, just like different faces of a dice are put in separate rows even if for the purposes of a given game they are equivalent.

They're not final outcomes, and they don't affect any of the car location outcomes, including the final outcome that they share in representing.

In the 4-line table presented by @cmb in post #1, the 2 'outcomes' that are acting as co-aliases for 1 car location outcome, the 1 such that the car is behind the door originally chosen, are listed not as sub-outcomes for 1 car location outcome, but instead are each presented as outcomes at the same level as the 2 other car location outcomes, and consequently they make the 4-line table misleading.

I think that using a second row for that single car location makes the table less readily useful for the above-illustrated purpose of enumerating the relevant probabilities by correctly counting all and only the 3 actually different car location outcomes.

It's true that there are 2 possible pre-revealed-location-sub-outcomes for that 1 behind-the-originally-chosen-door 1/3 probable car location outcome, but the difference between those 2 non-car-location-sub-outcomes in no way affects the car location outcome, so giving procedural cognizance to them, in my view, is at best embracement of a distraction.

I think that even listing them as 2 sub-outcomes with 1/2 each of the 1/3 probability for the 1 possible car location that they share in representing, prescribes too much attention being given to which door is opened when the car is behind the door originally chosen.