B The Monty Hall paradox/conundrum

[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.

 

[cmb]

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.

Thank you for your supportive position, and later comment. Yes, this is how I see my style as a constructive contribution. If everyone is always nice about ideas, then they don't really get tested. I put forward something, it was tested, and has (per other threads here on the same topic) thoroughly reinforced the conclusion, having been shown to be wrong. I think that's OK, it's what science should do, for fear of ending up with 'group-think' and cognitive biases.

Personally, I think part of the issue of (particularly my style, but) questioning something head-on by putting forward the counter-proposition is that people get confused about why they think a proposition is 'aggressive'. I mean, yeah, I may well get 'aggressive' when I am trading one proposition with another, I want those propositions to smash into each other like head-butting goats! But what's fascinating at the level of human interaction is that people feel that 'you' are being aggressive to 'them' if you put their ideas into a battle arena with their ideas. I mean, in what possible way is it logical to ascribe such an anthropomorphism to 'an idea' when one is attacking it?

People are often very precious of their ideas and thoughts, and that's OK but it is ludicrous when people feel that means 'they' are being 'attacked'.

My OP 'attacked' no-one, and it is not possible to be 'aggressive' to an idea, it is a silly anthropomorphism.

On your subsequent post, thanks for progressing the rationale, but I am totally comfortable with the situation now and am happy to leave my OP and declare it misinformed and wrong, having 'seen the light' (or 'seen the goat', as the case may be! ;) ).

 

[DaveC426913]

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.

I saw a distinction between the loaded dice and the penny toss in that loaded dice are usually not known to all parties, thus it is typically a "cheat". The penny toss game, in my mind, had the rules explained to all (including the edge as a valid outcome) and thus is fair.

But it's reaching a state of mootity, since they made be made equivalent with just a few implicit conditions being made explicit (eg: "BTW players - these dice are loaded!").

I see now that, with whatever examples one wants, it can be shown that some problems don't have ideal same-probability-per-row table solutions.

 

[Dale]

My OP 'attacked' no-one, and it is not possible to be 'aggressive' to an idea, it is a silly anthropomorphism.

To be clear, I also did not view your OP as aggressive. That was @lavinia’s word. I don’t think that your OP was aggressive, but I do think that aggressive speech of any sort is inappropriate, especially in a classroom.

I also don’t believe that any of the replies that you received were particularly harsh. Your question was addressed factually and professionally. This meta-discussion is more general than directly about your OP or the subsequent responses, at least for my part.

 

[Dale]

when the 2 numbers are different, (An,Bn) is a different outcome from (Bn,An)

Why is it right to consider the dice distinguishable but not to consider the goats distinguishable? You are using different rules for constructing the different tables.

With the dice you have to consider them distinguishable to get an equal probability table, and with the goats you have to consider them indistinguishable to get an equal probability table. So just to construct the table you already need to consider probabilities.

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.

But as I just pointed out a table inherently cannot do that without external information regarding probabilities. You have to know a priori what events have equal probability so that you can put those events on individual lines in the table.

In one case you treat dice as distinguishable and in another you treat goats as indistinguishable. Why? Because if you don’t then the probabilities on each line are different. So you already need to have the basis of the information that you are purporting to obtain from the table before you can even construct the table.

So since you must have probability information to begin with you may as well include it in the table. That allows you to set the rows as you like without being forced into an ad-hoc structure and it also permits you to analyze scenarios that simply cannot be analyzed as an equal probability table. So you are not putting in any information that is not already required for an “ideal” table, but putting it in explicitly allows you to do more and do it more easily and accurately.

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

The same can be said about the two dice. The distinction between (An,Bn) and (Bn,An) in no way affects the summation outcome. In one case embracing the distraction leads to an equal-probability table and in another case embracing the same distraction leads to an unequal probability table. It is not possible to determine if a particular table is a “embrace the distraction” table or not other than by examining the probabilities.

 

[pbuk]

Do we still want a table for Monty Hall? I think it looks best with two tables, one for each strategy for the contestant. I refer to the doors by what is behind them, any other labelling is redundant.

Stick strategy:

I pick​	Probability​	Host reveals​	I stick with​	Probability of win​
G1​	1/3​	G2​	G1​	0​
G2​	1/3​	G1​	G2​	0​
Car​	1/3​	G1 or G2​	Car​	1/3​
Total stick​	1​			1/3​

Swap strategy

I pick​	Probability​	Host reveals​	I swap to​	Probability of win​
G1​	1/3​	G2​	Car​	1/3​
G2​	1/3​	G1​	Car​	1/3​
Car​	1/3​	G1 or G2​	G2 or G1​	0​
Total swap​	1​			2/3​

The swap strategy is therefore twice as good as the stick strategy.

 

[lavinia]

Thank you for your supportive position, and later comment. Yes, this is how I see my style as a constructive contribution. If everyone is always nice about ideas, then they don't really get tested. I put forward something, it was tested, and has (per other threads here on the same topic) thoroughly reinforced the conclusion, having been shown to be wrong. I think that's OK, it's what science should do, for fear of ending up with 'group-think' and cognitive biases.

Personally, I think part of the issue of (particularly my style, but) questioning something head-on by putting forward the counter-proposition is that people get confused about why they think a proposition is 'aggressive'. I mean, yeah, I may well get 'aggressive' when I am trading one proposition with another, I want those propositions to smash into each other like head-butting goats! But what's fascinating at the level of human interaction is that people feel that 'you' are being aggressive to 'them' if you put their ideas into a battle arena with their ideas. I mean, in what possible way is it logical to ascribe such an anthropomorphism to 'an idea' when one is attacking it?

People are often very precious of their ideas and thoughts, and that's OK but it is ludicrous when people feel that means 'they' are being 'attacked'.

My OP 'attacked' no-one, and it is not possible to be 'aggressive' to an idea, it is a silly anthropomorphism.

On your subsequent post, thanks for progressing the rationale, but I am totally comfortable with the situation now and am happy to leave my OP and declare it misinformed and wrong, having 'seen the light' (or 'seen the goat', as the case may be! ;) ).

You are welcome. And thank you for your thoughtful post.

 

[sysprog]

Why is it right to consider the dice distinguishable but not to consider the goats distinguishable? You are using different rules for constructing the different tables.

The 2 door-opening possibilities don't have any effect on the door-originally-chosen possibilities, the car location possibilities, or the stick or switch possibilities. Not only do I not distinguish them; I don't enter them in the table at all. The fact that a door is eliminated has an effect, but other than the constraints that it has to be a not-chosen non-car-concealing door, it doesn't matter which door.

In one case you treat dice as distinguishable and in another you treat goats as indistinguishable. Why? Because if you don’t then the probabilities on each line are different. So you already need to have the basis of the information that you are purporting to obtain from the table before you can even construct the table.

I have to know that there are 6 equally likely outcomes for a throw of a single die, and that for a throw of a pair if dice, there are the 6 possibilities for the 1 die for each of the 6 possibilities for the other, but I don't have to know in advance how many of those pairs of possibilities sum to each of the possible sums, and that can be shown by constructing the table.

With the dice you have to consider them distinguishable to get an equal probability table,

There's no legitimate reason to not consider them to be distinct, except when they show the same number, and if I don't, my 6×66×6 table will have 15 empty cells.

So since you must have probability information to begin with you may as well include it in the table.

This doesn't distinguish between input possibilities and outcome probabilities. With the dice throw, we use the 6 equiprobable input possibilities for 1 die to determine the columns, and the 6 equiprobable input possibilities for the other die to determine the rows. In the Monty Hall game, we use the 3 equiprobable possibities for the door choice to determine the columns, and the 3 equiprobable possibilities for the car location to determine the rows.

Also in the Monty Hall game, we count the original outputs to get the original probabilities and the probabilities for sticking, which are (obviously) the same, and their inverse for the probabilities for switching.

In both the dice throw and the Monty Hall game tables the probabilities are established by counting the outputs of each kind; although they could be entered in another table as inputs after they are determined, they are not determined in advance and then entered in a separate row or column in a table that is being constructed and filled in for the purpose of establishing what the probabilities are by counting the outcomes of each kind.

On looking at the blank table below, after the rows and columns are established but before any content cells are filled in, it's obvious that the (An, Bn)(An, Bn) cells are not the same cells as the (Bn, An)(Bn, An) cells, except when the AnAn column number is the same as the BnBn row number:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline \phantom {\mathtt{A1}}\ \ &\ \ \mathtt {A1}\ \ &\ \ \mathtt {A2}\ \ &\ \ \mathtt {A3}\ \ &\ \ \mathtt {A4}\ \ &\ \ \mathtt {A5}\ \ &\ \ \mathtt {A6}\ \ \\
\hline\ \ \ \mathtt {B1}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B2}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B3}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B4}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B5}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B6}\ \ &\ \ \phantom {\mathtt{(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline
\end{array}

The distinction between (An,Bn) and (Bn,An) in no way affects the summation outcome.

That's of course true for each of the sums individually, but the distinction does affect how many instances of each sum are represented on the table, except when each of the 2 dice shows the same number, and that's what we need to know in order to establish the probability for each sum. We know that there are 6 possible outcomes for each die, so we construct a 6×66×6 table.

In the following 3 tables, I've entered input pair identifiers (corresponding to how many pips are on the up face of each die) rather than the sums of the 2 dice. The 1st 2 tables are incorrect, and the 3rd is correct.

The 1st table shows the result of incorrectly deciding that when a non-equally-numbered (An, Bn)(An, Bn) pair is entered in a cell, the corresponding (Bn, An)(Bn, An) pair represents the same possible outcome, so it should not be entered in its cell.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline\ \ \phantom {\mathtt {A1}}\ \ &\ \mathtt{A1}\ \ &\ \ \mathtt {A2}\ \ &\ \ \mathtt {A3}\ \ &\ \ \mathtt {A4}\ \ &\ \ \mathtt {A5}\ \ &\ \ \mathtt {A6}\ \ \\
\hline\ \ \ \mathtt {B1}\ \ &\ \ \mathtt {(1,~1)}\ \ &\ \ \mathtt {(2,~1)}\ \ &\ \ \mathtt {(3,~1)}\ \ &\ \ \mathtt {(4,~1)}\ \ &\ \ \mathtt {(5,~1)}\ \ &\ \ \mathtt {(6,~1)}\ \ \\
\hline\ \ \ \mathtt {B2}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \mathtt {(2,~2)}\ \ &\ \ \mathtt {(3,~2)}\ \ &\ \ \mathtt {(4,~2)}\ \ &\ \ \mathtt {(5,~2)}\ \ &\ \ \mathtt {(6,~2)}\ \ \\
\hline\ \ \ \mathtt {B3}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \mathtt {(3,~3)}\ \ &\ \ \mathtt {(4,~3)}\ \ &\ \ \mathtt {(5,~3)}\ \ &\ \ \mathtt {(6,~3)}\ \ \\
\hline\ \ \ \mathtt {B4}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \mathtt {(4,~4)}\ \ &\ \ \mathtt {(5,~4)}\ \ &\ \ \mathtt {(6,~4)}\ \ \\
\hline\ \ \ \mathtt {B5}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \mathtt {(5,~5)}\ \ &\ \ \mathtt {(6,~5)}\ \ \\\hline
\hline\ \ \ \mathtt {B6}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \mathtt {(6,~6)}\ \ \\
\hline
\end{array}
Similarly, the 2nd table incorrectly eliminates the non-equally-numbered (An, Bn)(An, Bn) possibilities:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline\ \ \phantom {A1} \ \ &\ \ \mathtt {A1}\ \ &\ \ \ \mathtt {A2}\ \ &\ \ \ \mathtt {A3}\ \ &\ \ \ \mathtt {A4}\ \ &\ \ \ \mathtt {A5}\ \ &\ \ \mathtt {A6}\ \ \\
\hline\ \ \ \mathtt {B1}\ \ &\ \ \mathtt {(1,~1)}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \ \\
\hline\ \ \ \mathtt {B2}\ \ &\ \ \mathtt {(1,~2)}\ \ &\ \ \mathtt {(2,~2)}\ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B3}\ \ &\ \ \mathtt {(1,~3)}\ \ &\ \ \mathtt {(2,~3)}\ &\ \ \mathtt {(3,~3)}\ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B4}\ \ &\ \ \mathtt {(1,~4)}\ \ &\ \ \mathtt {(2,~4)}\ &\ \ \mathtt {(3,~4)}\ &\ \ \mathtt {(4,~4)}\ &\ \ \phantom {\mathtt {(0,~0)}}\ \ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B5}\ \ &\ \ \mathtt {(1,~5)}\ \ &\ \ \mathtt {(2,~5)}\ &\ \ \mathtt {(3,~5)}\ &\ \ \mathtt {(4,~5)}\ &\ \ \mathtt {(5,~5)}\ &\ \ \phantom {\mathtt {(0,~0)}}\ \ \\
\hline\ \ \ \mathtt {B6}\ \ &\ \ \mathtt {(1,~6)}\ \ &\ \ \mathtt {(2,~6)}\ &\ \ \mathtt {(3,~6)}\ &\ \ \mathtt {(4,~6)}\ &\ \ \mathtt {(5,~6)}\ &\ \ \mathtt {(6,~6)}\ \ \\
\hline
\end{array}
The 2 incorrect tables show that if the dice are not distinct, then there are only 36−15=2136−15=21 possibilities, which is inconsistent with there being 6 rows×6 columns⇒6×6=366 rows×6 columns⇒6×6=36 possibilities.

The 3rd table correctly shows the 6×66×6 possibilities:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline\ \ \phantom {A1} \ \ &\ \ \mathtt {A1}\ \ &\ \ \ \mathtt {A2}\ \ &\ \ \ \mathtt {A3}\ \ &\ \ \ \mathtt {A4}\ \ &\ \ \ \mathtt {A5}\ \ &\ \ \mathtt {A6}\ \ \\
\hline\ \ \ \mathtt {B1}\ \ &\ \ \mathtt {(1,~1)}\ \ &\ \ \mathtt {(2,~1)}\ \ &\ \ \mathtt {(3,~1)}\ \ &\ \ \mathtt {(4,~1)}\ \ &\ \ \mathtt {(5,~1)}\ \ &\ \ \mathtt {(6,~1)}\ \ \\
\hline\ \ \ \mathtt {B2}\ \ &\ \ \mathtt {(1,~2)}\ \ &\ \ \mathtt {(2,~2)}\ \ &\ \ \mathtt {(3,~2)}\ \ &\ \ \mathtt {(4,~2)}\ \ &\ \ \mathtt {(5,~2)}\ \ &\ \ \mathtt {(6,~2)}\ \ \\
\hline\ \ \ \mathtt {B3}\ \ &\ \ \mathtt {(1,~3)}\ \ &\ \ \mathtt {(2,~3)}\ \ &\ \ \mathtt {(3,~3)}\ \ &\ \ \mathtt {(4,~3)}\ \ &\ \ \mathtt {(5,~3)}\ \ &\ \ \mathtt {(6,~3)}\ \ \\
\hline\ \ \ \mathtt {B4}\ \ &\ \ \mathtt {(1,~4)}\ \ &\ \ \mathtt {(2,~4)}\ \ &\ \ \mathtt {(3,~4)}\ \ &\ \ \mathtt {(4,~4)} \ \ &\ \ \mathtt {(5,~4)} \ \ &\ \ \mathtt {(6,~4)}\ \ \\
\hline\ \ \ \mathtt {B5}\ \ &\ \ \mathtt {(1,~5)}\ \ &\ \ \mathtt {(2,~5)}\ \ &\ \ \mathtt {(3,~5)}\ \ &\ \ \mathtt {(4,~5)}\ \ &\ \ \mathtt {(5,~5)}\ \ &\ \ \mathtt {(6,~5)}\ \ \\
\hline\ \ \ \mathtt {B6}\ \ &\ \ \mathtt {(1,~6)}\ \ &\ \ \mathtt {(2,~6)}\ \ &\ \ \mathtt {(3,~6)}\ \ &\ \ \mathtt {(4,~6)}\ \ &\ \ \mathtt {(5,~6)}\ \ &\ \ \mathtt {(6,~6)}\ \ \\
\hline
\end{array}
That each of the cells is equally probable is a condition of fair dice thrown fairly. I didn't arbitrarily ex ante assign them a 1/36 probability each. What determines their 1/36 each probability is that there are 6 equiprobable faces of 1 die, and 6 equiprobable faces of another die, and the result of throwing 1 of them is independent of the result of throwing the other, so there are 6 equiprobable results for 1 die, for each of 6 equiprobable results for the other die, which means that there are 6×6=366×6=36 equally likely possibilities for a throw of the pair. That there are 6 equiprobable faces on each of the 2 dice and that their throw outcomes are independent of each other is necessary for determining that the table will be a 6×66×6 table, but it doesn't pre-establish the content of the cells.

Filling in the table with a sum for each cell requires only adding the row number to the column number:

\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}
That's the same table as the 1st 1 that I presented in post #120. In that post I also made the following remark about why (An, Bn)(An, Bn) is distinct from (Bn, An)(Bn, An)

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, (An, Bn)(An, Bn) is a different outcome from (Bn, An)(Bn, An), just as on a planar (x, y)(x, y) graph, when x=yx=y, then only the same point is defined as when y=xy=x, not 2 distinct points, as when x≠yx≠y.

At no time in the constructing or filling-out of the table for the dice throw was a probability value entered in a label or a cell.

From examining that table we can verify that the probabilities are not the same for all the sums. For example, there is only 1 instance of 2, and only 1 instance of 12, but 6 instances of 7, so we now construct a table that shows the number of instances for each sum, and from that, we can then emplace a 3rd row that shows the per-sum outcome probabilities, which result from dividing each number of sum-instances by the total number of possibilities, which is 36.

Also from post #120:

It can be seen at once that the sums in the cells along each of the (An→Bn)(An→Bn) diagonals, i.e., those diagonals such that (A1→B1…A6→B6)(A1→B1…A6→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 An→BnAn→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 probabilities listed in the last row are not used ex ante, and are not in the first 2 tables, by which they are established.

Regarding the distinguishability of the goats in the Monty Hall problem, they are both non-cars, and we know Monty is going to show a non-car, regardless of whether there are 2 available to him to show, or only 1, so there is no reason to be concerned about which non-car is shown; the possible outcomes that matter before a door is opened, and the differences between them and their inverses that matter before a choice is made whether to stick or switch, are only the ones that vary depending on either the car location or which door is chosen, or on which choice is made. That Monty opens 1 of the 2 non-chosen doors matters, but given the game rules requiring that he not open the originally chosen door or the door that conceals the car, which door he opens does not matter, and we don't have to know the chances in advance to establish them from the tables.

The opening of a door in accordance with the rules, reduces from 2 to 1 the number of members of the 2/3 probability original 2-door non-chosen subset of the 3 doors. Before Monty opens a door, the 2/3 probability of that 2-door subset was distributed over 2 doors; opening 1 of the doors collects the 2/3 probability for those 2 doors into the remaining door. Whether the opened door was the only 1 available or not, and if not, which 1 Monty randomly chose, does not split 1 of the final outcomes for the car location. The car stays behind 1 door from start to finish.

The following tables and accompanying commentary suffice to establish the probabilities for sticking and for switching, without any table or its construction including or depending on which door Monty opens.

There are 2 original input variables that matter:
1. which door conceals the car (3 possibilities)
2. which door is initially chosen (3 possibilities)

So a table with 3 rows for the possible door locations, and 3 columns for the door chosen by the contestant is constructed:

\begin{array}{|c|c|c|c|}
\hline \mathtt {~car~is\\behind} & \mathtt {chooses\\~~door~1} & \mathtt {chooses\\~~door~2} & \mathtt {chooses\\~~door~3} \\
\hline \mathtt 1 & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} \\
\hline \mathtt 2 & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} \\
\hline \mathtt 3 & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} \\
\hline
\end{array}
If the game stops here, then if the 'car is behind' numbered row matches the 'chooses door' numbered column, the contestant wins the car. Otherwise he gets a non-car. He has a 1/3 chance of winning the car, and a 2/3 chance of not winning the car.

Then a door is opened and the contestant is offered the option to either stick with the door he originally chose, or switch to the remaining unopened not-chosen door.

That 2-way stick-or-switch decision leads to 2 sets of outcome possibilities, so that there are now 18 possible outcomes: 9 for sticking, 9 for switching, so now we can use 2 9-cell tables; 1 for sticking and 1 for switching.

If the contestant sticks, his chances are (obviously) the same as they were at the outset, so all that is needed to construct the possibility table for sticking is to copy the at-the-outset table and relabel the columns:

\begin{array}{|c|c|c|c|}
\hline \mathtt {~car~is\\behind} & \mathtt {sticks~to\\~~~door~1} & \mathtt {sticks~to\\~~~door~2} & \mathtt {sticks~to\\~~~door~3} \\
\hline \mathtt 1 & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} \\
\hline \mathtt 2 & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} \\
\hline \mathtt 3 & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} \\
\hline
\end{array}
If the contestant opts to switch, the table cells will no longer contain the same 9 possibilities. The contestant will no longer win the car if he originally picked the car-concealing door, but those 3 cases are the only ones in which he will not win; in the other 2 cases he will win the car.

We know that one of the doors cannot be switched to, because Monty will open it before the contestant is given the option to switch, and obviously we can't switch to the door originally selected, because that wouldn't be switching.

We don't know which door Monty will open, but we do know that whichever door he opens it will not be the one originally chosen, and it will not be the one that conceals the car.

If the contestant switches, of the 3 original possibilities, the only possibilities that are consistent with those rules, are the 1 possibility such that the contestant originally chose the door that concealed the car, or the 2 possibilities such that the car is behind whichever of the 2 doors it is that remains available for the contestant to switch to it.

When sticking would result in the car, switching results in a non-car, and when sticking would result in a non-car, switching results in the car. The result of sticking is never the same as the result of sticking. Each cell is switched for a cell of the other kind.

The door number column headings have the same door numbers as before, but are changed to indicate that the column now shows the 3 possible results for switching from the door from which the contestant switched:

\begin{array}{|c|c|c|c|}
\hline \mathtt {~car~is\\behind} & \mathtt {switches\\~~from~1} & \mathtt {switches\\~~from~2} & \mathtt {switches\\~~from~3} \\
\hline \mathtt 1 & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} \\
\hline \mathtt 2 & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} \\
\hline \mathtt 3 & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} \\
\hline
\end{array}
There is 1 possible door that the contestant could switch from and not win the car, and there are 2 possible doors that he could switch from and win the car. There is 1 possible car location that the contestant could switch from and not win the car, and there are 2 possible car locations that he could switch to and win the car.

We entered the cell contents by listing the 3 possibilities for each of the 2 inputs. As with the game conditions at the outset, to determine how many rows and columns are required, we needed to know only the number of inputs and the number of possibilities for each input, and for the cell contents, we needed further to know only how the game rules constrained the inputs.

 

[Dale]

That's of course true for each of the sums individually, but the distinction does affect how many instances of each sum are represented on the table,

This is my whole point. You must know (from information obtained outside of the table) that this would result in unequal probabilities in order to construct the table. The probability information fundamentally does not come from the table. The probabilities are an input to constructing the table, not an output.

At no time in the constructing or filling-out of the table for the dice throw was a probability value entered in a label or a cell.

Obviously not, it is this omission that I am recommending against.

 

[sysprog]

Why is it right to consider the dice distinguishable but not to consider the goats distinguishable? You are using different rules for constructing the different tables.

That's quoted and fully answered in my post. Here's the brief immediate response:

The 2 door-opening possibilities don't have any effect on the door-originally-chosen possibilities, the car location possibilities, or the stick or switch possibilities. Not only do I not distinguish them; I don't enter them in the table at all. The fact that a door is eliminated has an effect, but other than the constraints that it has to be a not-chosen non-car-concealing door, it doesn't matter which door.

The rest of the post includes greater detail on that concern.

This is my whole point. You must know (from information obtained outside of the table) that this would result in unequal probabilities in order to construct the table. The probability information fundamentally does not come from the table. The probabilities are an input to constructing the table, not an output.

The sample spaces are inputs. The outcomes permute the inputs. The probabilities result from counting the outcomes of each type.

 

[sysprog]

This is my whole point. You must know (from information obtained outside of the table) that this would result in unequal probabilities in order to construct the table. The probability information fundamentally does not come from the table. The probabilities are an input to constructing the table, not an output.

The number of instances of each sum number is not an input to the table; it's the result of counting the instances in the table.

Obviously not, it is this omission that I am recommending against.

I was showing that it wasn't necessary to use the outcome probabilities ex ante, pursuant to supporting my contention that it doesn't make sense to prescribe entering the outcome probabilities in a table you intend to use to establish the probabilities; you only need to enter the outcomes for each input pair into the table, and you can then determine the probabilities for each type of output by counting the number of each type, and then dividing by the total number of possibilities.

 

[Dale]

The sample spaces are inputs. The outcomes permute the inputs. The probabilities result from counting the outcomes of each type.

Not all sample spaces consist of equiprobable events.

Considering the goats distinguishable results in a perfectly valid input sample space, with non-equiprobable events. Considering the dice indistinguishable also results in a perfectly valid input sample space with non equiprobable events.

Counting outcomes in the table only gives the probability of the outcome if the input sample space is equiprobable. That is information that does not come from the table itself, so it is a good idea to include it as a column in the table.

Why are you so resistant to adding a column for the probabilities?

 

[sysprog]

Not all sample spaces consist of equiprobable events.

The ones for the inputs to the 2 games under consideration here do.

Considering the goats distinguishable results in a perfectly valid input sample space, with non-equiprobable events. Considering the dice indistinguishable also results in a perfectly valid input sample space with non equiprobable events.

In each case, that emburdens the table maker with the need to know ex ante what the probabilities are, which is not given in the game conditions, and not having established which, is the motivation for constructing the table.

Counting outcomes in the table only gives the probability of the outcome if the input sample space is equiprobable. That is information that does not come from the table itself.

In each of the 2 games under consideration, the equiprobability of the different possible values for each input is given in the outset conditions. The number of instances of each cell type is not specified in the outset conditions. It is the inclusion of those that I'm saying is not necessary, and not reasonable as a prescription, if you want to use the table to establish the probabilities for each cell type.

 

[Dale]

that emburdens the table maker with the need to know ex ante what the probabilities are

That doesn’t emburden the table maker. The table maker already was emburdened with the need to know ex ante what the probabilities of the sample space are.

Again, not all sample spaces are equiprobable.

The ones for the inputs to the 2 games under consideration here do.

Not necessarily. I gave valid counter examples for each game. But regardless, the use of tables to calculate outcome probabilities is a common tool so it should apply to other games also. So you cannot avoid this fact:

Not all sample spaces are equiprobable

 

[sysprog]

That doesn’t emburden the table maker. The table maker already was emburdened with the need to know ex ante what the probabilities of the sample space are.

It's not onerous to be told the input conditions.

Again, not all sample spaces are equiprobable.

The reasonable and necessary sample spaces for the inputs to these 2 games are.

Not necessarily. I gave valid counter examples for each game.

You could add the distractant that the 2 dice are thrown 1 at a time, and that the first die is thrown with a prior random choice of the left hand or the right hand, and that the 2nd was always thrown with the right hand. That, you could insist, would mean that you could use a $\mathtt {12 \times 6}$ table instead of a $\mathtt {6 \times 6}$ table, with 1/12 probability for each of the 12 outcomes for the 1st die, but that would be unnecessary and unreasonable, just as in the Monty Hall game, counting 1 of 2 non-cars as if were meaningfully different from the other is.

But regardless, the use of tables to calculate outcome probabilities is a common tool so it should apply to other games also. So you cannot avoid this fact:

Not all sample spaces are equiprobable.

When they naturally are equiprobable, as in the 2 cases under consideration here, there's no sufficient reason to make them artificially otherwise.

 

[Dale]

It's not onerous to be told the input conditions.

Then it isn’t a burden to write it down in a column.

The reasonable and necessary sample spaces for the inputs to these 2 games are.

How do you know if a given sample space is “reasonable and necessary”

You could add the distractant that ... but that would be unnecessary and unreasonable

How do you know if a given sample space is “unnecessary and unreasonable”?

counting 1 of 2 non-cars as if were meaningfully different from the other

How can one know that it is wrong to count the goats as meaningfully different but right to count the dice as meaningfully different? In both cases the distinction does not affect the outcome of the game. So in constructing the table the determination about whether to treat something as meaningfully different is not based on whether or not it affects the outcome. So what is it based on?

When they naturally are equiprobable, as in the 2 cases under consideration

I disagree that they are “naturally equiprobable”, but these aren’t the only 2 cases for which you might want to build a table. So again:

Not all sample spaces are equiprobable

Your persistent avoidance of the issue of non-equiprobable sample spaces is quite telling. I think that you realize it is completely damning to your position. Since they exist and cannot be represented by entry counting in a table that implies in general that probability cannot come from the table. Even for equiprobable spaces you need to know in advance that they are equiprobable before you can determine that the probability column can be dropped and replaced by counting. So the probabilities fundamentally must come from outside the table.

 

[sysprog]

Your persistent avoidance of the issue of non-equiprobable sample spaces is quite telling. I think that you realize it is completely damning to your position.

If we're given the condition of equiprobability, we don't need to weight the rows or columns to reflect input probabilities, and if the inputs aren't equiprobable, and only then, we need to model the table differently.

Since they exist and cannot be represented by entry counting in a table that implies in general that probability cannot come from the table.

Only for non-equiprobable input possibilities is it necessary to weight the inputs. A table listing the possibilities can be used for establishing probabilities of types of outcome. Those probabilities can be established by counting the instances of occurrence of the types in the table.

For a table of 2 fair 6-sided dice, the probability for any type of outcome can be determined by counting the outputs of that type on the table, and then dividing the result by the number (36) of entries in the table.

The fact that the probability of getting a prime number on a fair throw throw of 1 fair die is 1/2 is readily available on looking at the equiprobable sample space $\mathtt {S=\{1,~2,~3,~4,~5,~6\}}$, and counting the number of primes. The numbers 2, 3, and 5 are prime; the numbers 1, 4. and 6 are not. We determine by counting that there are 3 numbers of the prime type. There are 6 possibilities for the number, and 3/6 is 1/2, so the probability that the number from a toss of 1 die is prime is 1/2.

The number of primes in the possible sums from a toss of 2 dice can be counted in a table of pair sums. The 11 possibilities for the sums are 2-12 inclusive. Of those numbers, 2, 3, 5, 7, and 11 are prime; 4, 6, 8, 9, 10, and 12 are not.

Here again is such a table:

\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}
Counting, we find 15 cells that contain primes, establishing $\mathtt {15/36=5/12}$ probability for a prime result from a throw of a pair of dice, which leaves $\mathtt {21/36=7/12}$ probability for a non-prime number. A row and column showing the 1/6 probability per number per die would not have assisted us in finding that, and counting the instances of primes in the table does allow us to establish that.

Even for equiprobable spaces you need to know in advance that they are equiprobable before you can determine that the probability column can be dropped and replaced by counting.

Again, you're failing to distinguish between the input possibilities and the probabilities of outputs of a given type. You could unnecessarily put the input value probabilities in rows and columns, but that wouldn't replace the need for counting the outcomes of each type for the purpose of establishing the probabilities for the output types. You still have to do the counting.

So the probabilities fundamentally must come from outside the table.

For equiprobable inputs, only the fact that they are equiprobable, the number of inputs, and the number of possibilities for each, are needed for the making of the table. The table allows the number of instances of an output type to be more easily counted. For the probability quotient of an output type, the number of equiprobable cells establishes the divisor, and counting establishes the dividend.

 

[Dale]

If we're given the condition of equiprobability

Exactly. The condition of equiprobability must be given. It does not come from the table itself. I am glad that we finally have agreement on this point.

if the inputs aren't equiprobable, and only then, we need to model the table differently

Yes, agreed. It is only necessary to include a probability column if the inputs are not equiprobable. However, it does no harm to include such a column in the equiprobable case.

When formulating general problem solving strategies (strategies that are intended to be broadly applicable to a large number of problems), if something is often necessary and never harmful, it is usually best to simply make it a habit. For example, in the case of the OP, where a non-equiprobable sample space was inadvertently chosen, the habit of including the probability column would have helped. Because the OP was not used to including the probability column they made two mistakes: one was to not include it where it was necessary and the other was to assume that all rows in any table are equiprobable. A habit of including it costs little and would have, at a minimum, benefited the OP in this specific case.

For equiprobable inputs, only the fact that they are equiprobable, the number of inputs, and the number of possibilities for each, are needed

Being given that there are N equiprobable events is the same as being given that their probability is 1/N.

Again, you're failing to distinguish between the input possibilities and the probabilities of outputs of a given type.

I don't think that we have any disagreement about how to calculate the output probabilities, just the best structure of these tables.

Only for non-equiprobable input possibilities is it necessary to weight the inputs. A table listing the possibilities can be used for establishing probabilities of types of outcome. Those probabilities can be established by counting the instances of occurrence of the types in the table.

For non-equiprobable inputs it is a sum rather than a count. Summation also works for equiprobable tables with probability columns as I recommend.

So, what is your remaining objection to including a probability column in general? It is sometimes necessary, as you recognize, and it is never harmful. It would have helped the OP. So why not use it as a general rule?

 

[FactChecker]

There have been so many threads on this that I hesitate to respond to this one. But I will tell you what made it clear to me. Suppose, instead of 3 doors, that there are 1,000,000 doors, with 1 car and 999,999 goats. You choose a door, so your odds of choosing the car are clearly 1/1,000,000. Then the host opens 999,998 doors which all contain goats. Do you really believe that the odds that your door has the car has increased to 50:50? The car didn't move, so how could your odds have increased? Think about it from this standpoint.

I have one concern about this. If you know that the host will intentionally avoid opening a door with the prize, then this is correct. However, if you know that the doors opened are random, then your odds actually did increase to 50:50 and you have just witnessed a bazaar streak of luck that the prize door was not opened. But that is not the Monte Hall puzzle. It is an essential part of the Monte Hall puzzle, that you know that Monte will never open a door with the prize.

 

[phyzguy]

It is an essential part of the Monte Hall puzzle, that you know that Monte will never open a door with the prize.

Of course. This is clear from the beginning.

 

[FactChecker]

Of course. This is clear from the beginning.

Right, but what is not so clear is that a random opening of doors, all without the prize, WILL change your odds to 50:50. And that makes all the difference.

 

[sysprog]

Right, but what is not so clear is that a random opening of doors, all without the prize, WILL change your odds to 50:50. And that makes all the difference.

In the original question, the chance goes from 1/3 to 2/3 -- your first chance doesn't change if you don't switch, and it changes to the complement if you do, and the door-opening stuff is meaningless sideshow. Monty's essentially letting the contestant swap his one door for two. If you know that Monty's not going to open your door, or the door that he knows to be the prize door, and you know that he's going to leave only one door other than yours unopened, then you know that the only remaining door has all of the chance of being the prize door that your first-pick door does not have.

 

[FactChecker]

It seems to me that if Monty picks the door at random then the odds are still 2/3.

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

 

[lavinia]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

I don't see that.

 

[lavinia]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

What if the new door does have the prize?

 

[FactChecker]

What if the new door does have the prize?

What do you mean by "new door"? If you mean the one that Monte chose to open, then the game is over and both your door and the remaining door odds go equally to zero. Of course, this is not how the game is actually played. Monte avoids opening a door with the prize.

 

[lavinia]

What do you mean by "new door"? If you mean the one that Monte chose to open, then the game is over and both your door and the remaining door odds go equally to zero.

Yes I meant the door that Monty opens.

Sorry. I didn't understand your point. You are saying what if Monty chooses randomly and the cases where he randomly picks the car are excluded from the game? I guess I don't understand what the game is.

 

[DaveC426913]

I guess I don't understand what the game is.

It's important that you understand the game first.
https://en.wikipedia.org/wiki/Monty_Hall_problem

1. The host must always open a door that was not picked by the contestant.
2. The host must always open a door to reveal a goat and never the car.
3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

 

[sysprog]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

Monty chooses at random only when your door conceals the prize. If the prize is behind one of the other 2 doors, he can open only the non-prize door. It's really as simple as this: if chosen has prize (1/3 chance) then switch loses, else switch wins (2/3 chance) -- there is no 50:50 chance $\dots$

 

[FactChecker]

Monty chooses at random only when your door conceals the prize. If the prize is behind one of the other 2 doors, he can open only the non-prize door. It's really as simple as this: if chosen has prize (1/3 chance) then switch loses, else switch wins (2/3 chance) -- there is no 50:50 chance $\dots$

In the real game that is true. My statement was intended to show the difference between the real game, in which the expected gain is better if you switch, and another variation, where the 50:50 odds result always holds. In that other variation, Monte is not influenced by any knowledge of where the prize is and makes a random pick of the door to open.

 

[sysprog]

In the real game that is true. My statement was intended to show the difference between the real game, in which the expected gain is better if you switch, and another variation, where the 50:50 odds result always holds. In that other variation, Monte is not influenced by any knowledge of where the prize is and makes a random pick of the door to open.

So in such a variation, Monty could reveal the prize before offering the switch, which would make the switch offer meaningless.