- #106
Ibix
Science Advisor
- 11,950
- 13,690
Not if the dice are biased.DaveC426913 said:Yes it will. One in 36.
Not if the dice are biased.DaveC426913 said:Yes it will. One in 36.
Red herring.Ibix said:Not if the dice are biased.
Dale said: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.
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.DaveC426913 said:Red herring.
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.DaveC426913 said:Yes they will. One in 36.
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)DaveC426913 said:Red herring.
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.lavinia said:Cultivating allies or enemies is irrelevant to pedagogy
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.lavinia said:No one earns an unkind response in a learning situation
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.lavinia said:Part of leaning is having the confidence and audacity to challenge.
If the coin lands on its edge, you just disregard that toss, and toss again.DaveC426913 said: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 said:Red herring.Ibix said:Not if the dice are biased.
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.Dale said: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)
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.sysprog said:If the coin lands on its edge, you just disregard that toss, and toss again.
Then you might as well include spurious winds and inclement weather in the table!Dale said:Not in general. That is only true for a pair of fair dice. Not all dice are fair.
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.DaveC426913 said: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.
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.sysprog said: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.
@cmbcmb said: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.
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.DaveC426913 said: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.
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.Dale said: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.
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.Dale said: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.
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.lavinia said: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.
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.sysprog said: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.
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.cmb said:My OP 'attacked' no-one, and it is not possible to be 'aggressive' to an idea, it is a silly anthropomorphism.
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.sysprog said:when the 2 numbers are different, (An,Bn) is a different outcome from (Bn,An)
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.sysprog said: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.
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.sysprog said: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 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 |
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 |
You are welcome. And thank you for your thoughtful post.cmb said: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! ;) ).
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.Dale said: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.
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.Dale said: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.
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.Dale said:With the dice you have to consider them distinguishable to get an equal probability 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.Dale said:So since you must have probability information to begin with you may as well include it in the table.
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.Dale said:The distinction between (An,Bn) and (Bn,An) in no way affects the summation outcome.
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.sysprog said: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.
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.sysprog said: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}
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.sysprog said: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,
Obviously not, it is this omission that I am recommending against.sysprog said: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.
That's quoted and fully answered in my post. Here's the brief immediate response:Dale said: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 rest of the post includes greater detail on that concern.sysprog said: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 sample spaces are inputs. The outcomes permute the inputs. The probabilities result from counting the outcomes of each type.Dale said: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.Dale said: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.
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.Obviously not, it is this omission that I am recommending against.
Not all sample spaces consist of equiprobable events.sysprog said:The sample spaces are inputs. The outcomes permute the inputs. The probabilities result from counting the outcomes of each type.
The ones for the inputs to the 2 games under consideration here do.Dale said:Not all sample spaces consist of 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.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 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.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.
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.sysprog said:that emburdens the table maker with the need to know ex ante what the probabilities are
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:sysprog said:The ones for the inputs to the 2 games under consideration here do.
It's not onerous to be told the input conditions.Dale said: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.
The reasonable and necessary sample spaces for the inputs to these 2 games are.Again, not all sample spaces are equiprobable.
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.Not necessarily. I gave valid counter examples for each game.
When they naturally are equiprobable, as in the 2 cases under consideration here, there's no sufficient reason to make them artificially otherwise.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.
Then it isn’t a burden to write it down in a column.sysprog said:It's not onerous to be told the input conditions.
How do you know if a given sample space is “reasonable and necessary”sysprog said:The reasonable and necessary sample spaces for the inputs to these 2 games are.
How do you know if a given sample space is “unnecessary and unreasonable”?sysprog said:You could add the distractant that ... but that would be unnecessary and unreasonable
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?sysprog said:counting 1 of 2 non-cars as if were meaningfully different from the other
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:sysprog said:When they naturally are equiprobable, as in the 2 cases under consideration
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.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.
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.Since they exist and cannot be represented by entry counting in a table that implies in general that probability cannot come from the table.
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.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.
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.So the probabilities fundamentally must come from outside the table.
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.sysprog said:If we're given the condition of equiprobability
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.sysprog said:if the inputs aren't equiprobable, and only then, we need to model the table differently
Being given that there are N equiprobable events is the same as being given that their probability is 1/N.sysprog said:For equiprobable inputs, only the fact that they are equiprobable, the number of inputs, and the number of possibilities for each, are needed
I don't think that we have any disagreement about how to calculate the output probabilities, just the best structure of these tables.sysprog said:Again, you're failing to distinguish between the input possibilities and the probabilities of outputs of a given type.
For non-equiprobable inputs it is a sum rather than a count. Summation also works for equiprobable tables with probability columns as I recommend.sysprog said: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.
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 said: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.
FactChecker said:It is an essential part of the Monte Hall puzzle, that you know that Monte will never open a door with the prize.
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.phyzguy said:Of course. This is clear from the beginning.