The Monty Hall paradox/conundrum

[sysprog]

I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

This is why, if you are interested in looking at outcomes of games (as opposed to probabilities), the proper tool is a Monte Carlo simulation, not a table. With a table or a decision tree you should calculate the probabilities for each row or path. In my tables I also included the impossible rows and calculated their probabilities which, as expected, was 0.

Tables or decision trees that show all the possible outcomes and how they are arrived at can be used to establish probabilities for each variable, and those probabilities can be verified experimentally by means of a Monte Carlo simulation; however, the rules for the simulation and the rules for constructing the table or decision tree are inter-derivable, and both are derivable directly from the rules of the game.

 

[FactChecker]

The critical issue is the screening process that Monte applies in deciding which door to open. He will intentionally avoid opening the door with the car. That changes the odds. If he had 100,000 doors and opened 99,999 goat-doors, one must ask why he avoided the remaining door.

To emphasize the screening process, consider another problem:
Suppose there is a bucket of sand containing one diamond. You take a random pinch of sand from the bucket. It has a tiny chance of containing the diamond. Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass. You clearly know that there is an overwhelming chance that the small pinch remaining in the bucket has the diamond, while your pinch of sand still has a tiny chance. You should switch.

 

[Orodruin]

Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass. You clearly know that there is an overwhelming chance that the small pinch remaining in the bucket has the diamond, while your pinch of sand still has a tiny chance. You should switch.

I do not think that this is true unless you know that somebody deliberately dumped sand that they knew did not contain the diamond. If you just dumped sand randomly from the remainder, the largest probability is that the diamond ends up in the screen. It would be the equivalent of a game show where Monty opens one of the remaining doors at random, ie, the Monty Fall problem - where the probabilities are known to be 50-50.

 

[Dale]

Tables or decision trees that show all the possible outcomes and how they are arrived at can be used to establish probabilities for each variable

The problem with that is that e.g. the first version of this table had four entries two representing wins for “stay” and two representing wins for “switch”. So simply counting entries in the table doesn’t give you probabilities. So the table itself must always be augmented with the probabilities. Even in the case where counting entries in the table gives you the probabilities it is because you know that the probabilities for each row are equal.

 

[FactChecker]

I do not think that this is true unless you know that somebody deliberately dumped sand that they knew did not contain the diamond.

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

 

[PeroK]

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

If you ended up on the Monty Hall show, I'll bet you'd bring a giant magnet and point it at each door and see which one bulges as the car gets pulled towards you. But, that's a different sort of screening process if you do something to find out exactly where something is!

That's nothing to do with probabilities and information. That's to do with physically isolating something.

 

[FactChecker]

If you ended up on the Monty Hall show, I'll bet you'd bring a giant magnet and point it at each door and see which one bulges as the car gets pulled towards you. But, that's a different sort of screening process if you do something to find out exactly where something is!

That's nothing to do with probabilities and information. That's to do with physically isolating something.

So I should have said "the sand was dumped through a screen that would not allow the diamond to pass if it is in the bucket. " Thank you for the English lesson.

 

[Orodruin]

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

This is irrelevant. Somebody would have to pre-screen the sand before the dumping. You would have to explicitly arrange for the diamond to be left in the bucket.

It would be the equivalent of a game show where Monty opens one of the remaining doors at random, ie, the Monty Fall problem - where the probabilities are known to be 50-50.

To put that into mathematics, consider the following:
You pick a door and Monty picks a door not to open at random among N > 2 doors. All the other doors are opened and none of them contains the car. Let A being the event that the car is behind your door, B the event that the car is behind Monty's door and C the event that the remaining N-2 doors do not contain a car. Clearly, P(A∪B∪!C) = 1 and A, B, and !C are exclusive. Both P(A) and P(B) are equal to 1/N because without further information to condition the probabilities on, they are both just choosing one out of N doors. This means that P(C) = 1-P(!C) = 1 - (N-2)/N = 2/N.

We seek P(A|C) and P(B|C).

P(A|C) = P(C|A) P(A)/P(C) = 1 * (1/N)/(2/N) = 1/2

Here, P(C|A) = 1 because if the car is behind your door then the remaining N-2 doors will not contain the car. P(A) = 1/N because there is originally a chance of 1/N of your door containing the car. The same logic can be applied to B.

 

[sysprog]

The problem with that is that e.g. the first version of this table had four entries two representing wins for “stay” and two representing wins for “switch”. So simply counting entries in the table doesn’t give you probabilities. So the table itself must always be augmented with the probabilities. Even in the case where counting entries in the table gives you the probabilities it is because you know that the probabilities for each row are equal.

That criticism seems to me to be apropos of either incorrect table construction or incorrect counting. It is clear from the rules of the game that there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many, and examination of the 3 car locations shows that 1 of them is repeated. The repeated car location line should be merged with its repetition, in the table construction or at least in the count, or the repetition should not be counted.

 

[DaveC426913]

...there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many

What? No.

'Possible locations of the car' is not what the number of rows is representing. Your idea that there is one too many rows is in error.

 

[sysprog]

What? No.

'Possible locations of the car' is not what the number of rows is representing. Your idea that there is one too many rows is in error.

In post #55 I included this table:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & \mathtt 1 & \mathtt 2 & \mathtt 3 & \mathtt {st\ 1} & \mathtt {sw\ 1} & \mathtt {st\ 2} & \mathtt {sw\ 2} & \mathtt {st\ 3} & \mathtt {sw\ 3}\\ \hline \mathtt 1 & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}}\\ \hline \mathtt 2 & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}}\\ \hline \mathtt 3 & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} \\ \hline \end{array}

There are no rows there that repeat a car location and thereby obfuscate the fact that there are exactly 3 different locations for the car. That the 3 locations are equiprobable at 1/3 each is immediately discernible from the stated game conditions. Counting of the 1 car and 2 goats in each of the stick columns, and of the 2 cars and 1 goat in each of the switch columns, shows that the chance for the car is 1/3 if you stick, and 2/3 if you switch.

On the more complicated tables that have more than 1 row per car location, there is still 1/3 probability per car location; not 1/3 probability per reference.

 

[mfb]

This is irrelevant. Somebody would have to pre-screen the sand before the dumping. You would have to explicitly arrange for the diamond to be left in the bucket.

I think the whole bucket is put through the screen which let's everything pass apart from the diamond and a small bit of sand.

 

[DaveC426913]

In post #55 I included this table:

Right. OK. That was your table, not the tables that have been mostly at the centre of the discussion. I might have picked that up if my knee hadn't jerked.

 

[Orodruin]

I think the whole bucket is put through the screen which let's everything pass apart from the diamond and a small bit of sand.

The initial statement makes it clear that not all sand left in the bucket is passed through the filter (my boldface):

Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass.

 

[sysprog]

Right. OK. That was your table, not the tables that have been mostly at the centre of the discussion. I might have picked that up if my knee hadn't jerked.

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

When the car is behind a door not initially chosen, there is only 1 door that Monty can open. There are exactly 3 doors and exactly 1 car, so there are exactly 3 possible outcomes that pivotally depend on where the car is. Listing 4 possibilities is therefore suspect.

If you ensure that you count exactly once per car position, you (presumably) won't wind up with the spurious 1/2 probability assessment that the ceremonious door-opening invites, and you'll (presumably) arrive instead at the correct 1/3 for sticking and 2/3 for switching probability distribution.

 

[Dale]

It is clear from the rules of the game that there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many

The rows of the table indicate more than just the location of the car.

Listing 4 possibilities is therefore suspect.

There are four possibilities. There is nothing suspect about that.

Not all the possibilities are equally probable, but there is no requirement that all entries in a table must be equally probable.

 

[FactChecker]

The initial statement makes it clear that not all sand left in the bucket is passed through the filter (my boldface):

Just to clarify what I meant: All the sand except a pinch is dumped out of the bucket through a filter. The remaining pinch of sand remains in the bucket along with any possible diamond.

 

[mfb]

Just to clarify what I meant: All the sand except a pinch is dumped out of the bucket through a filter. The remaining pinch of sand remains in the bucket along with any possible diamond.

Then you don't have the Monty Hall problem. You have the Monty Fall problem, and both pinches of sand are equivalent.

 

[FactChecker]

Then you don't have the Monty Hall problem. You have the Monty Fall problem, and both pinches of sand are equivalent.

That is not true. If the diamond remained in the bucket after you removed a pinch of sand, it will remain in the bucket after the majority of cand has been filtered out. That gives a very high probability that the remaining bit in the bucket has the diamond.

I have to think that we are not talking about the same thing since what I am saying is trivially true, IMHO, but people are arguing about it.

EDIT: I see what was ambiguous in my statements. I meant that as most of the sand is screened and dumped out, any diamond that hits the screen remains in the bucket. It is not removed.

 

[sysprog]

The rows of the table indicate more than just the location of the car.

True, but using an extra row for 1 of the car locations is misleading to many, who would, as you pointed out, then have to include a column, as you did and as @cmb did not, to show the fact that each of the 2 rows has only half of the probability for the 1 car location that they have in common. With only 1 row per car, you don't need a probability column for this problem, because the rows are simply 1/3 likely each.

There are four possibilities. There is nothing suspect about that.

The fact that there are more than 3 possibilities listed, when all that matters are the 3 possibilities for where the car is, in my view, makes the listing of 4 possibilities suffice as grounds to suspect that something that doesn't matter may be being given undue emphasis. Cursory investigation immediately makes apparent the fact that 1 car position is represented by 2 rows, and to my mind, that reinforces the suspicion. I think that the best correction of the overemphasis is to merge the 2 rows for the same car position into 1 row, but I recognize that your correction of including a probability column was also valid, and was well suited to the explanation that accompanied it, although I liked your earlier terser explanation:

Another way to look at it is as follows: the stick strategy wins if you initially pick the car. The switch strategy wins if you initially pick a goat. The probability of initially picking the car is 1/3 and the probability of initially picking a goat is 2/3.

I was about to post something similar, but I was still thinkig about how to word it, and you beat me to it.

Not all the possibilities are equally probable, but there is no requirement that all entries in a table must be equally probable.

I think that using 4 rows when there are only 3 possible car positions, without noticing that 2 of the rows share the same car location possibility and recognizing the implications of that, maps out exactly what misled @cmb in post #1.

 

[Dale]

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

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

 

[gmax137]

I think you should always do it

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

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

 

[DaveC426913]

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

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

 

[DaveC426913]

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

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

 

[Dale]

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

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

 

[DaveC426913]

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

The point is to crate a table of ideal results.

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

 

[Dale]

The point is to crate a table of ideal results.

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

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

 

[lavinia]

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

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

 

[Dale]

Some people aggressively challenge things as a way of learning

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

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

 

[mfb]

The point is to crate a table of ideal results.

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

 

[Dale]

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

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

 

[DaveC426913]

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

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

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

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

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

 

[DaveC426913]

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

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

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

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

 

[Dale]

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

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

 

[DaveC426913]

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

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

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

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

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