I don't think it has anything to do with some lack of specificity of the definition of the problem. The whole reason the "Monty Hall Problem" is a brain teaser is that it is counter-intuitive.
I would then say it is counter-intuitive because we automatically assume "additional information" is "useful additional information". Here, the revelation of the goat behind Door No. 3 is additional information which is not actually useful. Let me illustrate....
Suppose, rather than revealing the goat behind Door No. 3, the host simply says, "Rather than keeping what's behind Door No 1, would you like to take the best of what's behind Door No. 2 and Door No. 3?"
You see, at this point, a goat is assuredly behind one or the other door. The revelation of which is entirely immaterial and therefore useless. To be more direct, if you switch now, you get the best of two doors, or a 2/3 chance at the prize.
This result, of course, matches up with the combinatoric analysis.
To me, it only changes if I understand why the door is being opened. These doors do not contain Schrödinger's car. both doors start out at 1/3 and drop to 1/2. Unless the host is attempting to sway me. In short, it depends on if the second choice is real, or a red haring. Which is not explicitly stated. (See direct calculations section)
The logic of the Pigeon switching also seems false. As The thinking of the pigeon is to continue choosing after failing, as it is not going to choose the same door twice, if it knows it needs to make a new choice.
In addition, it feels like I am being told to accept an absurdity, as a new base logic. And after learning of this on the front page of reddit, I do not trust this to be a truthful logic. It seems they are skewing the base logic by, concealing a fundamental rule that changes the results.
If I may reiterate, the knowledge of the host and his revelation are both entirely useless. That's precisely what throws off our intuition.
Remove those from your imagination completely. The host knows nothing and will show you nothing. You pick #1, then the host says, "Do you want to keep #1, or take the better of either #2 and #3?"
Well, not much of a question in that formulation, is it? I could not believe how long the wiki page was on this.
Look at it this way. There is no winning in the first round. There is no losing in the first round. Therefor there is never a 1/3 choice. Switching or not switching in the second round will always give you a 1/2 chance.
I'm also noting that the subversion agent u/free-will-of-choice is AstroTurf-ing this thread. 3 comments with zero interaction. Two of which are not to an active reply. Odd thing for him to do.
Again, please examine the diagram in the "Conditional probability by direct calculation" section.
Note: the diagram shows 8 probable outcomes 1/2 of which are car. This aligns with my "First round doesn't count logic"
At a high level this is confusing because of the idea of informative and conditioning. As humans we're naturally used to explaining things based upon events that have happened this is a hack that enables us to process information. Unfortunately it backfired in cases such as these.
The probability of it being behind your door (1/3) or the other doors (2/3) doesn't change, it's independent of the events. When a door is opened we know that it is not behind one door so the 2/3 probability now only has a degree of freedom of 0 whereas before it has a df of 1. So the offs haven't changed but the df has.
Here's another way to look at it. The average probability does change however (1/3 + 2/3)/2 --> 1/2. This differs from what it was at the beginning (1/3+1/3+1/3)/3-->1/3 because there are fewer degrees of freedom.
So notice there at first there were N-1 or 2 degrees of freedom for the entire context. And N-2=1 for the other doors. Opening a door removes the only degree of freedom for the subgroup.
It is this added constraint that increases the rate at which you can infer about this subgroup, especially upon the last guess. It has no more degrees of freedom to vary.
I don't think it has anything to do with some lack of specificity of the definition of the problem. The whole reason the "Monty Hall Problem" is a brain teaser is that it is counter-intuitive.
I would then say it is counter-intuitive because we automatically assume "additional information" is "useful additional information". Here, the revelation of the goat behind Door No. 3 is additional information which is not actually useful. Let me illustrate....
Suppose, rather than revealing the goat behind Door No. 3, the host simply says, "Rather than keeping what's behind Door No 1, would you like to take the best of what's behind Door No. 2 and Door No. 3?"
You see, at this point, a goat is assuredly behind one or the other door. The revelation of which is entirely immaterial and therefore useless. To be more direct, if you switch now, you get the best of two doors, or a 2/3 chance at the prize.
This result, of course, matches up with the combinatoric analysis.
To me, it only changes if I understand why the door is being opened. These doors do not contain Schrödinger's car. both doors start out at 1/3 and drop to 1/2. Unless the host is attempting to sway me. In short, it depends on if the second choice is real, or a red haring. Which is not explicitly stated. (See direct calculations section)
The logic of the Pigeon switching also seems false. As The thinking of the pigeon is to continue choosing after failing, as it is not going to choose the same door twice, if it knows it needs to make a new choice.
In addition, it feels like I am being told to accept an absurdity, as a new base logic. And after learning of this on the front page of reddit, I do not trust this to be a truthful logic. It seems they are skewing the base logic by, concealing a fundamental rule that changes the results.
If I may reiterate, the knowledge of the host and his revelation are both entirely useless. That's precisely what throws off our intuition.
Remove those from your imagination completely. The host knows nothing and will show you nothing. You pick #1, then the host says, "Do you want to keep #1, or take the better of either #2 and #3?"
Well, not much of a question in that formulation, is it? I could not believe how long the wiki page was on this.
Look at it this way. There is no winning in the first round. There is no losing in the first round. Therefor there is never a 1/3 choice. Switching or not switching in the second round will always give you a 1/2 chance.
I'm also noting that the subversion agent u/free-will-of-choice is AstroTurf-ing this thread. 3 comments with zero interaction. Two of which are not to an active reply. Odd thing for him to do.
Again, please examine the diagram in the "Conditional probability by direct calculation" section.
Note: the diagram shows 8 probable outcomes 1/2 of which are car. This aligns with my "First round doesn't count logic"
At a high level this is confusing because of the idea of informative and conditioning. As humans we're naturally used to explaining things based upon events that have happened this is a hack that enables us to process information. Unfortunately it backfired in cases such as these.
The probability of it being behind your door (1/3) or the other doors (2/3) doesn't change, it's independent of the events. When a door is opened we know that it is not behind one door so the 2/3 probability now only has a degree of freedom of 0 whereas before it has a df of 1. So the offs haven't changed but the df has.
Here's another way to look at it. The average probability does change however (1/3 + 2/3)/2 --> 1/2. This differs from what it was at the beginning (1/3+1/3+1/3)/3-->1/3 because there are fewer degrees of freedom.
So notice there at first there were N-1 or 2 degrees of freedom for the entire context. And N-2=1 for the other doors. Opening a door removes the only degree of freedom for the subgroup. It is this added constraint that increases the rate at which you can infer about this subgroup, especially upon the last guess. It has no more degrees of freedom to vary.