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.
"As humans we're naturally used to explaining things based upon events that have happened"
I am explaining this by ignoring events, not by including events that have happened. Because we always move onto the second round, as the first round doesn't count, and we will always be shown a goat at that time. Therefor, there is no actual choice in the first round. It is always a 1/2 choice, in the second round.
Again, I must insist that you examine the first diagram in the "Conditional probability by direct calculation" section, and give me your thoughts on this matter.
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 am explaining this by ignoring events, not by including events that have happened. Because we always move onto the second round, as the first round doesn't count, and we will always be shown a goat at that time. Therefor, there is no actual choice in the first round. It is always a 1/2 choice, in the second round.
Again, I must insist that you examine the first diagram in the "Conditional probability by direct calculation" section, and give me your thoughts on this matter.
https://en.wikipedia.org/wiki/Monty_Hall_problem#Conditional_probability_by_direct_calculation
As it boils down to 3 choices at 1/3, with one being given a false choice dividing it to two choices of 1/6.
Or 3 equally probable conclusions, from 3 doors.
Note: the diagram shows 8 probable outcomes 1/2 of which are car. This aligns with my "First round doesn't count logic"