
The Aha of Monty Hall - having_datum
https://medium.com/@data.is.in.domain/the-aha-of-monty-hall-661ff743ac8b
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ksaj
The real trick is in the wording. You choose a door. Monty chooses a door that
is NOT be the car, leaving one that may or may not be the car, depending on if
the contestant was wrong.

In _words_ Monty pretends to be removing ONE door from the pool of choices,
but he's not doing that - he is removing ALL possible doors except for the one
that is a car if you were wrong, or a goat if you were right, and then saying
(truthfully) that every single other one is a goat. When there are 3 doors,
that is one that gets discarded. But that's the tricky word play.

Now, if you did it with 100 doors, Monty isn't simply removing 1 option to
give you the appearance of a 50/50 chance. He is removing ALL of them except
for 2 - one that he shows is not a car, and the other that you have to decide
whether your first choice was right or not.

Following the same exact sequence, he removes _97_ of them from the pool
because they are all goats, leaving the one you chose, the one he says is NOT
a car, and one that is the car if your original choice was wrong, or a goat if
your original choice was correct. Suddenly it is super obvious that this is no
50/50 choice.

That's how the Monty problem actually scales. Looking at it from a game with
100 doors makes it extremely obvious that the person's first bet is more
likely to be wrong than right, and that changing is the obvious choice to make
since the odds are 1/100 that your first choice was right, and 99/100 that the
one Monty left behind is the car.

Hence in the normal way of playing the game, it is a 2/3 chance the contestant
is initially wrong, and should always choose to change their choice.

------
jlongr
I often see people try to "prove" the solution to this problem as the author
did, using a large number of trials to approximate the long-term
probabilities.

But I don't think this method really proves it, nor does the proof require so
much work.

Consider that C = car and G = goat. There are three possible configurations of
cars and goats, represented by rows of the matrix, and the three doors are
represented by the columns:

    
    
      C G G
      G C G
      G G C
    

Suppose you chose door #1 (it doesn't matter which door); you then have 1/3
chance of winning:

    
    
      (C) G G
      (G) C G
      (G) G C
    

The host reveals another door with a goat. That door is X'd out:

    
    
      (C) X G
      (G) C X
      (G) X C
    

Thus, if you stay with your original choice, you retain a 1/3 chance of
winning, shown by the set [C, G, G].

If you switch, you have a 2/3 chance of winning shown by the remaining set [C,
C, G].

~~~
jamie_homs
I don't think the simulation is intended as a proof (it never is), more like
the coding clarified the solution.

It is true that your method is simple and good enough, just that the most
people might find it difficult to think in terms of abstract symbols.

