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On 17/03/2013 20:17, Tom Sult wrote:
> ------------------------------------------------------------------------
> This explanation is a commonly held view but I think it is in error.
> These are really two independent events. The first was a choice of
> three. The second was a nonrandom elimination  of one of the choices.
> This makes the second event an independent choice of one out of two.
> Therefore keeping your current curtain or changing to the other curtain
> results in the same 50-50 probability.

I still remember the shame of getting this wrong in front of a hundred
people at university, so I'll take a stab at explaining it.

Imagine you do this experiment many times. You always choose door A as
your initial choice. How often will you be correct if you stick with it?

Obviously, since the prize is placed randomly, over many iterations door
A will be the correct 1/3 of the time. So not switching doors has a 1/3
chance of winning the prize. Monty doesn't move the prize around!

Now, imagine you choose door A at first, but switch to the remaining
door after Monty opens B or C.
- One third of the time, the prize is behind A, Monty opens an empty
door and you switch to the other empty door. Bad luck.
- One third of the time, the prize is behind B, Monty opens C, and you
switch to B to win the prize.
- One third of the time, the prize is behind C, Monty opens B, and you
switch to C to win the prize.

So in conclusion, by switching you win 2/3 of the time, whereas by
sticking with your initial choice you win 1/3 of the time.


By the way, hello everyone! I recently got interested in celestial
navigation on land. Most of the discussion so far is slightly over my
head (badoom tish!).

Luc Van den Borre

   

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