|
|This 3-door Monty problem was a thread that lasted six weeks on
|the Usenet newsgroup: sci.math.num-analysis in the Summer of 1995.
|It is unusual to have a topic generate that much discussion..
|so be prepared!
|
|At that time, the participants had divided themselves into
|"two-thirdsers," and "one-halfers" and had vigorous debates.
|
|I was so amused by the whole thing that I wrote an article about
|this long thread for an Internet newsletter that I was writing for
|at the time, and my spouse (now ex) wrote a simulation and put it on
|the Web to show that the result had to be 2/3. I don't think that
|simulation exists now, but it would be easy to write to prove
|for yourself.
For the 3-door Monty problem, most humans intuitively compute 1/2.
Since the correct answer computes to 2/3, this problem's answer
goes against intuition (aka common sense).
You do not need any simulation, as the following discussion shows:
A B C
+---+ +---+ +---+
| | | | | |
| 0 | | 0 | | 1 |
| 0 | | 1 | | 0 |
| 1 | | 0 | | 0 |
| | | | | |
+---+ +---+ +---+
Assume you choose door A.
Then Mr.Gameshow opens a door from the set {B,C},
namely the one with a zero behind it:
A B C
+---+ +---+ +---+
| | | | | |
| 0 | | X | | 1 | X = taken (equal to zero).
| 0 | | 1 | | X |
| 1 | | 0 | | X | (in last case, Mr.Gameshow could equally
| | | | | | well choose 'B' as 'C'. 'C' taken here)
+---+ +---+ +---+
Now compute the number of '1's behind A
to the number of '1's behind B and C.
A contains 1/3 and B+C contain 2/3 (total = 3/3 = 1, naturally).
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