Come forth and be confused!

**aperson** · 01-10-2004, 02:25 PM

Yes, though it's not exactly correct, it is pretty close.

Here's an exhaustive proof:

There are three initial possibilities. Each have equal probability:

Code:

code:--------------------------------------------------------------------------------
Door 1   Door 2   Door 3   Probability
                  Prize      1/3
         Prize               1/3
Prize                        1/3
--------------------------------------------------------------------------------

You picked a door. Since you had no knowledge about what was where, this choice is essentially random. Thus, for each position of the prize there are three equally probably choices you could have made.

Code:

code:--------------------------------------------------------------------------------
X denotes your choice
Door 1   Door 2   Door 3   Probability
  X               Prize      1/9
           X      Prize      1/9
                  X, Prize   1/9
  X      Prize               1/9
         X, Prize            1/9
         Prize      X        1/9
X, Prize                     1/9
Prize      X                 1/9
Prize               X        1/9
--------------------------------------------------------------------------------

These are all possibilities and they are equally probable.
Now, when the host chooses which door to pick, if you chose wrong, there's only one door he can open that does not have the prize.
If you chose right, he can choose one of two doors to open; assuming he chooses randomly, two new cases are created, each with 1/2 the probability of the initial case:

Code:

code:--------------------------------------------------------------------------------
X denotes your choice
O denotes opened door
Door 1   Door 2   Door 3   Probability
  X        O      Prize      1/9
  O        X      Prize      1/9
  O               X, Prize   1/18
           O      X, Prize   1/18
  X      Prize      O        1/9
  O      X, Prize            1/18
         X, Prize   O        1/18
  O      Prize      X        1/9
X, Prize   O                 1/18
X, Prize            O        1/18
Prize      X        O        1/9
Prize      O        X        1/9
--------------------------------------------------------------------------------

Now count the total probabilities of the cases in which you should stay. They are 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 = 6 * 1/18 = 1/3.
The total probabilities of the cases where you should switch:
1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 6 * 1/9 = 2/3.

01-10-2004, 02:25 PM	#11
aperson FFR Hall of Fame Join Date: Jul 2003 Location: Houston Posts: 3,431	Yes, though it's not exactly correct, it is pretty close. Here's an exhaustive proof: There are three initial possibilities. Each have equal probability: Code: code:-------------------------------------------------------------------------------- Door 1 Door 2 Door 3 Probability Prize 1/3 Prize 1/3 Prize 1/3 -------------------------------------------------------------------------------- You picked a door. Since you had no knowledge about what was where, this choice is essentially random. Thus, for each position of the prize there are three equally probably choices you could have made. Code: code:-------------------------------------------------------------------------------- X denotes your choice Door 1 Door 2 Door 3 Probability X Prize 1/9 X Prize 1/9 X, Prize 1/9 X Prize 1/9 X, Prize 1/9 Prize X 1/9 X, Prize 1/9 Prize X 1/9 Prize X 1/9 -------------------------------------------------------------------------------- These are all possibilities and they are equally probable. Now, when the host chooses which door to pick, if you chose wrong, there's only one door he can open that does not have the prize. If you chose right, he can choose one of two doors to open; assuming he chooses randomly, two new cases are created, each with 1/2 the probability of the initial case: Code: code:-------------------------------------------------------------------------------- X denotes your choice O denotes opened door Door 1 Door 2 Door 3 Probability X O Prize 1/9 O X Prize 1/9 O X, Prize 1/18 O X, Prize 1/18 X Prize O 1/9 O X, Prize 1/18 X, Prize O 1/18 O Prize X 1/9 X, Prize O 1/18 X, Prize O 1/18 Prize X O 1/9 Prize O X 1/9 -------------------------------------------------------------------------------- Now count the total probabilities of the cases in which you should stay. They are 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 = 6 * 1/18 = 1/3. The total probabilities of the cases where you should switch: 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 6 * 1/9 = 2/3. __________________

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