Quote:
Originally posted by punkworld
Wrong.
If he *randomly* (!!!) selects which box to open of the boxes you didn't pick, and the one he selects happens to be empty, the ones that are left both have a 50% chance of containing the prize.
You see, in the original problem, you don't get any new information, because he will never reveal the box containing the prize, and there is always a box to open which doesn't contain a prize.
However, if a box is randomly selected, there is a 1/3 chance that it's the real one. If the box happens to be empty, that does not magically transfer 1/3 chance to the other box you didn't pick. It's just eliminated as a genuine possibility.
If a genuine possibility gets eliminated, chances get redistributed entirely.
The actual problem is:
X = A (1/3 chance) or Y (2/3 chance)
Y= (B or C)
Y will always contain at least 1 empty box, and an empty box will always be revealed. No genuine possibility will get eliminated by revealing a box.
The problem when used with randomness is:
X = A (1/3) or B (1/3) or C (1/3)
if X = B or C it might get revealed. A genuine possibility is confirmed or eliminated.
Non-random revealing: 1/3 vs 2/3
Random revealing: 1/2 vs 1/2
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If the host randomly chooses which of the 2 boxes to open, then you get tons more information. Here's the possible outcomes, presuming for this example the cash is in C and you always select a different box (if the host selects C that doesn't eliminate it):
(1) You select A, host shows B, you choose C = WIN
(2) You select A, host shows C, you obviously choose C = WIN
(3) You select B, host shows A, you choose C = WIN
(4) You select B, host shows C, you obviously choose C = WIN
(5) You select C, host shows A, you choose B = LOSE
(6) You select C, host shows B, you choose A = LOSE
Your odds of winning by always switching, even though the host RANDOMLY chooses between the 2 remaining boxes = 66.7%
It's because you get the additional information when the host randomly chooses C that alters your odds.
Maybe you just didn't understand my original statement: You aren't randomly selecting the boxes...only the host is randomly selecting which of the 2 remaining boxes to show you in round 2. Sometimes, he will show you exactly which box to choose. Now if the host revealing a box eliminates it, then your odds of winning drop to 33.3% no matter what you do (because instances #2 and 4 will then be loses). So the end result is your odds are never 50% if the host randomly reveals one of the 2 remaining boxes.