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Old 12-17-2004, 05:30 PM  
ModelPerfect
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Join Date: Nov 2003
Location: USA
Posts: 2,862
Ok. I'm not reading 4 pages of conflicting views, but let me state for the record your odds are the same whether or not you switch.

ROUND 1: Your odds are 33.3% that you'll choose the correct box. Every box has a 33.3% chance of being the right box

ROUND 2: Your original odds were indeed 33.3%, and that's what you selected from. But now each box has a 50% chance of being the right box: There's only 2 boxes and 1 of them has the cash. Providing everything is random, then no box can possess a higher probability of being right: one has it; the other doesn't. You're odds don't differ from this. If the original box has a 50% chance of being right, then your odds are the same, because it's the same box. It's only your odds that have changed.

Look at it this way. The original post stated that if you switched, the odds of choosing the correct box is 50%, but if you don't switch, it is 33.3%. A LAW of statistics is the sum of all probabilities must equal 100%. This reasoning only totals 83.3% and has no other choices. Therefore, that logic cannot be correct. If your odds of choosing the other box is 50%, then your odds of staying with the original box has to be 50% as well.

Another way to look at it: The choice in round 2 is completely independent of round 1. You're asked to make a choice between 2 boxes. Being that you previously selected one of the boxes has no bearing on this decision. It is just as if a random audience member, who was taking a leak during round 1 and didn't witness anything that happened before, was asked to choose between just those 2 boxes. Obviously, his chances (and yours) are 50% to select the right box.
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