You are shown three boxes, and told that one of them contains a prize. You are then asked to pick one box, and if that box is the one with the prize, you will win it. After picking a box, you are shown that one of the other two boxes is empty, and offered a chance to change your selection.

Should you do this? Would changing your choice to the other remaining box affect your odds of winning? Why or why not?

From Levik's solution.<o:p></o:p>

So let's say the box revealed to be empty is (B). If the chance of the prize being in either (B) or (C) is 2/3, and we know that it's definitely not in (B), that means that we can say that the probability is 2/3 that it is in (C) - and 1/3 that it is in (A).<o:p></o:p>

No.  That is wrong.  There is still only 1/3 chance that the box is in C.  Eliminating an unchosen  box does not make it twice as likely that it is in one of the other boxes.  If so, we would have to also say that  it is twice as likely to be in the chosen box as well.  Which makes the possibility 2/3 that it is in both boxes.  That is 4/3, a mathematical impossibility.  Therefore an invalid argument.  That does not make sense.  The extra 1/3 probability that the box is in b or C is discarded when box b is discarded or eliminated from the possibilities.<o:p></o:p>

Or say we now know it is not in B.  Now there are only two boxes it can be in.  So now it is 1/2 probability that it is in either box (A or C) Whether to switch or not, is an independent choice with different odds. Switching the box does not increase the likely hood that it is in the chosen box.  Yes 1/2 are better odds than 1/3, but changing the box does not give us those better odds.  Eliminating a box does.<o:p></o:p>

Posted by john on 2005-03-03 18:15:56