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Switching your choice would actually double your odds, and would be the smart thing to do. Here's why:
Let's say you start off by picking an arbitrary box. Let's call it box (A). Let's also call the other two boxes (B) and (C). Since the chance of the prize being in a particular box is 1/3, your selected box, (A), will be the winner one out of three times.
Which means you will lose the other two times, when the prize will be in either box (B) or box (C). Now the problem states that you are shown that one of these boxes is empty. But of course one of them has to be empty since the prize cannot be in both of them. And of course the odds of the prize being in either (B) or (C) are not affected by revealing which box the prize is not in.
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).
Thus changing your choice raises your chance of winning from one out of three to two out of three.
You can see the statistics of this here.
Update: Here's a more intuitive explanation...
At the time you make your initial guess and pick one box, the prize is either contained in that box, or not contained in it.
If the prize is in the box you picked, changing your choice is obviously the wrong thing to do.
However, if you were initially wrong, then the prize is in one of the other two boxes. After one of those boxes is revealed to be empty, the prize will most definitely be in the last remaining box, so you should definitely switch
So you stay with your choice if you are initially correct, and switch if you were wrong. Now the chance that you were right with your first guess is 1/3, and that you were wrong - 2/3, so in
two out of three cases a switch makes sense. |
