*Originally posted by Yongliang *

**Wow!**

Interesting problem and solutions.

Well, I'm no maths freak, but theoretically, I guess Roygoh and Toasty are correct.

However in reality, simply put, the situation has changed.

You are now faced with 2 chests, and you need to choose one.

Of course, the probability is still 1/2. It doesn't make practical sense that one chest shd have a probability of 1/3, and the other 2/3.

Just my personal views. =)

The situation only changed after the intial choice is made, so it cannot change the probability that the intial choice is right or wrong.

If, before the player made his initial choice, the host opens an empty chest and let the player choose from the remaining 2, then yes, the probability of winning is an obvious 0.5.

The twist is that the empty chest is opened after the player has made the initial choice. So it cannot change the probability of winning from 1/3 to 1/2 automatically.

Here's yet another way to look at this.

Say the player is allowed to pick 2 out of 3 chests. What is his chances of winning?

2/3, right?

So after he has picked the 2 chests, the host opens one of the 2 that he picked and it is empty. Does it increase his chances of winning? No. Because we already know that at least one of the chests his picked will be empty, and the host knows exactly which one is empty. The host merely picks the one that he already know is empty and opens it. The player's chances of winning is still 2/3, and it does not drop to 1/2 because now there are only 2 unopened chests.

In this case, if the hosts offers the player a chance to switch with the box he did not pic initially, is it more obvious that he should not switch?

Reversing the logic back to the original scenario, where the player picks one chest, his chances of winning is 1/3 if he does not switch, and his chances does not automatically increase from 1/3 to 1/2 after the host opens an empty chest from the two that he did not pick.

Convinced yet?