Logic Puzzles

[Doug31]

Yea, and this other approach is probably something that I completely don't understand you're allowed to do.

If there were 2 people and 2 boxes, and each person got to pick one box, then the first person tries a box, and has a 50-50 chance. Then the second person would know if they were still alive that it must be the second box, so that would be 50%.

Now, if there were 3 people, 3 boxes, and each got 2 choices, then the first person could try the first 1, so they'd be at 33.33~%. But, then the remaining two would know they both had to be in the last 2, which would be a 100% chance, so I can see how it could be done like that, as well.

Now, if there were 4 people, 4 boxes, and 2 choices each, I can't even see how to do that. What can the first person do? They have to pick 2 boxes or it's already less than 30%, so they pick 2 boxes...50%. Then the next person now knows the first person's was in one of the first two, so theirs is much more likely to be in the second 2. So the only way they can stay about 30% from that point is to pick the second 2. Even still, there was only a 2/3 chance that it would have been in one of the second 2, so they're down to 33.33~%. Now, the third person would only have a 50% chance, so it's over.

What am I missing here?

Edit: And ~HentaiXXX~, that would give the first person a 50/100 chance, and the second person a 50/99 chance, which is already only 25.2525~%, so it's already too low of a chance after only 2 people, just like all the ways I can come up with.