k-man

Since the Angel cannot turn ON a lantern that is currently OFF (only the Devil can do that), as long as there is at least one lantern in the OFF state from the beginning, the Angle cannot win by having all lanterns ON. This would require the Devil to turn ON the final lantern that will deliver the win to the Angel. Assuming the Devil wants to win, this will never happen, so the only way the Angel can win in practice is when all the lanterns are OFF. These are some trivial and uninteresting examples and I'm not sure how they help.

I understood the OP that you can only remove X coins from ONE chest and deduce the contents of all chests by observing removed coins. What's the lowest X?

Highlight text below to see my theory While the average wait time should be 15 minutes, there are longer and shorter wait times during the day. A random placement of a point in time is more likely to land on a longer wait time period. For example, if we look at a 1 hour period that has two 10-minute waiting intervals and two 20-minute waiting intervals, then an average wait time is 15 minutes, but a random point in time selected within that hour is twice more likely to be within the 20-minute interval.

...or more in general...

New York. And they all work as cab drivers.

I built the probability tree using PowerPoint. It's not freeware, but it's easy to learn and use. If you don't have PowerPoint you can try Google Docs, but I don't know if they have the same widgets that PowerPoint has or not.

I've been following this thread without comment as there wasn't anything I could contribute (didn't have the aha! moment like harey did). I'm convinced that harey's approach is correct. Probably the easiest way to see it is to draw the probability tree of complementary events (see below). The green track leads to the complete annihilation with total probability established earlier, the red track is the "no collision" track with total probability of 1/n! and the yellow tracks are all the partial annihilation tracks. This tree also explains jasen's finding that the most likely outcome is that 2 bullets remain. Half the leaf nodes of this tree will have that outcome and the total probability will be the sum of those leaf node's probabilities.

cheating...

My thoughts...

I would agree with jasen's answer. I think you're reading too much into details that are merely artifacts of the quality of the test. In this second puzzle there are 9 ways to combine 2 sticks of 3 types. 8 of them are given and the 9th is missing.

Well, I don't have the context that you have as I don't see the rest of the questions, but looking at this question alone I wouldn't call it particularly complex and my solution/answer seems reasonable to me. BTW, I've never seen a test where two of the possible answers were exactly the same as in this test.