bonanova

What ... my description of selecting random points does not overwhelmingly convince you? How about this.

OK, and I have a suspicion that my answer is too easy and there's more to it.

Totally! And, nice find on that last one.

@bubbled, I would interpret it this way: Round 1: Four games => 4 winners. Round 2: Two games => 2 winners. Round 3: One game => 1 winner. Seven games in all, and there are no "middle two" rounds. If that's right, my take is

Looks like

@jasen, check your 3rd iteration.

First thoughts

@jasen, In an earlier post you said to switch boxes. (A) of course that is not permitted. (B) you can succeed even if the warden randomly re-positions them before each visit! @bubbled, if the statement is incorrect, that you say is incorrect, the puzzle has no 30% solution. We agree on that. So it's not a guessing game. It's a strategy-driven game. And the strategy does group the successes and failures to advantage for the prisoners. We're both traveling the high road, trust me. @bubbled, let's not discuss my last point right now. This is because to clarify what I was getting at would mean to restate it in "much too revealing" terms. Perhaps I'll do that as a later clue if needed. For now I'd rather not retract it "with explanation." Let me simply retract spoiler #3. That is, your objection to it is valid, but also your objection to it does not preclude a solution. And if I say more, I say too much. Thanks.

I guess I misunderstood then. It sounds like the OP asks how to do it with 11, and then ask what the minimum length would be (even shorter.) What was confusing was trying to do it with 11. Good puzzle.

Thalia may have it, but ... seriously? twice your mother's age?

Yes, it is not a gimme that you win. The best strategy only means that among all possible strategies, it gives the best shot at winning.

@harey, we have language barriers here, and I think "in the middle" in the OP is safe to take as meaning "between the groups of white and black balls." @BMAD, did you mark my previous post as the correct answer? It only found a solution for length 12, and there was an earlier indication that a shorter chute was possible.

Here are three nudges. Not exactly clues. More like clarification of various points covered in the OP And here is a slight "Oops." Did I forget to mention that the boxes are externally numbered 1-100? They are, and now the OP says so.

More good advice (R rated)

Number 5

Each prisoner can open up to 50 boxes.

Bingo. Can you determine the expected number of flips needed for p(H) = 1/π ?

You have a coin which is known to be fair: p(H) = p(T) = 0.5 exactly. Devise a method for simulating a biased coin where p(H) = 1/π exactly. Edit: A gold star awaits a method that requires on average only two flips of the fair coin to produce an arbitrary bias.

I'm not getting this. Can you do the first couple numbers step by step? How to you decided on n? Just starting with 1, 2, 3, ... ? Where does "addition" come in? Thx.

I think that gives the opposite of the first flip. Pick the (slightly) lower, or bigger, hand.

Lets try it together. Hidden Content Again, to obvious to be true. But what is wrong?

One way