plasmid

Ah, thanks for catching the requirement about integer weights. If you multiply the terms in that example by 2 to make them integers, it would be a counter to the second point above where n=1. Regardless, based on the first point, I believe you're very close to a solution that's far more elegant than the sort of approach I initially thought this problem was going to require.

You're probably on the right track, but that's not an airtight proof. Here's a counterexample. (It uses four weights per pan instead of six in order to make it easier to see what's going on, but it would still apply to the case of having six gears per pan.)

My riddles are always about something concrete, but riddling about a concrete thing can involve and even be dominated by references to things the subject does or other things it interacts with. That is indeed the case with this riddle. Regarding your answer, if I were riddling about a chef then I would consider it a perfectly good answer to say that I'm describing making dinner because you would clearly have understood the point and imagery of the riddle. But I'm not a chef, although you're in the right ballpark.

Apologies if this seems a bit rambling, but it is an analytic solution unless I goofed somewhere. I wonder if there's a more elegant approach with less brute force.

I like how a lot of the imagery from that answer could apply to this riddle, but I'll have to say that it leaves a little too many clues hanging around without a nice explanation.

Does a big-shot mafioso have a Since the Arabian clue is more spouse than reception, and moms and dads are generally spouses Or how about this twisted Byzantine

How about:

Just wondering: how did you find the theoretical probability to be 50%? I'll spoilerize this since it references the previous incarnation of this problem.

Thanks for kicking it off, Wilson. That's a surprisingly nice fit for the first half, but quite different from what I have in mind.

Patagonia costume jewelry Arabian spousal reception Byzantine with a twist

A wooden deck with rawest ore Contains the seven seas Opposing sword leads arms galore A hundred through the breeze On anvil crash and grab thy share With thieving hand now masked Unite the victims should you dare If subtle be thy task Then sail away and loot deplete Across our flattened earth May Atlas stand on all his feet For two would be a dearth

..........

For the sake of kicking this off, and without proof of optimality, I'll say this is the best I can come up with for now.

Araver's right.

For part 1: If you have (a+b+c+d)2 >= Kbc for all such reals,

If you want to solve for the probability that any cards from card N to card (N-M) are in the left deck, I guess you could try: That looks too complicated for me to attempt without a more Ah-ha approach.

Correction to my previous post:

I can calculate the probability that at least one of card N-1 or card N-2 falls in the left pile. I think this might be extendable to larger numbers, albeit with the equations becoming somewhat unwieldy.

Oops, my mistake