CaptainEd

good point, Curr3nt!

Just a question about the conditions. "The difference is small relative to the weight of the coins". I'm guessing that means something like "the difference between the lightest and the heaviest coin is less than the weight of the lightest coin." Is that sort of what you mean? (What an interesting weighing puzzle!)

Some day, when I look back on this, I'll laugh. But right now, my ability to do word problems is not very satisfying. I believe I did exactly what you advised, I clearly graphed some points randomly generated according to this approach, I saw a graph in which the probability is exactly 1/3, yet my Excel spreadsheet continues to show numbers larger than 1/3, as you said (and presumably, as brifri238 says). Thanks in advance for my education in mathematical thinking!

Indeed, more than one way. Bonanova, what is the goal of this puzzle? a) show us that the notion of "randomly chosen" depends on how you define it? b) have us show lots of ingenuity in defining truly different ways to define it? c) elicit from us the really-really-best way to define it? d) all of the above e) none of the above f) other

I think I just learned something from this clarification. In Boggle, the tableau ABC DEF GHI has E adjacent to each of the others, and each of the others adjacent to two others, arranged in a ring. But I believe you are willing to relax the ring requirement on the neighbors of E; your only requirement is that AE <-> EA. All that is required is the graph. Thanks! (Another interesting puzzle from superprismatic!)

Thank you, phil1882, for finding and extending such an interesting puzzle! Now that I view the video a few times, I understand what you were saying about how the diagram is created. Overcoming my disbelief that such a simple mechanism could create a chart representing presence or absence of common factors was a pleasurable and enlightening experience. Thank you, curr3nt, for attacking this puzzle and sharing your progress with us.

I've got to confess that I couldn't relate what you just said to the analysis curr3nt and I have been doing. I fear we are off target. Here's a question. Did you originate this problem? (For example, when curr3nt spotted the surprising color on 4x24, you agreed that it should have been black. That makes me think you created this.) Or did you find it elsewhere (for example, you said you had removed the outer shell. That made me think someone else originated it).

If we assume this coloration rule is correct, we are asked to consider what is now the collection of NW-SE diagonals, and answer where there's one with more blacks than reds. At a snail's pace, I point out that the diagonal that intersects (k,1) consists of all the n1, n2 where n1+n2 = k+1. And I find it fascinating that there are some that are all white (never mind black and red!) k = 4, 6, 10, 12, 16, for example. There must be an obvious mathematical reason why all pairs summing to those numbers are mutually prime... (blushing) yes, if K is prime, and n1 and n2 add to K, then they don't share any factor, otherwise, K would share that factor.

Yes, sorry curr3nt, I was slow catching up to your observation about (4,24), and his corroboration.

Another apparent oddity: the "diagonal" for 4 has a BLACK square for 4, 8, 12, 16, 20, but RED for 24 ALso, the "diagonal" for 9 has Black for 9, but Red for 18 Of course, these wouldn't be oddities if we understood the pattern...but I don't

False alarm

So now I have to figure out the OP--randomly break the stick, then randomly break each piece. Discard one piece. Next time...

Thank you, Superprismatic! (And thank you, Y-san. Such a simple question, with so little intuition available for guessing how it should go, and such a tiny, but repeatable effect. Great fun!) I'm quite surprised how much more important the ability to hit 20/21 is than the ability to be the survivor. Over 80% of the wins due to 20/21! And yet, even so, the first position is only very slightly stronger than the other two... Fascinating!

Thank you, bonanova. I've been feeling dumb and behaving lazy. You've prodded me, so I have an answer for the first of the two.

Would you say the winning seat benefited more from 20/21 or from the other two going bust?

"infinite stack of decks"...One shuffled deck offers sampling WITHOUT replacement, which is what makes it possible for someone to benefit from card counting. If we stacked another shuffled deck on top, and another, etc. producing an "infinite stack of decks", there would still be sampling without replacement. (Indeed, with only three players, you're sure to exhaust the players before the first deck of cards.) Is the purpose of this phrase to imply sampling WITH replacement?

Is an ace either 1 or 11 at the player's choice?

Must be Bonanova is right. Cards are dealt one at a time, and completion criterion is tested each time. As soon as the first person hits 20 or 21, or the second person busts, the hand is over.