CaptainEd

Assuming we need to find the number in each cell, here’s a variant that will give the others a bit more information: “is the total either 9 or 21?” this shows the others that Kleene has 4, 5, or 6. I don’t see that that’s enough yet. But maybe y’all do.

Thanks Thundercloud, that makes sense. Ok I’m on board with two empty boxes.

Ok I believe Al is empty...

But...but...but, the OP points out that after N events, there will be N coins in the boxes! I always found infinity confusing, but how will they become empty?

Do we all agree on what the task really is? is it (1) ensure someone can correctly state the total number of coins —seems like that is solved by Molly Mae, OR (2) ensure someone can correctly identify the number of coins in each cell?

Does this do it?

I think for part (b), I’ve got the right answer for the wrong question.

Good job!

For @plasmid, your argument sways me for part b, but I’m certain of part a.

Oh, when you calculate them out...

Cute puzzle, Bonanova, as always. (1)

I can do it in 8

As always, we hope for some communication. The prisoners can see each other. It’s not clear they can hear each other (after all, if they shout simultaneously, they can’t benefit from hearing the others). Are they allowed to turn their bodies to face in a variety of directions, or some such thing? You said “no communication”, and I fear you mean it, but just askin’...

Here’s a tiny observation about what the next to last step looks like.

much more clearly stated than my babbling. I’m tickled that I’ve shown that it can be evaluated one flip at a time, based merely on the parity of contiguous Hs. here is my argument, expressed by plagiarizing your expression: Let e be the expected number of flips from the initial (even) state There are two states that are easy to analyze and cover all the possibilities: E is the initial state, it also represents the state of having seen an even number of H (including zero), since the beginning or the most recent T. O is the odd state, representing the fact of having seen an odd contiguous run of H. State E requires e more flips. In this state, H changes to state O, while T remains in state E State O requires o more flips. In this state, H changes to state E, while T terminates with a win. That allows us to write an expression for x as the sum of these terms, weighted by their respective probabilities, all 1/2. e = 1/2{1+e} + 1/2{1+o} o = 1/2{1+e} + 1/2 substitute o into e e = 1/2{1+e} + 1/2{1+1/2{1+e} + 1/2} = 1/2{1+e} + 1/2+1/4+e/4 + 1/4 = 3/2 + 3e/4 e/4 = 3/2 e = 6

Gardner sets high standard in many ways. I was a child reading Childrens Activities and a few years later I was enjoying hexaflexagons and later mathematical games. I was kneeling behind you in worship. i enjoy the puzzles here, and sometimes I don’t understand something that is obvious to anyone else. I think I may have a touch of ambiguity flu. Keep on puzzling, Bonanova!

Thank you Bonanova, sorry to be so dense

I want to be sure I’ve got this right. There are six (presumably distinguishable) dice. I want to demonstrate that each is capable of showing all six faces. Paul lets me roll all six (and tabulate which individual dice showed which numbers) charges $1 for the combined roll, and pays $50 when the tabulations show each die has shown all six faces. Peter has me roll one die at a time, charges $1 for the individual roll, tabulates the results, and pays $20 when each die has shown all six faces. OR... my Termination condition is seeing all six numbers on the table at once. Paul has me roll all 6 dice each time, and only pays me if I roll a full straight (123456). Peter lets me roll one die at a time, once I’ve rolled all six dice, he lets me improve my hand by rolling a single die that duplicates another one, and pays once all six numbers are showing.

Perhaps

Good job slashpuzzler! And neat puzzle BMAD!