k-man

Need more time for the 3/4 case.

Thanks, plasmid. It's clear now.

Can you gain as much by switching back? Using the same reasoning? There is a paradox: A gain for switching can be anticipated. Yet, there is a preferred envelope, and if we initially chose it we should not switch. Not sure what you mean by "switching back". The 50/50 comes from randomly picking one of 2 envelopes from which we know one has double the money of the other. I don't see a paradox.

Sorry, TSLF. I don't think I understand your comment. Are you saying that it's possible that a liar gets jolted twice by pressing LT followed by LR? It's either a T or R to his left, so both of these cannot be true.

It's your puzzle, so your interpretation prevails. However, you mentioned that it's a 'classic' and the prisoners are in a circle, which, in my opinion, strongly suggests the classic interpretation of the problem known as the Josephus problem. The reason they are in a circle is that you don't stop at the end and start the next round with the 'next first person' as if they were in a row. That distinction is what makes this problem interesting, IMHO.

correct answers Respectfully disagree. Let's test this solution for 5 prisoners. Execution order will be 1,3,5,4 with #2 surviving. For 13 prisoners: 1,3,5,7,9,11,13,4,8,12,6,2 with #10 survivng.

" by " is not " into " It isn't indeed. And English is my second language, so all this time I've thought that it meant the same thing when applied to division. So, BobbyGo has it then...

I think there's an error here (see part highlighted in red above) You're right. I made a wrong assumption here and that's what amounts to the difference in the answers. I'm convinced that Bonanova's solution is correct.

Nice! Can't beat the shoe lace approach. The total path comes to

I think you should mark this as solved. There is no way to place 6 points on a surface in a way that the minimum distance between any 2 points was m and the maximum distance between any 2 points was m*sqrt(3).