k-man

I'm not going to spoiler this as I'm not providing any specific solutions. I've been fiddling around with this problem for some time and still couldn't find a generic formula for digital scale that would determine a number of weighings necessary for a given number of eggs. However, I can definitely solve the original problem with 12 eggs in 4 weighings using a digital scale. If limited to 3 weighings, then only up to 9 eggs (36 pairwise combinations) can be solved. 10 eggs (45 pairwise combinations) requires 4 weighings, but if we can reduce the number of combinations by 1 (down to 44) then it can be solved in 3 weighings. So, if anyone is interested in solving this variation then try this: Among 10 eggs two are switched, but you know that it's definitely not 1 and 2 that are switched. Find the switched pair using only 3 weighings with a digital scale.

I started working out a similar solution for 10, but didn't have enough time to finish it yet. I'm trying to do it in 3 weighings using a digital scale, but so far, looks like it will require 4. Maybe a combination of digital/balance could be optimal. Is it allowed?

That's a cool solution, but a small correction is required...

Maybe this wording would be easier to understand (not that it changes anything): 4. If you see two cards that are not A and K, move the card that has an empty space to its right (wrapping around) into that empty space.

witzar beat me to it

Some more... I also think MC looks more like...

So much for a "proof in a hurry" LOL

Thanks for clarifying, but it seems that something is still missing... Also, if prisoner #1 guessed correctly and wants to save prisoner #2, who guessed incorrectly or refused to guess what happens to prisoner #2?

I don't have the time to put together a complete proof, but here is how I would go about it...

Interesting puzzle. I have some clarifying questions: 1) Do the prisoners know the number of their own cell? I would guess yes, but better ask than guess. 2) When you say that "they can only send it once to other cell and received it once from other cell" do you mean there is only one pass of dice allowed? Or each prisoner can do a single pass for a total of 6 passes of the dice? 3) Can the prisoner with the dice specify the cell number of the prisoner number that shall receive the dice from him? 4) What is the significance of the blindfold? They obviously need to be able to set the dice to something specific and read the dice. Some people may be able to feel the numbers on the dice using their fingers, but not all people can do that reliably.

plasmid, you're right. I didn't have the time to do the math and it seemed like this approach was worth a shot. Now, that I had some more time...

Welcome back, Wolfgang!

Hmm... when I read the puzzle, I interpret 1, 2, 3, ... 7, 8, 9 as all digits from 1 through 9 including 4, 5 and 6. Phil's solution appears to include only 6 numbers 1, 2, 3, 7, 8, 9. Also, it's not clear from the OP that going over 15 means losing the game, but if so, then...