EventHorizon

Some unnecessary code:

Yeah, that's not a very interesting definition of the state. Try the one I described. So if the angel and devil are approaching lantern 5, the state would be the state of lantern 5, followed by the state of lantern 6,..., ending with the state of lantern 4. This would be equivalent to cyclically bit shifting the binary number so the next lantern the angel and devil will visit is at the front. Say there are 5 lanterns: 1 is on, 2 is on, 3 is off, 4 is off, 5 is off. If the angel and devil are approaching lantern 4, the state is 00110. This is because looking forward from the pair's position they see off,off,on,on,off.

Was there an easier method? Concave

Good. Yeah, I overthink everything and can get rather paranoid.

And by "put the devil in his place", I didn't mean you @harey. I reread that and remembered you said "As I am a nice guy, I let you the role of the angel." So if you found that belittling, I apologize, as that was not my intent. I am sure you see things I don't and vice-versa.

First, let's nail down the definition of configuration so we know exactly when the devil wins. It cannot simply be the state of the lanterns, since not changing a lantern on a given step would then repeat the configuration. So the location of the devil and angel need to be part of it. There are two ways I see of incorporating this. Include which lantern number the devil and angel are at and the string of lantern states. Which could be notated by a location marked in a string of lantern states (e.g., "01101;001") or a number and the string (6,01101001). The lantern numbers don't matter, just look forward from the lantern the angel and devil are at (e.g., ";00101101"). And then you don't need the location marker ";" because it is always at the front. Option 1 has more states, lanterns x 2^lanterns compared to just 2^lanterns. Let's choose option 2 since less states means more of a chance of repeat and gives the devil a better chance. That is the notation I used for my code. I notice that you use the word "turn" to sometimes mean one whole stroll around all the lanterns. That is something that threw me off a little. I'd call that a revolution or a full rotation. Alright, time to put the devil in his place.

I couldn't follow that, harey.

Just saw your post harey.

recreating the results of the code...

sorted configurations with 6 lanterns (and added devil best choices I omitted before) 5 lanterns 4 lanterns 3 lanterns Edit: Another thing of note. The devil and angel both choose between the same 2 options once each (except the choice between 000...00 and 000...01 for angel and 111...10 and 111...11 for devil, due to angel win conditions).

On the "ugly strategy"

Edit: Oops. I guess switch to 1 meaning off and 0 meaning on to match the story. Who wins? Ugly Strategy with not much intuition as to how/why it works. (Obviously, I don't consider this as the answer...) Yay code Code output for 4 lanterns 5 lanterns 6 lanterns

Which angles correspond to which sides? There are two sides that create a given angle and two away.

Proof 3: Factors of x^5-x

Proof 1: Modulus 10 Proof 2: Modulus 5 I have a third proof idea... I'll see if I can flush it out later. Are you a math professor/researcher? If not, where do you find / get inspiration for these?

Constant fraction of current money to maximize the median/mode: Code output (assumes infinitely divisible money): A little more math:

It seems my initial guess and idea about how the probabilities would end up were... umm, bad. Yay code... My best approach so far... It seems we calculate the expected value for the day the rat is found differently.

Is this just spam? The link at the bottom of that page seems shady.... installyourfiles dot com? No thanks. It looks another like this was posted recently too. "can anyone out there solve this puzzle?" -- that is also hosted on sites.google.com and goes to getafilenow dot com.

Hey Bob, welcome to the den. It's been a while since I've checked this site. This puzzle, at first glance, seemed similar to "Groundhog in a Hole." -- http://brainden.com/forum/topic/11943--/ I have posted a couple variations, too. -- http://brainden.com/forum/topic/12010--/ and http://brainden.com/forum/topic/18524-groundhog-in-a-hole-yet-again/ Anyway, I thought I'd ask for some clarification. If the rat is in hole 10 on day 1, does that mean on day 2 it has an equal chance (20% each) of having gone to holes 8, 9, 10 (stayed there), 11, or 12? If the rat is in hole 2 on day 1, on day 2 is there an equal chance (25% each) of being in holes 1, 2, 3, or 4? And a third chance each for holes 1, 2, and 3 if starting right on the edge in hole 1? And does the rat have an equal chance of being in any given hole on day 1 (1% each)? Just thought I'd nail down the way the probabilities move. Here's a quick guess before checking... anything really...

Alternatively,