EventHorizon

Yup...

Ahh... Chicken. If you could, rip the cooperate card so the opposing player can choose to either lose 1 or lose 2. If your opponent thinks you are crazy (so you will only play defect)... the rational choice is to play cooperate to avoid the worst possible outcome. But if the opposing player is thinking the same, you'll both owe the Grand Master 100 beers. So, who will try and stop the bleeding first? It depends on how risk tolerant the players are and what they believe about the other player. Neither mutual cooperation nor mutual defection is an equilibrium of the single-shot game (of the repeated game, the folk theorem comes into play... but mutual defection cannot be an equilibrium (without "knowing" incorrect information about the opponent)). CD and DC are the only pure strategy Nash equilibria of the single-shot game (which both happen to be stable). There's a mixed strategy NE in there as well (though it is unstable... like all mixed strategy equilibria) as Octopuppy pointed out (50-50). As long as both players believe the other will play according to it (and both players know it), there is no incentive to deviate from the mixed strategy equilibrium (e.g., it can last more than the first round given the right information). There is no best strategy without knowing the opponent's strategy (just like I have proven for the prisoner's dilemma in a previous post). So I still believe there is no "best"/logical strategy without extra information (knowledge about the other player, or some history of play, etc). There are definitely some bad (dominated) strategies, but there are lots of good ones too. And as Neida pointed out, Masters of Logic are definitely not myopic... so they'll look at the big picture. Of course, some extra information won't necessarily result in a best strategy (like knowing the other player is also a Master of Logic). Logically? (sorry... I had to ) Does it not just depend on the information available to them? If both players know that they both will act according to a specific logic, and that logic includes knowledge that the other player has the exact same available information and will act according to the same logic, why would it be illogical to use that information to know that mutual cooperation would be best (due to the obvious symmetry)? It seems to me that superrationality is simply acting rationally when in the presence of certain extra information (though perhaps with inhibited free will mentioned previously to keep the assumption "everyone will play the same" correct). What if both players are told that the other is superrational and that both players have exactly the same information, but one retains his free will? Wouldn't the best strategy be to cooperate until the last round (letting the other continue to believe incorrectly that he will always cooperate), and then defect on the last round?

The same logic in my original post... Essentially you want to be the one to defect first if there is going to be defection, but defecting early lowers total reward... I thought this paragraph included the 10th round by "look at the second to last round as the last round and continue the pattern of mutual defection," and didn't think it was worth mentioning specifically. Essentially, this shows the circular logic at the end of my first post. Optimal with respect to what? If one person is going to cooperate all the time, isn't defecting on the last time (without time for retaliation) a better strategy (you end up with a higher score than otherwise). Perhaps only Super Masters of Logic +5 would do that (sorry for the sarcasm). I don't see the reason for the "Woo yay!" No disrespect to unreality, I'm sure he came up with that on his own. But as I see it, we are basically in agreement (even with the somewhat arbitrary decision of having them cooperate on the last round). The final two rounds become something like a paper rock scissors game (creating the circular logic I mentioned originally) CC - Just cooperate... it's better than the two rounds of mutual defection. CD - Well, if he's going to cooperate, I should defect on the last round to get the 51. DD - I'll still get 50 If I defect a round earlier than him. rinse and repeat. There are a few pareto optimal solutions, included is the 50-50 (Nash bargaining solution / the fair outcome). I argue there is no best (without qualification) solution, but I am partial to the fair pareto optimal one.

It won't effect the answer, but can the points go negative?

though the sequence in The Simpsons was consistent with the symmetry (I think the left half was always the number and the right was the reflection)... I'm not sure how you are doing yours (it seems inconsistent).

Season 9, the episode titled "Lisa the Simpson" From what I remember, Lisa finds a similar sequence. All her classmates know the answer immediately, but she cannot figure it out. She then stumbles onto knowledge of the "Simpson curse," which is a quick drop in intelligence after a given age. In the end, Homer gathers a bunch of relatives to try and cheer her up. He runs around to a bunch of them, asks them what they do, and their jobs don't inspire confidence. Homer thinks he's failed to cheer her up and most of the relatives leave. A female relative says something to her (can't remember what), but then Lisa notices her name tag. It has the title "Dr.". She then learns that the "Simpson curse" only affects the Simpson men. Relieved, she goes back to the sequence and solves it quickly.

That's a great lemma. I just figured out how to use it using the following links: http://mathforum.org/library/drmath/view/56240.html http://blog.plover.com/math/polya-burnside.html

superprismatic is right. I didn't get an answer that agreed with any of the ones posted, so I searched online and found this nifty formula.

Nice little exercise. I'm positive there's an easier way to do this, but here's my proof.

ECA is a perfect cube, not ECD.

I just stumbled on this puzzle at http://tierneylab.blogs.nytimes.com/2009/03/30/the-god-einstein-oppenheimer-dice-puzzle/ I thought it was pretty good, so I'm copying it here to let braindenners have a try: Instead of simply asking who is favored to win, my question for you is this: What are the best odds for the dice war game that Oppenheimer can get given that Einstein will minimize it by his die choice? In other words, I want to know how much the winning player will win by, and not just who wins. Good luck...

yup... that was quick. Good job.

Sort the following Boxes (described by their axis aligned perimeter measurements) by increasing volume. Box 1 - 10, 18, 20 Box 2 - 14, 14, 26 Box 3 - 14, 14, 16 Box 4 - 12, 16, 24 Have fun!

Exactly. If you take one section (one of each color) away you'll end up with my first posted solution. You can still remove quite a few more, but it requires a slight change in structure.

So, it looks like you can force 4...

It's not pretty, but...