@Killer, assuming ASCII or unicode, "Remember to drink your ovaltine"

@Diplomat, good job. Bonus question, find a solution that takes less time.


Now for a new one, this is barely a riddle and really just a spacial reasoning/math problem, but:

There are three spiders and a fly moving along the edges of a regular tetrahedron (edges, not faces). The spiders are given a completely deterministic set of orders for catching the fly, and the fly is given access to that set of orders ahead of time. The fly is completely impossible to detect by the spiders until caught, and can choose its starting location given the starting locations of the spiders, but moves slightly slower than the spiders (how much slower is irrelevant). What set of orders ensures that the spiders catch the fly?

@abge,

(1) Pyramid: the ruts were deeper from the mine to the pyramid.

(2) Sentence: "You won't sentence me to four years in prison."

Here's one for you (or anybody). Due to Raymond Smullyan. Somebody offers to let you play either of two games.

Game 1: You get $20.

Game 2: You make a statement. If it's true, you get $10. If it's false, you get some undetermined amount.

Which game should you play and why?

About the prisoner one. How long would it take on average for that to happen? It seems it would take many years, possibly much longer than the lives of the prisoners. Not sure if that's a reasonable solution.

If you're not a gambler you should play game 1 because you will get a predetermined amount of money for certain

Diplomat's solution should take an average of about 100/3 years.

You get to chose the statement? And, the undetermined about is positive?

If so, I'd chose Game 2 and make a false statement. The chances of getting a number from [20,infin) is greater than from [0,20).

what does THIS binary code translate into with ASCII:
01101011 01101001 01101100 01101100 01100101 01110010 00110001 00110011 00110101 00100000 01101001 01110011 00100000 01100001 00100000 01100010 01100101 01100001 01110011 01110100 00100000 01101101 01101111 01100100 01100101 00100000 01101101 01100001 01100011 01101000 01101001 01101110 01100101

I found it ghug. The prisoners should use their lotting time to plot how to escape and overpower the warden, and then do so. :-)

That's not the solution, abge, and also, your probability is wrong. There is no uniform probability distribution on the real numbers, and knowing nothing about what the probability distribution may be, you don't know whether the chances from [20, infin) is greater than from [0,20).

@ killer your binary statement is so utterly wrong it is funny. everyone knows "killer135 is a beast mode machine" is completely inaccurate :-)

diplomat, no, because then they will be fugitives and either A. Found or B. forced to live much of their lives in hiding or risk serving more time

I would do two and say "killer135 is a beast mode machine" which would be false and hope the monetary value is a lot based on what abg said which is correct.

That's a good point. I (wrongly) assumed the distribution would be uniform.

In that case, I suppose I'd have to go with Game 1.

and no diplomat, I am a machine made in a factory and am set on beast mode

I was kidding killer.

I would still take the risk in game 2 regardless.

Killer was made in china! :-)

Also wrong, abge. ;-)

lol

I'll freely admit I hate probability.

How many times can I play? Just once and I can't imagine Game 2 is better. Many times, maybe there's some Gaussian thing on large samples I'm forgetting about.

I kill the person offering the games and take all of their money off of the body.

I think Skyrim has altered my morality.

and that is why in my response to the game question, which was completely ignored for some reason, said "if you're not a gambler"

Just once, abge.
And I'll give you a hint. It's not a probability problem. It's a logic problem.

Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

This is called by some the hardest logic puzzle ever. Figure it out if you can.

Sorry killer, this thread is so interwoven.
Your answer is wrong, though, you should take game 2 even if you're the most risk averse person in the world and only playing once.

@semck

Can you say something like: I will be getting less than $10 by playing Game 2?

Oh!

I've heard this before and have always thought in terms of probability and it always trips me up. I'll PM you my answer.

That's getting close, uclabb, but it's not there. Game 1 might still be better.

@ghug

I haven't thought of a way to make the lightbulb one much, much faster, but you could definitely not have the counter predetermined and then just have the person on the second day become the counter. This would save about 100 days on average.

Hmmm..Mine's still wrong, but I'm on the same line of thought at uclabb. Just can't figure out how to word it right.