Let's take the situation where he flipped the 37 heads. Then there's no tails left on the table, so the 37 he flipped contain 0 tails, and so do all the others, right?
Now, take the situation where he flipped 36 heads (now tails) and one tail (now a head). That's his first group (one head). Since he flipped only 36 heads and appearantly missed one, that group still has the unflipped head it started out with.
Etc., this goes for any number of heads he flips.

But let's all go to my other thread where we discuss the merits of centrist parties :)

OK well my answer is correct if you aren't allowed to remember which ones you flipped

(desperately trying to justify all that wasted effort)

3rd person because you have a 75% to get 1euro and 25% to lose 2euro.

You should consider a personal attack on me to justify your failure. It often works on webdip.

Done and done.

oh god that was awful by me! shikari is right, i was thinking that 2x25 was equal to 1x75 :|

hey uclabb there is a better way to your prisoner puzzle than the one I chose?

you said I could do !99 times better :P

although if you are capable of flipping the coin beyond any finite number of times it should be the team

or were you just trying to say I should have written 1/100! using that silly mathspeak ?

The answer I gave to uclabb's problem gives a 100% chance of surviving. Maybe it was skipped over.

The prisoners *open* all 99 boxes but do so with their eyes closed so they never find their name.

Draug, the original wording of the first problem makes it pretty obvious that your overlap solution doesn't work.

Whoops, stupid phone.

@YJ- 99! is mathspeak for 99*98*97*96*...*5*4*3*2*1. So yeah, you can do a lot better. ghug figured it out, so presumably at some point he will pst a solution. If he doesn't some time this morning, though, I will.

@Socrates/shikari- Wrong.

@Yonni- My question is not meant to be a trick question. So while your answer is clever, it is not the one that is being looked for. I don't think the evil king is going to buy "I kept my eyes closed the other time!"

@uclabb - i'm i right in saying the BEST stratagy is still very, very likely to result in death of all the slaves? In which case prisoner 1 doesn@t open the left hand box and all those that follow will know left hand box has prisoner 1's name in it. Prisoner 2 doesn't open 2nd from left and therefore if he survives all those that follow know what names are in first two boxes. The odds of them all surviving are quite small, but this is a better stratagy than all prisoners acting randomly.

I've been called.

First, the 1/100! solution has been posted three times, as a new insightful solution. How did that happen?

Second, start by assigning each prisoner and each box a number from 1 to 100. Each prisoner starts with the box that corresponds to him, and then opens the box that corresponds to the name inside that box. He repeats, following the trails and boxes until he has opened 99. If he doesn't die, the numbers form a compete loop and therefore all of the other prisoners will be assured success by starting at the numbers that correspond to them.

Apologies if that makes no sense, phone typing makes me lazy.

yeah uclabb is'nt being very clear. I think those of us who said 1/100! are right.

I actually think the question I answered accidentally (where each prisoner only opens 1 box) is a more interesting one, because you're actually increasing your chances in a meaningful way, rather than improving on infinitesimals infinitesimally.

@YJ- Literally, ghug just posted the answer, which makes you 99! times more likely to succeed. Again, 99! is a much, much larger number than the total number of molecules in the universe, so your solution may as well be a zero percent chance.

If what ghug posted isn't clear, I'm happy to try to explain the solution, but what ghug posted is the answer.

lol, OK I'm just blind - I didn't see his answer. Good thinking Ghug!

1/100 chance of survival isn't nearly so bad.

Apparently I'm terrible at logic puzzles.

I'm going to try to upload a visual one for you guys later - this one isn't too hard, but its fun.

Anyone got the spider/fly one figured out yet?

@uclabb, for your coin flipping one: I choose to be on the flipping team, and agree with my partner beforehand that we'll always guess the value of our own coin. This changes the odds to 50/50 and leaves us with a net gain.

There it is, ghug. :-) I knew I could count on you.

Here's an easy one:
You have two ropes of different lengths, and a lighter. Each rope will burn for one hour total, but not in a uniform manner (i.e. the length that burns every minute will vary). Figure out a way to time 45 minutes.

Light both ropes, one of them on both ends. When the one you lit on both ends is done, 30 minutes have passed. Light the other end of the one that is left, and it will last another 15 minutes for a total of 45 minutes.

Nice thinking uclabb

Nice. This one might stump you (I'm really running low on these, this puzzle gets boring very quickly, hence the stumping):

A circular cake is frosted on one side. You pick an arbitrary angle x, cut a slice of cake that large, and flip it in place. You then cut another piece starting at the end of the first piece, and continue repeating this process, always going in the same direction. Which values of x will lead to the entire frosted side being up again, and how many slices will that take, relative to x?

I have the spiders and fly solution but won't answer unless ghug says OK.