What's your favorite lateral thinking puzzles?

There’s only one iteration? They each get one opportunity to guess, and if they all pass on that first and only opportunity, they all die?

In that case, I can only note that the puzzle says to maximize their chance of survival, not to guarantee it.Therefore,They pick one designated guesser, and he guesses at random. Knowing what hats the others are wearing is a red herring, without knowing something about the distribution of hats (i.e., on the one hand maybe the kingdom only has a maximum of two hats of each color, but on the other hand maybe the kingdom only has one color of hat at all, and the existence of the other color is a bluff, or maybe they have plenty of each, and each prisoner’s hat is chosen randomly and independently). 50% sucks when your life is on the line, but it’s the best we can do.

@ Tom Scud: Correct! For a 75% chance of survival. Obviously a bit easier than I thought. To me, it seemed intuitive that because merely seeing the other hats doesn’t tell you anything about the colour of your own hat, how could they do better than just one guy guessing at random?

…Wait. That’s true, isn’t it? Assuming, at least, that we’re not in the “the kingdom only owns one color of hat” territory.

No, we can do better than 50%, as Tom Scud notes. His strategy has a 75% winning probability, which is indeed the maximum.

ETA: Damn, I’m slow on the draw. This was in response to Chronos’s post which has already been responded to by both Ximenean and Chronos himself…

Or “The king is a malicious bastard who has worked out the optimum strategy for himself, and always gives his victims 3 same-colored hats”. In which case your only consolation is the knowledge that he’s sure to be run over by a runaway trolley car at some point in the near future.

To me, the intuitive explanation for why they can do better than just guessing at random is because, while it’s true that no individual person, when asked to guess their own hat, can do any better than random, we still have the option of arranging how their failures match up collectively: do different people tend to fail in different cases, or do they all tend to fail in the same case? The more we can bunch up their failures, the less distinct cases with anyone failing.

Well, in which case, it’s not really the optimum strategy, is it? It’s only the optimum strategy relative to a probability distribution which, perhaps, the prisoners should not have rashly chosen to model the situation…

You know, I originally had something about the hats being selected randomly with equal probability, but I thought it was a bit wordy :smiley:

Randomly and independently with equal probability: Otherwise, the method could be that the king flips a single coin, and if it’s heads, he puts white hats on all of them, and if it’s tails, he puts black hats on all of them. Details like this are always important in probability puzzles.

I know I’m a couple days behind this, but this answer clearly doesn’t work. If they are assumed to be perfectly conforming to the Norton circuit, the current source is ideal and can’t be dissipating heat. If the two boxes are electrically equivalent (as stated), the power source in both must be dissipating the same amount of heat.

In short, the Thévenin & Norton equivalent circuits do not model power correctly, so you can’t use it to solve that problem.

Upon consideration, I think I misunderstood the problem. The circuits are intended to be exactly what is described, not different circuits modeled as Thévenin or Norton.

Either way, the statement “Electrically, the two circuits are equivalent” is misinformation; I expect it’s meant that they will interact with a load the same. I thought that somehow these were two black boxes of unknown circuits.

Also I failed to read ‘constant current source’ syndecdochally, which made me think you saying it was what was dissipating the power.

Random thoughts on Islanders:

A: Everyone sees 99 or 100 blue eyed people. Subject A and B exist. A is blue-eyed, B is brown-eyed.

On day 0-98, nobody kills themselves. This is because everyone knows there are at least 98 blue islanders, so everyone’s safe. On day 99… everyone learns there are at least 99 blue islanders. B is not surprised by this. A is. Therefore, A kills himself, because he must logically be blue eyed.

This would be true for all A and all B. So it’s going to be a mass suicide on day 99. (or 100, if they have to wait for the next day.)

B: These people are islanders. What, they never looked at their reflection in the ocean?

… My bad. On day 100, everyone learns there is at least 100 blue islanders. A is surprised, etc. Because if there were only 99, as A thought, then they would have killed themselves on day 99.

We don’t need this complicated ‘X thinks that Y thinks that Z thinks’ chain at all.

However you word it, the chain of reasoning all the way down to one blue islander is necessary, because the knowledge that if there were only 99 blue islanders, they would have killed themselves on day 99 is based on their reasoning about what 98 prisoners would have done, and so on.

Actually, that’s a good point. Everyone knows there’s at least one blue eyed person, and not been telling him, all along.

If they’re that kind of person, who says they’re going to count 100 days?