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This is a bit of a stretch and I'm not sure it'd even work, but...

I think this would work? Sorry if it's confusing. I sort of confused myself when I came to this conclusion. So the next guy should then be able to determine the color of his hat by: Then the 9th person would look ahead and: But the first guy has no way of knowing his own hat color, so he's got a 50/50 chance of dying either way. If anyone has trouble ...

It will be Both A and B can see, what C sees, and that's why they both know that From there, the problem reverts to the earlier one:

A simple way would be that Either there is, or there isn’t, but since everybody guessed the same, everyone will be right, or everybody will be wrong. No-one can be wrong by more than one, so whenever the guesses are a different number, the correct number is guaranteed to be the number in between. EDIT: On closer inspection, this seems to be exactly what’s ...

Here is another strategy, which requires looking at just 1 other person's hat:

The players can achieve $15/16$ win probability, which is optimal. Associate the 15 players with nonzero vectors in $\mathbb{Z}_2^4$ like $(0,1,1,1)$. Let $S$ be the sum (entry-wise XOR, or Nim sum) of the players with black hats. Each player doesn't know $S$ because they don't know their own hat color, but they know the two possible values of $S$ depending ...

Sure, it's perfectly doable.

Answer: First, some observations Alright. Now, let's take a look at different possibilities: Generalizing this, we get the rule above: Do note however that the actual answer is actually slightly more complicated, thanks to the effect of asking them in order:

Before the "game", the prisoners decide on this strategy: Assign an integer 1, 2, ..., n to each prisoner arbitrarily and call this the "prisoner number". When the numbers are shown, each prisoner determines how many swaps (swapping adjacent prisoners) it would take to change the others from "forehead order" to "prisoner order". If it takes an even number ...

The strategy is: Why it works: For example:

I'll try another explanation(with same result): Here are the steps: Next step: So: Who talks?

How about this strategy Reasoning

First observations Since the puzzle is symmetric in white and black hats, I assume that the first man has a black hat. Then only three cases remain: (1) BWWB; (2) BWBW; (3) BBWW. ($\ast$) If a man sees two hats of color white, then he knows that his hat is either red or black. If a man sees two hats of color black, then he knows that his hat is either ...

Generically, what do you do if there are $c$ colors, and $n$ people? Imagine further the aliens say the following: You can plan your strategy now, but we are only going to tell you what the colors are going to be (including their hex representation) and how many there are after you are no longer allowed to communicate. How well can the humans do? It's ...

The players can win with probability This is optimal because: A strategy that achieves this is:

No such algorithm exists. Suppose Bob sees 0's on 99 shirts. He must call out 0, since it is possible that 0 is the only represented number. Suppose Bob sees 0's on 98 shirts, on everyone except Alice's shirt. If Bob's shirt were 0, then Alice would see 99 zeroes, and announce 0 (by the previous bullet point). This means Bob must announce 0, else there ...

I will describe a lower bound and an upper bound on the probability of success the dwarves can guarantee. These bounds are: Lower bound: Imagine that there is a stack of 100 shirts in the corner of the room, one of each number. Then each dwarf can see a total of 199 shirts. The dwarves can use the following strategy: a random integer $n$ between 1 and 199 (...

Everybody walks out free and happy.

I think this is a classic (and probably a duplicate): To separate the two groups cleanly

Create infinite groups of finite size with $$size = (2^k)-1 ; k=2,3,...,infinity$$ Each of the groups will handle themselves as per the optimal strategy in this answer by xnor Thus each group has a probability of correctly guessing equal to : $$1- \frac{1}{2^k}$$ And the probability that all of the groups guess right is the infinite multiplication of ...

Call the logicians A, B, C, and D, in the order they speak. WLOG, say A is wearing a white hat. Note that each logician sees 2 black, 1 white if he's wearing white or 2 white, 1 black if he's wearing black, so he always knows his own hat is either [the colour it actually is] or red. A is wearing white and sees 2 black, 1 white. As far as he knows, he could ...

This is a fantastic puzzle, one of the best I have had to solve, so give yourself time to think it over. It took me a month to come to the solution. To make you guys understand the solution, I will first demonstrate it in a particular case where there are just two colors. I will then expand it to N colors. Preliminary logical deduction: Particular case: ...

This is a proof that it is impossible to beat $\frac{7}{9}$. There are 27 equally likely situations for the choice of colors. Each person will guess correctly in 9 situations. However, for any two people, they will both be correct in 3 situations (three possible colors for A, B has the color A guesses and C has the color B guesses). By inclusion-exclusion, ...

One possibility is It is not possible for the wizards to do better. There are $16$ possible combinations of hat colors. Of these, $7$ of them have at least one wizard with no black hat, so cannot be won. Of the remaining $9$ possibilities, in $6$ of them Alice will have one black hat and one white hat, and no matter what Alice's guessing strategy, she will ...

You're all making this much more complicated than it needs to be. :) Let's take an example:

I think they can get all the way to probability of getting all three guesses right. There's probably a possible explanation that utilises Galois fields, modular exponents and discrete logarithms, (well, more likely just modular subtraction and averages are enough) but let's instead give each kid one of these: Kid 1 then observes the other hat colours, ...

This puzzle has a computer science interpretation¹. The humans must provide 10 bits of information in total (each hat color is one bit of information). Participant number $k$ (with #0 at the back and #9 at the front) will hear $k$ bits of information from the humans behind him and will see $10-k-1$ bits of information on the heads in front of him. So there'...

If the PhDs were allowed to plan a strategy together ahead of time, then the following would work: Unfortunately, the king forbids them from collectively planning. Each couple may come up with a plan that would work if everyone followed it, but how do we know everyone will come up with the same plan? Proof:

Here is a way to guarantee 500 correct guesses. Edit: The above is optimal, because the expected value of random guessing is 500, and with no extra information given to the logicians about how the hat colours are chosen, this cannot be raised. The only thing their strategy can change is the variation, and a variation of zero is optimal. For the second ...

So I just wanted to solve an amazing extension of this puzzle that I happen to know of. Suppose that there were more than two colors. In fact, let's suppose that there were an uncountably infinite number of colors (so we're effectively writing a real number on each hat). And of course, we still have a countably infinite number of people in line. Is there ...