71 votes
Accepted

Guessing hat colors. 4 prisoners

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

6 prisoners, 2 colors, one mute

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:
  • 70.8k
25 votes

Guessing hat colors. 4 prisoners

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 ...
  • 70.8k
24 votes

Infinitely many hats

Here is another strategy, which requires looking at just 1 other person's hat:
  • 13.5k
23 votes
Accepted

Cooperative guessing game: no incorrect guesses

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) ...
  • 24.1k
22 votes
Accepted

Numbered hats, Warden and Maths

Sure, it's perfectly doable.
  • 142k
21 votes
Accepted

3 Numbers on Hats, A = B + C

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 ...
  • 4,617
19 votes
Accepted

Alternating Hat Colors

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 ...
  • 6,554
19 votes
Accepted

10 prisoners and 10 lists of numbers

The strategy is: Why it works: For example:
  • 33.8k
18 votes

6 prisoners, 2 colors, one mute

I'll try another explanation(with same result): Here are the steps: Next step: So: Who talks?
  • 420
18 votes
Accepted

Infinitely many hats

How about this strategy Reasoning
  • 128k
16 votes

Five hats and four logicians in a circle

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 ...
  • 45k
14 votes
Accepted

Hat Guessing Game in Vegas

The players can win with probability This is optimal because: A strategy that achieves this is:
  • 24.1k
13 votes
Accepted

100 Dwarves and a tiny room

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 ...
  • 31.7k
13 votes
Accepted

100 Dwarves vs. the Evil Elf

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 ...
  • 13.9k
13 votes

Numbered hats, Warden and Maths

Everybody walks out free and happy.
  • 5,501
13 votes
Accepted

8 Prisoners wearing hats

Solution: Edit:
  • 391
13 votes

Sorting kids by hat colours

I think this is a classic (and probably a duplicate): To separate the two groups cleanly
  • 2,757
12 votes

Infinite wizards and hats

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 ...
11 votes
Accepted

Five hats and four logicians in a circle

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 ...
11 votes
Accepted

Surviving Captain Nefarious

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 ...
  • 33.4k
11 votes
Accepted

A hat puzzle involving wizards

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 ...
  • 13.9k
11 votes

Numbered hats, Warden and Maths

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

Hat Puzzle with 5 different colours and 3 people

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, (...
  • 70.8k
10 votes
Accepted

MORE Prisoner Hats!

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 ...
  • 31.7k
10 votes
Accepted

Guess simultaneously, color of your hat, no passing allowed

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 ...
10 votes

Infinitely many hats

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 ...
9 votes
Accepted

Hat-guessing with a spy

To start: Then: Total losses: Example: Same example, no spy: Either way, the spy only affects the outcome for themselves and the person after them in the line This only works for N > ...
9 votes

100 Dwarves and a tiny room

I think f'' basically has the right answer, but here's a simpler explanation of why it's not possible. Assume there is an algorithm. Consequently there must be an output i when all the shirts have ...
  • 683
9 votes

Surviving Captain Nefarious

This solution gives a 7/9 probability of survival. As JonTheMon has explained, agreeing on a cycle of A $\to$ B $\to$ C $\to$ A is necessary to ensure that the same person won't be chosen twice. This ...

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