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...
HugoBDesigner's user avatar
49 votes

Red and Blue in the Chocolate Fountain Room

The prisoners agree that: Each prisoner with a blue nose will: Each prisoner with a red nose will:
humn's user avatar
  • 21.8k
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:
Bass's user avatar
  • 75.8k
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 ...
Bass's user avatar
  • 75.8k
24 votes

Infinitely many hats

Here is another strategy, which requires looking at just 1 other person's hat:
phenomist's user avatar
  • 13.6k
22 votes
Accepted

Numbered hats, Warden and Maths

Sure, it's perfectly doable.
Deusovi's user avatar
  • 145k
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 ...
votbear's user avatar
  • 4,637
19 votes
Accepted

Red and Blue in the Chocolate Fountain Room

QBrute's user avatar
  • 1,178
19 votes
Accepted

10 prisoners and 10 lists of numbers

The strategy is: Why it works: For example:
athin's user avatar
  • 34k
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?
Tode's user avatar
  • 420
18 votes
Accepted

Infinitely many hats

How about this strategy Reasoning
hexomino's user avatar
  • 133k
17 votes
Accepted

Again! 6 prisoners, 2 colors, one mute

Reasoning:
EightAndAHalfTails's user avatar
13 votes

Numbered hats, Warden and Maths

Everybody walks out free and happy.
Evargalo's user avatar
  • 6,153
13 votes
Accepted

8 Prisoners wearing hats

Solution: Edit:
Mazement's user avatar
  • 401
13 votes

Sorting kids by hat colours

I think this is a classic (and probably a duplicate): To separate the two groups cleanly
Braegh's user avatar
  • 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 ...
Anonymous Coward's user avatar
12 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 ...
f'''s user avatar
  • 33.6k
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 ...
Julian Rosen's user avatar
  • 14.2k
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:
Trevor Powell's user avatar
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, (...
Bass's user avatar
  • 75.8k
11 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 ...
Jaap Scherphuis's user avatar
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 ...
Mike Earnest's user avatar
  • 32.1k
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 ...
greenturtle3141's user avatar
9 votes
Accepted

Three applicants, six hats

The king has to be nondiscriminatory for each person applied to this job so putting one black on one of them and two red on the others would make the game unfair! So The only way to make this game ...
Oray's user avatar
  • 30.2k
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 ...
lightbulb's user avatar
9 votes
Accepted

Guessing game with infinity coin flips

A possible strategy is for them to To get the probability of success,
ffao's user avatar
  • 21.6k
9 votes

6 prisoners, 2 colors, one mute

alternately, for lateral thinking:
Destructible Lemon's user avatar
8 votes
Accepted

One hundred and one hats

Over all possible situations, each prisoner will guess right at most half the time. If we want to maximize the probability of all guessing right, we must make it so that all prisoners are correct in ...
f'''s user avatar
  • 33.6k
8 votes

A hat puzzle involving wizards

Each hat is independently black or white with probability one-half So... Which brings us to...
Ryan Smith's user avatar
8 votes
Accepted

Five Hats and Three Logicians

I claim that: Reasoning:
Dennis Meng's user avatar
  • 1,531

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