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...
- 1,488
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:
- 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:
- 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 ...
- 75.8k
24
votes
Infinitely many hats
Here is another strategy, which requires looking at just 1 other person's hat:
- 13.6k
22
votes
Accepted
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 ...
- 4,637
19
votes
Accepted
19
votes
Accepted
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
17
votes
Accepted
13
votes
13
votes
Accepted
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 ...
- 221
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 ...
- 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 ...
- 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:
- 5,701
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, (...
- 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 ...
- 50.8k
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 ...
- 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 ...
- 9,892
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 ...
- 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 ...
- 91
9
votes
Accepted
Guessing game with infinity coin flips
A possible strategy is for them to
To get the probability of success,
- 21.6k
9
votes
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 ...
- 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...
- 531
8
votes
Accepted
Only top scored, non community-wiki answers of a minimum length are eligible