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
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
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
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
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
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
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 ...
- 115k
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:
- 5,631
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 ...
- 46.6k
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 ...
- 9,792
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 > ...
- 8,588
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 ...
- 91
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