Guess simultaneously, color of your hat, no passing allowed

Question

1000 logicians standing in a circle are blindfolded and a hat of either black or white color is placed on each person's head. The blindfolds are then removed and each person can now see everybody's hat color but his own. After exactly one minute, they all have to simultaneously guess their own hat color.

No communication in any form is allowed after the hats have been placed. However, before the hats are placed, the logicians are allowed to gather and devise a strategy for maximising the number of correct guesses.

Questions :

1. What should be their strategy to get the maximum guaranteed correct guesses?

2. Had the total number of different hat colors been three instead of two, then what strategy should they have devised to get the maximum guaranteed correct guesses?

P.S: I do not know the answer of this puzzle but I have been wondering about the answer for very long . It is based on a similar well known puzzle with infinite hats where the aim is to make only a finite number of incorrect guesses. The infinite version and it's solution can be read here : https://tinyurl.com/infinitehats

Answer 1

Here is a way to guarantee 500 correct guesses.

Beforehand they split into two groups of 500. One group will assume that the number of black hats is odd, the other that the number of black hats is even. After the hats are placed they determine whether their own hat is black, given their assigned assumption. One of the groups will use a correct assumption and will all get their colours correct, the other group will be all wrong. This method therefore gives exactly 500 correct guesses every time.

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 question, using three colours of hats:

A guaranteed 333 correct guesses is possible.
Label the three colours by the numbers 0,1, and 2.
They split into three groups of 333 (and one left over who doesn't take part). One group assumes that the sum of all the hat colours is divisible by 3, one group that it gives a remainder of 1 when divided by 3, and the last that it has remainder 2 when divided by 3.
Exactly one of those three assumptions must be true.
Every person can work out their own hat colour given their assumption, so the group whose assumption is true will give all correct guesses. The other groups will all guess wrongly (and the person left over might guess wrong too).
So there are guaranteed to be 333 correct guesses.
This is optimal for the same reason as the two-colour version.

Edit 2:

As @Acccumulation pointed out in a comment, in practice it is much easier to split the 1000 people into smaller groups, each of which then apply the solution strategy outlined above. That way they only need to look at the hats in their own small group instead of at all the hats.

So for two colours we have 500 pairs of people, and in each pair one assumes their hats will be the same colour (i.e. that there are an even number of black hats), the other that their hats will be different (i.e. an odd number of black hats).
Similarly, with three colours of hats there could be 333 groups of three people, each person in a group using a different assumption w.r.t. the sum of their trio's hat colours modulo 3.

Answer 2

First, I'd note three things:

1. The usual solution to the hat puzzle cannot be used, since the guesses must be made simultaneously, meaning nobody could pass on any information to the other players their guess.
2. With only a minute to count 1000 hats, you wouldn't have enough time to work out a definite answer by elimination, even assuming you knew the totals in advance.
3. You would have enough time to count a fairly large sample which, if selected sufficiently at random, ought to give you a fairly accurate estimate of the overall populations. Hopefully, being logicians, they'd know how to do this.

Bearing that in mind, my strategy would be:

They could all give the answer as the hat they can see the most of. That way, in aggregate, they'll at least beat random guesswork. For instance if there are 600 black hats and 400 white ones, they'll be correct 60% of the time rather than just 50%, with the accuracy rising to nearly 100% if there's a high proportion of one over the other.

The worst case is when it's 500 of each, where everyone will see slightly more of the hat that isn't their own. However, its unlikely that everyone could count to such accuracy so the results should still be about as good as guesswork in that limit.

The strategy remains the same if there are more kinds of hat added; just keep going with the one that has the highest count.

Answer 3

Assuming:

The colour of each hat is picked randomly from the available colours with equal chance of it being either colour (e.g. 50% for 2, and 33.3% for 3 different colours).

Then:

Once they've counted the total of hats of each colour on all other heads, then there is a higher chance of them having the colour that has the least amount.
E.g if there are 499 white and 500 black, then they are more likely to have a white hat.
In the case of the 3 colours this breaks down if the total is exactly 333 in each direction, as one colour will need 334 to round it up, but if the outcome is not actually evenly spread (e.g 331, 333, 335) then the strategy still holds for being one of the lowest counted colours.

However:

I believe this falls into the common misconception that if you have flipped 10 coins in a row and they've all been heads, then the next one has a higher chance of being tails. Which is obviously wrong as it will continue to be a 1 in 2 chance.
In this case, the chance of your hat being a specific colour is still that 1 in 2 or 1 in 3, despite what everything else is.

Therefore:

I believe there is no such strategy that can maximise this, given my assumption.