Aleph-Naught Wizards

I know an infinite number of wizards. If I add our own beloved Wizard to the group, that makes a grand total of, well, an infinite number of wizards! I asked them (nicely, because you don't want to piss off even a single wizard if you can help it) to prepare for a contest. They must derive a method that guarantees a winning probability of at least 90 percent given the following rules:

Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?

Quote: dblanch256

I know an infinite number of wizards. If I add our own beloved Wizard to the group, that makes a grand total of, well, an infinite number of wizards! I asked them (nicely, because you don't want to piss off even a single wizard if you can help it) to prepare for a contest. They must derive a method that guarantees a winning probability of at least 90 percent given the following rules:

Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?

Here are two trivial solutions given the literal statement of the problem:

At the count of 1, each wizard takes off their hat and looks at it.
At the count of 2, each wizard replaces their hat, just because they're wizards and like wearing hats.
At the count of 3, each wizard announces their own hat color.

At the count of 1, each wizard takes a selfie with their smartphone.
At the count of 2 they look at it.
At the count of 3 they announce their own hat color.

Here's one that's probably closer to what you intended, assuming the wizards are perfectly truthful (and perfect logicians, etc, etc.):

Quote: dwheatley

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The wizards agree on some common pointing strategy, such as every wizard finds two nearby wizards with the same hat, and points at both of them. Any wizard being pointed out can then announce the colour of their hat by looking at where the pointer's other hand is pointing. With an infinite number of wizards, it is not hard to see there will be an infinite number of wizards being pointed at, guessing correctly. And 0 wrong guesses.

I like Pascal's triangle but don't see how my solution is related.

The wizards consider pointing (and for that matter) any other gestures to be rude. They must use only their perfect vision and enormous minds to decide their answers.

At the risk of being overly-enumerative, the wizards may not (collectively or individually):

(1) Remove their hats and look at them.
(2) Point at other wizards hats or body parts.
(3) Use any form of pantomime.
(4) Use their wPhones to collaborate.
(5) Melvin each other for giggles.
(6) Need I go on?

Quote: MathExtremist

Quote: dblanch256

I know an infinite number of wizards. If I add our own beloved Wizard to the group, that makes a grand total of, well, an infinite number of wizards! I asked them (nicely, because you don't want to piss off even a single wizard if you can help it) to prepare for a contest. They must derive a method that guarantees a winning probability of at least 90 percent given the following rules:

Each wizard will be assigned a random hat, black or white, with a probability of 1/2 for each choice. The wizards can see everyone's hat, except for their own. At the count of three, each wizard must either guess the color of his hat or abstain from guessing. The wizards collectively win if, among them, there are an infinite number of correct guesses and zero wrong guesses. What strategy do they use?

Here are two trivial solutions given the literal statement of the problem:

At the count of 1, each wizard takes off their hat and looks at it.
At the count of 2, each wizard replaces their hat, just because they're wizards and like wearing hats.
At the count of 3, each wizard announces their own hat color.

At the count of 1, each wizard takes a selfie with their smartphone.
At the count of 2 they look at it.
At the count of 3 they announce their own hat color.

Here's one that's probably closer to what you intended, assuming the wizards are perfectly truthful (and perfect logicians, etc, etc.):

Show Spoiler

At the count of 1, the wizards pair up. For convenience they may move to stand adjacent to their partner wizard.
At the count of 2, each wizard in the pair holds up 1 finger if their partner's hat is white, and does not hold up anything if their partner's hat is black.
At the count of 3, every wizard correctly announces his or her own hat color.
This has 100% chance of success

Funny! But sorry, no fingers. Please see reply above to other "would be" finger pointer.

Hopefully that answer conforms to the rules. I'm worried it violates the "count of three" rule but maybe the wizards are fast.

Am I missing something?

Quote: dblanch256

Quote: dwheatley

Show Spoiler

The wizards agree on some common pointing strategy, such as every wizard finds two nearby wizards with the same hat, and points at both of them. Any wizard being pointed out can then announce the colour of their hat by looking at where the pointer's other hand is pointing. With an infinite number of wizards, it is not hard to see there will be an infinite number of wizards being pointed at, guessing correctly. And 0 wrong guesses.

I like Pascal's triangle but don't see how my solution is related.

The wizards consider pointing (and for that matter) any other gestures to be rude. They must use only their perfect vision and enormous minds to decide their answers.

At the risk of being overly-enumerative, the wizards may not (collectively or individually):

(1) Remove their hats and look at them.
(2) Point at other wizards hats or body parts.
(3) Use any form of pantomime.
(4) Use their wPhones to collaborate.
(5) Melvin each other for giggles.
(6) Need I go on?

I interpret those rules (plus the original statement) to mean that the Wizard's can't communicate at all, even through other Wizards' guesses. That means whatever ordering or arrangement the wizards are in, they can't use it to sequentially perform their guesses. All guesses are simultaneous. Is that right?

Edit: can't communicate after the counting has started; the OP clearly says they need to derive a method/strategy to play the game so they're obviously talking beforehand.

Quote: DJTeddyBear

Am I missing something?

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According to the rules, every wizard can abstain, thus using zero wrong answers.

For the part of the problem which mandates "there are an infinite number of correct guesses and zero incorrect guesses" I think you have covered the second part, but not the first. I don't see how you can have an infinite number of correct guesses if no one is guessing. Sorry.

Quote: MathExtremist

Quote: dblanch256

Quote: dwheatley

Show Spoiler

The wizards agree on some common pointing strategy, such as every wizard finds two nearby wizards with the same hat, and points at both of them. Any wizard being pointed out can then announce the colour of their hat by looking at where the pointer's other hand is pointing. With an infinite number of wizards, it is not hard to see there will be an infinite number of wizards being pointed at, guessing correctly. And 0 wrong guesses.

I like Pascal's triangle but don't see how my solution is related.

The wizards consider pointing (and for that matter) any other gestures to be rude. They must use only their perfect vision and enormous minds to decide their answers.

At the risk of being overly-enumerative, the wizards may not (collectively or individually):

(1) Remove their hats and look at them.
(2) Point at other wizards hats or body parts.
(3) Use any form of pantomime.
(4) Use their wPhones to collaborate.
(5) Melvin each other for giggles.
(6) Need I go on?

I interpret those rules (plus the original statement) to mean that the Wizard's can't communicate at all, even through other Wizards' guesses. That means whatever ordering or arrangement the wizards are in, they can't use it to sequentially perform their guesses. All guesses are simultaneous. Is that right?

Yes that is right.