dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 12th, 2014 at 3:30:08 PM permalink
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?

The key can be found along the edge and adjacent values of Pascal's triangle.

David C Blanchard
dwheatley
Joined: Nov 16, 2009
• Posts: 1246
January 12th, 2014 at 4:37:10 PM permalink
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.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
January 12th, 2014 at 5:33:21 PM permalink
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.):
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
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 13th, 2014 at 6:42:29 AM permalink
Quote: dwheatley

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?
David C Blanchard
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 13th, 2014 at 6:44:42 AM permalink
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.):
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.
David C Blanchard
Wizard
Joined: Oct 14, 2009
• Posts: 25448
January 13th, 2014 at 7:08:59 AM permalink

The Wizards define a certain order. It doesn't matter how, but every wizard must know his place. No matter what, only one wizard will hazard a guess. If one hazards a guess then all others will abstain. Here is the strategy:

1. If the first Wizard sees every other wizard wearing the same color hat then he will guess that same color. He will have a 50% chance of being right.
2. If the first Wizard abstains, then the second wizard must know the first one sees a mixture of colors. If the second wizard sees all the same colors (not counting the first wizard's hat) then he will know his hat will be the opposite color, and will guest that. Otherwise, the second wizard will abstain.
3. If the first two wizards abstain, then the third wizard must know the second one sees a mixture of colors. If the third wizard sees all the same colors (not counting the first two hats) then he will know his hat will be the opposite color, and will guest that. Otherwise, the third wizard will abstain.
4. Keep repeating this process until somebody hazards a guess, which must be right. It must eventually lead to a guess. If it gets down to the last two wizards then second to last one will guess the opposite color of the last one, and must be right.

The only chance of failure is if every hat is the same color and even then there is a 50% chance of being right.

This is similar to the "cheating husbands" riddle.

Hopefully that answer conforms to the rules. I'm worried it violates the "count of three" rule but maybe the wizards are fast.
Extraordinary claims require extraordinary evidence. -- Carl Sagan
DJTeddyBear
Joined: Nov 2, 2009
• Posts: 10903
January 13th, 2014 at 7:22:47 AM permalink
Am I missing something?

According to the rules, every wizard can abstain, thus using zero wrong answers.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/  Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
MathExtremist
Joined: Aug 31, 2010
• Posts: 6526
January 13th, 2014 at 8:00:26 AM permalink
Quote: dblanch256

Quote: dwheatley

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.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 13th, 2014 at 8:04:35 AM permalink
Quote: DJTeddyBear

Am I missing something?

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.
David C Blanchard
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 13th, 2014 at 8:07:34 AM permalink
Quote: MathExtremist

Quote: dblanch256

Quote: dwheatley

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.
David C Blanchard