Quote:WizardI'm still waiting to hear if my second solution violates the infinite right answers rule. I could always break down my method into infinitely many subsets, but some of them would be wrong. Just any one subset would have a probability of success over 95%.

I'm sure this isn't optimal, but will get the probability of success over 95%. It is based on ME's solution.

1. Create 5*(2^1000/2002) groups of 1000 wizards each. Everyone else is to abstain no matter what.

2. Instruct each member that if he sees all hats of the same color to guess the opposite way.

3. Here are the probabilities for any given group:

Correct guess = 2*1000/2^1000

Incorrect guess (everyone making it) = 2/2^1000

Probability group is correct given it guesses = 2000/2002 = 1000/1001.

Probability guess tendered = 2002/2^1000

4. The expected number of guesses tendered will be 5.

5. An approximation to the number of incorrect guesses is 5*(1/1001) = 5/1001 = 0.50%.

6. An approximation to the probability of no guesses is exp(-5) = 0.67%.

7. Total probability of failure = 0.50% + 0.67% = 1.17%.

You're not actually making an infinite number of guesses, so doesn't that automatically fail?

Quote:MathExtremistYou're not actually making an infinite number of guesses, so doesn't that automatically fail?

Ungh! I guess so. I'm about ready to throw my hands in the air and give up -- speaking for myself only.

Quote:WizardUngh! I guess so. I'm about ready to throw my hands in the air and give up -- speaking for myself only.

I have an inkling there was something slightly amiss with the problem statement; I don't see how you can have an infinite number of guesses and zero wrong ones unless the probability of being right is always 100%. But I admit to not having studied the various flavors of infinity too closely.

Quote:MathExtremistI have an inkling there was something slightly amiss with the problem statement; I don't see how you can have an infinite number of guesses and zero wrong ones unless the probability of being right is always 100%. But I admit to not having studied the various flavors of infinity too closely.

Its possible as long as the limit of the probabilities of being wrong go to 0. Remember you only want a 90% chance of being right not 100%. So if you can get an infinite set of wizards first set at 9% chance of failure, next at .9%, and so on it would work. Since you'd have 9.999 repeating percent chance of missing at least one. Not sure how to construct these sets though just know that the question isn't necessarily wrong from the start.

Edit: Oh do feel the need to point out the 9% .9% and so on must be both percent chance to either miss or not answer otherwise you may run afoul of the infinite number of people answering criteria. Hope that helps someone but I am really failing right now at building appropriate sets and criteria for them to answer.

Really, if none of them can tell individually, but all of them must somehow be able to tell simultaneously through reasoning, if the problem is to have a solution, then all of the hats must be of the same color, or it would be unsolvable.

Quote:beachbumbabsWell, they are Wizards, so why can't they cast a magical spell that, whatever they answer for themselves, it will be correct (and the hat will change to reflect their answer if necessary)?

Really, if none of them can tell individually, but all of them must somehow be able to tell simultaneously through reasoning, if the problem is to have a solution, then all of the hats must be of the same color, or it would be unsolvable.

That was exactly my reasoning but I didn't want to fly in the face of conventional wisdom and posters MUCH smarter than me who are actually trying to figure this thing out :)

Quote:TwirdmanSince you'd have 9.999 repeating...

Uh oh...

Quote:MathExtremist

a) The wizards divide themselves beforehand into a first group of N, a second group of N^2, a 3rd group of N^3, etc. Something exponential, and something such that product[1-1/2^(N^M)] as M->infinity converges to more than 90%.

b) One wizard in each group is selected to be the spokesperson for that group.

c) If the spokeswizard in each group sees an entire group of white hats, they guess black, and vice versa. Otherwise, they abstain. This is done simultaneously at the count of 3.

There are an infinite number of groups and a much smaller (but still infinite) number of guesses. If the product in (a) converges to more than 90% they will (in total) make one or more wrong guesses less than 10% of the time. If I did that right, N=5 (groups of 5, 25, 125, etc.) converges on a 96.875% success rate.

A group of size N has been shown by MathExtremist and the Wizard to have a probability of (1 - 1/2

^{N-1}) of not failing. The probability of global win is the product of all (infinite) group probabilities. (No communication means no correlations needed to be computed.)

This product must be higher than 90%, or its logarithm higher than ln 0.90 = -0.1053.

ln(1-x) can be approximated by -x; it is necessarily higher, anyway. So we are looking for N

_{K}'s such that the SUM of ln(1-1/2

^{NK-1}) >

SUM(-1/2

^{NK-1}) > -0.1053.

It is known that the geometric series yields a + a² + a³ + ... = a/(1-a) if a²<1. With a<=0.0953178 you get a/(1-a) <= 0.1053, so you define 1/2

^{N1-1}<= 0.0953178, or N

_{1}-1 >= 3.39111. The first group must be at least of size N

_{1}=4.39111 and MathExtremist was right to choose 5. The next groups need not be N

_{K}=N

_{1}

^{K}. It is sufficient to define N

_{K}=(N

_{1}-1) K + 1 . Or even N

_{K}>=3.39111 K + 1.

Note: not every francophone is French, exactly like Canadians are not United-Staters. Is Blanchard not a French name?