MathExtremist
MathExtremist
Joined: Aug 31, 2010
  • Threads: 88
  • Posts: 6526
January 13th, 2014 at 7:12:35 PM permalink
Quote: Wizard

I'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?
"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
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1423
  • Posts: 24361
January 13th, 2014 at 7:14:40 PM permalink
Quote: MathExtremist

You'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.
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist
MathExtremist
Joined: Aug 31, 2010
  • Threads: 88
  • Posts: 6526
January 13th, 2014 at 7:21:46 PM permalink
Quote: Wizard

Ungh! 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.
"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
Twirdman
Twirdman
Joined: Jun 5, 2013
  • Threads: 0
  • Posts: 1004
January 13th, 2014 at 7:31:45 PM permalink
Quote: MathExtremist

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.



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.
beachbumbabs
Administrator
beachbumbabs
Joined: May 21, 2013
  • Threads: 99
  • Posts: 14234
January 13th, 2014 at 7:35:17 PM permalink
Well, 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.
If the House lost every hand, they wouldn't deal the game.
24Bingo
24Bingo
Joined: Jul 4, 2012
  • Threads: 23
  • Posts: 1348
January 13th, 2014 at 9:08:30 PM permalink
I would go so far as to say that Cayonero has proven that this problem, as presented, has no solution.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
Tomspur
Tomspur
Joined: Jul 12, 2013
  • Threads: 28
  • Posts: 2019
January 13th, 2014 at 9:42:47 PM permalink
Quote: beachbumbabs

Well, 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 :)
“There is something about the outside of a horse that is good for the inside of a man.” - Winston Churchill
Buzzard
Buzzard
Joined: Oct 28, 2012
  • Threads: 90
  • Posts: 6814
January 13th, 2014 at 10:36:16 PM permalink
I came up with DOOR #3. Am I close ?
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet
paisiello
paisiello
Joined: Oct 30, 2011
  • Threads: 21
  • Posts: 546
January 13th, 2014 at 11:00:03 PM permalink
Quote: Twirdman

Since you'd have 9.999 repeating...


Uh oh...
kubikulann
kubikulann
Joined: Jun 28, 2011
  • Threads: 27
  • Posts: 905
January 14th, 2014 at 7:57:21 PM permalink
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 - 1/2N-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 NK's such that the SUM of ln(1-1/2NK-1) >
SUM(-1/2NK-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 1/2N1-1 <= 0.0953178, or N1-1 >= 3.39111. The first group must be at least of size N1=4.39111 and MathExtremist was right to choose 5. The next groups need not be NK=N1K . It is sufficient to define NK=(N1-1) K + 1 . Or even NK>=3.39111 K + 1.

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

  • Jump to: